GEOMETRY CALCULUS

GEOMETRY & CALCULUS

SC-102

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Directorate of Distance Education

SWAMI VIVEKANAND SUBHARTI UNIVERSITY MEERUT-250 005 UTTAR PRADESH

About Author

Bhupendra Singh, the author, is presently working as an assistant professor in the department of

Mathematics, Meerut College, Meerut. His more than a decade long experience at both

Undergraduate and Postgraduate level has earned him a reputation of a learned and a dedicated

individual. More than 30 students of P.G. have submitted their dissertation in different disciplines

e.g., Fluid Dynamics, Operations Research, Number Theory, Mathematical Biology etc. under his

supervision. He has also authored more than 35 books of different levels. To upgrade himself and to

keep the good work going he is doing his Ph. D. in the area of Micropolar Fluid.

Reviewed by:Mr. Rishabh Raj Mr. Neeraj Kumar

Assessed by:

Study Material Assessment Committee, as per the SVSU ordinance No. VI (2)

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Copyright © Geometry and Calculus, Pragati Prakashan, Meerut

DISCLAIMER

No part of this publication which is material protected by this copyright notice may be reproduced or transmitted or utilized or stored in any form or by any means now known or hereinafter invented, electronic, digital or mechanical, including photocopying, scanning, recording or by any information storage or retrieval system, without prior permission from the publisher.

Information containing in this book has been published by Pragati Prakashan, Meerut and has been obtained by its authors from sources believed to be reliable and are correct to the best of their knowledge. However, the publisher and its author shall in no event be liable for any errors, omissions or damages arising out of use of this information and specially disclaim and implied warranties or merchantability or fitness for any particular use.

Published by : Pragati Prakashan, 240 W.K. Road, Meerut Tel. : 2643636, 2640642, 6544643, E-mail: [email protected]

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EDITION : 2017

mailto:[email protected]

1. Successive Differentiation2. . Expansion3. Indetermination4. Tangents and Normals5. Curvature6. Asymptotes7. Singular Points and Curve Tracing8. Rectification and Curve9. Area of Bounded Curves10. Surfaces and Volumes of Solids of Revolution11. Differential Equation of First Order and First Degree12. Differential Equation of First Order and of Higher Degree13. Linear Differential Equations with Constant Coefficients14. Homogeneous Linear differential Equations15. Wronskian16. General Equation of Second Degree17. System of Conics18. Confocal Conics19. System of Co-Ordinates20. Direction Cosines and Direction Ratios21. The Plane22. The Straight Line23. The Sphere24. The Cone25. The Cylinder

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SyllabusCALCULUS & GEOMETRY

SC-102CHAPTER ISuccessive differentiation, Leibnitz theorem, Maclaurin and Taylor series expansions. Indeterminate form, Tangent and Normal (Cartesian curve), curvatures, Asymptotes singular points and curve tracing (only cartesian curve).CHAPTER IILength of curves, Area of Cartesian curves, Volumes of revolution and surfaces of revolution.CHAPTER HIDefinition of differential equation, Order and degree of differential equation. Differential equation to first order and first degree. Exact differential equations, First order higher degree differential equations, Clairaut's form and singular solution. Linear differential equation with constant coefficient. Homogeneous linear ordinary differential equations. Linear differential equation of second order Wronskian.CHAPTER IV

Geometry (2-dim) : Coordinate system, General equation of second degree, System of conics, Confocal conics.Geometry (3-dim) : Coordinate system, Direction cosines and ratios, The plane, The straight line, Sphere, Cone Cylinder.

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UNIT Successive Differeniiautm

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SUCCESSIVE DiFFERENTIATIONSTRUCTURE

• Introduction• Successive Differentiation of Standard Forms

• Test Yourself-1• Leibnitz’s Theorem

• Test Yourself-2• Summary• Student Activity• Test Yourself-3

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LEARNING OBJECTIVES

After going through this unit you will learn :• How to differentiate the given functions upto finite number of times• Leibnitz’s rule which is applicable for the product of two or more functions

• 1.1. INTRODUCTIONLet y =f{x) be a function, then the differential coefficient off[x) denoted by./'(*) is defined

as follows,f(x + &x) - fix) _ dyf'(x) = lim

&V-+0 5* dxIf the limit exists (i.e., limit is finite and unique), then f'(x) is called fiht differential coefficient

off{x) witfrrespect to x. Similarly, if f[x) is differentiable twice, it is denoted by f"{x), if it is differentiable thrice, it is denoted by /"'(*), i.e.,

d f _ d2y "f"(x)=~ dx2dx dx

dx dx2 ~ dx3f"'(x) =

If y =f(x) be a function of x, then we adopt the following notations.

£(/W). ^=/"w=0=d2/w = d2^ = Df(x) =y'=f'(x) = dx -rim)dx/

y*=rxx) =/

dn=Dnm=^nm)This process of finding the differential coefficients of a function is called successive

differentiation.

Similarly, yn=/"(x) =dxn

• 1.2. SUCCESSIVE DIFFERENTIATION OF STANDARD FORMS(i) Ify =/(*)=*" then Dn (xn) = n).

(ii) If y =f[x) = xm, then Dn (x*) = j xm ~ " (m > n)

1 l (-iyVn! *ax + b

1 , then Dn(iii) If y =f{x) = {ax + b) m + n{ax + b))

Self-Instructional Material 1

J LJLL __

Calculus & Geometry (- 1)”a’n ![^Tl]=^ then D"(iv) Ify=/(x) = m + n(ax + AF'

(v) If y = f(x) = sin (ax + A), then D” = [sin (ax + ft)] = a71 sin ^ + ax + ft

(m - 1)! (ax + ft))

(vi) Ify =</(x) = cos (ax + ft), then if ~ [cos (ax + ft)] -an cos ~ + ax + ft

Xyii) Ify ^f(x) = e" + fr, then Z>fl = (e" + *) = ctf* + b:

(viii) If y =f(x) = log (ax + ft), then If ~ [log (ax + ft)] - (- l)" "1 an(n -1) I(ax + ft)"

(ix) y =/(x) = e** sin (ftx + c). Here, we have

y, = /'(x) = ae™ ■ sin (ftx + c) + be™ cos (ftx + c) = e™ [a sin (ftx + c) + ft cos (ftx + c)]

a = r cos 9, ft = r sin 0 => r2 = a2 + ft2 tan 0 = ft/a i.e., 0 = tan”1 ft/a.

Putand

y\ =/'(x) = r- e™ sin (ftx + c + 0)( ft'• e^ sin ftx + c + tan-1— .

Therefore,

= (a2 + ft2)1/2a

“ sin (ftx + c + tan”1 ft/a + tan-1 ft/a)Similarly, y2 = /"(x) = (a2 + ft2)1/2 (a2 + ft2),/2= (a2.+ ft2)272- sin (ftx + c + 2 tan” 1 ft/a)

>>3 = f"(x) = (a2 + ft2)3/2 a® sin (ftx + c + 3 tan 1 ft/a)

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Similarly, yn = /"(x) = (a2 + ft2)n/2 sin (ftx + c + «tan 1 ft/a)d" e" sin (ftx + c + n tan 1 ft/a)= ft (ftx + c)] = (a2 + ft2)n/2ax

(x) ^ = /(x) - cos (ftx + c).Similarly, we may obtain

=>

yn = 7^ [«“ COS (ix + c)] = (a2 + i2)”/2 ax

• e™ cos (bx + c + n tan"1 ft/a)

SOLVED EXAMPLESExample I. Find the nh differential coefficient of log (ox + x2). Solution. LetDifferentiating n times, we get

y = log (ox + x2) = log [x(a + x)] = log x + log (a + x)

— (,0gx)+ —log(a+x) ax ax

_ (- i)n~1 («-1)! «pi5; eir-1

x" (x + af1 .1

~ (“ 1)"" 1 • ('t “ 1) • ■T +V (x + a)fl

Example 2. Find the n'h differential coeffcieints of (t) e™ sin ftx cos cx.(ii) e2* sin3 x.

y = eax sin ftx cos cx = | e'1* [2 sin ftx cos cx]

Solution, (i) Let

~je<lx ts'n + c*) + sin (ftx - cx)]

= ^ sin (b + c)x + e™ sin (ft - c) x] ...(1)

Since, we know that<r-^7 [c“ sin (ftx + c)] = (a2 + ft2)"72 e“ sin (ftx + c + n.tan”1 ft/a)

dlx"Therefore, by differentiating (1) n times, we get

2 Self-Instructional Material

SAtteitkive Differentiation. H^+.^ + c)2)""2 ■“e sin {(b + c)x

+ n tan"1 (b + c)/a} + {a2 + {b - c)2}^2 e sin((6-c)jc +rttan“ (b-c)/a\\.(ii) Let y-^ sin3 x.Now using the result

sin 3* = 3 sin * - 4 sin3 x. ■ *14 sin3 jc = 3 sin x - sin 3xWe have

lsin3 x = ^ (3 sin x - sin 3x).

.y = ^ e2* [3 sin x - sin 3x] = ^ e2* ~\ef‘ s’n ^Therefore,

Now, differentiating n times, we get.J| [(22 + l2)1/2f e^sih [x + n tan"1 1/2]

1. ^ [(22 +l32)1/2r e2* sin [2x + n tan"1 3/2].

Example 3. Ify = sin nix + cosmx, prove that yn = mn [l +.{- 1)" sm 2mx]Wn.Solution. We know that

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[sin (ox + A]] = n” sin - ~ + ox + ft

(F (n—- [cos (ox + ft)] = a” cos n • — + ox + ftand

d1Therefore, )'n = ^(Sin ^ + (C°S ^\ •

n • n= m sin mx + n— + m cos mx + n — £ &

■ Y>2-|1/2' ' ‘ ?t'\" jsin mx + n ^ + cos mx + n —= m2

1/2m" 1 + 2 sin mx + n^ n• cos mx + n —

= mn [1 + sin(2mx + /i7i)]1/2 = m" [1 ± sin 2mx]l/2 = m” [1 + (- l)" sin 2/nx]1/2.

Example 4. differenial coefficient of log [(ox + ft) (cx +Solution. Lety = log [(ax + ft) (cx + if)] = log (ax + ft) + log (cx + d). We know that Z/1 log (ox+ ft) = (- ly-1 (n - 1) ! an (ax + ft)"”.

>. = (- ir' (« - 1) ! a" (fix + *)-" + (- l)"- 1 (n - 1) !

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d)Y * f

y (cx+co "t n

+(ax + ft)" (cx + <0\

Example 5. Find the n,h derivative ify = cos* x.y = cos4 x = (cos2 x)2 = [ 1 /2 (1 + cos 2x)]2

= 1/4 (1+2 cos 2x + cos2 2x)= 1/4 [1 + 2 cos 2x + 1/2 (1 +cos4x)]= 1/4 [3/2 + 2 cos 2x + 1/2 cos 4x]= 3/8 + 1/2 sin 2x + 1/8 cos 4x.

D" cos (ax + b) = an cos (ox + ft + /m/2).

Let

Now1 1 11yn = 0 + —. 2" cos 2x + — nn + — • 4” cos 4x + — /m 2 2 8 1 2 / •;

1 1- 1 2n-3= 2” .. cos 4x + — /I7t ..cos Tx + ^nn +2 &

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Calculus <fe Geometry • TEST YOURSELF-1Find the nth derivatives of (i) sin3 jc

(ni)e cos jcsin^:(v) sin2j:sin2x.

2. Show that the value of the ntl1 differential coefficients ofand is - n !, if n is odd and greater than 1. * - i

cfy 13. (i) If * = fl(r-sin 0 and y = a (1 + cos r), prove that —7 = —cosec

dx~

1.(ii) cos x cos 2x cos 2x (iv) sin nr cos bx

x3for r = 0, is zero if n is even

4 t2 'v /

&(ii) If r = a (cos 0 + 0 sin 0j]‘_y = a (sin 0-0 cos 0), finddx2

ANSWERS

1 /*\ 3 n7t.I.- (1) ^ ~sm x + ~f ^ ^ '

(ii) >„ =■ 6'1 cos 6jc + ^-/m +4" cos 4r + ^ +2" cos 2r + ^• ^ v / \ / \

(ih) Jn= ^ [(«2 + f)"72 eax sin [x + n tan-1 1/a} + (a2 + 9)”/2 em sin (3r + n tan 1 3/f()]

(iv) yn~^ (a + i>)w sin ■ (a + fr) r + ^- /m + (a - £>)" sin |(a - .b) * +an

(v) _yn = 2" “ 1 sin 2r + ^/17t

1 —n o an— • 3 sin 3* + —

1

1-1- 4" sin 4r + — an

1 sec 03. (ii) ^2 = -; 0 'a'

• 1.3. LEIBNITZ’S THEOREM

This theorem help us to find the n‘h differential coefficient of the product of two functions in terms of the successive derivatives of the functions.

Statement Jf u, v be two functions of x, having derivative ofn11' order, then

Dn (uv) = unv + "Cj un _ jV! + "C2 H„ -2 V2 + ... + X un~rvr+...+ "C,, uv,,

where suffixes of u and v denote differentiations w.r.t. x.SOLVED EXAMPLES

Example 1. Find the n'h derivative of x’ sin x.Solution. Let

2u = sin x and v = r .\

nnThen, Un - sin x + —

= sin x + (n-l)~s ’ ■

uh - 2 - sin x + (n~2)~ .

1 Un — l

Also V! = 2r v2 " 2 v3 = 0.

Now, by Leibnitz theorem, we have .

!

d'‘ («v) = . V + "Ci«n _ , - + nc2 Un

(r2 sin r) = sin r + ^~ r2 + X sin r + (n - 1) ^ 2r+''C2sin j: + (/i-2)“ 2 dx11 2 .2 2

-2 • v2


Stt ccess iv# D iffe ret i iia tio 11= A:2 ,sin x + ^ + 2«j: sin x + (n-l)^ +n(n-

Example 2. If y = a cos (log x) + d sin (log x), show that

x2y2 + xy] +y = 0andx2yn±2 + (2n+ i)^’n + i+(»2+ l)yn=0.

Solution. Here, we have

I) sin x + (n-2)~ .

y = a cos (log x) + b sin (log x). Differentiating (1) with respect to x, we have

— sin (log *) + — cos (log *)X Xyi = -

xyx=-a sin (log x)+b cos (log jc). Again, differentiating w.r.t. x, we get __

xy2 + yi = -

=>

“.cos’(log x) - — sin-(-log j:) .

+ *y\ = -a cos (log x)-b sin (log x) = -y2 •*—

x y2 + xy2 + y = 0.Now, differentiating (2) both n times by Leibnitz theorem _

DV>2) + D"to1) + D,,Cy)=0(D"3-2)/ + ri(D"-1y2)(^2) + ”C2(D"''2yJoD24 . •

Proved ...(2)=>

=>

!JOV rtrr «•(Dfyfox +-?C, Dn~ 'ydJPx) + Dny = 0

£ . \ A

-^1+2 n(n -1) . ^

=» y,! + 2 + 2™yn + 1 + — 2y„ + xyn + l+ ny„ +yn = 0

=> x2yn + 2 + (2n + 1) xyn + \ + (n2 + i) y„ = 0. Proved.

SUMMARYn 'y=xr => yn =

(sin (ax + b)) = a sin ^ + ax + b)

(cos (ax + b)) = an cos ^ ^ + nx + 6

an («-!)!(ax + bf

(cCA sin (bx + c)) = (a2 + b2)n/2 eax sin*! bx + c.+ h tan

xm~n (m > n)(m - n) !

<r -1— (log (ax + b)) = (- l)ndx<f (b-idx acf bW— (eax cos (bx + c)) = (a2 + b2)n/1 dx"Liebnitz’s Theorem : If m, v be two functions of x having /1th order derivatives, then

D" (wv) = UnV+nCIU„_,V, _ 2V2 +(yn)o = value of yn (x) at x = 0.

-i Icos 6x + c + n tan aj /

+n

• STUDENT ACTIVITY1. Find the nl derivative of sin ax cos bx.

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Seif-InsiruciioTiaTMateridi &/

Calculus & Geometry

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If y = [* + Vl + x2 ]m, then find (yn)o-2.

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• TEST YOURSELF-2Use Leibnitz’s theorem, to Find yn in the following cases :

(ii) xV .

(vi) ex log x

1.(i) A"(iv) x*\o$x

(iii) a:3 sin ax (vii) xn log a:

"V

(v) x2^ COS x

2. If ~ (xn logx), prove that fn = nfn_ { + (n - 1)! and hence show that<r «dx *:

i i i/„ = n! logA:+l+- + ~+ ...+ —

3. If = [jc + V(i + x2) ]m, prove that(1 +Jt2)y»M:2 + (2n +

4. If y]/m + y~I/m - 2x, prove that (x2 -l)yn + 2 + (2n+l)xyn+I+ (n2 ~ m2) yn = 0.

+ (n2 ~ m2) yn = 0.n + 1

5. If y = cos (log x), prove that .x2yn + 2 + (2« + l) +1 + («2 + I) y„ = 0.

ANSWERS

(i) e^a"”3 [a3^ + 3na2A:2 4- 3n(n - l) ax + n(n- l)(n -2)](ii) ex[x2 + 2nx + n(n-l)]

r / n ,(iii) a"-3 a3*3 sin ax+~ +3na2x2sm ax + (n-l)^ + 3n (n - 1) ajr sin ■ cw + (n - 2) ^

1.

\n+ n(n - 1) (n - 2) sin ax 4- (n - 3) ^

y-1•»\ •-1(-1)" nil 3 3 1(iv) -■ +-3 n n-1 n-2 n-3xn

j

(v) 2n/2A:2cos x + ^~ +2(n i)/22nj:cos -«+ («-'!)“

\“) 7tn(n- 1) cos x + (n -2) —(n-2V2+ 2J


/

/-1 -nC2x~l + nCl2\x~l - ...+nC„ (- l)n Successive Differentiation(vj) ex [\o$x + nCiX

(viOyn+l =~

-1 (/J - 1) Ijc*"]

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OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. D" (log x) is equal to..........2. To find the ntl1 derivative of the product of two functions we use3. If y = sin (ax+ b), then [f sin'(cu: + b) = ...4. If y = (ox + b)~ \ then Lf(ax + ft)" 1 =......> TRUE OR FALSE :

theorem.-«

oWrite ‘T* for True and ‘F, for False :To find the n,h derivative of the product of two functions we use Leibnitz’s theorem. r.~(T/F) If we observe that one of the two functions is such that'all its differential coefficients after a

(T/F) (T/F)

1.2.

certain steps, become zero, then we should take this function as second function. „ 3. If y = a cos (log x) + ft sin (log x), then x2y2 + .xyi = y.> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :

ax + b1. [f(e ) is equal to :ax + b(a)

(c) it'baSx + b 2. Dn log x is equal to :

(n-1)!

(b)en_ax + b(d) a'‘e

- 1 (-l)V-!(~ l)n(n-l)! (-1)" (n - 1) ! (j;(b) (c)(a)xnx"

ANSWERS

Fill in the Blanks :j, (-ir'tn-D! 3. a" sin ax + ft +2. Leibnitz’s 2x" . /

-i4. (- l)rt • n ! ft” (ax + ft)"^ True or False:

l.T 2. T 3.F Multiple Choice Questions:

l.(c) 2. (c)

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Self-Ihsimctiotial Material 7I

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Calculus & Geometry UNIT

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EXPANSIONSSTRUCTURE

• Taylor’s Theorem• Maclaurin’s Theorem• Failure of Taylor’s and Maciaurin’s Theorem

• Summary• Student Activity• Test Yourself

LEARNING OBJECTIVESAfter going through this unit you will learn :

• Some standard results like Taylor’s and Maclaurin’s Theorem# Expansion of given function in a power series of its variable.

• 2.1. TAYLOR’S THEOREMStatement Let f(x) be a function of x which can be expanded in powers of x and let the

expansion be differentiable term by term any number of times, then _1' ffAa + h) =m + hf(a) + |^/'(ajV..

(i) Writing x for a in (2), we have

J{x + h) =f{x) = hfXx) + ~/'%*) + ... + —Ax)+ ...

(ii) Putting a + h = borh=b-a,m (2), we get(b-a)2,,, (b -a)H fn(a) + .-M) =f{a) + (b-a)fXa) + f"(a)+...+

2 ! n !(iii) Changing a + hto x i.e., h tox ~a in (2), we have

(x ~ a)2 ^ (x - a)n .„\/» + ...Ax)=Aa) + (x-a)f/(a) + f"(a)+...+2 ! n !

• *2.2. MACLAURIN’S THEOREMLet f{x) be a function ofx which possesses continuous derivatives of all orders in the interval

[0, x] and can be expanded as an infinite series in x, then2Ax) = ao)+xf'(o)+lyno) +... + ^r(o) +...

SOLVED EXAMPLESExample 1. Expand

ex («)(!+a-)”(0 (iii) sin x.(v) ax.(iv) log (1 +x).

Solution, (i) Let A*) = ^ AO) = e° = i ,f'(x) = =>/'(0) = 1 /n(0) = 1.Put all these values in Maclaurin’s series

2 3

Ax) = ao)+a/'(0)+—no)+hrw + -3 !x X2 Xz JC4 Xne -1+a:+ + + +...+ +....2! 3! 4! nlWe get

This is known as exponential series.


(ii) Here, we have Expansionsm=a+x)n m = if'(x) = n(l+x)n-l^f(0) = n f"(x) = n(n - 1) (1 + x)n-2=*f"(0) = n(n - 1)

fm(x) = n(n - 1) (» - 2) ... (n -m + 1) (1 /"(O) = n(« - 1) ... (n - m + 1).

Put all these values in Maclaurin’s series2m -xo)+^(o)+fy+no)+...

n(n -1) n{n - 1)... (n - ot +x1 +...+We get (1 + *)" = 1 + n* + 2 ! rm :This is known as Binomial series. (iii)Here, we have

/(x) = sin x =>/(0) = 0 f'(x) = cos x =>f'(0) = f"(x) = - sin X =>/"(0) = 0 f"'{x) = - cos X =>/"(0) = - 1

fn (x) = sin x + —

0, when n ~ 2m (- l)m when n = 2m + 1.

f\0) =

Putting all these values in Maclaurin’s series, we get2/n +1. x3 xs x7'-'" ; x

s.n = jc - — + — - — + 1) — + ... .(2m +1)!

(iv) Here,y(^) = log (1 + x) =>f{0) = 0 (-ir1 (»-!)!

(x+irPut all these values in Maclaurin’s series, we get

-1AO) = (-!)"rw= (n - 1) ! n = 1,2, 3, ... .

2 43-l («-!)! + ...

. .. . X2 x3log (l+x)=x-y+y x4J + ... + (-!)"=> n !

This is known as logarihmic.series, (v) Here, we have

^0) = a°=l => /'(0) = log a=. /"(0) = (log a)1 ==■ /"'(0) = (log a)3

f'(x) = ax log a /"(x) = (log a)2 /'"(x) = (log a)3

Putting all these values in Maclaurin’s series, we. getx2a* = 1 + x log a + — (log a)2 +

Example 2. Expand log sin (x + h) in power ofh by Taylor’s theorem. Solution. Let/(x + /i) - log sin (x + h)

J(x) = log sin x

|^(loga)3 + ....

1rw=- • cosx = cotxsinx

f"(x) = - cosec2 x/'"(x) = 2 cosec x cosec x cot x = 2 cosec2 x cot x

...(1)


Calculus & Geometry Now by Taylor’s theorem, we have2 3

A* + *0 =Ax) + ¥'{x) + Yif"{x)z+y7/"'(x) + ...

Putting all the values from (1) in (2), we geth2

log sin (x + h) = log sinx + h cot a: - -y

f 71^Example 3. Expand sin x in powers o/ x - — by using Taylor's series.

Solution. Let Ax) - sin x.r /

We may write/[x) =/ ^ + x - ^ v

• - / y { Tt 'Now, expanding/ — + x - — by Taylor’s theorem in powers of x - — , we get

£ z* z •L \ /J/y \ /.l 7C^) =/ 2 71 + ^ " 2

...(2)

A3« 2 ft 2— cosec x + ycosec xcotx + ....

» - /i V fc

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3 . / \ Tt r,Jtl2 J 2

\2/ \ «]rfiE| + _L 2 ; 2 ’2 !

1 ,v=/- ,/ 2

5 /■" — 2 J 2+ x- + 3T x~ + ...

V / \ /. ...(1)

A*) = sin x =>/y = sin —\ // \

71 —/'(x) = cos x =>/' — = cos — =

r(x)=.-sinx^r J V A

/,,-,(x) = — COS X f'" ~ = - cos ^ = 0

//'v(x) = sinx=^/'v ^

Now,

• 71 i= -s,nI = -li

V /

71 1 = sin — = 1

Putting all these values in (1), we get\2

sin x = 1 - 1. 7t » '+ 4!-x"227^2

• SUMMARY• Taylor’s Theorem : Let/(x) be a function of x which can be expanded in powers of x and that

the expansion be differentiable term by term any number of times, then :2

/(a+*)=■/(*)+*/'(a) + |y/'>) +

Other form of Taylor’s theorem\ . (x-af

/(*) =/(*) +(*-«)/'(*) + /'» +2 !• Maclaurin’s Theorem : Let/(x) be a function of x which possesses continuous derivatives of

all orders in the interval [0, x] and can be expanded.as an infinite series in x, then,;>'<»•......2

/ (x) =/(0) + xf' (0) +#t/"(0) +) ■+2 !

• STUDENT ACTIVITY1. Expand sin x in powers of ^ x - ^ J by using Taylor’s series.


Expansions

V" (-•

2. Apply Maclaurin’s theorem to prove that■ 1 2log sec * = — * + 1 v-2 a. 1 4 . 1 62^12*

• TEST YOURSELFExpand the following function by Maclaurin’s theorem

(ii) ex sec x J (iii) log (1 + sin x)1.

(i) sec x^x2 + “/ + x6 + ... .Apply Maclaurin’s theorem to prove that log secx = -

x = aQ + a\X + a2x2 + ... Prove that (n + 1) (n + 2) a„ + 2 = n2a„.

2.-13. If y - sin

4. Expand the following(n }

(i) sin ^ ® *n powers of 0.

(iii) sin-1 (x + /i) in power of x.

(ii) 2x3 + 7x2 + x - 1 in powers of x - 2.

(iv) log sin x in power of (x - a).\ ANSWERSs

, 2x2 4x3Oi) 1+x+2! + 3|+'"

x2 5x4 (1) 1 + 2T+4T

X2 X3 X (»0*-T + 7

3 lx61. +..+

\6 l54

X12 + 24

1 + 0 2! 3! 4! 5!‘"1

4. (i)V2

(ii) 45 + 53(x - 2) + 19(x - 2)2 + 2(x - 2)3 + ...

(iii) sin"' A + x(l - h2)-1/2 + ~h(l -h2)~3/2 + [(1 -h2)~5/2 (l + 2h2)} + ...

{x-af 2 , (x - a)3v —cosec a +

3

2 !

2 cosec2 atpta + ...(iv) log sin a + (x - a) cot a - 2 ! 3 !OBJECTIVE EVALUATION > FILL IN THE BLANKS :1. If y = tan x then y5(0) is......2. If>, = ^smAtheny3(0)is 3. By Maclaurin’s theorem


Calculus & Geometry 2n/1 sin (/m/4)x ■ ' 2 2 l •y = e sin^=^ + A: +tta:'+.. xn+.. + .. t3 ! /n !

Then y3(0) =..........> TRUE OR FALSE :Write ‘T’ for True and ‘F* for False : _ _ _

We get the Maclaurin’s series by putting a = 0, h-x \n Taylor’s series. If^.r) =ctx, then /"(0) is equal to (log a)'1.If y = log sec a: then yd(0) is 3.Second term in the expansion of log (1 + tan x) is - ~x2.

> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :

If./(•*) = then fn(0) is equal to :(b) 1

1. (T/F)(T/F)2.

3. (T/F)4. (T/F)

1.(a) 0

dog nf (d) (log a)".(c) n2. If y = tan x then y5(0) is :

(a) 4 (b) 8If y = log sec x then y4(0) is :(a) 0 .Expansion of e* sec x is equal to :

(c) 12 (d) 16.3.

(b) 1 (c) 2 (d) 3.4.

2x2 2(b) 1 + x + —7 + ...(a) 1+x+ — + ...

2 ! 2 !2 2x2

(c) 1 -X + —7 (d) 1 - x + — .2 ! 2 !

ANSWERS

Fill in the Blanks :1. 16 2. 0

True or False :1. T 2.T 3. F 4. T

Multiple Choice Questions : .1. (a) 2. (d) 3. (c) 4. (a)

3. 2/

. □□□

12 Self-Instructional Material /

UNIT Indetermiriaie Forms'V,

3INDETERMINATE FORMS

I

STRUCTURE• Indeterminate Forms• L’Hospital Rule for the indeterminate form 0/0• L’Hospital Rule for the Indeterminate form «>/«>

• Test Yourself-1• The Indeterminate form 0 x <*>• The Indeterminate form

•t

• The Indeterminate forms 0*, 1~• Student Activity• Summary• Test Yourself-2

— oo


# About all determinate and indeterminate forms witnessed while-evaluating the limit of the given functions.]

• 3.1. INDETERMINATE FORMSWhen a function involves the independent variable in such a manner that for a certain assigned

value of that variable, its value cannot be found by simply substituting that value of the variable, the function is said to take an indeterminate form.

. The most common cases occuring is that of a fraction whose numerator and denominator both vanish for the value of the variable involved.

As f(x) —»0 and g(x) —»0 when * —> a, then the quotient m is said to have attained the*(*)

indeterminate form ^.

fix)Similarly if lim f(x) = =* and lim g(jt) = <», then the fractionx—*a

is said to have attainedg(x)X ->„fl

the indeterminate form — .CO

The other important indeterminate forms are 0 x -

CO

/oo, 0°, 1“ akd oo°.

l

• 3.2. L’HOSPITAL RULE FOR THE INDETERMINATE FORM 0/0

lim f{x) = 0 = lim g (x)Ifx-* a x—*a

fix)fix) providedlim = limthen*'(*)g(x)X —0 x—* af' (x)J^ exists.fix)provided lim = lim

.r-* « gf(x)g (x)x a

• 3.3. L’HOSPITAL RULE FOR THE INDETERMINATE FROM oo/oo

If lim fix) = oo and lim g (x) = «>, thenx—» <7x-*a

r mfix)lim limSix)Six)


Calculus & Geometry f (x)provided limX—»«

Solved Examples

exists.S'to

Example 1. Find limx-*0 x - sin x

sinxex - e x-sinx ’

S-e*“.cosx1-cosx

e* - [cosx . es,nx. cosx + esinA (- sinx)]sinx

g* - eiin J [cos2 x - si n x]sinx[2 cos x (- sin x) - cos x] - [(cos2 x- sin x) esi" * cos x)

0Solution : limx-» 0 . aform

,olimx-»o

again —formv0

limx-» 0

0limx-*0

again — form

sinxex - e- limx-»o COS X

sinx [- sin 2x - cos x + cos3 x- sinx cos x]ex - elimx-»0 cosx

1 - 1 (- 1 + 1) 11 1

= I.xcosx - log (1 +x)Example 2. Find lim

x-»02

Xl

x cosx-log (1 +x)Solution : We have limx-»0

2X

/x2 x4 ) f x2 X3 2 ! + 4 ! ■■ “r " 2 + 3 "

0limx-»0

r form0

''x*

= limx-*0 x2

^ ^ x + terms containing x jlimx-»0

12 ‘

cosh x - cos xExample 3. Find limx-*0 xsinx

cosh x-cosx 0Solution : We have limx-+0

t formxsinx 0

cosh x - cos x x= limx-*0 x2 sinx

cosh x - cos x 0= limx 0

t form2 0x

sinhx + sin x 0limx-*0

— form2x 0

cosh x + cos x 1 + 1 .lim x —> 0

= 1.2/ 2

• TEST YOURSELF-11. Find the following limits :


Indeterminate Forms.1 -cos.x ■ ■... x-sinx(i) lim ------ r

x —^ 0 X

cf-b*

(ii) lim •2X

log-y(iii) lim *->0

(iv) lim x-*\ x - 1x

(1+*)"-! x^- log (1 +a:)(v) lim jr-»0

(vi) lim x^O

(viii) lim - x->a -

2x xx atan x - x - x(vii) lim ,

*-»0 'jr tanx xK-aa

sin x sin 1 x-x22. Find lim 6xx —* 0

1(1 + x)l/x - e + i- ex23. Find lim 2Xx—tO

ANSWERS

i * 1 3 . ... 12 VU 3(iii) log ~

2 A.1-0)f(viii)

®2 (iv) 1 (v) n (vi)

3.~log a - 1 ■ A-

24 ‘18log a + 1

•' 3.4. THE INDETERMINATE FORM 0 x ~.To find lim [f(x).g (a:)], when lim / (at) = 0 and .lim g (a:) = ■».

x-ta. x-*a0 CO

To determine this limit, the produce may be transformed into the form

of the following relations

— or —, using any one

fix) _*(*)f(x) g(x) = or,, fix)- 8 (x)1 1■■■Six)

and then apply previous method-____Solved Examples

Example 1. Evaluate lim (atlog.*).■ x-»0+

fix)

0fofmlog a:Solution : lim (* log *) = lim

x-»0+ 1/xx-»Of

1/X 2 - lim (-at) = 0.— 1/x X—Mk

limx—»Of

Example 2. Evaluate lim x log sin x.x—»0

| from 0 x colim x log sin x.x->0

Solution :

log sin x= limx —> 0 1/x

(1/sin x). cos x ooform—lim

x—>0 - 1/x2 . eo

-x2.cosx ooform—lim

x->0 oosm x

x2 sin x - 2x cos xfonn0lim

x—»0 cosx

= 0.


Calculus & Geometry • 3.5. THE INDETERMINATE FORM — ooftft •To determine lim (/C*:) - g (*)], when lim f(x) = °°= lim g (at)

x-*a/ o

Here, this can be reduced to/the form — by the relation

/ x—*a

1 1Z(X) f(x)

f{x)-g{x) =1-

m.g{x)and they aply previus method.

Working Procedure.1. Change all trigonometric-ratio into sin x and cos x (if T-ratio ae present).2. Take LC.M.Now the indeterminate form is reduced into jj form.

Solved Examples

■i

t*1-- —2 sin2* JExample 1. Evaluate lim

.r-»0 V*

1-x1 sin2 x )Solution : lim I' form oo-oo

•2 2' i form ^sin x-x

limj-»0

v ; J.a2 sin2 x

23X 2

-X

limx -> 0 ,23X2' - + ~ .rX - *-3!

2x- -r— + terms containing higher powers of i3 !lim

x-» o■'4

x4 + terms containing higher powers of x

terms conta*ninS x 'n numerator, : 4 ;

1 + terms containing x in the numeratorlim

x—»0•v

3 ! “ 3 '

Example 2. Evaluate lim (sec x - tan x).X-Ml/2

Solution : We have lim (sec x - tan x)X-»7l/2

| form oo — OO

1 form jjsmxlimX—»ti/2 COS X COS X

1 - sin xlim

cos xX—» 7t/2

- COS Xlim lim cotx = 0.

x—*n/2-sinxx-» n/2

• 3.6. THE INDETERMINATE FORMS O6, 1~, oo°gW when the limit is of the form 0—, 1°°, oo°.To determine lim [/(x)]

x—*aSix)Let

logy = g(x) log/(x).Taking logs;


indeterminate FormsThe R.H.S. assumes the indeterminate form 0 x <» in each of these above cases. The limit can, therefore, be dtermined by the method used in the article (4).

Supposelim [f(x)\ogf(x)] = l (say)x-* a

lim log y = /ix—*a

lim y = llimx—*a <- x —ia

lim y = el=>

lim (f(x)jx-*a

'w.Working Procedure1. Let the given limit - y.2. Take logs onboth sides to get the forms 0, <» and proceed by the method of the type

i0X«>.

Solved Examplesi/,2 •tan*Example 1. Find lim xx-*0

l/x"' tan-j: | form 1“, for x = 0Solution: Let y-X

11, 1 . tanj:tog y = ^ log —

1lim log y = lim -t log

^ ♦ 0 x-»0 X x

, tan a: leg —

tanx=>

pi

0form tlim*~»o ■2 0x

1x sec2 x - tan xtan x

2X Xlim*->o 2x

x sec2 - tan x tan*v lim * —»o x

= 1limx-»0 2x3

x. 2 sec x sec x tan x + sec2 x - sec2 xlim

x-»0 6x2

tanx sec2x2x tan x sec2 x= lim

x-»0- lim

x-»0 bx2 3x

11 , 2 = -x 1 xseci tanjc3 x

2= limx-»0

. sec x : 31/X2

tan xlim y = e1/3x-»0

1/3limx-»0 x

l/x’sinxExample 2. Evaluate limx—>0 X

l/x2smxSolution : Let y = lim xx-»0


3 -5Calculus & Geometry x x1 . sinx

~2 l°g-----X1 X

1- lim —zx—*0 X

log y = lim xjr-> 0

lim ^l0gf 1_f^ + 4^ + ...

X2 X41lim —; 1 - — • x-m x2L

lim -^log(l-z) x->0 x

+ ...120/ 2 4

where z~x + ...

6 12021 Z= lim —z

x —> 0 X Z 2 'rj**- * *'•

_1^ X^ x4 2.1 .6-” 120

2 41 X X= lim ~2 - j:—* 0 X -

+...6, 120.

x2 4 4 \1 X X- lim —z x-+o x

+ ...6 120 722 x41 X= lim -

x-»o x -+ ...

6 . 180

6 180limX -4 o

i+ ...

16 ' n- ■

• SUMMARYIndeterminate Forms : Some indeterminate forms are

0 A — , — , 0 X oo oo -0 oo

oo, 0°, 1“ and oo°.

: fix)If lim f (x) = lim g (x), then limx->0

If lim /(x) = oo - Hm g (x), then limx —> 0

limx —) 0

8 (a) * 0.OO =g'ixYsix)x —* 0 x—»0

f'jx) f' (x)̂ exists.fix) limx—»0

provided limx-» 0*'<*) g' (X)8ix)x->0 x-» 0

fix)If lim /(x) = 0 and lim g(x) = °o, then lim /(x)g(x)= limi/8 (x) 'X —4 0x—»a x —> (1

r i ig jx) fix)If lim |/(x)(x)] = lim\ 1x—>a x —) flfix) g ix)

g(x) is of the form 0°, 1*° and oo0, then log y - lim g (x) log (fix) = l (say)If the lim [/‘(x)]x—»0 • X —> o

/y = e.

• STUDENT ACTIVITYlog (x - 7t/2)1. Find lim

x—»ji/2 tan x


Indeterminate Forms

l/x2sin x

2. Evaluate limx~>0 X

•.Jit v. /

4

• TEST YOURSELF-21. Evaluate the following limits :

(i) lim x log tan x x-*0

(iii) lim2xsin —' x —» oo 2X

2. Evaluate the following limits

(i) lim ~ - “T log (1 + x)x->0'-x x

n(ii) lim tanx —»0

- -a-

1(ii) lim

x -»2 x-2 log (a: - 1) 1 2—r - cot X1 2 (iv) lim 2

x -> 0 V Jt(iii) lim —r - cosec x2x—»0 ^ x

3. Evaluate the following limits :\tanx l/x

1 n -i(ii) lim(i) limx->0

— - tan x2XJ

l/xtanx

(iii) limx 0 V x J

OBJECTIVE EVALUATION Fill in the Blanks :

limx-»l 1

x2 + 2x . 5 - 3x2

sin ax .is 2. limx-» o

1. issin bx

tanx is4. limx-» o x

3. lim

True or False :

0ooThe indeterminate form — can be converted into the form —1. O' (T/F)OO

i0 is not an indeterminate form.2. (T/F)


Calculus & Geometry Multiple Choice Questions :tanx .lim1. is :x

(d) -1(b) oo (c) 1(a) 0lim (1 + nx)1/x is : x—»o

2.

(c) e2' (d) en(b)(a) 1

’

ANSWERS

1. (i)0

2- ® I3. (i) 1

Fill in the Blanks :

(ii) ~ (iii) a (iv)

(iv)f1<a)-2 (iii) 1

(iii) 1(ii) 1

12 - 2'b1. 1 3-3 4. 1. v *IVue or False

1. TMultiple Choice Questions

2. (d) ‘

U12. T 3. F

1. (c)

□□□

S! lI

, .itI

:


UNIT Tangents am! Normals

4TANGENTS AND NORMALS

«*-•' i , i t.

STRUCTURE &

• Tangent• Equation of the Tangent• Normal• Equation of a Normal


•4


# Determine the tangents and normals to the given curve at the given points. • # How to determine Angle between two Curves.j

• 4.1. TANGENTLet P be any point on a curve y =/(*) and Q any other point

on it such that Q is very close to P. The point Q may be taken on either side of P.

As Q tends to P, secant line PQ, in general tends to a definite straight line TP passing through P. This straight line TP is known as the tangent to the curve y =f{x) at the point P.

(x + 5xr y + 5x)y = f (X)

(X, y)

X0Fig. 4.1

• 4.2. EQUATION OF THE TANGENTLet y =/(*) be a given curve. Let P be any given point (x, y) on this curve and Q be any other

point (x + Sx, y + 5y) on it such that Q is very closed to P. Let (X, Y) be a i arbitrary point on the secant line PQ, then the equation of the secant line PQ is

(X-x)y-y = x + dx-x

As Q tends to P, 5x —> 0 and PQ tends to the tangent at P. Equation (1) tends to an equation

... (1)y-y =or

.. & = L ' dx

4i lim{X-x)Y~y = dx SxJ&r-*0

Hence, the equation of the tangent to the curv at P (x, y) is given by

REMARKS&If we are to find the tangent to the curve y =/(x) at (xj, yi), we first find

the equation of the tangent to the curve at (x^yj is given by

at (xi.yj). Thendx

Self-Industrial Material 21

Calculus & Geometry di (X-Xi)y-y\~ V dx px^x)

y-yi _| dvX-X\

If 0 be the angle which the positive direction of the tangent at P makes with the positive direction of the x-axis, then

ordx/Xx.y,)

0Fig. 4.2

tan 0 =V dx-p>.yt)

& - 0. which means, the tangent is parallel to *-axis.If 0 = 0, then , dx XT»-yJdx = 0, which means, the tangent is perpendicular to x-axis or parallelIf0 = 9O*, then k dy Xxx.y,)

to y-axis.Two curves y ~f(x) and y = g (x) are at right angle, if /?i| x nt2 ~ - I where /«( = slope of the

tangent to y-f(.x) at common point and wj = slope of tangent to = g (x) at common point.... Two curves)’—/(x)andy = g (x) touch each other if they have the same tangent at the common

point.';,.If 0 be the angle between the curves y = /(x) and y = g (x), then

mj - m2

\ - m\m2Result 1. The tangent to a curve y = /(x) at a point />(X|,yj)'is parallel to x-axis if and only

tan 0 =

ifdx = 0., dx M.yJ

Result 2. The tangent to a curve y =/(x) at a point Pfo.yj) is parallel to y-axis if and onlyif

dx= 0,dy J(s,.y,)

• 4.3. NORMALThe normal to a curve y =/(x) at any point P on it, is the straight line passing through P and

perpendicular to the tangent to the curve at P.

* 4.4. EQUATION OF A NORMAL

Let y =/(x) be a given curve and FCxi.yi) be any point on it.Since, the normal to y =/(x) at f* (x!t yj) is perpendicular to the tangent at P.

- 1So, slope of normal -

*vy\)

The equation of a normal at P (xj, yi) is given by

- 1 (x-x,)y-y\ =

-1y-y\orX — X]

($L). dx tox-fi)

(y ~ yi)= o(x-x,) +or

Solved ExamplesExample 1. Find the slope of the tangent of the following curves :

22 Self-Industrial Material

Tangents and Normalv(i) y2 = 4ax at (a2, 2a) (ii) _y = *3 -* at (2, 6)

y2 = 4axSolution : (i) We have

dy2yTx=4a '

dy 2adx y

2'So, the slope of the tangent at (a , 2a) =

dx )(fl2,2<i)

2a 2a■y J(n.2fl) 2a

y = x3

dx '

(ii) We have -x

4lSo, the slope of the tangent at (2, 6) = dx h 6)

= (3x2 - 1)(2.6)

= 3 (2)2- 1 = 11.

Example 2. Find the equation of the tangent and the normal to the c jrve y = x3 -2x +7 at(1, 6).

y=x3-2x + 7

^L = 3xz-2

Solution : We have

dxd£ = 3 (l)2—2= 3-2=1.So, dx J(l. 6)

The equation of the tangent to the given curve at (1, 6) is

(x-l)dx p,6)y-6=l(x-l)

j:-y + 5 = 0.The equation of the normal to the given curve at (1, 6) is

oror

dx (x-l)j-6 = - dx 1(1.6)

- 1y-6 = — (x-l)or 1x + y - 7 = 0.

Example 3. Find the equation of the tangent to the curve .x2 + 3;y = 3, line ;y-4x + 5 =0.

Solution: We have

or

which is parallel to the

+ 3y = 3

2* + 3^ = 0dxdx__2x

3 ‘dx

Let (xi, y\) be a point of contact.

4 +3^1=3 ■Now the slope of the tangent at (xh to the given curve is

-d)So,

Self-Industrial Material 23

Calculus A Geometry 2*1— —^ dx /Xi,yt)

and, the slope of the line y - 4* + 5 = 0 is 4.Since the tangent is parallel to the given line.

_^ = 4. 3 .

X] = - 6

Putting xj = - 6 in (1), we get yj = - 11./. Point of contact is (- 6, - 11).Thus, the equation of the tangent at (-6, -11) is

y + 11 ■= 4 (x + 6)

4x -y + 13 = 0.2 2Example 4. Find the angles of intersection of the curves y = 4 - * and y = x .

Solution: We have

3

or

or

y = 4-x2

y-x2

Solving eqns. (I) and (2), we get, 2 y=4~xt

x2=4-x2

...(1)

... (2)and

[From eq. (I) j

[using eq. (2)]

2x2 = 4

x2 = 2 x = ±ll2

Putting x = ± "VT in eq. (2), we get y = 2.Thus the common points are (V2", 2) and (- VT, 2).

Now from eq. (1), we get

=>

-2x ...(3)dxSo, mx ~ slope of the tangent to eq. (1) at (V^, 2).

Jdl = -2'l2 [using eq. (3)]dx A>/2,2)From eq. (2), we get

^- = 2x ... (4)dxAnd, m2 = sope of the tangent to the curve eq. (2) at ('ll, 2)

= (J)V2.2) = 2VIIf 0 be the angle between the curves at ('ll, 2) then,

[using eq. (4)j

•* - .mi - m2

tan 0 =1 + mjm2 -I'll-I'll

1-8- 4'll 4'll=-7 7

4'll-i0 = tan7

24 Self-Industrial‘Material

Tangents and Normals• SUMMARY

dx• If y = /(*), then the slope of the tangent to y =f (x) at (jcj , y,) is m

• Equation of the tangent to the curve y = f(x) at the point(x1)>'1) is. \

• If y = f(x), then the slope of the normal to the cuve y =f(x) at (j;b yi) is given by* ✓ . i

mx--■ {dy fayd

• Equation of the normal to the curve y=f{x) at the point (*i, yO is(dx

dxt

(x-Xi)y-y] = - dy k.jj)J!• STUDENT ACTIVITY

2 .21. Find the equation of the tangent to the curve — + .

a22L = 1 at the point (a.cos 0, b sin 0).

2 22. Find the angles of intersection of the curve y=x-x and y = x.

• TEST YOURSELF

1. Find the slope of the tangent to the following curve :(i) y = (2x2 4- 3sin x) at ^ = 0(iii) y = sin2 x at x = n/4

2. Find the equation of the tangent and the normal to the given curves at indicated point:

(b) xy = c at (c, c)

x2 v2(ii) — - ^ = I at (a sec 0,6 tan 0).

a* b

(ii) y = (sin 2* + cot .x + 2)2 at x = tc/2

, v 2 A *( 2fl(a) y = 4ax at —r, — * tny m

-+^2 1.2 a b

4. y2 = 4ax at (at2,2at)

3. (i) ^ = 1 at (a cos 0, b sin 0)

Selfrlndustrial Material 26

25. Find the equation of the normal to the curve y = (sin 2x +cot r+2) atx = K/2....

6. . Show that the tangents to the curve y = 2x - 4 at the points x = 2 and .v = - 2 are parallel. Show that the curves xy^a2 and x2 + y2 = 2a2 touch-each other.

8. Find the angles of intersection of the curves x -y =a andx +y =<3 \2.

Calculus <£ Geometry

7.

ANSWERS

1. (0 3 (ii) - 12 (iii) 12. (i) m2x - my + a = 0, m2x + rny - 2am2 - a = 0 (ii) x + y- 2c = (,y = x3. (i) i»x cos 9 + ay sin 0 = ab, ax sec 0 - by cot 0 = a2 + 62

(ii) hx sec 0 - ay tan 0 = ab, ax ax cos Q + by cot Q~a2 + b24. x-ty + at2 = 0, tx + y = ati + 2at5. 24y - 2x + 7t-96 = 0 8. 71/4.

□□□

26 Self-Industrial Material

'CurvatureUNIT

5CURVATURE

STRUCTURE• Definition of Curvature• Formula for Radius of Curature (Cartesian form)

Radius of Curvature at the Origin• • Summary

Student Activity# Test Yourself


# How to calculate the radius of curvature of the given curve at a given point

• 5.1. DEFINITION OF CURVATURE . 4, I P

Let P, Q be two negihbouring points on a curve AB. Also, let AP = s, arc AQ = s + Ss and arcPQ = 8s. T-

Let the tangent to the curve at points P and Q makes angle and \[/ + 5\|r respectively with a fixed line say X-axis, then

B T

(i) The angle 8\\f through which the tangent turns as its points of contact travels along the arc PQ is called the total bending or total curvature of arc PQ.

A E

§1 is called the mean or average00 The ratio „05

curvature of arc PQ.{Hi) The limiting value of the mean curvature when

Q tends to P is called the curvature of the curve at the point P. Therefore, the curvature K at point P is

■\

X' >X0Y'

Fig. 2S\|/ <Aj/ 85 dsUrn Urn

q^p 8s(iv) The reciprocal of the curvature of the given curve at P, (provided this curvature is not

equal to zero), is called the radius of curvature of the curve at P. This is denoted by p1 ds

P~K~d\v'

& 0

• 5.2. FORMULA FOR RADIUS OF CURVATURE (CARTESIAN FORM)Let y ~f{x) be the equation of curve. Then the radius of curvature P is given by

^ VT372 idy 11 + dxP =

dx2

• 5.3. RADIUS OF CURVATURE AT THE ORIGINLet the curve y =f{x) passes through the origin. Then, we may use the following methdos, to

find the radius of curvature.dy d~y(i) Method of Direct Substitution. Since y =fx) be given. Calculate the values of -f and —-

dxAat origin and then use the following formula


\2-|3/2Calculus <& Geometry dl1 + dx\/P = d2y/dx2

(ii) Method of Expansion. Let y =f{x) be the equation of curve. Since, it passes through the origin, therefore/(O) = 0. In this case we use the following formula :

a+ftV'2p=P2

where pi = ^ and p2 = ^

(iii) Newton’s Method. If a curve passes through the origin, and axis of x is the tangent at the origin, then radius of curvature p at origin

x2lim*-*0 tyy —^ 0

REMARK

>• If a curve passes through the origin and axis of y is the tangent, then radius of

i.curvature at the origin is given by = limX-*Q 2* y—>0

SOLVED EXAMPLESExample 1. Find the curvature of the curve x3 +y3 = 2>axy at the point (3a/2, 3a/2). Solution. Here, we have the equation of the curve is

x3 +y3 = 3axy.Differentiating w.r.t. x, we get

3x2 + 3y2^=3ay + 3ax^- dx dx

'SIUCVV"dy

-(ii)=> = ay + ax , dxdx

dy _x2 - ay=s>

dx 2ax-y(i \dy = - 1=>dx h 3 '

From (ii), we havedy *£i=ady .dy2x + 2y +y + a + ax

dx2 dx2dx dx dx

ii dy dy{ax-y2) = 2x+2y -2a ...(.iii)=>dx2 dx dx\ /

_ 3a 3a .Putting x = —, y = — and dy = - 1, we getdxha 3n

2’ 2

h 32 }_ 3 adx2 3a 3a

2' 23^ 3^^

Hence, the radius of curvature p at —, — , is given by

/ , \213/2_,1+fel r idx

(1 + 1)3/2 3a\ /p = 32 i_

3 a8V2"

dx2 (3a 3a-Mt-t% Self-instructional Material

1 8V2 CurvatureTherefore, the curvature = — (By ignoring the negative sign)

Example 2. Apply Newton’s formula to find the radius of curvature at the origin for thep 3a

curvex3 - 2x2y + 3jcy2 - 4y3 + 5x2 - + 7y2 - 8y = 0.

Solution. Since, the curve passes through the origin. Equating to zero, the lowest degree terms, we may find y = 0

=$x axis is the tangent at the origin.

Therefore, by Newton’s formula, p at (0, 0) = lim —•x~*o ty y-*Q

Dividing the equation of the curve by 2y, we get2 I 2 ' 7x-^-~x2+^xy~2y2 + 5-~-3x + ~y-4 = 0.

2y 2 ’ * 2y 2yTaking lim x —» 0 and y 0, we get

x2 45 lim ~4 = 0=>5p-4 = 0=>p = ^x o - ^y 5y-»0

• SUMMARY• If y =/(x), then the radius of curvature is given by

2-3/21 +

dxp = -

dx21Curvature = —.P

• If a curve passes through the origin and x-axis is the tangent at the origin, then the radius of curvaure p at the origin is

limP = 2y(x.X)-»(0,0)• If a curve passes through the origin and y-axis is the tangent at the origin, then the radius of

curvature at the oriin is given by l2 \2LlimP = 2x /fry)-*(0.0)

• STUDENT ACTIVITY. 2a ^1. Find the curvature of the curve x3 + y3 = 3oxy at the point I — , —

/

2. Find the radius of curvature at the origin to the curvex3 - 2x2y + 3xy2 - 4y3 + 5x2 - 6xy + 7y2 - 8y = 0.


Calculus & Geometry

'1 *1*4

• TEST YOURSELF1. Find the radius of curvature of the following curves :

(0 v/2+,,/2 1/2mJ*--3” -2/3

. (v) •n£T+ Vy = 1 at .

(vii) ay2 = jc3.(viii) y = ^ at the point where it cuts the y-axis.(ix) x2/3 + y2/3 = a2/3 at (a cos3 6, a sin3 0).

71(x) y = 4 sinx - sin 2x at x = —•4->

2. Find the radius of curvature at the origin of the following curves :(ii) y = x3 + 5x2 + 6x

(iii) 5x3 + 7y3 + 4x2y + xy2 + 2x2 + 3xy + y2 + 4x ^ 0(iv) a (y2- x2) -x3

(ii)a2y = x3 - a3.- a(iv) x”1 + ym = 1.+y = a

(vi) s~4a sin \|/ at (s, \|/).

(i) x3 + y3 = 3axy

ANSWERS

2(x + y)3/2 (a4 + 9x4)372 (iii) 3a1/3x1/3-1/31. (0 (ii) y1/2 6a4xa2m-2 2m - 20/2)(PC 1 1+ y (vii)-f-(4a + 9x)3/2x1/2

oa(iv) (v)(1 - m) xm~2ym~2

5V5(viii) V§" (ix) 3nsin0cos0 w -r-„ ,.s 3a 37 ^!W2‘ (l> T (u)

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. The curvature of the curve at any point P is defined as the

at P.2. For a curve y =f(x), the radius of curvature p =...........> TRUE OR FALSE :Write ‘T’ for True and ‘F’ for False :1. The curvature of the curve at any point P is defined as the reciprocal of the radius of curvature

(T/F)'2. If y axis is the tangent to the given curve at the origin, then radius of curvature at the origin

(iii) p = 2 (ignoring negative sign) ' (iv)2aVJ

of the radius of curvature

of P.

2)L.is equal to lim

x->0 2x3. The curvature of the circle and circle of curvature, both are the same.

(T/F)

(T/F)



The radius of curvature of the curve the point where it crosses the y-axis is :(b) V2 (c) 2-fl

2 For the curve xy = a2 the radius of curvature at (2, 2) is :(b) 16

Curvature

(a) 2 (d) 1.i

(a) 4 (c) 10 (d) None of these.

ANSWERS

Fill in the Blanks :(i +y\2)2/2 iX. reciprocal 2. ^

True or False:l.T 2. T 3. F

Multiple Choice Questions : l.(c) 2. (d)

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. if

Self-Instructional Material ^31

U N I T' r;Calculus & Geometry

6ASYMPTOTES

STRUCTURE• Asymptote• Determination of Asymptotes• Asymptotes of General Equation• Number of Asymptotes of a Curve• Asymptotes Parallel to Co-Ordinates Axes

• Test Yourse!f-1• Test Yourself-2• Summary• Student Activity• Test Yourself-3

v •

LEARNING OBJECTIVESAfter going through this unit you will loam :

# How to determine the asymptotes of the given curves, which are helpful to draw the given curve

• 6.1. ASYMPTOTEIn calculus, there are some curves whose branches seem to go to infinity. It is not necessary

that there always exists a definite straight line for all such curves which seems to touch the branch of the curves at infinity but more or less there are some certain curves for which this type of definite straight line exists, this straight line is therefore known as asymptote.

Definition. A definite straight line whose distance from branch of the curve continuously decreases as we move away from the origin along the branch of the curve and seems to touch the branch at infnity, provided the distance of this line from origin should be finite initially, is called an asymptote of the curve.

More than one asymptote of a curve. Suppose in the equation of a curve, two or more than two values of y exists for every value of x, then we obtain different branches of the curve corresponding to these distinct values of y. If each branch have its own separate asymptote, then we can say that a curve may have more than one asymptote.

• 6.2. DETERMINATION OF ASYMPTOTESLet us consider a curve

Ax, y) = 0and also consider that there are no asymptotes parallel to y-axis. Thus we shall take the equation which is not parallel to y-axis, in the form of

-d)

...(2)y = mx + c.Let us take a point P(x, y) on the curve (1), therefore this point as tends to infinity along the

straight line (2), x must tend to infinity. Now find the tangent to the curve fix, y) = 0 at the point P(x,y).

The equation of tangent at P(x, y) is

= & dy v ( dy X+ y-x-^-(X~x) or ...(3)^ &v j

The equation (3) is of the form y = mx + c so in order to exist the asymptote of the curve there

must both

dx

di tend to finite limits as x tends to infinity. Therefore, if the equationand y-Xdxdx


ti

(3) tends to the straight line given in (2) as * tends to infinity, then the line (2) will be an asymptote of the curved, y) = 0 and also we have

m = lim x-*-

c = lim\

Since c is finite, then we have

Asymptotes

dz (slope of the asymptote)" dxdiand y-x dx

it

diy-Xdx ■ 2-f=0

^ x dx.= lim - = 0 orlim limxX —^ 00

lim ^ = lim or. lim ^- = m.«» x

or.«o dx -rXx—*°° x—>\ /

di\ ■it

Also c= limX —> » y~xdx i

c = lim (y - mx).X—» »

Hence, if y = mx + c is an asymptote to the curve 7(x, y) = 0, then we obtainlim * I

oo X \m = lim

oo dxIc = lim (y - mx).and

x »

• 6.3. ASYMPTOTES OF GENERAL EQUATIONLet/(x, y) = 0 a general equations of a curve of degree n. To find theiasymptotes y = m + c,

we proceed as follows : ■' (i) Put y = m and x = 1 in n* degree terms (highest degree) of/(x, y) = 0 and we obtain an

expression (|)n (m).(ii) Find the values of m by solving equatin 4>n (w) = 0.We have two cases:Case I : If all the values of m are distinct, then we calculate the values of c by using the

formulac([>'/,(m)+(l)n-1(/i)=0

Case II : If r values of m are identical, then we calculate the values of c for these identical values of m by using the formula

-1crcr -hv.C-I («) +77 <('") + (m) = 0.(m) + (J)n_r+ -r+ 11 !(r-1)!r!

• 6.4. NUMBER OF ASYMPTOTES OF A CURVESuppose the degree of an algebraic curve is n, then we find a polynomial (J)n(m) by putting

y = m and x = 1 in the n'h degree terms of the curve. Thus the equation (f>„ (w) = 0 is of degree n in m and which gives almost n values of m real as well as imaginary. These n values of m are nothing but the slopes of the asymptotes, which are not parallel to y axis. If there are some asymptotes, parallel to y-axis, then the degree of (J)n (m) will be smaller than n by the same number of parallelasymptotes. Suppose all the roots of <t>„(m) = 0 are distinct and real, then to each value of m we obtain one value of c. Hence, we obtain n asymptotes. In case, there some roots say r (out of n) of <t>n(m) = 0 are same, then we can find the values of c for these same roots by the following equation

c'-177 < (*«) + <fc_j(m) + ... + <|>B_r(»0 = 0.

r-1!This equation in c is of degree r so we get r distinct values of c for the same roots, hence,

again we obtain n asymptotes. Therefore we can say that the total number of asymptotes of a curve are equal to the degree of the curve. These asymptotes are real as well as imaginary but we have required only real asymptotes so we ignore all the imaginary asymptotes, j

Self-Instructional Material. 33

1

Calculus & Geometry • 6.5. ASYMPTOTES PARALLEL TO CO-ORDINATES AXES(a) Asymptotes parallel to *-axis. Let the general equation of an algebraic curve in

decreasing powers of x be

/ (})(y) + / -+ / - 202(y) + ... = 0 -.(I)where <J>(y), foCy) • • • are the function of y only.

To find the asymptotes parallel to *-axis we equate the coefficient of highest power of x in the given curve to 0.

(b) Asymptotes parallel toy-axis. Similarly, we may obtain the asymptotes parallel to y-axis by taking the coefficient of highest power Of y in the equation of the curve equal to zero.REMARK

>• If the coefficient of highest power of x or y or both are constant, then no asymptotes parallel to either x or y or both axis exists respectively.

SOLVED EXAMPLESExample 1. Find the asymptotes of the curve x3 + y3 - 3axy = 0.Solution. Obviously, the degree of the curve is 3, so it will have 3 asymptotes real as well

as imaginary. Here the coefficient of highest degree in x and y are constant so no asymptotes parallel to co-ordinate axis exist. Let

-(1)y = mx + cbe the asymptote of the curve.

So putting y = m and x = 1 in the highest degree terms of the curve, we get

03(m) = 1 + m3.

Solving the equation 03(m) = 0 i.e., 1 + m3 = 0

(1+m) (m2 ~ m + 1) =0 or m = - 1

is only real root and other two roots are imaginary so ignore them.Next, putting y = m and x = 1 is second degree terms in the equation of the curve (1), we get

$2(m) = - 3am.Now we find value of c by the following equation

+ or c <t>3'(m) + ^ 0c [3m2] + (- 3am) = 0

or

[v b3(m) = 1 + m3, <J)3'(w) = 3m2]orwhen m = - 1, then

c[3(-l)2] + [-3a(-l)] = 0 3c + 3a = 0 or c = ~a.

Hence, the asymptote isy = -x-a or x + y + a = 0.Example 2. Find all the asymptotes of the curve x3 + x2y - xy2 - y3 - 3x - y - 1 =0.Solution. The degree of the curve is 3 so it has 3 asymptotes which are real as well as

imaginary. Since the coefficients of highest degree i.e., 3rd degree of x and y are constant so there are no asymptotes parallel to co-ordinate axes. Thus there are oblique asymptotes of the form y = mx + c.

Now putting y~m and x = 1 in the third degree terms of the curve, we get2 3<J)3(m) = 1 + m - m - m .

Solving the equation <>3(m) = 0 i.e., 1 + m - m2 - m3 = 0, we get

(1 + m) (1 -m2) = 0 or m = 1.

Determination of c. For m = .1, we use the following equation

c tyn' (m)+ - \(,n) - 0 or c ^'(m) + 4>2<m) = 0-

Putting y = m and x = 1 in the second degree terms of the equation, we get

Mm) = °-From (1), we get\


c [ 1 - 2m - 3m2] +0 = 0 Asymptotes

at m = 1

c(l-2-3) + 0 = 0 or -4c = 0 or c = 0.Thus one of the asymptote is y = *.Determination of c for m = - 1, - 1. Since two out of three roots of the equation ^(m) = 0

are same, then we use the following formula to determine c2 •

^7 ^"("O + T7 {t)2,(^) + <t> i (m) = 0. ...(2)1 !

Putting y = in and a: = 1 in the first degree terms of the equation we obtain <|>i(m) = - 3 - m. From (3), we have

2“ (- 2 - 6m) + ~ ■ 0 + (- 3 - m) - 0

at w = - 12

I~ (— 2 + 6) — 3 + 1 = 0 or 2c2

Thus other two asymptotes are y = -*+l,y = -;t-'l. •Hence, all the asymptotes of the given curve arey = ;t, * + y- l = 0, jt + y+ l=0.

-2 = 0 or c = ± 1.

TEST YOURSELF-1Find all the asymptotes of the following curves :1. a2/x2 - b2/y2= 1.2. a2/x2 + b2/y2=l.3. y2 (a2 - x2)= x*.4. *y = n2 (.v2 + y2).

- j:2y - A:y2 - y + 1 = 0.6. 3x3 + 2x2y - Ixy2 + 2y3 + 14:cy + 7y2 + 4* + 5y = 0.7. 2-t3 - A'2y - 2:cy2 + y3 - 4.v2 + 8-xy - 4x + 1 = 0.8. jt3 + 2jy2y + xy2 — X2 xy + 2=0. - - ■9. y3 - 5xy2 + &x2y ~ 4x_ - 3y2 + 9xy - 6x2 + 2y - 2x + 1 = 0.10. y3 - x2y - 2;ty2 + 2x3 - 7*y + 3y2 + Tx2 + 2x + 2y + 1 = 0.

ANSWERS

2 2. 5. * y

2. x-±a,y = ±b 4. x = ±a\y = ±a6. a: + 2y = 1, 2jc - 2y = 7, 7a: - 6y = 15

1. x-±a 3. x = ±a5. y = 0; y = 1; a: = 0; a: = 17. jr + y- 2=0;A:-y + 2 = 0;2A:-y-4 = 08. A: = 0;A:+y = 0;x + y- l= 0 9. A:-y = 0;2A:-y+2 = 0;2x-y+l=010. A:-y-l=0;x + y + 2 = 0;2A:-y = 0

t

• SUMMARY• If y = mA: + c is an asymptote to the curve f(xt y) = 0, then

& = lo dx

c = lim (y - mx)

• The asymptotes parallel to Ar-axis are obtained by putting the coefficients of highest power of x in the quation of the given curve f(x, y) = 0 to zero.

• The asymptotes parallel to y-axis are obtained by putting the coefficients of highest power of y in the equation of the given curve/(A:, y) = 0 to zero.

1m = lim lim xJT —» «and


Calculus & Geometry • If .y = wjc + c is in asymptote to the given curve f(x, y) = 0, thenTo dtermine m : Put y = m and * = 1 in the highest degree terms say n of the curve so we get <[)„ (m). The roots of <)>„ (m) = 0 gives the values of tn.To determine c : For r identical values of m we use the following formula to find c

<tCl("0 + ---+4>n-r('«) = 0-1cr^7 K (m) +

(r-1)!r!

• STUDENT ACTIVITY1. Find the asymptotes of the curve x +y - 'iaxy = 0.

• i

2. Find all the asymptotes of the curvey3 - ry2 - x2y + + x2 - y2 - 1 = 0.

• TEST YOURSELF-2OBJECTIVE EVALUATION > FILL IN THE BLANKS :1. If y = mx + c is an asymptote of the curve^x, y) = 0, then m =

• 2. The equation <}>/i (m) = 0 gives theand c =

of the the asymptotes.If one or more values of m obtained from <j>/i(m) = 0 are such that = 0 and.<])/,- i(/h).3.then the asymptotes .

> TRUE OR FALSE :\Write ‘T’ for True and ‘F’ for False :1. The line y = mx + c is an asymptote of the curve

ABCy = mx + c + — + ^r + —r+... .X X2 X3

2. The. polynomial 4>«(m) is obtained by putting y = m and x = m in-the degree terms-of the.(T/F)

\

(T/F)

curve.31 If y = mx + c is an asymptote of the curve/(x, y) = 0 then

Zin ~ limx—» oo X (T/F)\ /

> MULTIPLE CHOICE QUESTIONS :Choose the most, appropriate one :1. If y = mx + c is an asymptote of the curve/(x, y) -0,.then lim (y/x) equals :

(b) m (d) -c.

(a) c(c) -m


Asympimcslim (y-m*) equals ::2. If y - mx + c is an asymptote of the curved*, y) = 0, then(b) -c (d). -m.

a: —»y/x -»m(a) m(C) C

(ANSWERS i

■rFill in the Blanks :1. Hm y/x, lim

y/x—* m(y - mx) 2. Slopes 3. Will not exist

X—»«•

True or False:1. T 2. F 3. T

Multiple Choice Questions: 1. (b) 2. (c)

1 *

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Calculus <fi Geometry UNIT **

£??:3(/gHA !i

-/SINGULAR POINTS AND CURVE TRACING•v. i

r

STRUCTURE: '• i

•itr * I• Concave and Convex Curves• Point of Inflaexion• Formula for Finding the Point of Inflexion

• Test Yourself-1Multiple Points and Singular Points

• Types of Double Points• Tangents at Origin• Position and Nature of Double Points

• Test Yourself-2• Curve Tracing

• Test Yourself-3• Summary• Student Activity• Test Yourself-4

tii

i

ii


About the singular points on the curve. # How to trace the given curve.

• 7.1. CONCAVE AND CONVEX CURVESDefinition. If P is any point on a curve and CD is any given line which does not passes

through this point P. Then the curve is said to be concave at P with respect to the line CD if the small arc of the curve containing P lies entirely within the acute angle between the tangent at P to the curve and the line CD and the curve is said to be convex at P if the arc of the curve containing P lies wholly outside the acute angle between that tangent at P and the line CD. Which are shown in figures below :

7 A

O oConcave Convex

Fig. 1i i

• 7.2. POINT OF INFLEXION y*

Definition. A point P on the curve is said to be the point of inflexion, if the curve in one side of P is concave and other side of P is convex with respect to the line CD which does not passes through the point P as shown in fig. 2.

Inflexion tangent. The tangent at the point of inflexion of a curve is said to be inflexion tangent. In the fig. 2 the line PQ is the O inflexion tangent. Fig. 2

St8 Self-Instructional Material

Singular Points and Curve Tracing• 7.3. FORMULA FOR FINDING THE POINT OF INFLEXION

= 0 but * 0. dx3

\We can have a point of inflexion at P, ifdx2

SOLVED EXAMPLESExample 1. Find the points of inflexion of the curve

x = {logy)3.Solution. Since the equation of the curve is

x = (log>’)3.

Differentiating (1) with respect to *y\ we get...(1)

O

Fig. 3

2 log y (log y)2 J 2 2

Again differentiating w.r.t. y

d2x...(2)

dy2 y yt.

Again differentiating w.r.t. y, we get

2 _ 4log y _ 2logy 2(logy)2. i “ ^ 3 .3 3 + ..3

•••(3)dy3 y y_y y

For the point of inflexion, we haved2x̂ = 0.dy2

2"2 log y - (log 3') J . 2 = 0 or 3 (log y) (2 - log y) = (

y . *logy = 0, logy = 2 or y=l,y = e2.

From (3) it is obvious that at y = 1, y = e2, ^4 ^ 0.dy

Hence, the points of inflexion are (0,1) (8, e ).

Example 2. Find the points of inflexion of the curve y = x (x + 1) . Solution. Since the equation of the curve can be written as

y = (x + 1) ^x.

Differentiating (i) w.r.t. ‘x\ we get

= 1 1/2dx 2

or

...(1)

1+2Vx

Again differentiating w.r.t. ‘x’d2y _ 3____dx2 4'ifx 4xV2

1 •••(2)

and again differentiating w.r.t. 'x\ we get

. dx* 8x3/2 8x5/2For the point of inflexion, we have

-(3)

& = 0.dx2

1 * ?>3 1 3-— =0 or x= 1/3.= 0 or4Vx 4xVx x

h 9t0.From (3) it is obvious that at x = 1/3 dt3


Thus, the point of inflexion are given by (1/3, ± 4/3^3).Calculus <& Geometry

. ;• TEST YOURSELF-11. Find the points of inflexion of the curve x = \og(y/x).2. Find the points of inflexion of the curve y (a2 + = x3.3. Find the points of inflexion of the curve y = (x - l)4 (x - 2)3.4. Find'the points of inflexion of the curve xy - a2 log (y/a).

ANSWERS

2. (0,0), ,

3, Point of inflexion at x = 2, (11 ± V2)/7. 4. {^ae~3/2,aeJ/2

41. (- 2, - 2/e2).

• 7.4. MULTIPLE POINTS AND SINGULAR POINTSDefinition. A point on the curve is said to be multiple points if through this point more than

one branches of a curve passes.Definition. A point on the curve is called a double point if through it two branches of the

curve passes.Definition. If three branches of the curve passes through a point, then this point is called

triple pointDefinition. If n branches passes through a point on the curve, then this point is called a

multiple point of nth order. ‘ JDefinition. The point of inflexion and multiple points are also called the singular points.

r;>

OrAn unusual point on the curve is basically called a.singular point.

♦ 7.5. TYPES OF DOUBLE POINTS(i) Node. A double point on a curve is said to be a node, if / f

through this double point two branches of the curve passes which are real and having two different tangents at that point. (Fig. 4)

->,VoFig. 4

(ii) Cusp. A double point on a curve is called a cusp if through this double point two real branches of the curve passes and have real coincident tangents at that point. (Fig. 5)

>

*.VoFig. 5

V (iii) Conjugate point. A point P on the curve is said to be conjugate point if there are no real points on the curve in the neighbourhood of that point and having no real tangent at that point. (Fig. 6)

oFig. 6

• 7.6. TANGENTS AT ORIGINThe nature of a double point depends on the tangents so we find the tangent or tangents there.

// a curve passes through the origin, then the equation of the tangent or tangents at the origin are obtained by equating to zero the lowest degree terms in the equation of the curve.


Singular Points and Curve Tracing •• 7.7. POSITION AND NATURE OF DOUBLE POINTS

Thus the necessary and sufficient condition for any point of the curve fix, >')-= 0 to be a multiple point are that

i£=0ss3£.3* dy

The double point will be node, cusp or conjugate point according as

<f£dx2 ' dy2>, = or <dx dy

REMARK^ d2f d2f d2fIf ^ ’ g g ’ are all zero, then the point P(x, y) will be a multiple point of order

greater that two.

SOLVED EXAMPLESExample 1. Show that the origin is a node on.the .curve x + y -3axy = 0.Solution. The tangent at the origin are obtained by equating to zero the lowest degree terms

i.e., second degree term in the given equation of the curve.- 3axy = 0 or x = 0, y = (X

Thus at the origin there are two real and distinct tangents. Hence (0, 0) is a node.Example 2. Find the double point of the curve (x - 2)2 = y (y - l)2.

Solution. Let /(x, y) = (x - 2)2 ~y (y - l)2 = 0.

Differentiating (1) partially w.r.t. x and y, we get

dx = 2<-X~2)

fy = -(y-l)2-2y(y-i).Since the necessary and sufficient condition for a double points are

^ = 0,^0,

-(y-l)2-2y(y-l) = 0.

...(1)

...(2)

•••(3)and

...(4)/. 2 (x - 2) = 03x 3y

...(5)

Now solving fix, y) = 0, = 0 and = 0 simultaneously.3x 3y

.•. From (4), we get x = 2 and from (5), we get -O-l)(y-l+.2yJ-0 - (y - 1) (3y - 1) = 0 or y=l,y=l/3.

Possible double points are (2,1) and (2, 1/3).But (2,1/3) does not satisfy/(Xy) = 0. Hence only double point is (2,1).Example 3. Examine the nature of the origin on the following curve i:

y2 = a2x2 + &x3 + cxy2.Solution. The given curve is

fix, y) -y1 ~ n2x2 - 6x3 - cxy2 - 0.Equating to zero the lowest degree terms in the equtaion of curve (1) we get

y2 -a2x2 — 0 or y = ±ax.Thus we have obtained two real and distinct tangent at (0,0). Hence (0, 0) is a node.

or

-(1)

• TEST YOURSELF-21. Find the equations of the tangents at the origin to the following curves :

(a) (x2 + y2) (2a - x) = b2x (b) a4y2 = x4 (x2 - a2)(c) x4 + 3x3y + 2xy~y2 = 0 (d) x3+y3 = 3oxy.

2. Examine the nature of the origin on the curve (2x + y)2 - 6xy (2x + y) - 7x3 = 0..


ANSWERSCalculus <4 Geometry

(d) x = 0, ,y = 0(c) y = 0, 2x — = 0(b) > = 0,}< = 01. (a) 'x = 02. Origin is a single cusp of first species

• 7.8. CURVE TRACINGCartesian Form of Equations :

To trace any curve of cartesian form we should apply following process :(a) Symmetricity. In order to find the symmetry of the curve we should apply following

rules :(i) If the powers of y in the equation of the curve are all even, then curve is symmetrical

about x-axis.(ii) If the powers of* in the equation of the curve are all even, then the curve is symmetrical

about y-axis.(iii) If the powers of x as well as y in the equation of the curve are all even, the curve is

symmetrical about both axes.(iv) If the equation of curve remains unchanged when jc is replaced by - ^ and y is replaced

by - y, then the cure is symmetrical in opposite quadrants.(v) If the equation of the curve remains unchanged when x and y are interchanged, then the

curve is symmetrical about the liney =x.(b) Nature of the origin on the curve. If the curve passes through the origin, then find the

tangent at (0,0) by equating to zero the lowest degree terms of the curve. If we obtain two tangent at the origin, then origin will be a double point and then find the nature of this double point.

(c) Intersection of curve with co-ordinate axes. We should check whether the curve cuts the co-ordinate axes or not, for this put y = 0 in the equation of the curve and find the values of x. then we get the points at which the curve cuts the x-axis. Similarly if the curve cuts the y-axis, then put x = 0 in the equation of the curve and obtain the points on the y-axis. Hence in this way we obtain the points of intersection of the curve with co-ordinate axis. Thereafter we should find the tangents at these points of intersection. For this first we shift the origin at these points and then obtain the tangent at these new origin by equating to zero the lowest degree terms in the new equation of the curve. On the other hand the value of dy/dx at these points of intersection can also be used to find the slope of the tangent at that point.

(d) Nature of y or x in the curve. We should now solve the equation of the curve either for y or for x whichever is convenient. Suppose we solve for y and see that nature of y as x increases from 0 to + oo. Similarly see the nature of y as x decreases from 0 to - <» and finally collect those values of x for which y = 0 or y ^ or - <*>.REMARK

>. If the curve is symmetrical about x-axis in opposite quadrants then there is no need to take the values of x of both positive and negative. We can take only positive values of x to see the variation in y.

(e) Regions in which curve does not exist. In order to find the regions where the curve does not exist we should solve the equation of curve for one variable in terms of the other. Therefore, the curve will not exist for those values of one variable which make the other variable imaginary.

(f) Asymptotes. Next, we should find all the asymptotes of the curve because the branches of the curve approach to the asymptotes if they exist.

(g) Sign of dy/dx. Next, we should find the value of dy/dx from the equation of the curve and find the points on the curve at which dy/dx = 0 or dy/dx = °o. Therefore at these points we obtain the nature of tangents, suppose in any region a<x <b, dy/dx remains positive throughout, then in this region y increase continuously as x increases. On the other hand if dy/dx remains negative, they y decreases continuously as x increases.

(h) Special points. If necessary, we should find the some special point on the curve.(i) Points of inflexion. If necessary, we should find the point of inflexion to know the position

of the curves at that point.Now taking all above considerations in mind, draw an approximate shape of the curve.

SOLVED EXAMPLES2 3Example 1. Trace the curve y (2a - x)= x .


Singular Points and Curve TracingSolution, (i) Obviously the given curve is symmetrical about x-axisj.(ii) The curve passes through the origin and the tangents at the origin are obtained by equating .

to zero the lowest degree terms i.e., 2ay2 in the equation of the curve.2ay2 = 0 or y = 0, y = 0.

Thus at the origin we obtained two coincident tangents y = 0, y = 0 i.e., x-axis. Therefore (0,0) is a cusp.

(iii) From the equation of the curve it is obvious that the curve does not cut the co-ordinateaxes.

(iv) Now solving the equation of the curve for y, we gety2 = x3/(2a - x)

when x = 0, y2 = 0 and when x = 2a, thus y2 —> ©o thus x = 2a is an asymptotes of the curve.It is observed that y increases as x increases from 0 to 2a.(v) When x lies between 0 and 2a, y2 will be positive and the curve v ill exist in this region.

When x > 2a, y2 will- be negative so the curve will not exist beyond the line x = 2a. When x < 0,’ |

again y^ will be negative and thus the curve will also not exist for x < 0. Hence we can say that thecurve only exists in the region 0 < x < 2a. J

(vi) In order to find the asymptotes, putting y = m and x = 1 in the third degree terms in the equation of the curve, we get

(J>3(m) = m2 + 1.Therefore the equation = 0 gives both its roots imaginary so ignore them. Consequently

x = 2a is only the asymptote of the curve.(vii) Differentiating the equation of the curve

yf

3/2Xy = x=2aV(2a - x)

dy _ (3a - x) xl/2 (2a - x)

Ox'

we get 3/2

yvd£In the region 0 < x < 2a, will be positive, sodx. Fig. 7therefore in this region y increase continuously as x increases. |

Now taking all above points of consideration in the mind and draw the curve whose shape is shown in fig. 7.

Example 2. Trace the curve y2 (1-x2) =x2 (1 + x2).Solution, (i) In the equation of the curve the powers of both x and y are all even so the curve

is symmetrical about both axes.(ii) The curve passes through the origin. The tangents at the origin are obtained by equating

to zero the lowest degree terms in the equation of the curve.

y2-x2 = 0 or y='±x.

Thus there are two real and distinct tangent at the origin so (0,0) is a lode.(iii) From the equation of the curve it is clear that curve does not cut £ ny co-ordinate axes.(iv) Solving the equation of the curve for y, we get

2 _ X (I + X2)” (i-*2).'

. Whenx = 0,y2 = 0 and whenx = ± 1, y2—> ©©sox = ± 1 are two asymptotes parallel toy-axis.2 2(v) When - 1 < x < 1, y is positive, so the curve exists in this region. When x > 1, y will

be negaive thus curve will not exist beyond the line x = 1. Also when x < - il, y2 will be negative so that curve will not exist for x < - 1. |

(vi) In order to find the asymptotes', putting y = m and x = 1 in the fourth degree terms of thecurve, we get (m) = m2 + 1.

Solving 0w(m) = 0, we are only two real asymptotes. .

(vii) Since, we have

y

get both values of m imaginary so ignore them, [consequently x = ± 1


VCalculus & Geometry 2\l+Xy=x1-x2 '

2 NxV(l+x2)yi^x2 Vl+X2 +-^

VT + a:2- x

V i - x2..(1-x2)

(1 - x2) (1 + X2 4- X2) - X (1 + X2) (- x) (l-x2)3/2(l+x2),/2

2x2 + 1 -x4(l-x^fl+x2)172'

/ I*

is negative this means •x=-lWhen - 1 < x < 0 x= 1x'4 >xthat when x decreases from - 1 to 0, y decreases.

When 0 < x < 1 ^ is positive this implies that when

x increases from 0 to \,y increases.Now taking all the above facts in mind and

draw the shape of the curve. We geti r

Fig. 8

• SUMMARY

h = 0 and * 0. dx3

We can find a point of inflexion to the curve y =/(x), ifdx2

% = 0 = %For multiple points in /(x, y) = 0 we must have ^dy'

( d2f >(& £lMultiple point {a, b) is node ifdx2dx dy J(n, b)

J&dx dy J(d. i>) ^ dx2 fa’ ^ t dy2 ^

dx dy J(a, t)<( 9a-2 ]{«. b) ^ dy2

To trace any curve/(x, y) = 0, we must remember the following points :(i) Symmetricity(ii) Nature of origin (if/(x, y) = 0 passes through the origin).(iii) Points of intersection of f(x, y) = 0 with coordinate axes.(iv) Nature of y or x in /(x, y) = 0.(v) Regions in which curve does not exist(vi) Asymptotes

Multiple point (a, b) is cusp if

$LMultiple point (a, b) is a conjugate point if(«. b)

&(vii)Sign of

(viii) Points of inflexion (if they exist).dx '

• STUDENT ACTIVITY3 31. Show that the origin is a node on the curve x +y -3axy = 0.


Singular Points and Curve Tracing2* 22: Trace the curve ay =x (a-x).

• TEST YOURSELF-3Trace the foUowihg'curves:

2' 3'ay =x.y=xix2-\).y2(a:+x)=x2Xa-x).X1- +-/ ~X.

2. cP'y =-jc3.4; xy2 - 4a2 (2a-x).6. x2(x2-4a2)=y2(x2-a2). .8. x2y2 = a2'(x2 + y2).

1.3i.5;. i

7..

ANSWERS

•yA-

- 6>x'+.0

y'if y'+2.1. \

y<>. 4 K

N.S \

N(2a:0)Ox'4 >Xx<r(1,0)

/» ’S

,x y=-x

yi

. 4:

4^ ✓s

- i -y s ✓1^^

✓s/✓ (-2a, 0) s (2a, 0j:= -a r<y\

jc'-^0x'<r VJ' vS4

• Vsr Is✓ ss \v.y ssX=a \✓✓ i * V,

v6.5;

Self-Instructional'Material 45

* ' yCalculus <S Geometry

“iy~a

1(1,0)

x'<r O

c-T

vy’ <5

s.7.

OBJECTIVE EVALUATION > FILL IN THE BLANKS :

d3yll1. A point P(x,y) on the curve j will be a point of inflexion if = 0 but —- is not equaldx2 dxto

A double point is a node if at this point there are.....A double point is a cusp if at this point there are two

> TRUE OR FALSE:Write ‘T’ for True and ‘F’ for False :

distinct tangents. .....tangents.

2.3.

(f d2For the point of inflexion of the curve y = 0 but # 0.1.dx (T/F)

2. If we obtained two real and distinct tangents at the double point, then double point is a cusp.(T/F)

> MULTIPLE CHOICE QUESTIONS : Choose the most appropriate one :/

cFx d3xAt the point of inflexion of the curve x =f(y), —r = 0 and —- is not equal to :

dy ' s1.

dy.(d) None of these.

2 22. The points of inflexion of the curve y = (x- a) (x - b) lie on the line :(c) 3x + b = 4a

(b) 0 (c)-l(a) 1

(d) 3x = a. '(a) 3x + a = 4b (b) 3x + a = b

ANSWERSFill in the Blanks :

1. Zero True or False :

1. T 2. F 3. T 4. T Multiple Choice Questions :

1. (b) 2. (a)

3. Coincident2. Two

□□□

• V’ * T


Rectification of CurvesUNIT

8RECTIFICATION OF CURVES

STRUCTURE-• Formulae for Finding the Lengths of the Curves


>

LEARNING.OBJ|CTIV^■ j-After going through this unit you will learn :

# How to find the whole length of the given curve between the given points.

• 8.1. FORMULAE FOR FINDING THE LENGTHS OF THE CURVES(a) Let the equation of a curve be y =f{x) and let A and B be two points on this curve between

A and B, the length of curve is to be required. Let s be the length of an arc from a fixed point on the curve to any point on it.

Therefore, we have

V \2~/ . \2-| dz dyds dx1 + , ds = ± 1 + ~i~dx dxdxJ ->

where positive and negative sign will have to take according as x increases and decreases as s increases. Thus the length of an arc between the points A and B where &tA,x = a and at B,x = b is given by __________

f . '\2'ld£ dx. {a < b).1 +s dx(b) If the equation of the curve is * = fly), then the length of an arc between C and D is given

by/ \2-dx dy, (c < d).1 + ^s dyc- v. y

(c) If the equation of the curve is in parametric form, i.e., x-flt), y = g(r), then we have

V \ J \ |

h \2 & .[(dx)2ds . dt.or ds =+ dtdtdt dt

Thus the length of an arc between A and B where at A, r = ^ and at 5, t = t2 is given by\2 / « \2“dx dz dt.+s dt dt

REMARK>. If the curve is symmetrical about some lines, then in order to fird the length of an arc,

first find the length of one of symmetrical part and multiply this length by the number of symmetrical parts.

SOLVED EXAMPLESExample 1. Find the length of the arc of the parabola y2 = 4ax cut off by its latus rectum. Solution. Latus rectum. A line which passes through the focus of the given parabola and

perpendicular to the axis of that parabola.Here the equation of the parabola is y = 4ax whose trace is shown adjoining figure and in

the fig. LU is the latus rectum, the co-ordinates of L and V are respectively {a, 2a) and {a, - 2d).

we


Calculus & Geometry Since y2 = 4ax is symmetrical about the line OX. Therefore the required arc length - 2 x arc lengthOL.

7Ay2 = 4ax. y ~ 2Va" yfx.

dy _ 'la dx~ ^

Since

(0,0)*A'<■

OX'

dxNow arc length OL = 1 + dx

\ / r('.• At point O, x = 0 and at point L,x = a)

Fig. 1

-JTV fJo1+^ dx - dx>CTX

fJo 1“2 Jo1 2a: + 2a • 'x + a

dx = T dxylx2 + ax V(x2+ ax)

f2 Vo

pa .f dx

2 Jo• 1 (2a: + a) dx a

V(a:2 + ax)

(2x + a) dx a

+ -^(x2 + ax)

f2 Jo 2 Jo1 dx+ -

V(x2 + ax) V 2 W21•/aa: + —2 2

ina'l-(2V(Aoa:)Jo + | log x + f + vx2 + axJo

= aV2 + ^ log (3 + 2V2) = a^2 + % log (1 + V2)2.

Arc length OL = aV2" + a log (1 + V2).Hence the required arc length = 2 x arc length OL

=■2^2 a + 2alog(l +V2).

Example 2. FtW r/re whole length of the astroidx^ ̂= a^x-a cos2 B, y = a sin3 0.

Solution. The equation of the curve is2/3 , 2/3 2/3x +-y -a .

Since, the curve is symmetrical about both the axis, i.e., curve, lies in all the four quadrants as shown in fig. 2.

Therefore the whole length of the given curve = 4 x arc length of the curve in first.quadrant.

Since the co-ordinates of A and B are (a, 0) and (0, a) ^ respectively. Thus in the first quadrant x varies from 0 to a.

Now differentiating (1) w.r.t. x, we get

or

-.(1)

a y

B 0=7i/2 ! !c ^0 = 0

■> Xo a{a. 0)

Dy *

Fig. 21/3/ \1or

dx x

Therefore, the length of the curve in the first quadrant\2~ ,2/3d£

1 + dx = 1 + dxdx 2/3x .V /

r


I.

i

- 1/3

H&Ctification of Cun>t'\

dx [using (1)]

J- -a1 r2/32

1/3 dx = axn~aJO

■i

= a1/3 =!- .

Hence, the whole length of the astroid = 4 x = 6a.

2°

<r• SUMMARY(1?.J• Length of an arc from .* = ato;c = 6ofth curve y = f{x) is

*j^ dx

• Length of an arc from y = a to y = 6 of the curve x =/(y) is given by

dx'?1 + dys =dy

• Length of an arc from t = tt to t = t2of the curve x = x (t), y = y(t) is given by

dx'? dy+s = dt dt

• STUDENT ACTIVITYex-\1. Find the length of the curve y = log from jt = 1 to x = 2./+1 i

2 2 22. Find the whole length of the curve x +y = a .


Calculus & Geometry TEST YOURSELF2 2 21. Find the whole length of the curve x +y = b .

2. Find the length of the arc of the curve x2 = 8y from the vertex to an extremity of the latus rectum.

1 2 13. Find the arc length of the curve y = - x - - log x from x = l to * = 2.

4. Find the length of the arc from 0 = 0 to 0 = 271 of the curve •x = a (cos 0 + 0 sin 0), jy = a (sin 0 - 0 cos 0).

ANSWERS

3. | + jlog2 4.. 2a7t22. 2[^+log(l +V2)]

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. The curvey =f(x) is rectified betweenx~a andx = b if.f(x) is............2. The whole length of the arc of the curve x2 +y2 = a2 is...........3. If x = J(y), then the formula for finding the length of the curve between y = a and y = h is> TRUE OR FALSE :Write ‘T* for True and ‘F’ for False :

1" 2nb

1. The length of the loop of the curve x -t2,y = t ls2. The whole length of the astroid x2/3 + y2/ = a2/ is la.

(T/F)(T/F)

> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :1. The whole length of the curve x2 + y2 - 2ax = 0 is :

(b) 27tfl(a) na (c) Tia2 (d) 37la.

2. The whole length of the curve x = a cos3 t, y = a sin31 is :(b) 8a (d) 2a.

The length of an arc of the cycloid x = a (0 - sin 0), y = a (1 - cos 0) is : (a) 6a (c) '3a

(a) 6a (c) 3a

3.(b) 5a (d) 8a.

ANSWERS

Fill in the Blanks : rv dxT dy2. 27ta 3.1. Continuous 1 + =dy

True or False :1. T 2. F

Multiple Choice Questions :1. (b) 2. (a) 3. (d)

□□□


UNIT Atelt nf'Bounded Curves

fa

9AREA QF BOUNDED CURVES/

STRUCTURE• Equation.of Curve in Cartesian Form

• Summary• Student Activity• Test Yourself if


# How determine the area of the bounded regions, bounded by the co-ordinate curves and the given cun/es.

• EQUATION OF CURVE IN CARTESIAN FORMLetj =f(x) be a continuous curve in cartesian form and let A be the area of the region bounded

by the curve y =f(x), the axis of x and the two ordinates x = a and x = b. Then

Ax) dx.A =, * ..

Similarly, the area bounded by the curve x = J{y), the axis of y and the ordinates y = a and y = & is given by

xdy.A =

REMARK► If the given curve is symmetrical either about .x-axis or about y-axis or both axes, then

find the area of one of the symmetrical part and multiply this area by the number of symmetrical parts, we get the whole area of the bounded region.

> Area bounded by two curves = | Area bounded by one curve - Area bounded by othercurve | .

y. .SOLVED EXAMPLESExample 1. Find the area of the region bounded by the

line x = 2 and the parabola y2 = Sx.Solution. Since the equation of the parabola is

y2 =which is symmetrical about x-axis and the line x = 2 intersects the parabola y2 = 8x in two points (2,4) and (2, - 4) as show in

- fig. 1. ■B

fJO

Fig. 1The required area = 2 y dx.1 t

,4-"

4’ i2

Vs7 dx = 4'!2 |x3/2_o

32= — sq. units.

Example 2. Find the whole area of the ellipse 2 2

a2 b2= 1.


2 2Calculus & GeometrySolution. Since the ellipse ^ ^ 1 is symmetrical about both axes and whose trace is

a oshown in fig. 2.

fJo y dxRequired area = 4

4 W 2''1-^ *

a\

2 ^ rl-~ dx.= 4b y'a

Fig. 2Let us put x = o sin 0.dx = a cos 0 dQ and 0 varies from 0 to ti/2.

•- -- -n/2" • i» tl/2

JoJo cos2 0cos 0 . a cos 0 ^0= 4ab= 46•v R. A.

(2-1) n 2 2

(By Walli’s formula)= 4a6

.. = 7ta6 sq. units.Example 3. Find the common area between the curves y2 = 4ax and x2 = 4ay.Solution. Since the curve y2 - 4ax is

symmetrical about x-axis and the curve x2 = 4ay is symmetrical about y-axis. Both curves intersects at two points (0,0) and (4a, 4a) which are obtained by solving both equations of curves. The tracing of the curves is shown in fig. 3.

The required area

4'y

M (4a. 4a)

*xO B

y^=4ax

= area of OBACO\ r

- area of OBADO. ...d)Fig. 3

|.4a

Jo y dx, where y2 = 4oxNow area of OBACO =

r4*Jo

-Aa

V4ox dx = I'la j x3/2

y dx, where y = ^

32 2 = Ta

JOr 2and area of OBADO -

-Aa

Jo 114a 3

16a23jo

From (1), we getRequired area =â2 --â2 = -â2 sq. units.

• SUMMARY• Area of bounded region, bounded by the curve y =/(x) and x-axis between x = a to x = 6, is

given byf> pb

A = I ydx - IJa Ja

f{x)dx

• Area of the region bounded by the curve y =/(x) and y-axis between y-a and y = 6 is givenby


-f•fa Area of Bounded Cun>esf(x)dy

• Area bounded by two curves = j Area bounded by one curve - Area bounded by other curve |

• STUDENT-ACTIVITY2 y2+ iT= 1.1. Find the whole area of an ellipse

a* b2

2. Find the area common to the curves y2 = Aax and x2 = 4ay.

• TEST YOURSLF

1. Find the area of the region bounded by the following curves, and the axis of x and the given ordinates:(i) y = log x;(ii) y = c cosh (x/c)\

x = bx = a, x = 0. x — a

2(iii)>’ = sin x\ x = 0, x = ti/2.Find the area of the region bounded by the parabola y2 = 4x and the line y = 2x. Find the area of a quadrant of the ellipse x2/<z2 + y2/b2 = 1.Find the area of a loop of the curve xy2 + (x + a)2 (x + 2a) = 0.Find the area of the loop of the curve 3ay2 = x(x- a)2.Find the whole area of the curve a2*2 = y3 (2a - y).

2.3.4.5.6.

ANSWERS

1. (i) b\og(b/e)-a]og(a/e) (ii) c2sinh(a/c) (iii) Jt/4

3. 4. 2a2(I-7t/4) 5. 8a2/(15VT) 6. na212* 3

OBJECTIVE EVALUATION > FILL IN THE BLANKS ;s:1. The integral fix) dx represents

Seif ■‘Instructional Material 58

2. The area of the ellipse x2/a2 + y2/b23. The area of the curve X2 + y - lax - lay = 0 is.......4. The area common to the curves y2 = 4ax, x2 = 4ay is> TRUE OR FALSE :Write 4T’ for True ‘F for False :

Calculus A Geometry = 1 is

f fty) dy represents the area of a plane bounded by the curve x =J{y) the1. The integral

y-axis and the lines y = a and y = b.If die curve is symmetrical about x-axis, then the area of bounded portion is equal to twice the area of the upper portion of the curve above x-axis.

>- MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :

(T/F)2.

(T/F)

r1. The value of the integral J{x) dx if fix) = “’ is :

(c) 0 (d) 1.(a) log 2 (b) log (2-1)

f2. The integral fx) dy represents the area of bounded region between the lines :

(b) x= \, x = b (d) y = a, y = b.

3. The area bounded by the curve y = sin2 x, x-axis and the lines x = 0, x = n/1 is : (b) n/4

(a) x = a, x = b (c) x = - a, y = b

(a) n/1 (d) ti/3.(c)7t


1. Area of bounded region 3. nab 4. Ina2True or False :

1. T 2. TMultiple Choice Questions :

l.(a) 2. (d)□□□


,1 Surfaces and Volume of Solids of Revolution

f fhUNIT r

SURFACES AND VOLUMES OF SOLIDS OFREVOLUTION

STRUCTURE>•• Surface of Revolution

• Revolution about x-axis• Revolution about y-axis• Revolution about any Line• Surface Formulae for Different form of Equations

• Test Yourself-1• Volume of Solids of Revolution• Volume of a solid of revolution when the evolutions of the curve are in different Forms

• Summary• Student Activity• Test Yourself-2

f

xV

LEARNING OBJECTIVESAfter going through this unit you will learn : ^

• How to calculate the surface areas and volumes of solids on revolution.

• 10.1. SURFACE OF REVOLUTIONDefinition. When a plane curve is revolved about a certain fixed line which is lying in its

own plane, a surface is generated. This surface is called a surface of revolution. Also the fixed line is called the axis of revolution.

• 10.2. REVOLUTION ABOUT x-AXISLet S be the surface area (curved surface) of a solid which is generated by the revolution of

the curve y=f(x) about x-axis between the ordinates x = a and x = b and let s be the arc length measured from the point a,f[a) to any point P(x,y). Then

y ?

QP,5sf {a,f{a\A-r ds27Ty ds - 2ny • dx.dx

dmds DCand 5 is theWhere — =dx O x-a —> M N x=bsurface area of the solid of revolution of the curve y =j{x) about x:axis between x = a and x = b.

Fig. 1

• 10.3. REVOLUTION ABOUT y»AXISLet S be the surface area of a solid generated by the revolution of

y-axis between y = a and y = b. Then Sthe curve x = f(y) about

.-f f ds2nxds = 2nx — • dydy

ds and s the arc length being measured from the point ifia), a).where =dy

Self-instructional Material 55

Calculus & Geometry • 10.4. REVOLUTION ABOUT ANY LINE• v

Let S be the surface area of a solid generated by the curve abouf any line between certain points. Let $ be the arc length of the curve measured from one of the two given points to any point P on the curve and let Q be any point very near to P such that PQ = 8s. Now draw a perpendicular PM from the point P to the line of axis of the revolution. Then

S = \%n{PM)ds.

Here the limits of integration are taken the given points between them a solid is formed by the revolution.

• 10.5. SURFACE FORMULAE FOR DIFFERENT FORM OF EQUATIONSEquation of a curve in parametric form. Suppose the equation of a curve is given in

parametric form x y - g(t), where t is the parameter, then the surface area of a solid generated by the revolution of the given curve about x-axis between the suitable limits is

( , \2-where

/ JSimilarly for y-axis as the axis of revolution we may find the surface area.

REMARK>. If we have to find the surface area of a solid generated by the revolution of the curve

about any line, then use the following steps :

1. Take any point P on the curve between the given points.

2. Draw the perpendicular from P to the line of axis which meets the axis at the point M (say).

3. Find the perpendicular PM.

4. And use the formula given is § Revolution about y-axis.

SOLVED EXAMPLESExample 1. Find the surface of a sphere of radius a.Solution. The sphere is generated, if a semi-circle is revolved about its diameter. I^t the

equation of a circle of radius a is

x2 + y2 = a2.

2x + 2y^ = 0 or dx

...(I)

£t = _x..dx y c

\2- V 2ds , x 1+^

/ JBA O<■ ».v

[using (1)]

Let A (- o, 0) and B (a, 0) be the bounding points of the semi-circle as shown in Fig. 2.

Here the diameter is taken as x-axis. Let S be the

i i

%yN

✓

c^ *

surface area of the sphere, then Fig. 2

S=L 2nydi-dxds

fJ- ads__a dx y

2ny. — dxy

M Self-Instructional Material

Surfaces and Volume of Solids of Revolutionfj-a H,= Ilia dx = 271a

S = 47ta2.Example 2. Find the surface of the solid generated by the revolution of the curve

x = a cos31 and y = a sin31 about the x-axis.Solution. We have 4k y

5(0,6)x = a cos31 and y = a sin31. This curve is symmetrical about both the axes. A' A<- ■> xAt A(a, 0) t = 0 and at 5(0, a) t = n/2.

dx 2 ■— = - 3a cos t sin t dtdy . 2

. = 3a sin t cos t

O (a. 0)

NowB'

dt

V 2 \2ds &dx Fig. 3+dt dtdt

= 3a Vcos41 sin21 + sin41 cos21 3l2

= 3a sin t cos 7 = ~ sin 27.

Let S be the surface of a solid of revolution of the given curve about x-axis. Then•'3 3a , _ J

a sin 7 . ~ sin 27 dtn/2 ,~ ds ,

2ny — dt = 4nn/2L JoS = 2 dt

ft/2n/2 2 sin5 7 IZna2IJo- 127ta2 sin4 7 cos 7 dt = 127ta

5 5Jo

• TEST YOURSELF-11. Find the curved surface of a hemi-sphere of radius a.2. Find the area of the surface formed by the revolution of the parabola y = 4ox about the x-axis

by arc from the vertex to one of the latus rectum.3. For a catenary y = a cosh (x/a), prove that

aS = Tta (ax + sy)where s is the length of the arc measured from the vertex, S is the area of curved surface of the solid generated by the revolution of the arc about x-axis.

4. Find the surface of the solid generated by the revolution of the astroid x2 /3 + y x-axis.

2/3 2/3 , .= a about

ANSWERS

(2^2- 1) 4.81. 27ta2 O 2— 7ia2* 3

• 10.6. VOLUME OF SOLIDS OF REVOLUTION(a) Revolution about x-axis. Let V be the volume

of a solid which is generated by the revolution of a plane y a area bounded by the curve y =J{x), the ordinates x = a, x = b and x-axis about the x-axis. Then

B

Q'

v-f ATty2 dx x=b

Px=a

where y =f{x) is a continuous and single valued function defined on [a, b], L

(b) Revolution abouty-axis. Similarly, the volume ^ of a solid formed by the revolution of a plane area bounded

DC M N-

Fig. 4

Self-Instructional Material ,57

Calculus & Geometry by the curve x = fty) and the lines y-a and y = b and y-axis about y-axis is

nx2 dy.fV-

(c) Revolution about any line. Thevolume of a solid formed by the revolution of a plane area bounded by the arc AB and the lines AC and AD and the axis CD about any line CD (different from .x-axis and y-axis) is

fJOC

where PM is the length of perpendicular from O any point P on the arc AB to the axis CD and O be any fixed point on the axis CD.

Y +

B. axis of rotation

A

ODNn (PM)2 d (OM)V = M

C■> A'

Fig. 5

• 10.7. VOLUME OF A SOLID OF REVOLUTION WHEN THE EQUATIONS OF THE CURVE ARE IN DIFFERENT FORMS

(a) Equation of curve in parametric form. Suppose the equation of a generating curve are in parametric form x =f(t) and y = g(t), then the volume of the solid generated by the revolution of the plane area bounded by the given curve, axis of x and the ordinates at the points where t - a and t = b, about x-axis is

-j: j* it2 dx\ ,Tty .dt~

dtdt

dxwhere

Similarly, the volume of a solid formed by the revolution of a plane area bounded by x =f(t), y = g(t) axis of y and the two absciassaes where t = a and t = b about y-axis is

f f *[/W]2f ■<*nx2\^V = . dt =dt

where

SOLVED EXAMPLESExample 1. Find the volume of a spherical cap of height h cut off from a sphere of radius a. Solution. The spherical cap is generated by the revolution of a plane area bounded by the ;i

curve x2 + y2 = a2, the axis of y and the liney = a- h and y = <2 in the first quadrant about y-axis as jshown in fig. 6.

Let V be the volume of this spherical cap, then’!

1°Ja-hCnx2 dyV=h

trn .Ja ~ h (a2-y2^dy ('.' x2+y2 = a2) 0 ■>x11 /1\ /✓

- (7 “A

^ r

' 3 , ..J 2 (a-hfa - (a ~h)i a - N 3= 7t Fig. 6U\

^ a^[3a2-(a-h)2)= n —


Surfaces and Volume of Solids of Revolution

. = ^ [2a3 -{a-h) {la2 -ft2 + 2ah}\

= j [2a3 - 2a2 (a-h) + (a- h) h2 - lah (a - ft)]

= ^[3a/i2-/i3]

y _ ^2 (a- —

V JExample 2. Find the volume of the solid generated by the revolutioi

x = a cos t+2 tan2 y = a sin t.

about its asymptote.Solution. We have

of the tractrix

Ay

(0. h)A

x = a cos ^ ^ log tan2 (f/2)

y = a sin t.

= - a sin / + -:

Oand2a___a cos t

sin t sin f Here x«axis is the aysmptote of the given

curve so that for x-axis i.e., y = 0,t = n/2 and at A(0, a), t-0.

Let V be the volume of the solid generated by the revolution of the tractrix. Then

dxNow

dt (0, - a)A

Fig. 7

fJji/21°Jn/a cos21dxny2 dt =2n a2 sin21V=2 dt

dt sin t-iO

1°Jn/

cos31cos21 sin t dt = Ilia3= 2na3

3 . k/2

2 3 = jTta .= 2na3\±-Q

• SUMMARY• Surface area of solid of revolution of the curve y = f(x) about .x-axis between .c = a to .x = 6 is

given by

dx.

• Surface area of solid of revolution of the curve y =/(x) about y-axis between y = a and y = b is given by

2nx\l+^dx}2S-fv ads dy2nx — dy =dy dy

about x-'axis between• Surface area of solid of revolution of the curve x = x(r) and y = y(r) t = t] and r = r2 is given by

-jMsHsJ dt

• Volume of solid of revolution of the curve y = f(x) about x-axis betweer x = a to x = 6 is givenby

f* a71 y2dx ■

Self Instructional Material 59

0 ,

Calculus & Geometry • Volume of solid of revolution of the curve y = f(x) about y-axis between y = a and v - bis given by

fbV = I n x2dy

* a

• STUDENT ACTIVITY1. Find the surface area of a sphere of radius a.

2. Find the volume of a sphere of radius a.

1 \

1

• TEST YOURSELF-21. Show that the volume of sphere of radius a is ^ fta3-

2. Find the volume of a hemi-sphere.3f The part of a parabola y = 4ax cut off by the latus rectum revolves about the tangnet at the i

vertex. Find the volume of a solid thus generated. ;4. Find the volume of the solid generated by the revolution of an arc of the catenary

y = c cosh (x/c) about x-axis.

4 .

ANSWERS

TIC2 c • JlxM — smh —2 c2. |ufl3 Tta3 4 i x +



Surfaces and Volume of Solids of Revolution

tegral Jo 2ny ds representsThe in of a solid generated by the revolution about -r-axis1.

of the area bounded by the curve y =J[x), the ordinates x = d,x = b and x-axis. 2. The surface area of a hemi-sphere of radius a is ............

> TRUE OR FALSE :Write 4T* for True and ‘F’ for False :1. The volume of a hemi-sphere of radius a is y tm3.2 The surface area of a sphere of radius a is 27ta2.3. The surface area of a solid generated by fhe revolution of the curve

^3+^3 = a2/3

> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :1. The surface area of a sphere of radius r is :

(a) 2^(c) Ttr2

2. The volume of a hemi-sphere of radius a is :. (a)

(c) |7ta3

(T/F)(T/F)

(T/F)

(b) 47tr2 (d) 37tr2.

(b) j™3(d) Tta3.

ANSWERS

Fill in the Blanks :1. Surface

True or False :1. T 2. F 3. T

Multiple Choice Questions : L (b) 2. (b)

2. 2na2

□□□



DIFFERENTIAL EQUATION OF FIRST

ORDER AND FIRST DEGREE• •

STRUCTUREDefinitions• Test Yourself-1Solution of the Differential Equation• Test Yourself-2Formation of a Differential Equation• Test Yourself-3Solution of Differential Equations• Test Yourself-4Homogeneous Differential Equations• Test Yourself-5 Linear Diffrential Equation• Test Yourself-6Equatin Reducible to Linear Form .• Test Yourself-7Exact Differential Equation• Test Yourse!f-8 Integrating Factor• Summary• Student Activity• Test Yourself-9

1

“ M

?

; ■


How to determine the solution of the given differential equation of order one and degree

• 11.1. DEFINITIONS1. Differential Equation. An equation involving one dependent variable and its derivatives

with respect to one or more independent variables is called a differential equation.\2n3/2(4lFor Examples : (\) exdx + e*dy = 0, (ii) y-x^- ■< (iii) 1 +

dx2dx dxJ -

(iv) are all diffrential equations.

2i Ordinary DiiTerential Equation. A differential equation which contains ..only one independent variable is called an ordinary differential equation

(dy'f _- •

For Examples : (i) a: djc + y dy = 0, (ii) 1 + are both ordinary differential equationsdx3dx

as they have one independent variables.3. Partial Differential Equation. A differential equation which contains more than one.

independent variables, is called partial differential equation.

i


„ , ... B2u d2u d2u _ .... ^ ..For examples : (i) ~z + —r + —T = 0, (n) —T =

ox oy dz oxd2 u 1 d2u

c By2as they have more than one independent variable such as x, y, z.

4. Order of Differential Equations. The order of a differential equation is the order of the highest derivative appearing in it.

5. Degree of Differential Equations. The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals and fractions as far as the derivative are concerned.

For examples : (i) ex dx + ey dy = 0, is of the first order and first degree.

Differential Equation of First Order and First Degreeare both partial differential equation

-Irxly = 0, is of the second order..and first degree.

\2-i3/2

- (ii) dx2

d£ can be written as(iii) The differential equation 1 +dx2dx

/ .2 2 djL

dx2

\2-3& , it is of second order and second degree.1 + dx

SOLVED EXAMPLESExample 1. Determine the order and degree of the differential equation

Jiff, ..feY*j+/=aSolution. The given differential equation containssecond order derivative which is the highest

derivatives, so the order of the differential equation is 2. The power of second order derivative is 3, so the degree of the differential equation is 3.

Hence, order = 2 and degree = 3.Example 2. Determine the order and degree of the differential equation

dy _ d y_ 7 [d*2J

Solution. Making the differential equation free from redicals, we ge

dz)21Jiz\2dx dx2

+ydx2

1/3

1 +

1 +/ ^

Now in this equation, the order of the highest derivative is 2, so its order is 2. The power of the highest derivative is 2, so its degree is 2.Hence, the given differential equation is of order 2 and degree 2.Example 3. Determine the order and degree of the differential equation

2\r\=y ■ &dxdx

Solution. Making the given differential equation free from fraction,

'-'2 afrf*)yUJ-Now, in this equation, frie ^rder of the highest derivative is one and ts power is 2. Hence, the differential equation of order 1 and degree 2.

we get

(dx x -f~ + 2-dx

• TEST YOURSELF-1Determine the order and degree of each of the following differential equation :

i)L dxdZ + 2y = 0.2. ^ + 3 dx2

+ 5>, = sinx.1. dxdx

2 f dz + 2xy - 6a:3 = 0.(a:3 -yl)dx + {xy2 - Ary) dy = 0. 4. *3. dx

ANSWERS

3. Ord :r = 1, Degree = 11. Order = 1, Degree = 2 4. Order = 1, Degree = 1

2. Order = 2, Degree = 1


Calculus & Geometry • 11.2. SOLUTION OF THE DIFFERENTIAL EQUATIONAny relation between the dependent and independent variables i.e., a function of the form

y ~AX) + G, which satisfies the given differential equation is called its solution or primitive.

cos a: be a given differential equation. Then a function >» = sin a: + CFor example : Letdx

is its solution.1. General solution. If the solution of nth order differential equation contains n arbitrary

constants, then it is called general solution or complete primitive.2. Particular solution. A solution obtained from the general solution by giving particular

values to the arbitrary constants in the general solution is called particular solution or particular integral.

Singular Solution. The solution which can not be obtainedfrom general solution by assigning jparticular values to the arbitrary constants, is called singular solution.

Example 1. Verify that y= A cos x - B sin x is a solution of the differential equationii + y = 0.dx2

Solution. We have, y ~A cos x - 6 s\n x. Differentiating (1) w.r.t. x, we get

dx_

...(1)

= -A sin;c-Bcosx.dxDifferentiating again, we get

<&_= - A cos a: + P sin a;

dx2= - (A cos a: - sin a:) = - y [using (1)]

Hence, y = A cos x ~ B sin a: is a solution of

Example 2. Show that y = Aex + Be~ x is a solution of the differential equation

-y = o.

£l + y = 0.dx2

i

iidx2

Solution. We havey=Aex + Be~x.

Differentiating (1) w.r.t. x, we get

~ = Aex - Be' x.

-.(1)

dx /Differentiating again, we get

ix = Ae* + Be x = y [using (1)]dx2

ii -y = o.dx2

Hence, y = Ad* + Be x is a solution of -y = 0.dx2

Example 3. Show that y = ae2* + be~x is a solution of the differential equationd2y _ d^

, - . - 2y = 0. dx2

Solution. We havey = ae2x + be x.

Differentiating (1) w.r.t. x, we get

— = 2ae2x-be~x.

...d)

-(2)dxDifferentiating again, we get

64 Self-Indtructiondl Material

^ = Aae2* + be~ x. Differential Equation of First Order and First Degree•••(3)dx2

Now, from (1), (2), (3), we get d2y _ dv dx2 dx

~2y= (Aae* + be~ x) - (Tae2* -be x)-2 {ae^ + be~ x) = 0.

Hence, y ~ ae2* + be ^ is a solution of the differential equation ^ -2y = 0.

• TEST^YOURSELF-2

1. Show that y = ae* is a solution of the differential equation ^ - y = 0.

2. Show that y = ae~ ^ is a solution of the differential equation ^ + y = 0-

dfd3. Show that .y = c (x-c)2 is a solution of the differential equation %y2 = Axy fr- j ■LlvK Cwv

2 14. Show that _y = 4 sin 3x is a solution of the differential equation + 9y

dx= 0. • -

• 11.3. FORMATION?OF A DIFFERENTIAUEQUATIONThe general solution of a first Order differential equation contains one arbitrary constant, which

is called a parameter. If this parameter "takes various values, then we get la family of curve of one parameter.

For example: The equation x2i+y2 = c2 represent a one parameter family of circles if c takes all real values.

Suppose there is an equation, representing a family of curves, containing n arbitrary constants. Then in order to find its differential equation we proceed as follows :

(i) Differentiate the given equation of family of curve n times to get n more equations containing n arbitrary constants and derivatives.

(ii) Eliminate all the n constants from all of these {n + 1) equat ons to>getqin equation containing a nth order derivativie, which is the required differential equation of the givenfamily of

!

curves.Example 1. Find the differential equation of the family of all straight lines passing through

the origin.Solution. The general equation of the family of all straight lines passing through the origin

isy = mx

where m is a parameter.Differentiating (1) w.r.t. x, we get

dy-f- = m. •••(2)dxEliminating m between (1) and (2), we get

d£y=Xdx

which is the required differential equation.Example 2. Find the differential equation of the family of all stra ght lines making equal

intercepts on the co-ordinate axes.Solution. The general equation of the family of all straight lines making equal intercepts on

the axes is

c c -= 1 or x + y = c •••(Dwhere c is a parameter.

Differentiating (1) w.r.t. x, we get^ = 0 dx

which is the required differential equation.

1 +


Calculus & Geometry • TEST YOURSELF-3Find the differential equations of the following families of the curves :y = 3x + c, where c is a parameter.(2x + a)2 + y2 = a2, where a is a parameter. x2 + y2 = a2, where a is a parameter. x2 + y = lax, where a is a parameter. y = a sin (x + b), where a and h are parameters. y = a cos (x + d), where n and 6 are parameters.

1.2.3.4.5.6.

ANSWERS

1. ^ = 3 2.2xy^~y2 + 4x2 = 0 Xx + y^ = 0 dx' ax

6& + y = 0 dx'

dx

5 A&4. 2xy +x2-y2 = 0 dx + y = 0

dx2

•. 11.4. SOLUTION OF DIFFERENTIAL EQUATIONS1. Method of Solving Differential Equation by Separation of Variables :If any differential equation can be expressed as ■

j{x)dx^g{y)dy.Then we say that variables are separable.In order to solve the given differential equation, using variable separable, we first reduce it

to the form (1) and then integrate both sides, we therefore obtain the required solution.

-(I)

Example 1. Solve -^=xy+x + y+l. .

Solution. We have

— =xy + x + y+ 1 dx&̂=(l+*)(l+y).

Separating the variables, we get

= (1 + x)dx.dy1+y

Integrating, we get2

log] 1 +y|=x + y+ C

which is the required solution.4lExample 2. Solve ^ = log (a: + 1).

Solution. We haved£ = ‘og (x+ l).

Separating the variables, we get.dy = log (a:+ 1) dx.

Integrating both sides, we gety = J log(A:+ l)dx+C

= log (^ + 1) J ^ - J (log (x + 1)) .

= j: log (a:+1)-

~x log (^ + 1) - J

J^} dx+ C

f—J a: + dx + C

x+l-l dx + Cx+l


4

Differential Equation of First Onier and First Degree

= * log (x + 1) - J *& + J -^r dx + C1

^ = jc log (j: + 1) - x + log | x + 1 [ + Cwhich is the required solution.

• TEST YOURSELF-4

Solve the following differential equations :dz x. 2. ^ + y=l.1. dx~^' dx

4. +dx

a ^_(l + y2) dx (i+j^'

ANSWERS

3.dx

5. &dx = (eX+l)y-

2. log | 1 -y\ +x- C 5. log | >> | = e-1 + a: + C

3. y = 2 sin (x + Q 6. tan”1 y [■ tan”1 x = C

l. y = Cx 4. ex + e-y = C.

• 11.5. HOMOGENEOUS DIFFERENTIAL EQUATIONSHomogeneous Function. A function fix, y) in x and y is said to be a homogeneous function

of degree n, if the degree of each term of fix, y) is n.For Example : fix, y) = ax2 + 2hxy + by2 is a homogeneous function of degree 2.

REMARK

^ In general, a homogeneous function of degree n can be expressed/ \

Ax,y) = xnF i .

as

Homogeneous differential equation. A differential equation of the form

dy J\(x, y) dx f2(x, y)

is said to be homogeneous, iff\(x, y) and fiix, y) are homogeneous functions of same-degree.For example = ^

dx xyMethod of Solving Homogeneous Differential Equation :In order to solve a homogeneous differential equation we proceed through the following steps: Step 1. Substitute y = v* (or * = vy).Step 2. Reduce the given differential equation in terms of v and x.Step 3. Solve the reduced equation by the method of separation of variables.

is a homogeneous differential equation.

yf x]Step 4. Replace v by ^ or v by — after integration.

Example 1. Solve (x2 + y2) ^ = xy.

Solution. The given differential equation can be expressed as dy _ xy dx x2 + /

Clearly (1) is a homogeneous differential equation.dv .= v + x— in (1), we get dxPutting y = vx and

2dv vx Vv + x — =dx x2 + v2x2 I + v2

3dv v - V— V =X dx 1 + v2 1+v2


Calculus <£ Geometry Separating the variables, we get1 + v2 , dx

------r— dv = —■v3 x

Integrating both sides, we get

=>

- log | v j = log 1*1 + log [ C | * 2v

.

1̂ = log | xvC |

\/2v

yC = e^xvC = e

which is the required solution.

TEST YOURSELF-5Solve the following differential equations :

j 2,22 dy _x + v ,

dx 2xy 5 dy _x + y

dx x-y

dy y2 - x2 dl1. 3* X + y't = 2y- - 6* (x_>;)^=-r+3-v-

dx 2xyx&4. dx=X+y-

ANSWERS

2. x = C (a-2 - y2) 3. log (y - x) = C + X

~1 ^ - X log (j:2 + y2) + C 6. log | +3' | +AT £

1. {x2+y2) = Cx

2x4. y = x\og,\x\ + Cx 5. tan = Cx + y

• 11.6. LINEAR DIFFERENTIAL EQUATIONDefinition. A differential equation in which the dependent variable and all its derivatives

are of first degree only and not multiplied together, is called a linear differential equation.

+ y = x is a linear differential equation.dlFor example : x ^

General Form of Linear Differential Equation : (i) The equation of the form

& + Py = Qdx

where P and Q are either constants or functions of x only, is most general form of linear differential equation.

(ii) The equation of the formdx— + Px = Q dy

where P and Q are either constants or functions of y only, is most general form of linear differential equation.

Solution of ^ + Fy = g : Multiplying by J

(i.e., I.F.), we get

Pdx , which is known as integrating factor

jpfix\ PdxPy = Qedx

=>dx

Integrating both sides, we getPdx dx+ C

y . (I.F.) = J g. (I.F.) dx + C, where I.E = Jy.e

Pdxor


Working rule for solving ^ + Py = 0.

1. Find I.F. =e^p dx.

2. The solution is y . (I.F.) = J Q . (I.F.) dx+ C.

SOLVED EXAMPLESExample 1. Solve (1 + X?) ^ + 2xy

Solution. The given differential equation can be written as dy 2x

+------- ry = -■dx l+x2 1 +xz

dyComparing (1) with ^ + Fy = Q, we get

Differential Equation of First Order ami • First Degree

= COS X.

COSJC••(1)

2x cos*p = » C =l+x2 1 +x2

I-z-2«I.F. = J

y (1 + x2) = j Q (1 + xz) dx + C

y(l+xz) = jy (1 + x2) = | cos xdx + C

y (1 + x2) = sin x + C. dy 2Example 2. Solve x -p - y = x .

e,oe(,+PdX - e l+x

Thus, the solution is

cosx (1 +x2) dx + C(l+x2)

=>

dxSolution. The given differential equation can be written as

dy 1-f - “ y = *. dx x -d)

Comparing (1) with ^ + = 2’ we Set

1P = — and Q = xx

J 1Pdx _ e~ log X _ i._I.F. =so, = eX

Thus, the solution of (1) is

y . ^ x dx + C = \ dx + c = x + C.

y = x2 + Cx.

=>

iTEST YOURSELF-6Solve the following differential equations :

2. ^ + y = <?.dy lx 3. ~-Ay~ e1*.1. lu'y = e • dx dxa dy -2x4‘ Tx+y = e ■

dx x

dy 6. ^ + 2y = 6ex. dx

9. ff- + - = xn. dx x

5. ?f + 2y = e~x. dxo dy _ 4x 8. -^ + 2y = xe . dx7.

ANSWERS

l 3.y = -±ez‘+Ceix[y = e2x + Cd* l.y = ^ex+Ce-x1.


S.y = e*+Ce'2i 6. y = 2/+ Ce'1'8. y = 4 xe* -4 ^ + Ce" ^ 9. (n + 2) Ay - a"

o 3o

Calculus & Geometry ~2x + Ce'x4. y = -e

7. y = a3 + Ca + 2 + (fi -r 2) C

• 11.7. EQUATION REDUCIBLE TO LINEAR FORM*4

A differential equation of the form% + ry = Qy”

is called an equation reducible to the linear form or Bernoulli’s equation. Solution of the equation. The given equation can be written as

-ndz -n+ 1 = Q-dx + P>y-/j+1Put y = v-ndl_dV'

dx dx(-n+i)y

Then (1), becomes- 1 dv + Pv = Q

(n-l)dxdv— -(n-\)Pv = Q(l-n)

which is a linear equation of first order, and can be solved in a usual manner.Working Procedure :

or

dy(i) Write the given equation in the form ^ + Py = Qyn.

(ii) Divide by yn, substitute y-n +1 = v and get the equation of the formdv~-(n-\)Pv = Q(l- n).

(Hi) Then, apply the method of solution of linear equation.SOLVED EXAMPLES

. Example 1. Solve a ^- + y = y2 log x.dx i

Solution. The given equation can be written as 1 ^y 1 1 1 ,

~2 —j~ ^ = log AyL dx Ay A_ \ dy dv

^ dx dx1Put — = v =>y y

Hence, the given equation becomesdv 1 1. - —v = — log a which in linear in v. dx x x

JPdx = e-l±dx = e-\ogx = }_IF. =A

The solution isj--^\ogxdx + c1—. v =

A

Jf1 1 1~ dx — log A + He A A Av = log a + 1 + ca or 1 = cyA + y + y log a 1 = y (1 + log a) + cxy.

= — log a -A

oror

Example 2. Solve ^ ^ /og y = 4 (l°s yf-a“

Solution. The given equation is of Bernoulli’s type because of presence of non-linear part2 ^

on right side, so dividing by y (log y) , we get

-.^ + 1.2 dx a log y ^2 .

11 1...(1)

.y (log y)

70 Self-Instructional MaterialIt «|

du1 1 , • ^ in (1), we get yOog).)2 *

Differential Equation of First Onier and First DegreeNow substituting u = logy °r dx

du l 1, du 1- —+ -« = -r — — u — —dx x x dx x

12 '• ■X

which is a linear equation in u, thereforel.F. = e'^/xt,x = --

xMultiplying equation (2) by I.F. and a(ter integrating, we get

« 1 *~ = —: + Cx 2x2

1 1, = —r + c, which is the required solution.x log y 2x2

=*

• TEST YOURSELF-7Solve the following ordinary differential equation :

dy y i dx x&+L=eL.dx x x2

3.^ x2. 4 = Wy".2. x1. +y=xy. + 2 ydx dx 1 -X

5. y{\+xy)dx-xdy = §.4.

7. (x3y2 + xy) dx = dy.6.

ANSWERS

= “ (l/x2) + c 2.4=^(3-cx)1L x5y5 2

3. ' Vy~(l - x2)^4 = - j (1 -x2)3/4 + c 4. 2xy2 + «V = 1y

x 1 2 .= -, +c /

7. 1=2>

6. 2x = (lex2 + 1) ^5.y

-x/2- eye

• 11.8. EXACT DIFFERENTIAL EQUATIONThe differential equation which can be derived from its primitive by direct differentiation

without any further transformation (such as elimination or reduction), is called an exact differentialequation.

For Example : The equation ydx + xdy = 0 is exact, as it is derived from it is primitiveyx = c.

Working Procedure :(0 Compare the given equation with Mdx + Ndy = 0 and find out M and N.

(it) Show that which conclude the exactness of the givei: equation.

(Hi) Integrate the coefficient of dx (i.e., M) with respect to x, regarding y to be constant.(iv) Omit the terms containing x is N and find the integral of the coefficient ofdy with respect

to y.(v) Add the above two results and equate this sum to an arbitrary constant, i.e.,

j* (terms in N not containing x) dy = cM dx +(y constant)

REMARK> It is clear from the definition that the exact differential equation is formed from its general

solution by direct differentiation and without any further operation of elimination or reduction.

Self-Instructional Material \71

Calculus <t Geometry > The necessary and sufficient condition for a differential equation (of first order and first degree)

Mdx + Ndy = 0,where M and N are the functions of x and y

to be exact equation is that

The equation = tt" is true whenever both sides exist and are continuous. Here we take ay oxthe hypothesis that all the functions, which we discuss are sufficiently continuous and differentiable to gurantee the validity of the operations we perform on them.

>•

SOLVED EXAMPLESExample 1. Solve (ax + hy + g) dx + (hx + by +f) dy = Q. Solution. The given equations

(ax +hy + g) dx + (hx + by+j) dy = 0. Compare with Mdx + Ndy = 0, we get

M = ax + hy + g and N-hx + by+f' dM , dN= ft, -r-dx=h equation is exact.dy

Hence, the solution is

i J (terms in N not containing x)dy = c

f (ax + hy + g) dx + j (by+f) dy = c

M dx +(y constant)

ax2 bl— + hxy + gx + -^-+fy = c

ax2 + 2kxy + by2 + 2gx + 2fy = 2c.

Example 2. (x2 - ay) dx-(ax- y2) dy = 0.

=>

Solution. Compare the given equation with M dx + N dy = 0, we get2 2 M = x -ay and N = y - ax.

Now, we havedM dN dM dN= -a and -r— = -a =>dy dx dy dx

=> The given equation is exact.Therefore, the solution of the given equation is

M dx + Ndy = cy is constant (terms which do not contains x)

| (x2 - ay) dx-y \ y2 dy-c

1 3 y 3 3— x - ayx + ^ =c => x +y - 3axy = 3c.

=>

• TEST YOURSELF-8Solve the following ordinary differential equation :

dy_ (2x - y) dx x + 2y-5

. , . , ixdy-ydx4. xdx + y dy-a —^—x2 + y2

1. (4x+3y + 1) <& +(3x + 2y + 1) dy = 0. 2.

3. xdx + yiySQ-yf-O.x +y

ANSWERS

1. 2x2 + 3xy + y2 + x + y = c.

3. x2 - 2 tan-1 (x/y) + y2 = c.

2. x2-y2-xy'+5y + c.

4. x2 + y2 + 2a2 tan-i I x = c.y


Differential Equation of First Order and First Decree• 11.9. INTEGRATING FACTOR

If the given equation Mdx + Ndy = 0 is not exact, then it can be made exact by multiplying some function of x and y. This multiplier is called an integrating factor (I.F).REMARKS

> The number of integrating factor for the equation Mdx + Ndy = 0, is infinite.

>• If ^ is an integrating factor. Then (Mdx + Ndy) = 0 is an exact differential equation. >• Although an equation of the form

Mdx + Ndy = 0always has integrating factor, there is no general method of finding them.

Here, we explain some rules for finding I.F. :(A) By Inspection. The following list of ^xact differentials should be

xdy - y dxnoted very carefully.

/ \ r.A .y dx-xdy (2)' d *(1M £ = 2 x2y Xy \ jv /-1 ZI_ xdy -y dx

x2 + /

ydx - xdy

y dx-xdy-i x(4) d tan(3) d tanx2 +y2yX

ZI_ xdy -y dx(5) d log- = (6) d log ^xy xyy

xdy+ y dx1(7) d(x, y)=x dy Fy dx (8) rf - = 2 2x y

2xy dx- x dy 2xy dy - y2 dx(9) d - Z_ _(10) d X2.2 .2y\ Xx y

V) lx2y dy - 2y2x dx (x2) 2xy2 dx - 2yx~ dy(12) d ^ =(11) d x2 4 4

X y y( 2xdx + 2y dyye dx - e dy(13) d - (14)d[log(x2 + /)] = t •? 2 * x +y j2y y

There are several methods which are usually given for determining integrating factors inparticular classes of equations.

M dxFN dy = 0.is the given equation, which is not exact, then we proceed as follows.

...(1)•If

is an integratingMethod (i); If Mx FNy*0 and the equation is homogeneous, then

factor of (1).Method (ii): If equation (1) is of the form

/i(x, y)ydx +/2(x, y) xdy = 0.

Mx^Ny

1 is an integrating factor, provide Mx - Ay * 0.Then Mx -NyMethod (iii): If ^ is a function of x alone say f[x) or a constant c, then

A L dy dx J

J/M* or JarI.F. = e

Method (iv) : If tt is a function of y alone say fly) or a constant, say c, thenM l dx dy JjofyI.F. =

Method (v) : If an equation is of the formxayb [my dx F nx dy] F xrys (pydxFqx dy) = 0

or e


where a, b, m, n, r, s,p and q are all constants, then the integrating factor will be where h and k are such that after multiplying by the integrating factor that the condition of exactness is satisfied.SOLVED EXAMPLES

Example 1. Solve (y2ex + 2xy) dx -x2 dy- 0.Solution. The given equation is

Calculus <fi Geometry

(y2ex + 2xy) dx ~ x2 dy = 0.

Compare with Mdx + Ndy = 0We get M - y2ex + 2xy and N =- x2

^ = 2yS,^ = _2xdy dxdM dN. dy *

The given equation is not exact.

=>

Now, if in the equation ex is multiplied by some other-function, then it must occur twice in the differential equation. But since it is occurring only once, therefore we should divide by y2.

2xy dx - x2dy lx'' 2f dy = Qex dx + = 0 or ex + — dx +

M = ex + — and N = ~ —

2 y y

Now2y y

dM 2x _ dN_ 2 ~ dx =» Equation is exact.=>

dy yHence, the solution is

M .dx + N .dy=cy is constant only those terms does not contain x

K 22x x X •=> ^ + — = c.ex + — dx = cy y

Example 2. Solve (x2y - 2xy2) dx-(xi - 3x2y) dy = 0. Solution. The given equation is

(/y - 2xy2) dx -(x3- 3x2y) dy = 0.

Compare withA/afr + Ndy = 0 We get M = xzy- 2xy2 and N = 3x2y - x2

dM 2-4xy and ^- = 6xy-3x2dy X dM dN dy * dx

=> The given equation is not exact.Here, we observe that M and N are the homogeneous functions of x and y and

dx

1 1 1* 0 therefore, I.F. =2 2 2 2Mx + NyNow, multiplying the given equation by I.F., we get

21, (3- dx+ -

xy x y

x')1 dy = 0.y x y y

In this equation

- and N = -~ —y y2

1M = —y x

dM 1 dNdy

=* This equation is exact.x y3

The solution is - + log ^ = c.

2 dxy

y X


Differential Equation of Firxt Otrier and First Degree• SUMMARY

• Differential equation : An equation involving one dependent variable and its derivatives with respect to one or more independent variables, is called a different equation.

4lFor example : x ^ = 0

• Order oiD.E : The order of the highest derivative appearing in D.E is called the order of D.E.• Degree of D.E : The degree of the highest derivative appearing in D.£|, after making D.E free

from radicals and factors as for as the derivative are concerned, is called the degree of D.E.• General solution of D.E : If the solution of n‘h order D.E contains n arbitrary constants, then

it is called general solution.• Particular solution of D.£ : The solution of D.E obtained from its gneral solution by assigning

particular values to the arbitrary constants in it, is called particular solution of D.E.• Singular solution of D.E : The solution of D.E, which can not be objtained from its general

solution as obtained particular solution, is called singular solution of D.E.

• STUDENT ACTIVITY1. Find the order and degree of the differential equation

Jlx 1/3V1 +dx2 ■dx

2. Solve, -^ = y - x tan . dx x


Calculus & Geometry • TEST YOURSELF-9iSolve the following ordinary differential equation :

(1 +^)y rfx + (l - xy)xdy = 0. ydx-xdy + (l'+x)dx + x sinyrfy = 0.(*y + x2y2 + xy + 1) y dx + (jr3?3 - A'V - xy+ l)xdy = 0. {y2e + 2xy) dx -x2dy = 0. x2y dx - (a-3 + y3) dy = 0. x dx + y dy + {x2.+ y^) dy = 0.{x2 .y2 + xy-‘r i)ydx + {x2y2-xy+ i)xc/y = 0.(a:2 + y2) dx - 2xy dy = 0.(xy sin xy + cos xy) y dx + (A'y sin xy - cos xy) xdy = 0.(2y dx + 3x dy) + 2xy (3y dx + 4x dy) = 0.

1.2.3.4.5.6.7.8.9.10.

ANSWERS

1. log (x/y) = c + (l/xy) 2. ^ x + cos y = c 3. xy + (-1/xy).-2 log.y = cJC X

5. y - ce*/3,yy

1 2 28. xy + logx - logy - — -c 12.x -y =cx

10. x2y3 + 2x3y2 = c

24. ex + - x2 + y2 = /-2>7.= c

9. x sec (xy) = cy

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. An equation, which contains, dependent, independent variables and different derivatives is.

called..................2. A differential equation, which involve only one independent variable is called...................3. A differential equation, which involve two or more independent variables and partial

differential coefficients, is called..................

> TRUE OR FALSE :Write ‘T’ for true and ‘F’ for false statement:1. An exact differential equation can be derived from its primitive by direct differentiation.f7'/Fj2. If is an integrating factor then \x{M dx + N dy) = 0 is not necessarily exact.3. If the given equation is homogeneous and Mx + Ny * 0 then 1/Mx + Ny is the I.F. of the given

(T/F)

(T/F) '!

equation.> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :1. The general solution of a first degree equation contains :

(b) two arbitrary constant (d) none of these.

(a) one arbitrary constant (c) three arbitrary constant

2. The equation (x2 + y2) = xy is :dx

(a) homogeneous (c) reducible to linear

(b) linear(d) none of these.

3. The number of integrating factor for a differential equation of the type M dx + N dy = 0\$ : (a) one (b) finite (c) infinite (d) none of these.

ANSWERS

Fill in the Blanks :1. differential equation


Multiple Choice Questions :1. (a) 2. (a) 3. (c) 4. (b)

2. ODE 3. PDE

□□□76 Self-Instructional Material

Differentia! Equation of First Order and of Higher DegreeUNIT

DIFFERENTIAL EQUATION OF FIRST

ORDER AND OF HIGHER DEGREESTRUCTURE.

• (A) Equations Solvable for p• Test Yourself-1

• (B) Equation Solvable for y• Test Yourself-2

• (C) Equation Solvable for x• Test Yourself-3

• Lagrange’s Equation• Clairaut’s Equation

• •Test Yourself-4• Summary• Student Activity• Test Yourself-5


How to determine the solution of differential equation of order ora and degree more than one

In this chapter we shall deal with some special types of differential equations of the first order and of degree higher than the one.

Definition. A differential equation of the formp‘ + A\ pn 1 + Ai pn ^ + A3pn ^ + ... + — 0

is called a differential equation of the first order and nth degree, wherep is

j

a symbol for dx'Some special types of equations are(a) Solvable for p. (c) Solvable for x.(b) Solvable for y.

• (A) EQUATIONS SOLVABLE FOR pConsider, the differential equation

pn+A\pn 1+A2pn 2 + ... + A„_ ] p + A„ - 0 ...d)where A!, A2 • •. A„ are functions of x and y.

Working Procedure :

(0 Put p formal) Solve the equation for p.(iii) Apply the method of solving the equation of first order and first c(iv) Write down the result in closed brackets in the form of product

equations equal to zero, by replacing C], c2,... cn by c.SOLVED EXAMPLES

Example 1. Solve p2 - 5p + 6 = 0.Solution. The given equation is

p2 - 5 p + 6 = 0 The linear factors of (1) are given by

agree.and put this resulting

...(1)


Calculus <4 Geometry (p-2)(p-3) = 0 => p = 2, p = 3p = 2=>,̂ =2 => dy - 2dx. dx

On integrating, we get y = 2x + c.Similarly p = 3 gives >» = 3x + c.Hence, the single combined solution of (1) is given by

iy-2x-c)iy-3x-c) = 0.Example 2. Solve xyp2 - {x2 + y2) p + xy = 0. Solution. Here, the given differential equation is

xyp2-{x2 + y2)p+xy = 0

which is the quadratic equation in p. Solving (1) for p, we get ' t

(x2 + y2) ± V[(x2 + y2)2 - 4 x2f] (x2 + y2) i (x2 - y2) ■P = 2xy 2xy

(x2 + y2) + (x2 - y2) _ x-..(2)so that P- 2xy y

(x2 + y2)-(x2-y2) _ i-..(3) •and P = 2xy x

Now from (2), we getp = ^~ = — =$ xdx = ydy

dx yOn integrating, we get

2 2 • •.(4)x -y = c.Now from (3), we get

dy y dy dxp = -f- = z- => -JL- — dx xOn integrating, we get log y - log x - log c

y x

Z -..(5)=> = £•xHence, from (3) and (5), the single solution of the given differential equation is

(r2 -y1 ~ c)(y- cx) - 0.

• TEST YOURSELF-1Solve the following differential equations :

(a) p2-Ip+12 = 0(x-yfp2-3y(x-y)p + 2y2 + xy-x2 = 0. 3. (p - xy) (p ~ x2) (p ~ y2) = 0. '

(b) p2 -9p+ 18 = 0.1.•2.

ANSWERS

1. (a) (y - 4* + c) (y - 3x + c) = 0 (b) (y - 3;c + c) (y - 6x + c) = 0.2n ,= logr+logc 3. (y - cex ){3y - x - c){xy+ cy + \) = Q

\2*i

1+f^x 2 xi2. tan

V /

• 12.2. (B) EQUATION SOLVABLE FOR yLet the given differential equation is

f(x,y,p) = 0

Solving fory, the equation (1) gives

y = F{x,p)Differentiating (2), with respect to x, we* get an equation of the type

-..(2)

42. -..(3)P = S x,p. dx

78. Self-Instructional Material

Differential Equation of First. Order and of Higher Degree

Equation (3), being an equation in two variables x and p, may possibly be solved. Let its solution is given by

...(4)(K*,p, c) = oThe elimination of p between equations (2) and (4) gives a relation involving x, y and c, which

is the required solution.Working Procedure :

(i) Solve the given equation for y and put it as F(x, p) only.(ii) Differentiate it with respect to x.

di(Hi) Put p for ^ and use the previous method.

(iv) Now eliminate p between (1) and (3) and get the required general solution.SOLVED EXAMPLES

Example 1. Solve y+px = p2xi.Solution. Here, the given equation is

2 4 ••.(I)y + px=p xDifferentiating (1), w.r.t. x, we get

4y = -r . ^ * & 2 1-p-x--+2px - + ipx.dpdx

diPut = p, we getdxp = - p - x^j- + 2pxA ^ + 4p2;t3

ax ax2p - 4p2 x3 +x~j^ (l - 2xz p)-0

2p (1 - Tx*p) + (1 - 2jc3p) = 0

(l-2x3p)(2p+x&) = 0

2p+x^- = 0 and 1 - 2x3 p = 0. dx

dp

=>

Now, we solve these two equations, separately. First consider

dR = Q ^2p+Xdx

On separating the variables, we get dp ~ 2dx

dx x-

xpOn integrating, we get

\ogp = ~2\of>x + log c2 7 Clog (px ) = log c =* px =c or p = ^f

x2Put this value in eq. (1), we get

/ \C C 4 c , 2

-X-x + -rX => y = - - +c X X

2y = X

xy = c2x-c, which is the required solution of (1).or2

Example 2. Solve y-x = .. dx

Solution. Here, the given differential can be written as

y-x = xp+p2, where P = ^

y = x(l+p)+p2.Differentiating (1), with respect to x, we get

dl

...(1)=>


Calculus & Geometry

i*i*.4.*-0 o, ‘‘ax ax

which is a linear differential equation of order one, with p as independent variable.lfW1 4- = /

...m ■ ■+ x=-2pordp

Hence, the solution is-1 - 2pe? dp + c => = -2 {p^ - + c

xe? = - 2^ (p - 1) + c => . x = -2 (p- 1) + ce~p. Putting this value of x in equation (1), we get

y = (l+p)[-2(p-l)-^ce~p]+p2 y — c (l + p) e~p + (2 — p2)

Hence, the equations (3) and (4) given the required solution of (1).

...(3)=>

..7(4)or


2 22. yp = x +p x. 5. y = 3x + log p.

1. y = 4p3 + 3xp. 4. y=xp2+pi.

3. y = x[p + V(1 +p2)}.

ANSWERS'o\- p3 + 3cp12 2

P +cp-3/2 - 1/21. X = ~ y = -i1 2 1/2-3P -cp

3p3 + Cpl/2

3. x2 + y2 ~2 cx = 01 (2C + 3p2-2p3)2 (P-D2

5. y = 3x+ log

+ p2 1 2 ,• JP +cp1/22. x = - y = p

2cp2 + 2p3 - p4y- 2(p-l)2

31 3^1 - c e

• 12.3. (C) EQUATION SOLVABLE FOR xSuppose the given differential equation is

pn+41p" + i42pn 2 + ... + An_l p + Ari = 0.Let us suppose, it can be expressed in the form

^Ay,p)Differentiating (1) with respect to y, we have

1 dx dp .-•(2)-t.='8 y>p,P dy dywhich is a new differential equation with variable y and p.

Let the general solution of (2) is of the form (J)(x,p, c) = 0.

Eliminating p between (1) and (3) we get, either F(x,y,c) = Q

as the required solution or in parametric form.SOLVED EXAMPLES

...(3)

Example 1. Solve y = 2px + p2 y. Solution. Here, the given equation is

j = 2px + p2 y.Solving, it for x, we get

2x = - py + y/p.Differentiating (1) with respect to y, we get


\

Differential Equation of First Order and of Higher Degreedy p dy

i+4

2= -p-y

p

-y{df1p + ~por

P2dy

dp (. 1• 1p 1+— +>- i+^ =0ordyP P '

1 dp1+— p + y-^ = 0.or dyP L

Now, since the first factor does not involve a derivative of p with respect to x or y, therefore it will be omitted (such factor always give the singular solution).

^ = 0.Now consider p + y , dyOn separating the variables, we get

^ + ^ = 0. p y

log p + log y = log c or py = c.Integrating

To eliminate p between (1) and (2), solve (2) for p, which gives.p = p-- Putting this value of

p in (1), we get

...(2)

2x = - c + y2/c =s> 2xc-y2 + c2 = 0.Example 2. Solve x = y+p2.Solution. Differentiating the given differential equation with respect

dy p

o y, we get

dy2d2 1dp = -2 p + l + dp.dy =or 1 -p p-l

Integrating, we have2

y = -2 ^-+p + log(p-l) +c

y = c - [ p2 + 2p + 2 log (p - 1)]. Putting this value of y in the given equation, we get

x = c~[2p + 21ogO - !)]•Relations (1) and (2) gives the required solution.

or

...(2)


1. p3 - 4pxy + 8y2 = 0. 2. y = 2px + p2y. 3. x=-py + ap2.4. 4 (xp2 + py) - y4. 5. x+ p/Vl+ p2 =0.

ANSWERS

2. y2 - 2cx - c2 = 0 p[c -a cosh"1 p]

1. y = c (x - c)2c — a cosh 1 p

3- y = - ap, x =

4. y = 4c (xyc + 1)V-1)

5. (x-a)2 + (y + b)2=l

• 12.4. LAGRANGE’S EQUATIONThe differential equation of the form

y = xF(p) +ffp)is known as Lagrange’s equation.Method of Solution :

Differentiating the equation (1) w.r.t. x, we get

Self-Instructional Material 81.

Calculus & Gepmetry . p=m+*r(p)%+m &̂

dx xF'(p) +f(p)dp—c- or^ p-m=m+m]dx .. dp p_m

which is linear in x and p and can be solved by the usual method.

• 12.5. CLAIRAUTS EQUATIONThe differential equation of the form

y^px+ffp)is known as Clairaut’s equation.

Method of Solution :Here, the given differential equation is

y = px+f{p).Differentiating (1) with respect to xt we get

d̂+mdfx

> «

-•(1)

d£= P=P+Xdx

^ [x +f(p)] = o

which gives either ^ = 0 or x +f'(p) = 0

dx

dx

~j~ = 0 => p = c dx -•(2)

Now, the elimination of p between (1) and (2) givesy = ex +f{c), where c is any arbitrary constant.

SOLVED EXAMPLESExample 1. Solve y = apx + bp2.Solution. Here, the given differential equation is

y = apx + bp2.Differentiating (1) with respect to x, we get

***■ *

...(1) .

2 dp_*L̂_=p-ap +axdx

p(l-a) = (ax + 3bp2)^

dx ax + 3bp2 dx dp (\-a)p ^

or

2>bpa . x +ordp (l - a) p U-fl)

which i^ linear in x and p.adp a .-rr'ogpI.F. = e =

Hence, the required solution is

27__ 36f 2fi- 1

“I Paa J

3bxpa p dp + c = dp + c1 - 1 -{(3fl-2)/(*^ 0)~__ Iffxpa + cor

(1 - a)' (3a - 2)/(a - 1) x = (3bp2)/(2 - 3a) + cp

The required solution is given by (1) and (2) in parametric form, being p as a parameter.

fl/l - n -..(2)or

Example 2. Solve y = px + —-\ P

Solution. The given differential equation is in Clairaut’s form Put p = c,we get

ay = CX + — y

cwhich is the required general solution of the given equation.

Example 3. So/ve (y - px) (p-l) =p.Solution. Here, the given equation is

1‘(y-px)(p-l)=p ...Cl)-


»

Differential Equation of First Order '' and of Higher Degree

which can be rewritten as

y=px +p-\

which is of Clairaut’s form. Hence, replacing p by c, the required general solution isC .y = ex +-----where c is any arbitrary constant.c — 1


y=px + ap(l -p). p2x = py-\.xp3 - (y + 3) p2 + 4 = 0. sin px cos y = cos px sin y + p.

2. y = px + p2.4. p = log (px - y).

1.3.5.

7. (y-px)2/(l+p)2 = a2.

ANSWERS

6.

12. y = cx+ c2

45. y = cx + - 3

3. y = cx + —1. y = cx + ac (1 - c) ciy = cx- ec

7. (y — cx)2/(1 + c)2 = a2

6. y = cx~ sin4. cc

• SUMMARY• Soluble for p :

&(i) Put p for in D.E.

(ii) Solve for p.(iii) Apply the method of solving the equation of first order and of iirst degree.(iv) Write down the result in closed brackets in the form of product and put this resulting

equations equal to zero, by replacing the constants ci, C2,... ci by c.• Soluble for y :

••... (!)•f^y.p) = oGiven D.E is Solving for y, we get

...(2)y = F(x,p)After differentiating eq. (2) w.r.t. x, we get the equation of the form

• ••(3)P = S x,P, dx I ■The solution of eq. (3) is given by

Q(x,p,c) = 0Now eliminating p between eq. (2) and (4), we get the required soluti Soluble for x :

Given D.E is Solving for x, we get

...(4)n.

-CD/(*. y>p) = o

• ••(2)x = F(y,p)Differentiating eq. (2) w.r.t. y, we get the equation of the form

1 -(3)

The solution of eq. (3) is given by(J) (x.p, c) - 0

Eleminating p between eq. (2) and (4), we get the required solution.

...(4)

Self-Instructional Material 83".

Calculus dt Geometry • STUDENT ACTIVITY './.V.

1. Solve /? (p-;y) =* (•K + y)-

2. Solve y = 2px + p2y.

• TEST YOURSELF-5Find the singular solution of the following differential equations :

y=px +a V(1 +/?2). y2 - 2pxy + p2 (x2 - 1) = m2.^:2 + >’2 + 2cx + 2c2 - 1 - 0.SpV' - pj: + 1 = 0.

1. 2. p = log (px - >).4. 4p2.x (x- a) (x-b)^ {hx1 ~2x {a + fc) + a6}2.£2,2,6. p +y =1.8. /(I +4p2)-2p^ - 1 = 0.

3.5.7.

ANSWERS

1. jc2 + y2 = a2 A. x-0, x-a = 0 and x- b = 0. 5. x2 + 2y2 = 2. 7. x2-Ue3y = 0.

3. y2 + m2 x2 = m2.(t.y = ± 1.

2. x + y-x log Jt = 0.

8. x2 - 4y2 + 4 = 0.

OBJECTIVE EVALUATION► FILL IN THE BLANKS :1. The differential equationp” + Z5] pn_ 1 +... +Fn = 0, where Fi, Pnare functions of x and

y, is said to be the differential equation of first order and...........degree.2. The general solution of a nth order differential equation contains...........independent arbitrary

constant.> TRUE OR FALSE :Write T'for True and F for False statement:1. The Lagrange’s equation is a particular case of Clairaut’s equation.2. The Clairaut’s equation is a particular case of Lagrange’s equation.3. The particular solution can be obtained by general solution.

JT/F) (T/F)

(T/F) (T/F)


Differential Equation of First Order and of Higher Degree


A solution, which can not be obtained from general solution by assigning some particular value of arbitrary constant is known as : ;(a) Particular solution (c) Integrating.factor

2. A differential equation, in-which both p and ^discriminant relations, are same, it is said to be:(b) Equation solvable for p *(d)'None of these. ;

.3. A solution'of a.differential equation, which is given by the-envelope of the family of curves represented by that differential equation is said to'be : •(a) Particular solution (c) General solution

4. To find the singular solution of a differential equation, we use :(a) p-discriminant relation (c) both (a) and (b)

1.

(b) Singular solution (d) None of these. ,

(a) Legrange’s equation(c) Clairaut’s equation

(b) Singular solution(d) None of these.

(b) c-discriminant relation (d) None of these.

ANSWERS

Fill in the Blanks :1. n

True or False:I.F 2. T 3.T

Multiple Choice Questions : 1. (b) 2. (Cj

2. n

□□□


Calculus. & Geometry UNITi; Wj.*

LINEAR DIFFERENTIAL EQUATIONS WITH

CONSTANT COEFFICIENTSSTRUCTURE

Linear Differential Equation Methos of Finding C.F.• Test Yourself-1General Method of Finding P.l.• Test Yourself-2Short Methods of Getting P.l.• Test Yourself-3• Test Yourself-4• Test Yourself-5• Test Yourself-6• Test Yourself-7• Summary• Student Activity• Test Yourself-8

}


• How to determine the solution of differential equation with constant coefficients.• About complementary function and finding particular integral of the given differential

equation

• 13.1. LINEAR DIFFERENTIAL EQUATIONin- 2d y ,2 ^+...+Any — 5

dxn~2having A\, ...,An and B either constant or function of x, is called the linear differential equation of the nth order.

If A0, Aj, ...,An are all constants and B may not be constant then equation (1) is said to belinear differential equation of nth degree with constant coefficients.

If we take fi = 0, then the corresponding equation is called homogeneous equation.d dnUsing the symbols D, D2, ...£>" for -r-, —... —- respectively in (1), then we getdx dx dx

D^ + Aj Dn~xy + A2Dn~2y+ ...+Any = B {if + Aj 1 4- A2Dn~2 + ... + A„) y = B => j[D) y~ B f(D) = Dn + A \ Dn~ 1 + A2 Dn ~ 2 +... +An.

Now, consider the homogeneous differential equation AD)y = 0

(obtained by putting right hand side i.e., B equal to zero).Working Procedure :

(0 Firstly, we find the general solution of (2), which is called the complimentry function (C.F.), contains as many arbitrary constants as is the order of the given differential equation.

-1dn yThe equation -..(1)+ A + Ai -1dxn dxn

...(2)=>

where

...(3)

86 Self-'Instructiohal Material

(ii) Next, find the solution of (1), with no arbitrary constant which is called the particular integral (P.I.).

(Hi) To find the general solution of (1), and C.F. & P.I., obtained in (I) and (2).y = u + v = C.F. + P.I. ;

Auxiliary Equation. Consider the differential equation (1) with B = 0 i.e.,(Dn + At D"-1 + A2D"-2 + ... + A„)y = 0 or f(D)y = Q.

I

Substitute y = enxx on a trial basis, then we get(rn + A \ + A2mn~2 + ... + An) = 0 <

Linear Differential Equations with Constant Coefficients

i.e.

which holds ifmn+A]mn l+A2tnn 2+...+A„ = 0 or f(m)=0.

Equation (2) is called the auxiliary equation.From (1) and (2), we observe that the auxiliary equation_/(m) = 0 will give the same value of

m as the equation ./{£>) = 0 gives of D.

...(2)

• 13.2. METHODS OF FINDING C.F.To find the C.F., the roots of the auxiliary equation (2) are to be cor Three different cases arises :

sidered.

(i) The roots of auxiliary equation (2) are real.(ii) The roots of auxiliary equation (2) are complex i.e., a + i (3 typ i.(iii) The roots of auxiliary equation (2) are surds i.e., a ± type.Case (i): (a) Suppose that the auxiliary equation (2) has n distinct foots mj, m2,.. ■ m„, then

C.F. is given bym.x ■\-CiemiX + ... + Cn em”xCj e

where Q, C2,..., C„ are arbitrary constants.(b) If the auxiliary equation having r roots are equal to mi (say) and remaining roots are

distinct, then the C.F. is given by[Cj + C2x + C3x3 + ... + Cy1]em'x + C, + 1 em' + I* + ... f

Case (ii) : If some of roots of the auxiliary equation are complex, then we shall follow the given procedure.

Let a ± ip be the roots of the auxiliary equation, then the corresponding part becomes(a +18) x . y-. (a - iB) x cu i&x , ^ ar -iRx= C| e +C2e ^ = C\ e .e +Cie e

= e™ [Ci cos px + iCi sin px] + e0* [C2 cos px - iC2 sin px]= e0* [(Ci + C2) cos px + (iC, - iC2) sin Px]

C.F. = etu [Bx cos Px + B2 sin Px] where B2 are arbitrary constants,

The expression (1) can also be written as (a) Bi e^cos (px + B2)

- -..(I)

(b) S, ^ sin (px + B2)-If, the equation has two equal pair of complex roots a + i'P and a - r'P, say, occur twice, then

the corresponding part of C.F. is written ase0* [(Bj + B2 x) cos Px + (fl3 + B4 x) sin Px].

In general, if a ± i p occur K times, then the corresponding part of he C.F. can be written as e™ {(B\ + B2x + ... + Bkxk ^ cos px + (Bx+i + B^+2x +... +Z?2a:X^ s*n P*1

where B\,B2,..., BK, BK+1,.... B2K are arbitrary constants.Case (iii): If a pair of the roots of the auxiliary equation involves surds, say a ± VjT, where

P > 0, then the corresponding part of C.F. in one of the following three forms(a) c0* [Bi cosh (x Vj3) + B2sinh (x Vp")](b) cosh (x Vp + B2)

(c) BiC^sinhfx Vp+jB2)- SOLVED EXAMPLES

Example 1. Solve [D3 + 6DZ + 11D + 6] y = 0.Solution. Here, the given differential equation is

[D3 + 6D2 + 1 ID + 6] y = 0 ...(1)


Calculus & Geometry To find the auxiliary equation, replace D by m, then (1) becomes, m3 + 6m2 + llm + 6 = 0

(m + 1) (m2 + 5m + 6) - 0 (m + 1) (m + 2) (m + 3) = 0 => m = - 1, - 2, - 3

i.e.. Roots are real and unequal. Hence, the general solution isy = Cl e~x+C2e~Zx+C3 e~3x.

Example 2. Solve [D4 + 2D3-3D2-4D + 4]y = 0.Solution. Here, the auxiliary equation is

m4 + 2m3 - 3m2 - 4m + 4 = 0 (m-.l) (m3 + 3m2 — 4) = 0

(m - 1) (m - 1) (m2 + 4m + 4) = 0 (m- 1) (m - 1) (m + .2)2 = 0

m = + 1, + 1, - 2, - 2

or

=>

=> Repeated real roots exist. Hence, general solution is

yi - (C| + C2 a:) ex + (C3 + C4 a:) e ^

TEST YOURSELF-1Solve the following equations :

A* dx2 + i dx

3. (D4 - D3 - 9D2 - 11D - 4) y = 0.5. (D3 - D2- 12D) y = 0.

2. (D3 - 9D2 + 23D - 15) y = 0.

4. (D2 + l)2 (D - l)2 y - 0.6. (D4 V 2n2D2 + n4) y = 0.

ANSWERS

+ 2y = 0.

1. y = Cle x+C2e~2x3. y = ex (Cj + C2 x + C3 x2) + C4 e4j:4. y = (Cj + C2 x) sin x + (C3 + C4 x) cos x + (C5 + Q x) ex5. y = C\ + C^e4x + C3 e~3x6. y = (C|^f'C2x) cos nx + (C3 + C4x) sin nx

Particular Integral:Consider the differential equation

AD)y = B => y =

2. y = C, ex + C2e3x + C3 e5x

1 . B.

1Let . B denote some function of x, which when operated upon by J(D) produces B.

P.I. = B.

AD)Hence, AD)

• 13.3. GENERAL METHOD OF FINDING P.I.Result If B is a function ofx, then

■1 B = en B e"™ dx.D-aSOLVED EXAMPLES

Example 1. Solve D2 - 5D + 6 = e3x.Solution. The given equation can be written as

(D-3) (D -2) y~ e3x C.F. = C| e3;f + C2 e2*

2x 3x e j e e11 13* '^dxand P.I.= D-3'D-26 D-3 1

D-36- 3x 1 3j:ax = xe .. e

>88 SelfJnstructional Material


Now, General solution = C.F. +P.I.y = Ci e3x + C2 e2* + x e3x.

Example 2. Solve (D2 + 1) y = sec1 x. Solution. Here, the given equation is

(Z?2+ I) y = sec2 j:To find the C.F. of (1).The auxiliary equation of (1), we get

m2-i- l = 0 =$ m = ±i C.F. = Cj cos x + C2 sin x

1 1 11 1sec2 x = 2 1sec * = — sec2*P.I. =Z>2+ l 2i D-i D + i{D + i){D-i)

[*\^ sec2 xdx - e u J ex sec2xdx12i

cos * + i sin *cos * -1 sin * dx - dxcos2* cos2*

= ~ (e1* l (sec* - i sec* tan*) dx- e ^ {(sec*+ i sec* tan*) dx]

= ~ {(e1*. — e~lx) | sec xdx - i (eix + e-w) J tan * sec * d*}

1 (2t sin * log (sec * + tan *) - 2i cos * sec *)2i- sin* log (sec* + tan*) - 1.

Hence, the general solution is.y = C.F. + P.I.y = Ci cos * + C2 sin * + sin * log (sec * + tan *) - 1.

• TEST YOURSELF-2Solve the following differentiaLequation : t1. (D2 + a2) y = sec ax. 2. (D2 + a2) y = tan ax. 3. (D2 + 1) y - cosec *.

ANSWERS

* 1y = C[ cos a* + C2 sin <2* + — sin ax + —: cos ax log cos axa a

1.

2, y = Cj cos ax + C2 sin a* —^ cos ax log tan f? +a V 2,

3, y = Q cos * + C2 sin * + sin * log sin * - * cos *

• 13.4. SHORT METHODS OF GETTING P.I.The general method for getting P.I., discussed above requires lot of calculations. In certain

cases the P.I. can be obtained by methods which are shorter than the general method.(1) To evaluate P.I., when B is of the form e®*, we use the following rule ’

1 ax _eax AD) e /(fl) ' provided f(a) * 0.

SOLVED EXAMPLESExample 1. Solve (Dz ~ 3D + 2) y = e5x. Solution. Here, the given equation is

(D2 - 30 + 2) y = e5*.Auxiliary equation is

m2 - 3m + 2 = 0(m - 1) (m - 2) = 0 =❖ m = l,2.

CF = Clex+C2e2x

JSelf-InstructionalMaterial -89

Calculus & Geometry 1 _____________J_ 5x25 - 3 x 5 + 2 '' 12

15xNow, P.I.= . eD2 - 3D + 2

Hence, the general solution isy = C.F. + P.I.y = ClS + C2e2* + ±.<?'.

Example 2. Solve (D3 + 1) _y = (ex + l)2. Solution. Here, the given equation is

(D3 + 1) y = (Z+l)2 ...(1)The auxiliary equation is

m3 + 1 = 0(m + 1) (w2 - m + 1) = 0

1^ 2

C.F. = C]e‘^ + ej:/2 C2 cos ^

[^+1]2 =

=> m = 2 .V3^', ^ • xy- + C3sin —

\[e2* + 2ex + 1]

therefore

1 1Now, P.I. =(D3 + l) (D3+l)

(e2* + 2c* + e0x) =1 11 1^ + 2 OxeA' + -----eD3+ 1D3+ 1 D3+ 1 D3 + l

2k + £?*+1.1 1 1____ Ox _ j.

0+1* 96e2* + 2 —23 + 1

Here, the general solution is y = C.F. + P.I.

y=C|C j: + cx/2 C2 cos

P+ 1

>/Til. 1+ C3sm — 2x + ex+L+ 9e=>7J

• TEST YOURSELF-3Solve the following differential equations :(D2 ~4D + 1) y = e2x - e~x.(4D2 + 4D ~ 3) y =

2. (D2 + 5D + 6) y - e2f.4. / - 2D + 1) y = 2e5x/2.

1.3.

ANSWERS

y = e2* (Ci cosh WT + C2 sinh xVJ) - T e2* - 7 e3 0

y = Cle'2" + C2e-3jc + ^e2t. Xy-Qe^ + Qe

e^2

-X1.

1 _2x- 3x/22. + 2fe84. y - (C, + C?*) + -

(2) To Evaluate P.I., when £ is of the form sin ax or cos ax :Case (I) : If/(D) contains even power of D :

, sin ax

Case (II): If/(D) contains odd powers of D : If/(D) = /, (D2) + Df2 (D2) 1 . /i(~a2) sin ax -fz(- a1)

mSl ' {fi(~ a2)}2 + a2 a2)}

—sin ax ~ad1)

a cosax

SOLVED EXAMPLES^-3^ + 2y = cos 3*. dx2 dx *Example 1. Solve

Solution. Here, the given differential equation can be written as (D2 - 3D + 2) y = cos 3x

To find C.F., the auxiliary equation is...(1)


Linear Differential Equations'with Constant Coefficients

m' - ~hm +2=0(m - 1) (m - 2) = 0

which gives m = 1 and m = 2.• C.F. = Cj ex+C2e2x.therefore,

1 1 (v D2 = - a2 = -9)cos 3xNow, P.I.= cos 3^ =D2 - 3D + 2 - 9 - 3D + 2

(7 - 3D)1 cos 3x = - 2 cos 3x(72 - 9D2)-7-3D

(7-3D)72 - 9 (- 9)

7 ^ 9 • i- ttt cos 3x - ttt sin 3a: =

1 [7 cos x - 3D cos 3x]cos 3x = -130

1 (7 cos 3x + 9 sin 3x).130130 130

Hence, the general solution of (1) is given byy = C.F. + P.I,

y = C] e* + C2 e2* - -j— [7 cos 3x + 9 sin 3xj.


(D2 + 9) y = cos 4x.(D2 - 3D + 2) y = sin 3x.

2. (D2 - 2D + 5) y = sin 3x.4. (D4 + 2D3 - 3D2) y = 3e2x + 4 sin x.

1.3.

ANSWERS

l1. y = C\ cos 3x + C2 sin 3x- — cos 4x

2. y = ex [Ci cos 2x + C2 sin 2x] + (3 cos 3x - 2 sin 3x)2613. y = Cj / + C2 (9 cos 3x - 7 sin 3x)

3 4 24. y = (Ci + C2 x) + C3 c* + C4 e” ^ + e2* + - sin x + - cos x

(3) To Evaluate P.I. when B is of the form of xm, when m is positive integer :

r.c., To evaluate xm, m G Z+

/(D) =A0D" +Aj D"_I + ... + A„, we apply working procedureand

l)xw~2m -1 m (m - 1) 2.11 1 mxm ,— X + •••(2).+... ++D-aX a 2 ama a

Here, we observe that (1) and (2) are the same.Working Procedure :

Take the lowest degree term from f{D) and remaining factor will be of the form [1 +/D)] or [1 ~f{D)}. Now, this factor can be taken in the numerator with a negative index, which can be expanded by Binomial theorem. Here, it shoula be noted that the expansion is to be carried upto

m + 2xm = 0 and all other higher differentialm + 1the term Dm, since, we always have D coefficients of xm are zero.

xm = 0, D

SOLVED EXAMPLESExample 1. Solve (D2 + D - 2) y = x + sin x. Solution. Here, the given equation is

(D2 + D - 2) y = x + sin x

To find C.R, the auxiliary equation is rn2 + m - 2 = 0

(m - 1) (w + 2) = 0 => m = 1,-2

...(1)

Self-Instructional Material -91

Calculus & Geometry C.E = C, ex + C2e~lx

P.I. = 1 1 1Now, (x + sin x) = x + sin x

(D2 + D- 2) (D2 + D - 2) (D2 + D- 2)1 1x + sin*

- I2 + D-2-2

2 2 2

i• (P + 3)P-3) (0 + 3)* + sin*5 — '—

11 1 P + 3) ■ D2 -9

D + 3 .1-2[1+^ + -

if A f i* + — - — [O (sin *) + 3 sin *]

. * + JC + T +sin * = - sin *22 - 1-9

2 2\ 7

= -|*-^--^(cos* + 3 sin*).

Hence, the complete solution is given by y = C.F. + P.I.y=Ci^ +C2^~2x~~x-^ - — (cos* + 3 sin*).

i

• TEST YOURSELF-^Solve the following differential equations :(O3 - D2 - 60) y = *2 + 1.(03 + 202 + D)y = eZr + *2+*. 4. (O3 - 30-2)y =*3.

2. (O4 - a4) y = *4.1.3.

ANSWERS

1. y = Ci + C2 e3* + C3 e“ 21 - 25 1__X-J-J + J-J108 18 36

4 242. y = C] e™ + C2 e '**•+ C3 cos a* + C4 sin;fl* - ^7a

3 2 /-* +4*n8

1 2t + ^*33. y - Ci + (C2 + C3 *) e ^ + Jg e

4. y = (C1 + C2*)e^ + C3e1,:-|*3 + |*2-~-*+ 153

~ A', where X is any function of * :(4) To evaluateAD)

1 1[em. X]=eax .X .AD) AD + a)

Working Procedure :Replace D by (O + a) and brought e™ before the operator —7—. After that, determine

AD)1 . X as usual.

AD + a)SOLVED EXAMPLES

Example 1. Solve (D2 + 4D - 12) y = (* - 1) e2*.Solution. Here, the given differential equation is

(D2 + 40 - 12) y = (* - 1) e2*To find C.F. of (1), the auxiliary equation is

m2 + Am -12 = 0 => (m-2)(m + 6) = 0

ii 5. ' ■ •••(I)

which gives m = 2 and m = - 6C.F. = Cj dtx + C2e~6x



1 1e2x(x-l)=e2xNow, P.I.= (■*-1)[(D + 2)2 + 4 (D + 2) - 12]

r(^-D .{D2 + AD - 12)

1 1 •(*-1)=^lx= e(D2 + 8D) ( D

8D \+~8i

8 8 D * 8

i_! 2*1 8 D

1 2* 1 = 8e

__L 2*f^ 9 )8e [2 gy

Hence, the general solution of (1), is given by y = C.F. +P.I.

2- 2* , -d* , 1 2t •* 9y = C|6 +C2e +ge ~2~sx '

11 D .. (*-l)1 -~D +8 8

(- IU9 1- 1 — ~ = &e 8-


(D2 -2D+l)y = ex. x2.(D1-l)y = e‘(l+x2).

2. (D2 - 5D + 6) y = x3. e2*. 4. (D2 - 4Z) + 1) y = e2^ sin x.

1.3.

ANSWERS'v.

2x X4y = (C1 + C2x)ej: + ^/./

y = Cj / + C2 e" x + vr ^ [9x + 2^ - S^2]

—+ x3 + 3^2 + 6^ 4

e(2-VT)x;

2. y = C, + C2 e3* - e

(2+VT)

1.

1 ^ sin jcX + C24. y = C!d3. 'A6120

(5) To evaluate e™, when/(a) = 0 :

Let us supposey(a) = 0.In this case (D - a) is at least one factor of AD).

AD) = (D-a)rg(D), where g(a)*0.Let11 1 11 ax ax axThen AD) 6 (D - a)r g(a) 6 g(a)'(D-aye

ax | <«« J *1 1 .e'^dxg(a) ' (D - a)r -1

1 1 1 1 ^.e-^dxaxzr\xe er- 2g(a) ' (D - a)r 1 1 _

g(a) ' (D - a)r~2 ' 2 ! 6 ‘

g(a) ' (D-a)2x ax

Proceeding in the same way, finally, we get

m** s(a) r!e<11SOLVED EXAMPLES

Example 1. Solve (D2 + D-6)y = e2*. Solution. Here, the. given equation is

(D2 + D - 6) y =.e:u.To find the C.F. of (1), the auxiliary equation is

m2 + 01 — 6 = 0(m +-3) (m - 2) = 0 => m = 2, - 3

C.F. = Ci e^ + Cre-**

...(1)v

=>

Self-Instructional Material 93'

Calculus & Geometry , . 1 12x 2xNow, P.I. = e .= (D + 3)(D-2)eD2 + D-61 12x ZX . 1(2 + 3) (D - 2) e 5 (D - 2) e

= ±-e2x5 (D + 2) - 2Hence, the complete solution of (1) is given by

y = C.F. + P.I.y=Clei,+ C2e^x + \xe2t.

o

;i !e^± l=Ire^ 5 e D 5. 1 =

• TEST YOURSELF-?Solve the following differential equations :

(D2 + 4D + 3)y = e~3x.(£>4 + D2 + D2 - D-2) y = ex.(D~ l)2 (D2 + l)2 y = e*.

2. (D2 + 6D + 9)y = 2 e~4. (D2 -9D + 18) y = cosh 3.r.

1.3.5.

ANSWERS■ \4 V

= Ci e~x + C2e~3x-~ - 3x 2. y = (Ci+C2x)e'2x + x2e~3x-

\[7~ x-C3 cos —at + C4 sin

1. 2" ^rvr \ ■ i-x/23. y = Ci e* + C2 e ■* + e

4. y = C| e1* + C2 e6x-^o

y = (Q + C2 J:) e',r+ (C3 + C4x) cos j: + (C5 + C6^:) sin;c + e*

+ 8^y J1 v3,+ -3j <

108 e1_*•

5.8

sin or, or cos or, when /(- a2) = 0, we use the following rules :(6) To evaluate

1 . x x \ • j. —r sin ax - - — cos ox = r sm ax ax D2 + a2 2a 2J

x . x f — smax = -J

and 1 cos ax dxcos ax =D2 + a2 2a

SOLVED EXAMPLESExample 1. Solve (D2 + a2) y = sin ax.Solution. Here, the given equation is

(£>2 + a2) y = sin ax.To find the C.F. of (1), the auxiliary equation is

m2 + a2 = 0 => m = 0 ± aiC.F. = [Cj cos ax + C2 sin ax] = [Cj cos ax + C2 sin ax]

2 sin ax

.-•(1)

1Now, P.I.D2 + a ir

fj xsin ax dx = - — cos ax.2aHence, the complete solution of (1) is given by

y = C.F. + P.I.Xy = C| cos ax + C2 sin ax - — cos ax.2a

• SUMMARYSolution of D.E

h &tf + P* + Q*‘* *

is of the form


1t

c


y = C.F+P.IC:F=clmm'x + c2em*(,

and mi and m2 are the roots of m2 + Pm + (2 = 0.

f(D) = D2 + PD + Q.

T ’•

where

RAlso, P.l =f{D)'

If mi = m2 = m, thenmxC.F. — (Ci + Cjx) €

• If R = eax, f(a)*0ax ax

p.i.=4. ef(D) f(a)

• If i? = fl',x,/(a) = 0, thenax 1PI —- e ----------

•• f(D) /p + fl) • If R = sin ax or cos ax, f(- a )*Q, then

sin ax

ax■r. •

cos axcos ax sin axP.I. = or ----- — z- or -/(D2) /(-a2) /(-«2)f(D2)

• If It = sin ax or cos ax,f(- a1) = 0, thenX cos ax xsin ax

2 2 — — /i cos ax,D2 + a2 2a• If /? = xm, me N, then

= — sin axD2 + a2 2a

xm = [/(D)] 1 xm [expanding [/(D)] 1 xm [expanding [/(D)]-1 by Binomial theorem]. PI =m• STUDENT ACTIVITY1. Solve (D4 + /:4)y = 0

f'

2. Solve 3 -+- + 2y = cos 3x. dxdx1


Calculus & Geometry • TEST YOURSELF-8V.)

2. (D2‘+4)y = cos2x. .4. (D2 + a2)y = sin ax.

■Solve the following differential equations :(D2 + a2)y = cos ax. .(D2 + 4) y = e* + s\n2x.

1.3.

ANSWERS

X '1. y = C{ cos ax + C2 sin ax + -—!£

2. y = Ci cos 2x + C2 sin 2x + — sin 2x. sin ax.2ay = C-i cos 2x + C2 sin 2x + ^- e-1 cos 2x.

x4. v = Ci cos ax + C2 sin ax - — cos ax.la

3.

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. If the order of the given differential equation is n, than C.F. of this equation contains

arbitrary constant.2. If the order of the given differential equation is n, then

any arbitrary constants.V TRUE OR FALSE :Write ‘T’ for True and F for False Statement:1. The particular integral of an n,il order differential equation contains n independent arbitrary

(T/F)2. : The-complementary functions of a differential equation of order n contains n independent

arbitrary constant..> MULTIPLE CHOICE QUESTIONS :

one : -

u

of this equation does not contain

*constants

(T/F)

' W 4

Choose, the.most appropriateThe solution of (£> + 1) y = 0 is y = :

A cos x- Bimx(a) (b) - A cos x - fi sin x(c) A cos x + B sin x

2. The general solution of the differential equation (D2 + 1) y = 0 is : (a) y = cos x (c) y = Ci cos (x + Ci)

(d) - A cos x + B sin x. ■I

(b) y = C cos x(d). Ci cos (Ci + C3 x).

ANSWERS

Fill in the Blanks :1. n 2. Particular integral

True or False :1. T 2. F

Multiple Choice Questions :1. (c) 2. (c)

□□□


Homogeneous* Linear Differential EquationsUNIT

HOMOGENEOUS LINEAR DIFFERENTIALEQUATIONS

J STRUCTURE• Homogeneous Linear Differential Equations• Solution of Homogeneous Linear Differential Equation

• Test Yourself-1• Summary> Student Activity

LEARNING OBJECTIVESAfter going through this unit you will learn : j

+ How to get solution of homogenous linear differential equations, j

• 14.1. HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONSDefinition. Any differential equation of the form

. 1

-14" y- \ + ...+Any-X]dxnis called a homogeneous linear differential equation of nth order, where •• •, are constantsand X is a function of a: or a constant.

dxn

For example, consider the following differential equations :

dx2 dc2^ + 2xdl

dx2 dxThe above two differential equations are linear as the dependent variable y and its derivatives

appear in their first degree and are not multiplied together.

+ 4y = /(i) jc

+ 2y = e*(H) x

i

• 14.2. SOLUTION OF HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONConsider the homogeneous linear differential equation

i n - 2

dxn 'where A,, A3, ...,An are constants and X is a function of x or constant.

Now, equation (1) can be transformed to an equivalent equation (linear differential equation) with constant coefficients by changing the independent variable by the relation.

x ~ ez i.e., z = logx I

-1_2 + ... +-A„y-X ...d)-1dx*dxn

Working Procedure ;(0 Put x-ez,x-~ = D

dx dzd2— = D (D - 1) and so on.dx2

(ii) Obtained the equation in terms ofD (linear equation).(Hi) To find the C.F. and P.1, used the usual method given in chapter 16.(iv) Find genera! solution by adding C.F. and PI.(v) Finally, substitute z = log x.

1

SOLVED EXAMPLESniz dx ...(1)Example 1. Solve x -4x + 6y=x.

dx2 " dx

Self-Instructional Material 97-

Calculus <& Geometry Solution. Putting . x = ez

z = log * and D = -—• •. •=>dz

Thus, the given equation becomes[D(D-\)-4D + 6]y = ez =* (D2 - SD Jr 6) y ~ ez ...(2)

which is a linear equation in y.To find the C.F. of (2), the auxiliary equation is'

m2 - 5m + 6 = 0(m - 2) (m - 3) = 0 which gives m - 2, 3.

C.F. = Cx elz + C2=>

11 1cz =Now, P.I. =

1-5 + 6D2 - 5D + 6 Hence, the general solution is given by

y = C.F. + P.I.y = C\e2z + C2 eZz + ^ ez.

Now put ez = x

y = CiX2 + C2x3

Example 2. Solve x2-3x^ + 4y = 2x2.

Solution. Here, the given equation is

^4-3** + 4, = 2>.dx2 dx ' ...(0

2^1Let x = ez, then z = logx, x^- = Dy,x = D (D- 1) ydx2dx

Put in (1), then (1) becomes{D (D - 1) - 3D +4) y = 2e2z

(D2 - 4D + 4) y = 2e2\To find the C.F. of (1), the auxiliary equation is

m2 - 4m + 4 = 0 =» (m - 2)2 = 0 => m = 2,2 C.F. = (Ci + C2 z) e2z = (Ci + C2 log x) x2

...(2)=>

1 12 26^ = 2 2^Now, P.I. =(D-2) (D-2)

. 1 = 2e2z 4.1= 2e2' .^- = z2e2i21= 2e22 = (log x)2. X2.

' 2[(D + 2) - 2]Hence, the complete solution of (1) is given by

y = C.F. + P.I.y = (Ci + C2 log x) x2 + x2 (log x)2.

D

=>

• SUMMARY• Homogeneous linear D.E

^ + A,/ dx11

• Solution Homogeneous Linear D.E(i) Put x = ez or z = log x, and x ^

etc. in the given D.E.(ii) The given D.E becomes

f(D)y = x,D = j^(iii) Find C.F and P.I as usual obtained in previous chapter.(iv) y= C.F + P.I(v) Put z = log x in y = C.F + P.I

d?~xh y-1x" + Any = x.+-1dxn

d x2d2 xV = D9D- l)(D-2)= D (D - 1),~ D = —rdz dx1 dx3

\


Homogeneous Linear Differential.Equationsi• STUDENT ACTIVITY r-ni •••-.

Solve x2 ^ -4x^- + 6y = x. dx2 d*

'■s i t1.

2 try dy + \2>y - log x.2. Solve d t + 7x , dx2 ^

I

• TEST YOURSELFSolve the following differential equations :

2 d2y _ dy dx2 dx

d^dx

\2. x2^

dx2

4.

&- 4x~f- + 6y = jc4. dx

34-^*+‘xy=Ldx

1 + y=x.

r + 2v . fit3 dx2

3.

ANSWERS

2. y = Clx2 + C2x3 + ^x4.1. y = X (C] + C2 log jc) + ^ (log x)2.

C 13. y = Cj x2 + —r + —a:2 log a:. x 4

14. y = (C] + C2 log a:) a: + C3 a:- 1 + -7- log x4x

OBJECTIVE EVALUATION>► FILL IN THE BLANKS :

In any homogeneous equation x .......... , only.

22 + x + v = X the X is either constant or a function of

dx v

2. The homogeneous linear equation can be reduced to a linear differential equation with constant coefficiants by putting x=...........

i

dx2

> TRUE OR FALSE :Write T for True and F for False statement:1. The homogeneous linear differential equation can be reduced to linear differential equation

with constant coefficients.2. The differential equation, in which the power of x in the coefficients are greater than the order

of its derivative, associated with them, is called the homogeneous linear equation.

(T/F)

(T/F)1

Self-Instructional Material 99/

Calculus & Geometry 3. The differential equation jn which the power of x in the coefficients are equal to the order of its derivative, associated with them is called the homogeneous linear equation. (T/F)

> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :1. The particular integral of (x2 D2 - 2xD) y = log x is :

(a) log x

(c) A log x2. The P.I. is (x2 D2 + 5xD + 4) y = x log .*• is given by :

(a) -^xlog*

, , 1 . 2 (c) -x\ogx- — x

(b) - - log * - - (log *)2

(d) none of these.

(b)|p(d) none of these.


1. *True or False :

2. ex 3. First 4. equal

3. Tl.T 2. FMultiple Choice Questions :

1. (b) 3. (a)2. (c)□□□

100 Self-Instructional Material\

WronskianUNIT •iiw uajtiijiiipa ifcoi

WRONSKIANSTRUCTURE

• Linear Dependence and Independence of Functions• The Wronskian

I Summary• Student Activity• Exercise


# About linear independent and dependent functions.# How to find the nature of the given function through Wronskian.

• 15.1. LINEAR DEPENDENCE AND INDEPENDENCE OF FUNCTIONS(a) The functions f(x), i = 1,2 ... n are said to be linearly dependent on an interval

a<x<b if there exist a set (a\, a2 ... an) of constants, not all zero, such thata\f\{x) + a2f2(x) + ... + anfn(x) =0ina<x<b

(b) The function f (x), i = 1, 2 ,...,« are said to be linearly independent if the only set of constants a\, a2, ■■■ an for which (1) holds, is the set = a2 = ■ ■ ■ = an = 0.Examples :

...d)

(i) sin x, 3 sin x and - sin x are linearly dependent on the interval - 1 < x < 2.(ii) x and x2 are linearly independent on 0 <x< 1 since

Q x + C2 x2 = 0 V x on 0 < x < 1 =* C| = C2 = 6.

• 15.2. THE WRONSKIANLety\, y2, ■■•,yn be n functions ofx such that each possesses at least (n - 1) derivatives. Then

the determinantCalculus 1

>*1 J2 •• y\ yi ■■

y,"-11 y?-’

ynyn

(«-l)Tnis called the Wronskian of the functions y\, y2 and denoted by Wiy^yi, ...,yn) or simply W.

• SOLVED EXAMPLESExample 1. Show that the solutions ex, e x and e2* of

S-S-s**-are linearly independent. Also, solve the given equation.

Solution. Since ex, e~x and e7* satisfy the given differential equation. So they are solutions of the given equations.

2x-xex e e

4^W[e,,e-',«2r]Now,

d2 i(e~X)T^) !dx1dx


Calculus & Geometry It/ e^ ~e-x le2* S e~x Ae2*

- X 1 1 1.e2* 1-1 2

1 1 4

e.juauuQt- = f.e-X

■i

After simplify, we get.I1 1 1

W^e2* 2 0 3 0 0 3

= e2* x (- 6) 0.

IHence, the solutions ex, e x, e2* are linearly independent solution of the given third order

differential equation, so it can possesses at most three independent solutions. Therefore, the complete solution is given by y = Cj e* + C2 e~x + C3 22*, with C2 and C? as arbitrary constants.

Example 2. Show that the Wronskian of the functions x~,x2logx is non-zero. Can these functions be independent solutions of an ordinary differential equation, if so determine the equation.

Solution. Let 2y\ =xy2 = x2 log x.and

Then, the.Wronskian isx2 x2 log x 2x 2x log x + x

yi yiy\ y*

= x**0.W(x) =

The functions yt and y2 are linearly independent solution of an ordinary differential equation.=>Now, we find the equation

y = Clyl + C2y2 = Ci x2 + C2 x2 log atLet

d£ = 2CiJt + 2C2 xlogx + C2x

and —2 — 2C| + 2C2 log x + 3C2. dx

Now, eliminating C\ and C2 from the above three relations, we get

dx2 dx+ 4>> = 0.

• SUMMARY• Linear dependent functions : Functions/i (x),f2 (*),/3 (x),

. be linearly dependent on [u, b] if these exist a2,an, not all zero such thata/, (x) + afc (x) + a3fi(x) + ...+ aj,, (x) = 0

• Linear independent functions : Function«i/i W + a/2(x)+/3(x), (x) + ... +anfri(x) = 0 =$ a] =0 = a2= ... = a„.

• Wronskian:

,/„ (x), x £ [a, b] are said to

yi yi yi y{ '^(y„y2) =

yi yi yi yi' yi y{ y\ yi' yi'

yi) = etc.

• STUDENT ACTIVITY1. Show that the functions x, x2, x3 are linearly independent.


Wmnskian2. Show that the functions sin x, sin 3x, sin3 x are linearly dependent.

i

• TEST YOURSELF1 Prove that the functions l,x,x are linearly independent. Hence, form jthe differential equation

whose roots are 1, ;t2. .;2. Show that if mi * m2, the functions emi x and emiX are linearly independent.3, Show that the following functions are linearly independent ex cos x, e* sin x.

Hence, form the differential equation of second order having these two functions as independent solutions.Which of the following sets of functions are linearly independent for all x.

(ii) sin x, sin 3x, sin x (iv) x2-x+1, x2 - l,3x2-x-l.

4.(i) sinx,cos*(iii) sin *, cos *, sin 2x

ANSWERS

3. ^-2^ + 2y = 0 dx2dx

4. (i) Linearly independent (ii) Linearly dependent(iii) Linearly independent (iv) Linearly independent.

1. ^ = 0dx2

□□□

1

\

:



GENERAL EQUATION OF SECOND DEGREESTRUCTURE

4Introduction Condition for a Pair of Straight LinesEquation of a Pair of Lines through the Origin and Parallel to the Lines/ix+ miy+ ni =0 and fcx+ m2y+ nz = 0Angle between the Pair of LinesCondition for Perpendicularity of a Pair of LinesCondition for Parallelism of the LinesCondition for Coincidence of the Lines ax2 + 2hxy + by2 + 2gx + 2fy+ c = 0 Point of Intersection of the LinesCondition for the General Equation of Second Degree to be a CircleCondition for the General Equation of Second Degree to be a ParabolaCondition for the General Equation of second Degree to be an EllipseCondition for the General Equation of Second Degree to be a Hyperbola• Summary• Student Activity• Test Yourself


• How the given general equation of second degree reduces to pair of straight lines circle, parabola, ellipse and hyperbola.

• 16.1. INTRODUCTIONThe general equation of second degree in x and y is given by

ax2 + 2lvcy + by2 + 2gx + 2fy + c = 0

where a, b, c,f g and h are six constants. If the equation (1) is divided by any one of six constants, then the-equation so obtained will have five constants, therefore in order to determine the values of these constants, there will required five equations between them. The solutions of these five equations will give the exact equation of degree two.

• 16.2. CONDITION FOR A PAIR OF STRAIGHT LINESTo find the condition that the general equation of second degree may represent a pair of

straight lines.Let the general equation of second degree in x and y be

ax2 + 2hxy + by2 + 2gx + 2fy + c-0. -.(I)Suppose the equation (1) represents a pair of lines and let these lines be

lix + m\ y + ni=0 Itf + m2y + n2 = 0.' ...(3)and

Then we have2 2(/jx + mj y + n{) (l^x Jr miy + nf) =ax + 2hxy + by + 2gx + 2fy + c = 0.

2 2Now comparing the coefficients of x , y , xy, x, y and constant term of both sides, we get


General Equation oj Second DegreeI\l2~a,m\m2 = b,n\n2 = c mj/12 + «im2 = 2/, + n2l\ = 2g ■ .

Iini2 + l2m\ = 2/i••.(4)

andNow eliminating mb nj and I2, w2, n2 from (4), for this we have Geometry and Vectors

/, l2 0mi m2 0ti\ n2 0

l2 lx 0m2 mi 0«2 ^1 0

= 0x

( .' the value of each determinant is zero)Multiplying both determinants row by row, we have

l\m2 + l2mx l\n2 + /2/ii min2 + m2n\

2nxn2

12/]/2

hm\ + hm2 2mim2 hn\ + hn2 ,n2nl +

= 0. ...(5)

Putting the values given in (4) in (5), we get 2a . 2h 2g 2h 2b 2/2g 2/ 2c

= 0

a h g h b f =0 8 f c

a (be ~/2) - h (he -gf) + g(hf- gb) = 0 abc + 2fgh - af2 - bg2 - ch2 - 0.

This is the required condition.

=>

=>

• 16.3. EQUATION OF A PAIR OF LINES THROUGH THE ORIGIN AND PARALLEL TO THE LINES hx+mi y+m = 0 AND hx+m2y+m = 0.

Let the equations through the origin and parallel to the lines Itf + m2y + n2 = 0bt

1* + /«] y + Ki = 0 and

l\X + nil y = 0 /^ + m2 y = 0

where /b l2, mb m2 are governed by the equation (4) in above section 2. Now the combined equation is

(lxx + mi y) (I2X + m2 y) = 0li[2x + xy (/im2 + l2m\) + m]m2 y =0.

Putting lil2 = a, mim2 = b, lim2 + l2mi = 2h in (2), we get ax2 + 2hxy + by1 = 0.

This is the required equation of a pair of lines.

...d)and

-(2)or

• 16.4. ANGLE BETWEEN THE PAIR OF LINEShx+m^y+n^=Q and kx+m2y+m-O.

The angle between the lines /i* + mi y + = 0 and l2x +m2y + n2 = 0 is

IVtf-abtan 0 =

a + b x*\%

%lAQ.2o.

//fiT*

1 '

$\

Figl


Calculus <£ Geometry • 16.5. CONDITION FOR PERPENDICULARITY OF A PAIR OF LINESa/ + 2hxy+b)f + 2gx + 2fy+c = 0.

Let 9 be the angle between the a pair of lines, then we have,2^h2 - ab

tan 0 - a + bFrom this equation, if a+ 6 = 0, then 0 = 90° i.e. the lines are perpendicular. Hence, the

required condition for the perpendicularity of the lines is given by ax1 + 2hxy + by2 + 2gx + 2fy + c = 0

a + 6 = 0.is

• 16.6. CONDITION FOR PARALLELISM OF THE LINESax2 + 2hxy +f>y2 + 2gx + 2fy+c = 0.

Since the angle between the lines is governed by the relation2^h2 - ab

tan 0 =a + b

If the lines are parallel, then tan 9 = 0, then we get h2-ab = 0 or h2=ab.

This is the required condition for parallelism.

• 16.7. CONDITION FOR COINCIDENCE OF THE LINESax2 + 2hxy + b/^2gx+2fy+c = 0.

2 2The condition for coincidence of the pair of lines ax + 2hxy + by +2g x+ 2 fy + c = 0 isgiven by

k2 -ab = 0, g2-ac = 0, /2 - 6c = 0.

• 16.8. POINT OF INTERSECTION OF THE LINESax2 + 2hxy+ b/ + 2gx + 2fy+c = 0.

The point of intersection of the lines represented by ax2 + 2hxy + 6y2 + 2g * + 2/y + c = 0 isgiven by

V tf-ab V n2- ac- ab

• 16.9. CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE TO BE A CIRCLE

Let the general equation of second degree in x and y be ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.

Definition. The locus of a point in a plane, which moves in such a way that its distance from fixed point is always constant, is called circle.Then by the definition of a circle, the equation of a circle is

...d)

{x-txf + (y - P)2 = .2 .••(2)!

where (a, |3) is the centre and r its radius.Comparing (1) with (2) and then we observed that the general equation (1) represents l\ circle

ifa = 6 + 0 and h - 0.

Therefore, this is the required condition for the general equation of second degree to be acircle.

• 16.10. CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE TO BE A PARABOLA

Let the general equation of second degree in x and y be


General Equation of Second Degreeax' + by1 + 2hxy + 2gx + 2fy + c = 0.Definition. The locus of a point, which moves so that its distance from a fixed point (called

focus) is equal to the distance of this variable point from a fixed straight line, is called a parabola.Let /* + my + n = 0 be a fixed straight line and (a, P) be a fixed point, then by the definition

of the parabola, the equation of parabola is fo: + my + n

(lx + my + n)2 = (l2 + m2) [(x: - a)2 + (y - P)2]l2x2 + m2y2 + n2 + 2lmxy + 2lnx + 2mny = (/2 + m2) (j:2 + y2 - 2ca - 2py + a2 + p2) m2x2 + /2y2 = 2lmxy + 2x (~ In - a/2 - am2)

...(1)

V(*-a)2 + (y-p)2

ororor

+ 2y (- mn - p/2 - Pm2) + (a2 + p2) (l2 + m2) - n2 = 0

takes the form :or(mx - ly)2 + 2Gx + 2Fy + C = 0. ...(2)

From (2) it has been observed that the second degree terms in the general equation of a parabola forms a perfect square. Therefore, using this conclusion on the general equation of second degree equation (1), we can say that ‘

ax2 + 2hxy + by" + 2gx + 2fy + c = 0reduces to a parabola if

h2 -ab = 0.This is the required condition.

• 16.11. CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE TO BE AN ELLIPSE

Let the general equation of second degree in x, y be

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. • ■ -(1)

Definition. The locus of a point which means in such a way that its distance from a fixed point is always less than its distance from a fixed line, is called an ellipse.

On the based as discussed in above postulates, the required condition for the general equation given (1) to be an ellipse is

h2 <ab.

• 16.12. CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE TO BE A HYPERBOLA

The general equation of second degree in x, y is

ax2 + by2 + 2hxy + 2gx + 2fy + c = 0.

Definition. The locus of a point which moves so that its distance from a fixed point is always greater than its distance from a fixed line, is called hyperbola.

Therefore, (1) becomes the equation of a hyperbola if

h2>ab. ' ’ .

...0)

This is the required condition.SOLVED EXAMPLES

Example 1. Show that the equation of second degree 5x2 - 2*y + 5y2 + 2* - lOy - 7 = 0

represents an ellipse.Solution. Compare the given equation with

ax2 + 2hxy + by2 + 2gx + 2/y + c = 0 ja = 5,b~5,h = ~\,f--S,g = \,c~~l. !

& = abc + 2fgh-af2 - bg2 - ch2= (5) (5) (- 7) + 2 (- 5) (1) (- 1) - (5) (- 5)2 - 5 (l)2 - (- 7) (- l)2 = - 175 + 10- 125-5 + 7 = - 288 * 0

we get


Calculus & Geometry h2 -ab = (~ l)2 - (5) (5) = 1 - 25 = - 24 < 0 h2 <ab.

and

Hence, the given equation is an ellipse.Example 2. Show that the equation

Zx1 + %xy + 2y2 + 26*+ 13>» + 15 = 0represents a pair of parallel lines.

Solution. Since the given equation is

Sx2 + 8^ + 2/ + 26* + 13? +15=0.

Compare (1) with the equationax2 + 2hxy + by1 + 2gx + 2fy + c = 0

a = M = 4, * = 2, * = 13, /= 13/2, c = 15.A = abc + 2fgh - af2 - bg2 - ch2

= (8) (2) (15) + 2 fy) (13) (4)- 8 lyj - 2 (13)2-15 (4)2

...d)

we get

= 240 + 676 - 338 - 338 - 240 = 0./i2 - = 16 - 16 = 0.Also

Hence, the given equation represents a pair of parallel straight lines.Example 3. Show that the following equation represents a pair of lines. Find, also the angle

between them :6X2 + 13*? + 6?2 + 8* + 7y + 2 = 0.

Solution. Since the given equation is6X2 + 13*? + 6y2 + 8* + 7? + 2 + 0. -d)

Comparing (1) with the equationa*2 + 2hxy + by2 + 2gx + 2Jy + c = 0

a = 6, h= 13/2, b = 6, g = 4,/= 7/2, c = 2. A = abc + 2fgh ~ of2 - bg2 - ch2

we get,

= 6.6.2 + 2.1.4.H_6. I - 6 . (4)2 - 2 . j y

147 169= 72 + 182 - -y- - 96 - -y-

58_ 316 = 116^316 =0 2 2

Thus the given equation represents a pair of straight lines.

f2^h2 - ab-1The angle between these lines = tana + b

' Vf^2 2 V 2 -365 'l-i -1- tan = tan12 '6 + 6

• SUMMARY• General equation of second degree in *, ? :

/(*, ?, z) - ax2 + by2 + 2hxy + 2gx + 2fy + c = 0

where a, b, c,f, g and h are constants.

• Equation ax2 + 2A*y + 6y2 + 2£* + 2/y + c = 0 represents a pair of straight lines if abc + 2fgh -af - bg2 + ch2 = 0.

• ax2 + 2hxy + fcy2 = 0 represents a pairs of straight lines passing through the origin.2 2• angle between the lines represented by a* + 2hxy + by = 0 is given by

f 2 ab '-!0 = tan a + b


2 2• If ax + 2hxy + by + 2gx + 2/y + c = 0 represents a pair of straight lines, then these lines perpendicular ifa + b = 0 and parallel to each other if h2 = ab, g2.= ac knd f2 = be.

• If the lines intersect each other then the point of intersection is

General Equation of Second Decreeare

V?y h2-ab y h2- ac-ab J

2 2• The general equation ax + 2hxy + by + 2gx + 2fy + c-0 will represent:(i) a circle if a = b*0 and /i = 0, A * 0(ii) a parabola if/i2 = ab, A*0(iii) a ellipse if h2 > ab, A * 0(iv) a hyperbola if h2 < ab, A ^ 0 where A = abc + 2fgh - af2 - bg2 - ch2.

• STUDENT ACTIVITY2 21. Show that the equation 6x +I3xy + 6y + 8x + 7y + z = 0 represents a pair of straight lines.

2. Show that jc2 + 2xy + y2-2x-l=0 represent a parabola.

• TEST YOURSELF1. Show that the equation 25j:2 - 120jty+144y2-25x +60y - 36 = Orepresents two parallel

straight lines.2. Show that the equation 3*2 + 3y2 - 2jM- 3y - 5 = 0 represents a circle.3. Show that the following equation llx2-4xy+ 14_y2 - 58x - 44y + 71 = 0 represents an ellipse.4. Show that the following equations :

(i) 16x2 - 24^y + 9y2 - I04x - I12y + 44 = 0 (i) 4^2 - 4^y + y2 - 8^: - 6y + 5 = 0. represents parabolas. j

Self-Instructional Material, 109

0 95. Show that the equation 3^: -5xy-2y + 5a: + 11^ - 8 = 0 represents a hyperbola.

OBJECTIVE EVALUATION> FILL IN THE BLANKS :

1. The general equation of second degree in x and y has at least nature of the equation.

2, The equation ax2, + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines ifabc + 2fgh - af2 - bg2 - ch2 =................

V TRUE OR FALSE :

Write T for true and F for false statement:2 2L The equation ax + 2hxy + by + 2gx + 2fy + c = 0 represents a rectangular hyperbola if

« + 6 = 0 and ab - h2 < 0. (T/F)2 22. The equation a: +2hxy + y + 2gA; + 2/y+ c = 0 represents a circle if/t = 1.

3 The equation x2 +Axy + 4y2 + 2gx + 2fy + c = Q represents a parabola.

> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate are :

2 2l' The.equation ax +2hxy + by +2gx + 2fy + c = 0 represents a rectangular hyperbola if h2 > ab and :

(d) b = 0.

(d) Two straight lines.

Calculus & Geometry

constants to find the

(T/F)(T/F)

(b) a- b = Q2. The equation 9x2 + \2xy + 16y2 + 2gx + 2fy + c = 0 represents :

(b) Parabola

(a) <3 + 6 = 0 (c) a = 0

(a) Ellipse (c) Hyperbola

ANSWERS

Fill in the Blanks :1. Five 2. 0

True or False :1. T . 2. F

Multiple Choice Questions : 1. (a) 2. (b)

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System of ConicsUNIT

SYSTEM OF CONICS* .* ,> .■ STRUCTURE

Standard Forms of ConicsReduction of General Equation of Second Degree into a Conic Determination of the Co-ordinates of the Centre of a Conic Equation of a Conic when the Origin is at the CentreDetermination of the Lengths, Posiions and Direction of the Axes of the Central Conic Determination of the Eccentricity of a Central ConicDetermination of the Co-ordinates of the Focus or Foci and the Equation of the directri) or Directrices• Summary• Student Activity• Exercise

LEARNING OBJECTIVES}

After going through this unit you will learn :What are conics in detail.How to calculate the centre, lengths, directionof its areas and its eccentricity.

• 17.1. STANDARD FORMS OF CONICS(I) Standard form of an ellipse. The standard form of an ellipse is

2 2X )L = i:.2 + ,2 a bCase 1. If a > b, then its major axis is x-axis and its minor axis is y-axis. Also the length of

semi-major axis is a and semi-minor axis is b. jCentre is (0, 0), focus are (± ae, 0) where b2 = a2 (l - e2). .The equations of directrix are x ± a/e = 0, also all the four vertices are (± a, 0) (0, ± b).Case 2. If 6 > a, then its major axis is y-axis and minor axis is x-axis of length 2b and 2a

respectively.Centre is (0, 0), focus are (0, ± be), where a2 = b2 (l - e).Also, the equations of its directrix are y ± b/e - 0, and all the four’vertices are (±a, 0),

(0,±6).(II) Standard form of a parabola.(i) y2 = 4ax. Its vertex is (0, 0) and focus is (a, 0).Also the axis is x-axis and the equation of its directrix is x + a = 0.

y2 = - 4ax. Its vertex is (0,0) and focus is (- a, 0).Also the axis is x-axis and the equation of its directrix is x - a = 0.(ill) x2 = 4ay. Its vertex is (0,0) and its focus is (0, a).Also the axis is y-axis and its directrix is y + a = 0.(iv) x2 = - 4ay. Its vertex is (0, 0) and its focus is (0, - a).Also the axis is y-axis and its directrix is y - a = 0,(III) Standard form of a hyperbola. The standard form of a hyperbola is

(ii)

>22X 2L = l.-

2 ,2. a bHere, the centre is (0,0) and focus are (± ae, 0), where, b2 = a2 (e2 - 1). Also the vertex are (± a, 0) and the equation of its directrix are x ± a/e = 0.

Self-Instructional Material HI

Calculus & Geometry • 17.2. REDUCTION OF GENERAL EQUATION OF SECOND DEGREE INTO A CONIC

Let the general equation of second degree in x and y be ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.

Change the equation (1) to the equation in which the terra xy is removed. This can be done

such that the x change to xcosG-^sinG and y to

...(1)

1 2h-iby putting G = — tan

* sin 6 -f-y sin G. Thus transformed equation becomes Ax2 + By2 + 2Gx + 2Fy + C = 0.

a~ b

.-(2)Now there arises some cases:Case I. If A = B, then the equation (2) becomes a circle. Case II. If neither A nor B is zero, then (2) becomes

Ax2 + 2Gx + By2 + 2Fy + C = 0f 2 G ] ( i F'lA x2 + 2^-x +B y +2Iry +C = 0

/\ LJor

/V Gi n 2 ~ F F2! G2 F2- A1 y +2By+B* ~AA 2 j- 0 ^A x +2~ A -^r+C = 0x +or B

= £ + £_C A + BG [ F

+By+a /-2 •. f + f-C = D'

A x + — Aor

Let2 \2<: G F\ + B y + — = D.A x + —A

^ * G F \Now by shifting the origin to the point - — ■ The equation (5) is transformed toA B

...(3)B

Ax'2 + By'2 = D/2 /2

X y = l. -.(4)+or D/A D/BTherefore the equation (6) represents an ellipse if both D/A and D/B are positive. Also the equation (6) represents the hyperbola if D/A and D/B are of opposite sign. Case III. If either A or £ is zero, let us suppose A = 0, then (4) becomes

fiy2 + 2Fy + 2G;c + C = 0 By2 + 2Fy = - 2Gx - C

rT 2-u2F ^B y +Ty

or

= - 2G.x - Cor

F2= - 2G^ + ^ - Cor B

( Ff F2 C_ 2BG + 2G

_2G F2 C | B X 2BG + 2G I

zL2BG 2G ’ B

= - 2G x-or> <2

FBy + z;or

, then above equation is transformed toNow shift the origin to

...(7)BThis equation represents a parabola. Similarly if we take £ = 0, then we shall again obtain a

parabola.

• 17.3. DETERMINATION OF THE CO-ORDINATES OF THE CENTRE OF A CONIC

Working Procedure for Finding the Centre :(1) First treating the given conic as a function 4>(-X, y) = 0.

112 Self-Instructional Materialv -

(H) Now differentiate (J)(jc, y) = 0 partially with respect to x and y respectively and System'of Conics

^and^-obtain dx dy^ = Oand^(HI) And solving ^

(x, y) of the conic.

= 0 for x and y. These value of x and y give the centredy

•17.4. EQUATION OF A CONIC WHEN THE ORIGIN IS AT THE CENTRELet the equation of a conic be

cue2 + 2hxy + by2 + 2gx + 2fy + c = 0The required equation of a conic referred to centre as origin is given by

ax2 + 2hxy + by2 + —r ab - h

where A = abc + 2fgh - af2 - bg2 - ch2.SOLVED EXAMPLE

Example 1. Find the co-ordinates of the centre of the conic whose equation is 32X2 + 52xy - ly2 - (Ax - 52y - 148 = 0 and hence, the equation of the conic referred to centre as origin.

-d).

= 0 ...(2)or

Solution. Let us assumey) s 32x2 + 52xy - ly2 -(Ax- 52y - 148 = 0.(1)

Differentiating (1) partially w.r.t. x and y respectively, -we get < •f’J'

90̂ = 64* + 52y - 64

90and = 52*- 14y —52.dy

Now solving = 0 and90 = 0, we get * = 1, y = 0.9y

Hence (1,0) is the centre of the given conic. Now the new constant d = ga +,/|3 + c

= (- 32) (1) - 26 (0) - 148 = - 32 - 148 = -Therefore, the equation of the conic referred. to centre as origin is

32*2 + 52*y-7y2-180 = 032 2 52* +

180.

7 y=i.or *y- 180180 180

• 17.5. DETERMINATION OF THE LENGTHS, POSITIONS AND DIRECTION OF THE AXES OF THE CENTRAL CONIC

Let the equation of a central conic beAx' + 2Hxy + By2 = 1.

Let us choose a concentric circle of radius r of the equation2,2 2* +y = r

2 •>* +r i •••(2)or

rThe equation of the lines joining the origin to the points of intersection of (1.) and (2) is

obtained by making.(1) homogeneous with help. of.(2), we. have( 2 , 2^* +yAx2 + 2Hxy + By2 - = 02r

-r *2 + 2Hxy + B -— y2 = 0. r r

•••(3)A-or

These two lines given in (3) will coincide if and only if the circle and central conic touch each other at the extremities of either axis of the conic and therefore, we have

1 1A 5 -- H2 = 0r r

| a

(v For a pair of coincident straight lines we know that -/i =0)

Self-Instructional.Materiql 113:

Calculus & Geometry ±-±(A + B) + (AB-H2) = 0. r r

(AS - H2) r4 - (A + S) r2 + 1 = 0Since the equation (5) is a quadratic equation in r2 so (5) gives two values of r2 Let these

values of r2 be r,2 and rj '.Now using (3) and (4), we get

.-(4)or

...(5) • 'or

i2

A -x + Hy =0

which is obtained by multiplying (3) by A - ^ and using (4).

Thus the equation of either axis is

ri).'

r

1A - ~ x + Hy = Q.

rJ

Since rt2 and r22 are the two values of r2 so that there arises some cases.

Case I. If both r| and r2 are positive and rj > r2“ then the conic will be an ellipse and the greater value of r i.e. rx will be semi-major axis while the smaller value of r i.e. r2 will be semi-minor axis. Hence the length of major axis is 2ri and the length of minor axis is 2r2.

Case II. If r2 > 0 and r2 < 0, then the conic will be hyperbola and the real value of / i.e. ri will be semi-transverse axis while the imaginary value of r i.e. r2 will be semi-conjugate axis. Hence, the lengths of transverse axis and conjugate axis are respectively 2r-[ and 2r2.

Now the equation (6) corresponding to r2 and r2 will give the equations of semi-axes of the conic as follows :

-(6)

1A - — A: + Hy = 0 ...(7)

r\1 'and A----- j x + Hy = 0.

I r2Direction of axes. The equation (1) can be rewritten as

ax2 + 2lixy + by2 + d = 0

...(8)

Awhere d =- —ab-h2

. vJ.; [IfCbrder to remove the term xy from above equation we rotate the axes through an angle 0 keeping the origin fixed. Then we get

1 2h-10 = - tana-b

2 tan 0or ------- =—=---- -1 - tan2 0 a-bh tan2 0 + (a - 6) tan 0 - /i = 0.

This equation (9) is quadratic in tan 0 so it have two values of tan 0, say tan 0], tan 02. These two values give the slopes of the axes of the given conic.

2h 2htan 20 =ora-b

-.(9)or

• 17.6. DETERMINATION OF THE ECCENTRICITY OF A CENTRAL CONIC

1. Eccentricity of an ellipse. Since we know that if r12>r22>0, then the conic will be ( ellipse and its eccentricity e is given by

r,222. Eccentricity of hyperbola. If rf and r2 are of opposite sign, then the conic will be

hyperbola and its eccentricity e is given by

e=ve-A r?1 + 2

'‘I


SOLVED EXAMPLESExample 1. Find the lengths and the equation of the axes of the conic whose equation is

3&C2 + 24ry+ 29>'2 -72r+ i26.y + 81 =0., ^ ■

System of Conics

■ Solution. Let us assume<Ka-, >■) = 36a2 + 24x>> + 29y2 - 72x + 126>> + 81 =r.QJ)

Differentiating (1) partially w.r.t. x and y respectively, we get94)2x = 12x + 24y - 72

9^ =^ - 24x + 58y + 126.and

Now solving ^ = ^ and ^

(2,-3). Therefore the new constant

0, we get x = 2, y = - 3. Hence the centre of this conic is

i

d = o.g +/}3 + c£ = - 36,/= 63, c = 81 and a = 2, P = - 3

d = 2(- 36) + 63 (-3)+ 81 =-72- 189 + 81 =- 180. Thus the equation of the conic referred to centre as origin is

36*2 + 24xy + 29y2 - 180 = 0 36 2 24 29 2180 X + 180 ^ + 180 ^

Here

= 1or

which is the form ofAx2 + 2Hxy + By2 = 1

1236 29A m'H 180’S -180

Now the lengths of the axes are given by±-±(A+B) + (AB-H2) = 0

where

\2n1 1 r 36 29 V r 36 29180 ' 180

i12 = 0/"r2 180+ 180 +or 180r ■

900 _r4 180 r2 + (180)2

1 ilVl.^O-Vor2 180 V 180 i

k / N /

r2 = 9 or 4r\2 = 9, r22 = 4 => r, = 3, r2 = 2.

Obviously r2 and r22 both are positive, therefore the given conic is an ellipse. Hence the lengths of major axis = 2r| = 2 (3) = 6

and the length of minor axisNow the equation of major axis referred to (2, - 3) as origin is

A - x + ffy = 0

or

or

ori.e.

= 2/2 = 2 (2) = 4.

r\^36__0

180 9\ JNow shift the origin back from (2, - 3) by putting x - 2 for x and y + 3 for y, we get

4 (x-2) + 3 (y + 3) = 0 or 4x + 3y+l=0.This is the required equation of major axis referred to (0,0) as origin.And the equation of minor axis referred to (2, - 3) as origin is

4--^ x + Hy-0

12x+my = 0 or 4A: + 3>' = aor

r2

(36 1 12180-4 X + TTOy = 0 °r 3x + 43' = aor

Again shift the origin back from (2, - 3) by putting x - 2 for x and y + 3 for y, we getSelf-Instructional Material 116

Calculus.& Geometry - 2r + 3y = 0.Shift the origin back from (2,1) by putting x - 2 for * and y - 1 for y, we get

-2(x-2) + 3(y - 1) = 0 - 2x + 3y + 1 = 0

2jc - 3y - 1 = 0.Hence the required major and minor axes are respectively

, 3x + 2y - 8 = 0 and 2jc - 3y - 1 = 0.Eccentricity of the ellipse. The eccentricity c is given by

or

ori.

orA ^

= V3/4 = VJ/2.1 —^'•j2

Latus rectum. The latus rectum of an ellipse is given by2r22

L.R. = — r\

= 1Q}!

('•' > >2)

= 1.2

Foci. The foci lie on the major axis 3x + 2y - 8 = 0. Therefore, the slope of this axis is given by

tane = -fsin 6 = 3/VI3", cos 0 = - 2/Vl3'

The distance of the foci from the centre (2,1) is •,

-±V3.

( .' 0 is obtuse)

r = ±ae = ±rle = ±2. —

/.The foci of the ellipse are(a + r cos 0, B + r sin 0) s 2 ± Vlf- |

vn '‘*^1

■[-i-si'Pi-i*Equation of the Directrices :

The equation of the directrices of an ellipse are given by (x - a) cos 0 + (y - (3) sin 0 ± rt/e = 0 (x - 2) (- 2/VTT) + (y - 1) (3/ViT) ± 2 (2/VI) = 0--L(x-2) + -|=().-I)±-p = 0vn vii vior

- 2 (x - 2) + 3 (y - 1) ± 4 0or

-2x + 4 + 3y-3 + 4or

and -2x + 4 + 3y-3-4

2x-3y- 1+4or

and 2x - 3y - 1-4

These are the required equation of directrices.

• SUMMARY• Equation ox2 + 2hxy + by2 + 2gx + 2fy + c = 0 is transformed to the equation

Ax2 + By2 + 2Gx + 2Fy + c = 0 by putting x cos 0 - y sin 0 for x and x sin 0 + y cos 0 for x ingiven equation, where 0 = ^ tan ( 2/t-j

a- b }'

r!18 Self-Instructional Material

„ 1-4

I

System of Conics• Coordinates of centre of the given conic :(i) Find ^ and ^

ox ay(ii) Solve ^ = 0 and ^ = 0, the values of x and y so obtained give the centre (x, y).

ox oy• Equation of conic with origin as centre

ax2 + 2hxy + by2 = —~—r ab - h

where A = abc + 2fgh - at - bg - ch .• Equation of central conic is Ax2 + 2Hxy + By2 = i• Equation of semi-axes of a central conic Ax + 2Hxy + By = 1 are given by

A-^-jjc + /yy = 0 and ^A-^-j^ + f/y = 0

2 2where rj and r2 are the roots of the equation

(AB - //2) r4 - (A + B) r2 + 1 - 0• If r2 > > 0 then the conic is an ellipse, inthis case length of major axis is 2rj and the length

| . If r? > 0 and r%<0, then the conic is a r2

hyperbola, in this case length of transverse axis is 2ri and the length of conjugate axis is 2r2

and its eccentricity is

= 0

*! -

*

of minor axis is 2r2 and its eccenticity is '^1 r?

HI)• STUDENT ACTIVITY

2 21. Solve the coordinates of the centre of the conic 32x +52Ay-7y -64v:-52y-148 = 0.

Show that the conic 36a:2 + 24A:y + 2ay2 - 12x + 126y + 81 = 0 is an el2. ipse.


Calculus & Geometry • TEST YOURSELF1. Find the co-ordinate of the centre of the conic whose equation is

36x2 + 24xy + 29y2 -12x+ I26y + 81=0.2. Find the centre of the conic section

13*2 - 18^ + 37;y2 + 2* + 14;y - 2 = 0.Also, find the equation of the conic section referred its centre as origin.

3. Find the lengths and equations of the axes of the conic$x2 + 4xy + 5yz-24x-24y = 0.^

4. Find the equation of the the directrices of the conic*155x2 - SOty + 39y2 - 4Qx - 24y - 464 = 0.

5. Find the centre, foci and the equations of the axes of the ellipsex2 + xy + y2-x + 4y + 3 = 0.

Also trace it.6. Find the lengths and the equations of the axes of the conic

5x2 -6xy + 5y2 + 26* - 22^ + 29 = 0.

ANSWERS

2- •\ J

3. Major axis : 2x + y = 4, length = 6; Minor axis : * - 2y + 3 = 0, length = 4.4. 3* + 5y = 36, 3* + Sy = - 28.

( - y[S Vil - • •5. Centre (2, - 3), foci 2 + — ’ - 3 ± —

{ ^3Major axis : * + y + 1 = 0,Minor axis : x - y - 5 = 0.

13*2 - 18*y + 37y2 -4 = 0.1.(2,-3)4

1

1

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t '


Confocol Conu sUNIT

CONFOCAL CONICSSTRUCTURE *

• Confocal Conics• Equation of Confocals to an Ellipse• Some Properties of Confocal Conics



# How we define confocal conics and what are their properties.

• 18.1. CONFOCAL CONICSDefinition. Those conics which have same two points as their foa

same focal points, are called confocal conics are also having same axes.or those conics having

• 18.2. EQUATION OF CONFOCALS TO AN ELLIPSELet the equation of an ellipse be

2,2 a b...d)= 1.

If a > 6, then the co-ordinates of its foci are (± ae, 0) where e = a(± ae, 0) = (± Vtf2 - b2,0).

Let us consider the equation2*2 y ...(2)= I.+

+ A. + A .---------------------- .Now the co-ordinates of the foci are [± 'v(a2 + A) - (fc2 + A), 0] o ■ _(± Va2 - b2,0), for all

Therefore, the foci of (2) and (1) are same. Hence, the equation (2) are the confocal conicsvalues of A.

of (I).

• 18.3. SOME PROPERTIES OF CONFOCAL CONICSProperty I. To prove that confocals cut at right angles. Let the two confocals be

22yX

•••(I)= 1+fl2 + A| ft2 + A i

22X y •••(2)and = 1.+a2 + A2 b2 + A2

Suppose both confocals (1) and (2) intersects at (a, |3) so that (a, p) will satisfy (I) and (2),we get

a2 £ = 1+a2+ X b2 + A,1

£a2and = i.+

fl2 + A2 fr2 + A2Substracting these equations, we get


_1£1^ + X2

(Xq — ^-l) Cc(a2 + X]) (a2 + k2) (b2 + Xl) (b2 + X2)

Calculus & Geometry 1 11a2 + P2 = 062 + ^.1 b -{-Xn

= 0or I •:y

a2 £ = 0 -(3)+or(a2 + X{) (a2 + Xq) (b2 + Xl)(b2 + X2)

(■-' X^Xq)

Now the tangents at (a, P) to the confocals (1) and (2), are given byPyax\

= 1 ...(4)+b2 + Xi

Jl=i.a2 + X i

axand •••(5)+q2 + A.2 b2 + X2

Slopes of (4) and (5) are respectively, m| and m2\ where a/tf + Xi) a(b2 + Xi)

M\=-P/(*2 + ^l)- a/(a2 + X2)

P (a2 + X|) a(b2 + X2)

P(a2 + X2) a(b2 + X2) -

P (a2 + Xq)

and m2 = -p/(62 + X2) a(62 + A1)''

P(a2 + ^1);C(2 (b~ + Aj) (b~ + X2)

m\ . m2= -

[using (3)]= - 1.p2 (a2 + A.]) (a2 + Xq)

This implies tangents at (cc,'P) to the confocals (1) and (2) cut at right angles. Hence confocal | cut at right angles.

Property II. Through a given point in the plane of a conic, two confocals can be drawn, one of them is an ellipse and other is a hyperbola.

Let the equation of a confocal to an ellipse be

* 1

22 yX= 1. -(1) •+

a2 + A. 62 + A.Suppose (a, p) is a given point in the plane of (1) so it will satisfy (1), we get

a2 £ = 1. -.-(2)+a2 + A. 62 + A.

The equation (2) is a quadratic in A.. Hence through (a, P) we can draw two confocals. Further, let us take £»2 + A. - p i.e. X = \i-b2.Now putting the value of X in (2), we get

a2 £ = 1a2 + p - 62 14

a2 £ = 1p + a2-Z>2%or

2 £a (y ci'e2 - a2 - b ) ii= 1+or2-2 \i (p + a e

p2 + {flV-a2-p2)p-p2aV“ = 0.Let Pi and p2 be the roots of (3), then we have

|4iP2 = -P

5or

2 2 2a e=> PJP2 is negative=> one of p = 62 + A. is positive and other is negative. Again, let us take a2 + A. = p i.e. A. = p - n2.Now put A. = p - a2 in (2), we get

—+(4 p + ^-a

£ 5=1


a2 £ . Confocal Conic\a2e2 = a2 ~ b2)~ 1

\i2 - (a2e2 + a2 + p2) + a2a2e2 = 0.Let jii and \i2 be the roots of (4), the we have

}ii + fi2 = «2g2 + cc2 + P2 •

^2 = a2a2ez

From (5) it is observed that both the roots of (4) are positive i.e. both value of a2 + X are

or

...(4)or

•••(5)and

positive.Hence one of the confocals is an ellipse and other is hyperbola.

SOLVED EXAMPLES i2 2

XExample 1. If the confocals through Pfpi, p) to the ellipse ~ = 1 are

b22 2 2 2

X y X y= 1 and - 1+b2 + X2a2 + X

a2 b2cc2 + P^ — a2 — £>2 ^ A,j + A,2.

X2 v2Solution. Let the equation of a confocal to the ellipse — + *—

a b

fc2 + X,

— XjXj

a ^ X2i

show that -1 =a2b2

and

= 1 be

2X = 1.+

a2 + l b2 + X This confocal (1) passes through (a, P), then we have

a2 £ = 1+a2 +A b2 + X

a2 (ft2 + A) + p2 {a2 + A) = (a2 + A) (62 + A) aV + a2A + pV + P2A = a2b2 + A (a2 + b2) + A2

A2 + A (a2 + £2 - a2 - P2) + a2b2 - aV - p2a2 = 0.This is a quadratic equation in A so it has two roots A[ and A2 (say) Hence, we obtain two confocals . .

oror

...(2)or

222 x2 yX y = 1 and = 1.a2 + A2 b2 + A2a2 + A b2 + Aji

Also we haveAi + A2 = - (a2 + 62 - a2 - p2)

AiA2 = aV - aV - pV.43)

andFrom (3) we obtain

a2 + p2 - a2 - fc2 = A, + Aj

2 ,2 n o

A,A2and - 1 =-

a2b2

• SUMMARY

X2 y2• Confocal conic of -^ + 2 = 1 are given by

a bx2 2

y + 1, A £ /?.+a2 + A /72 + A

• Confocal conics cut each other at right angles.• Through a given point in the plane of a conic, two confocals can bejdrawn one of them is an

ellipse and the other is a hyperbola.


Calculus & Geometry • STUDENT ACTIVITY:

1. Prove that confocals cut at right angles.

i

i

i|■ -r* v ; l

■>* \ .1

i

S l 2 2X y2. If the confocals through P (a, p) to the ellipse .a b2

= 1 are = 1 andq2 + Ai b2 + X\

2 2x, ^— = 1 the prove that Xi + X2 = a2 + 62 - (a2 + b2).or + X2 o + X'l

*’124 Self-Instructional Material f•J

I

’

Confocal Conics• TEST YOURSELF !

1. Prove that the locus of the point of contact of the tangents from a given point to a system ofconfocals is a cubic curve which passes through the given point and ti rough the foci. .

2. Prove that the two conics a*2 + 2hxy + by2 = 1 and ax^ + Ih'xy + b'y1 ■ -1 can be placed so asto be confocals if

(a-b)1 + 4/z2 {a -b'f + 4h'2 (ab - h2f

OBJECTIVE EVALUATION > FILL IN THE BLANKS :1. Confocal have the same two points as their foci as well as same

(a'b'-h'2)2

2 2xConfocals to an ellipse — + 2

o' o2. = 1 are

3. Confocals cut one another at................> TRUE OR FALSE :Write ‘T’ for true and ‘F’ for false statement:1. Confocal have same foci but do not have same. axis. (T/F)2. Through a given point in the plane of a conic, two confocals can be drawn, one of them is an

(T/F)ellipse and the other a hyperbola.2 2

The foci of the confocal —r-— + —^— - 1 are [± ^a2 -b2 - 2X, 0]. a2 + \ b2 + X

> MULTIPLE CHOICE QUESTION :Choose the most appropriate one :

2 2 x y1. The centre of the confocal -----+ —r— = 1 is :a+X bz + X

(a) (O.X)(c) (0,0)

3.\ (T/F)

(b). (K0) (d)

2 2X2. The foci of the confocal -t- y = 1 are :

a2 + X b2 + X(a) (±^la2-b2,0)(c) (± Va2 -b2 -2\, 0) (d) None

(b) (0, ±Vtf2 ->)

.of these..3. Confocal cut at an angles :

(a) rc/4 (c) n/3

(b) 7t/2.(d).k/6...

ANSWERS

Fill in the Blanks :2 2

2 * + >a2 + X + b2 + k1. Axis = 1 3. Right angles'

True or False : 1. F 3. F2. T

Multiple Choice Questions :1. (c) 2. (a) 3. (b).

Self-Instructional Material«. 125.}


SYSTEM OF CO-ORDINATESSTRUCTURE

• Co-ordinate System in Space• Co-ordinates of a Point in Space• Octants• Change of Origin (Shift of Origin)• Distance between Two Points• Division of the Join of Two Points• Application of Division of Join of Two Points

I Summary• Student Activity• Test Yourself

LEARNING OBJECTIVESAfter going through this unjt you will learn :

• About the 3-dimensional co-ordinate system How to locate a given point in space.

• How to find the distance between two given points in space.

• 19.1. CO-ORDINATE SYSTEM IN SPACEIt has been observed from review of

two-dimensional geometry that an object having some volume can not be described by two dimensional geometry, because any object occupied some space will have some points which are not in a same plane. Therefore to describe the location of the points of any volumetric object, we consider a space system in which there are three mutually perpendicular lines called the axes of coordinates and the geometry having three dimensions also known as solid geometry.

Let XOX', YOr and ZOZ' be three mutually perpendicular lines shown in fig. 1. The point O is known as origin and the lines are called rectangular axis, if these axes are taken in a pairs i.e., YOZ, ZOX and XOY, these from co-ordinates planes known as YZ, ZX and XY-planes respectively.

Zi i

y71

X’<- »A'O

> tZ'

Fig. 1

• 19.2. CO-ORDINATES OF A POINT IN SPACELet P be any point in space. To find the location

of this point P, let us draw the planes through P parallel to the co-ordinates planes as shown in fig. 2.

These planes cut the co-ordinate axis at points A, B and C on X-axis, K-axis and Z-axis respectively. Let OA =x, OB = y and OC = z. The co-ordinates of P be (x, y, z). Hence we can say that the co-ordinates of any point are the perpendicular distances with proper signs from co-ordinates planes to that point.

4 V

Q

sA cp2

A

aY

Fig. 2


Syxlems of Co-ordinates• 19.3. OCTANTSThe co-ordinate planes i.e., YZ, ZX and XY planes divide the whole space into eight parts,

called the octants. These octants contain negative axes also. htKXOX, YOY and ZOZ* be three'mutually perpendicular axes.Then the octants are :

1. OXYZ,5. OXYT, 6. OX'YZ, l.OXYZ,

• Here dashes represent the negative side of the respective axis. In the first octant OXYZ all the co-ordinates are positive.Let these co-ordinates for any point P be (x,y. z), then theco-ordinates of this point P with respect to other octants are ___

z), Or,-y,z), (*, y,-z), (x,-y,-z), (-J:,y,-z), A" 2(- x, - y, z) and (- x,-y,~ z) respectively.

To know about the octants, consider a sphere of any radius and let the centre be assumed the origin O of the co-ordinates axis and consider a great circle of this sphere. Now taking two perpendicular diameters of this circle as two axes XOX' and YOY and a line through 0 perpendicular to the plane of great circle is taken as third axis ZOZ as shown in fig. 3. In this way the whole sphere is divided into eight surfaces as written above in the beginning of this section. These surfaces are called the octants.

Examples 1. Find the position of the following points :(0 (4,5,7)(iv) (-3,-1,-6)(vii)(7, -1,-1)(A) (1,0,0)Solution, (i) (4,5, 7) is in the octant OXYZ.(ii) (1, - 3, 2) is in the octant OXYZ.(iii) (-2,1, 5) is in the octant OX'YZ.(iv) (- 3, - 1, - 6) is in the octant OX'YZ.(v) (0,0,1) is oh the z-axis.(vi) (3,2, - 1) is in the octant OXYZ.(vii) (7, - 1, - 1) is in the octant OXYZ.(viii) (- 5,2, - 2) is in the octant OX'YZ.(ix) (- 6, - 6,6) is in the octant OX'YZ.(x) (1,0, 0) is on the X-axis.(xi) (0,7, 0) is on the T-axis.

" iYZ

- 7\\\2. OX’YZ, 3. OXYZ, 4. QXY%,\h'

8. ^n^./Ni

/

p VX I\

\ I\ /r

Fig. 3

HO (l;-3,2)(v) (0,0,1)(viii) (- 5,2, - 2) ( (xi) (0,7,0).

(Hi) (-2, 1,5) (W) (3,2,-1) ix) (- 6, - 6,6)

• 19.4. CHANGE OF ORIGIN (SHIFT OF ORIGIN)Let OX, OY and OZ be three mutually

perpendicular axes and let P be any point whose co-ordinates with respect to these axis be (a, b, c). Now we want to shift the origin O at P. For this purpose draw three lines through P parallel to OX, OY and OZ respectively and thus obtained new co-ordinates axes.Let these new axes be denoted by PX\, PY\ and PZ\ as shown in fig. 4.

Let Q be any other point whose co-ordinates with respect to OX, OY and OZ be (flh bt, C|)- Now we shall 2

find the co-ordinates of Q with respect to new axes

PXh PYj and PZ,. Let r be the position vector of a point P with respect to the position vector of Q with respect to O. Then

------* ^ AAA

OP=r=ai+bj+ck------> .AAA

OQ - rj = a j i + bij + Cj h

i > ,2/,/!

P

*1t

Fig-4

the origin O and r\ be

andAAA 1

where i, j and k are the unit vectors along X-axis, K-axis and Z-axis respectively.

In AOPQ, OP + PQ = OQ(By triangular property of vector)AAA AAA

(a 11 + bi j + clk)-(ai + bj + c k)PQ = OQ - OP =7?-r*=

Self-Instructional Material

I I IV -

Calculus A Geometry SOLVED EXAMPLESExample 1. P is a variable point and the co-ordinates of two points A and B are

(- 2,2, 3) and (13, - 3, 13) respectively. Find the locus of P if 3PA = 2PB.Solution. Let the co-ordinates P be (x, y, z). Then the distance between P and A\ P and B

respectively are ______________________PA = 'V(x4-2)2 + (y - 2)2 + (z- 3)2PB = ^(x - 13)2 + (y + 3)2 + (z - 13)2 ,

3PA = 2PB________3 V(x + 2)2 + Cv-2)2 + U- 3)3 = 2 V(^-i3)2 + (y-+3)2 + a- I3)3.

Squaring of both sides, we get9 [(x + 2)2 + (y - 2)2 + (z - 3)2] =4 f(x- 13)2 + (y +3)2 + (z - 13)2] '9 (x2 + 4 + 4x + y2 + 4 - 4>' + z2 + 9 - 6z]

■ = 4 [x2 + 169 - 26x + y2 + 9 + 6y + z2 + 169-26z]9(x2 + y2 + z2 + 4x-4y-6z+ 17) = 4 (x2+y2 + z2 - 26x + 6y-26z + 347)'9x2 + 9y2 + 9z2 + 36x - 36y - 54z + 153 = 4x2 + 4y2 + 4z2 - 104x + 24y ~ 104z +.1388 Sx2 + 5y2 + 5z2.+ 140x - 60y + 50z - 1235 = 0 x2 + y2 + z2 + 28x - 12y + lOz - 247 = 0

which is the required locus.Example 2. Show that the points (1,2, 3), (2, 3, 1) and (3, 1,2) form an equilateral triangle. Solution. Let ABC be triangle and the co-ordinates of A, B and C be (1, 2, 3), (i‘. 3, i) and

andSince

or

orororor

(3,1,2).AB = V(1 - 2)2 + (2 - 3)2 + (3 - l)2

= ^(~ l)2 + (~ l)2 + (2)2 = Vl + 1 + 4 = V6 ' BC = V(3 - 2)2_+ (1 - 3)2 + (2 - l)2

CA = V(1 - 3)2 + (2 - l)2 + (3 - 2)2= ^(-2)2 + (i)2 + (i)2 = Vi+4 + i = Ve.

and

Thus/4fl = flC = G4. Hence A/tBC is an equilateral.Example 3. Find the co-ordinates of the point that divides the join of two points (2. - 3, 1)

and (3,4, ~ 5) in the ratio \ : 3.Solution. Let P and Q be two points whose co-ordinates are (2, -3,1) and (3,4, - 5) and let

R be that point which divides the join of P and Q in the ratio 1:3.Let m : n = 1 : 3 and co-ordinates of R be (x, y, z).

mx2 + nx, 1(3)+ 3 (2) 3 + 6_94 ”4

x =1+3m + n

my2 + nyi 1 (4) + 3 (- 3) 4-9 5m + n

tnz2 + 1 (- 5) + 3 (1) - 5 + 3

y- 4 4 42=_ !_ 4 2z = 4 4m + n

(9 5 lVHence divides the join of (2. - 3, 1) and (3, - 4, 5) in the ratio 1 : 3.

Example 4. The mid-points ofthe sides ofa triangle are (1,5, - 1), (0,4, - 2) and (2, 3, 4). Find its vertices

Solution. Let ABC be a triangle and let the co-ordinates of A,B and C be (x,.y,,Z|), (x^y^Zi) and respectively. Then we have

• x, + x2____i=1 yi+yi2 *'2

X] +x2 = 2,y, +y2= 10,Z| +z2 = -2.' ’ 2

= -l

•:•(!)or BSimilarly,

x2 + x3 = 0, y2 + y3 = 8, z2 + z3 = - 4 x3 + X] = 4, y3 + y] = 6, z3 + zi = 8.

Adding (1), (2), (3), we get

...(2)

..,'3)Fig. 9

130 Self-Instructional Material-

2 fa +x2+ xz) = 6,2(yl+y2+ y3) = 24,2 (^i + z2 + z3) = 2

+ ^ + ^3 =y\+y2+y3 = 12, zl+z2 + zi = i. Subtracting successively (1), (2), (3) from (4); we get

X{- 3,X2 = - l,x3 = 1, yi =4,y2 = 6,y3 = 2 z\=5,z2 = -l,zi = 3.

Hence the vertices of the triangle areA (3,4,5), B (- i, 6, - 7), C(l,2,3).

Systems of Co-ordinates

f ••.(4)ori

and

• SUMMARY• In 3-D geometry there are eight octants. __________________________• Distance between (xb yi.Zj) and (^,y2, z2) is given by v(x2-xt)2 + Cva -yi)2 + (z2-Zi)2-• If a point P {x, y, z) divides a line segement AB with A (jcj, yb Z\) and B (Jc2, y2, z2) in the X : 1

Xz2 + Z\\Xy2 H-y]Ax2 + X\

"X+FX < 0 then the division is externally.

• Coordinates of centroid of the triangle ABC with A (*i, yi, zj), B (x2, y2, z2) and C (*3, y3, z3) is

If X>.0, then the division is internally and ifthen x = , z =, y = X + 1X +1

x{+x2 + x2 yi +y2 + y3 Z]+Z3 + zA33 3

*2). C (x3, y3, z3) and• Co-ordinates of centroid of tetrahedron ABCD with A yj, zj), B (x2, y2, xi+x2 + x3 + x4 y\+yi + y3+y4 zi+z2 +

D (x4, >>4, Z4) is

• x = r cos 0, y = r sin 0, z = z are the transformations for cylinderical polar coordinate systemj,. “i ywhere 0 = tan ,L

44 4

• Lx

• ^ = r sin 0 cos (|), y = r sin 0 sin <|) and z = r cos 0 are the transformations for spherical polarFcFy2 ' . ' ‘ .y

co-ordinate system, where tan <J> = , tan 0 = z

• STUDENT ACTIVITY1. Show that point (1, 2, 3), (2, 3, 1) and (3, 1, 2) form an equilateral triangle.

Show that the points (3, - 2, 4), (1, 1, 1) and (- 1, 4, -2) are collinear.2.


Calculus dt Geometry• TEST YOURSELF1. A variable point P moves in such a way that its distance from two fixed points A (3,4, 5) and

5 (- 1, 3, - 7) are always same. Then find the locus of P.2. Find the locus of a point P which moves in such a way that its distance from the point

A(a, b, c) is always equal to r.3. A, 8, C are three points on the axes of x, y and z respectively at distances a, b, c from the origin

0\ find the co-ordinates of the point which is equidistant from A, B, C and O.4. Find the incentre of tetrahedron formed by planes x = 0, y = 0, z = 0 and x^ + y + z = a.5. Show that (0,7, 10), (-1,6, 6), (- 4, 9, 6) form an isosceles right angled triangle.6. Find the co-ordinates of the point which divides the join of (2, 3, 4) and (3, - 4, 7) in the ratio

2:-4.

Ff1.,

ANSWERS

1. 8x + 2y + 24z + 9 = 0 2. x2 + y2 + z2 ~ lax - 2by - 2cz + a2 + b2 + c2 - r2 = 0 . fa b c

I 2 ’ 2 ’ 2

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. (V2", - V2", - V2) lies in the octant.........2. (- 1, - 2, - 3) lies in the octant...........3. In the octant OXY'Z, the y-co-ordinate is> TRUE OR FALSE :Write ‘T’ for true ‘F’ for false statemet:1. The number of octants are 6.2. (1,0,0) lies on the X-axis.3. (0,3,4) lies on the XK-axis.> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :1. The point (- x,-y,- z) lies :

(a) OXYZ2. The number of octants are :

(b) 73. The distance between (1,2, 3) and (3,2,1) is_j

(b)VT4. The middle point of the join of (1,2, 3) and (- 1, - 2, - 3) is :

(a) (1,-2, 3) (b) (1,1,1)

. a a a4‘ 4’ 4’ 4 6. (1, 10, 1)

(T/F)(T/F)(T/F)

>•(d) OX'YZ. "(b) OXTZ' (c) ox'rz

(a) 6 (c)8 (d) None of these.

(a) V8 (c) 3V2 (d) 1.

(c) (0,0, 0) (d) (-1,-2, 1).

ANSWERS

Fill in the Blanks :1. oxrz

True or False :1. F 2. T 3. F

Multiple Choice Questions :1. (b) 2. (c) 3. (a) 4. (c)

2. OXTZ 3. Negative

□□□\• \


Direction Cosines and Direction RatiosUNIT

m

DIRECTION COSINES

AND DIRECTION RATIOSSTRUCTURE

• Direction Cosines of a Line• Co-ordinates of a Point in Terms of D.C.’s• Relation between Direction Cosines• Direction Ratios• Direction-Cosines of a Line through Two Points• Projection of a Line through Two Points on Another Line whose D.C.’s are /, m, n• Angle between Two Lines• Angle between the Lines whose Direction Ratios are given• Condition for Perpendicularity and Parallelism



9 How to calculate the direction cosines and direction ratios of the gi How to determine the angle between the / given lines.

/en line.

• 20.1. DIRECTION COSINES OF A LINELet AB be a line which makes the angles a, (3 and y with the positive axes of X, Y and Z

respectively. Draw a line OP through O and parallel to the line AB as shown in fig. 2.The line OP makes the angle a, with positive axes of X, T, Z respectively provided

a + |3 + y * 27t. Thus the non-dimensional quantities cos a, cos p, cos y are called the direction cosines of the line AB.

The direction cosines of any line are usually denoted by the letters /, m and n respectively, i.e., I = cos a, m = cos P, n = cos Y-

B

Direction Cosines of Co-ordinate Axes.(a) Direction eosines of X-axis. Since we know that X-axis makes the angle ct = 0o, P = 909,

Y = 90° with positive axes so its direction cosines are cos 0°, cos 90°, cos 90° i.e., 1,0,0.


Calculus & Geometry (b) Direction cosines of y*axis. K-axis makes the angle a = 90°, P = 0°, Y - 90° with positive axis respectively so that its direction cosines are cos 90°, cos 0®, cos 90° i.e.\ 0,1,0.

(c) Direction cosines of Z-axis. Likewise z-axis makes the angle a = 90°, P = 90°, y = 0° with positive axis X, Y,Z respectively. Thus direction cosines of z-axis are cos 90°, cos 90°, cos 0° i.e., 0,0,1.

• 20.2. CO-ORDINATES OF A POINT IN TERMS OF D.C.’sLet P be any point in a space and let OX, OY and

OZ be rectangular axes. Suppose OP = r. Draw a perpendicular from P to X-axis which meets at M. Let the co-ordinates of P be (x.y.z). Then in tsOPM, Z.PMO = 90®,

Z4 k

P {x. y. z)

OM= cos a

OP .OM - OP cos a 0/ ».Y

‘Mx

(Y OM = x,OP=r) (y / = cos a)

x = r cos aY\x=lr

Similarly,Hence the co-ordinates of P are (Ir, mr, nr) where

/, m and n are direction cosines of OP.

y = mr, z - nr. Fig. 3

• 20.3. RELATION BETWEEN DIRECTION COSINES

Let OX, OY and OZ be the system of co-ordinate axes mounted mi/tually perpendicular. Now draw a line OP through O and parallel to the given line AB. Let OP be r. Then the co-ordinates of P are (Ir, mr, nr) where /, m, n are direction cosines of AB. Suppose that co-ordinates of P are assumed (x, y, z). Then

OP = r = 'lx2 + y2 + z2 and we have x = lr, y = mr, z = nr squaring and then adding, we'get

x2+y2 + z2 = (Ir)2 + (mr)2 + (nr)2 x2 + y2 + z2 = r2(f- + m2 + n2)

Z

O

or 'r2 = r2 (/2 + m2 + «2) = x2 + y2 + z2)(vor

r/2 + m2 + n2 = 1.Hence proved the result. Rg. 4

• 20.4. DIRECTION RATIOSDefinition. The three real numbers a, b, c which are proportional to the direction cosines

l, m, n of a line, are called direction ratios of that line.I _ m_ n a b c

---- , ' m-±

Thus we have — =:

b ca:. l = ± , n = ±V*2 + *2 + c2Va2 + *2 + c2%2 + b2 + c2

• 20.5. DIRECTION-COSINES OF A LINE THROUGH TWO POINTSLet P and Q be two points through which a line is passing. Let the co-ordinates of P and Q

be (x],yi, zj) and (x2, y2» *2) respectively, and let the direction cosines of PQ be l, m, n. Now draw two perpendicular PP1 and QQ' from P and Q to X-axis respectively. Suppose the line, PQ makes the angles a, p and y with the positive X-axis, K-axis and Z-axis respectively; then

/ = cos a, m = cos P, n = cos y.


Calculus &. Geometr)' 2. The axis of the cylinder yfx, y)=ax2 + 2hxy + by2 + 2gx+ 2fy + c-0 is..........

3. The equation of a right circular cylinder of radius r having its axis as z-axis is ...4. The equation of the right circular cylinder of radius 5 whose axis is the y-axis is

> TRUE OR FALSE : jWrite ‘T’ for true and ‘F’ for false statement:

1 2 2The axis of the cylinder_/(y, z) s ay +2/yz + cz + 2vy + 2h>z + c = 0 is x-axis.2. The intersection of the curve_/(*,y, z) = 0 and <J)(jr,y, z) = 0 gives the equation of a cylinder

with its axis z-axis.2 2 23^ The equation y +z =a ,x-Q represents a cone.

(T/F)

(T/F)(T/F)

> MULTIPLE CHOICE QUESTIONS : Choose the most appropriate one :

= ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is : (b) y-axis (d) None of these.

2. The generators of the cylinder fly, z) = 0 are parallel to the axis :(a) *-axis\.ii‘i (c) z-.axis

The axis of the cylinder^, y) (a) x-axis (c) z-axis

(b) y-axis(d) None of these.

\ANSWERS

Fill in the Blanks :1. Cylinder 2. z-axis 4. x2 + z2 - 25

True or False :1. T 2. F 3. F

Multiple Choice Questions :1. (c) 2. (a).

■ i i 2 , 2 23. x +y =r

□□□

\


The Cylinder• STUDENT ACTIVITY

x y z1. Find the equation of the cylinder whose generators are parallel to the line — = = — and the2 2 1 Z ^

guiding curve is the ellipse x +2y =l,z = 3.

2 2 2Find the equation of the cylinder which intersects the curve ax +by.+cz = 1, lx + my + nz= p whose generators are parallel to the axis x.

2.

■j

• TEST YOURSELFX1. Prove that the equation to the cylinder whose generators are parallel to the line — =

and passing through the curve x2 + 2y2 = 1, z = 0 iso 2x2 + 2y2 + z2-~xz + ^yz-l=0.

2. Find the equation to the surface generated by a straight line which is parallel to the line y = mx, z-nx and intersects the ellipse

JL = L -2 3

8

22X = l,z = 0.a2 +b2

3. Find the equation of the cylinder with generators parallel to z-axis and passing through the curve ax2 + by2 = 2cz, lx + my + nz~p.

4. Find the equation of a cylinder whose generators are parallel to the line ;t = y/2 = -z and passing through the curve 3x2 + 2y2 = 1, z = 0.

AANSWERS

'I2. b2(nx - z)2 + a\ny - mz)2 = a2b2n2.3. anx2 + bny2 + 2c (lx + my) - 2pc = 0. ,4. 3x2 + 2y2 + 1 lz2 + 8yz + 6z* '--1=0.

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1 The equation fix, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a


\

Calculus & Geometry d.r.’s of this line 2, 3, 1.2 3/. d.c.’s of this line are -i—, ——,---- •

Vl4 VIT Vl4"Let /4(1,0, 3) be any point dn the axis and let 0 be a point on the axis a such that PQ = 2.

The projection of AP ori the axis = QAor Q/t = (* - 1)-L + O - 0) •+ (z - 3) •

PA2 = (x- l)2 + y2 + (z - 3)2.

1

1 1 (2x + 3y + 2 - 5)VI?’ VI?

andP (x. y, z)

2A (1,0, 3)

Q

Fig. 2In kPQA, Z.PQA = 90°, then

PQ1 = PAA -QA14 = [(-t - l)2 + y2 + (2 - 3)2] - ■? (2x + 3y + ^ - 5)2

56=l4[(x-i)2 + y2i(z-3)2]-(2x + 3y + z-5)214 (x2+y2 + z2- 2x-'6z + 10) - (4x2 + 9y2 + z2+ 25 + I2xy + 6yz

or

oror

+ 4£c-20,y-30v- 10£) = 56IQx2 + 5y2 + I3z2- 6yi - 4zx - 12xy - 8x + 30y - 14z + 59 = 0.

Example '3. Find the equation of the cylinder which intersects, the 222 1ax + by + cz = l, lx + my + nz = p whose generators are parallel to the axis of x.

Solution. The equations of guidingax2 + by2 + cA = 1

orcurve

curve are...(1)

lx + my + nz p pSince the generators of the cylinder are parallel to .Y-axis, therefore the required equation of

the cylinder will not contain the terms of x. So eliminating x between (1) and (2), we get

and ...(2)

p - my - nz + by2 + cz2=\al

a (p2 + m2y2 + n2z2 - 2pmy - 2pxz + 2nmyz) + bl2y2 + cP’z2 = C (am2 + bl2) y2 + {an2 + c/2) z + 2amnyz - 2apmy - 2apnz + («p2 - /2) = 0,

which is the required equation of the cylinder.

oror

• SUMMARY

* —ct y-3• Equation of a cylinder whose generator is

ax2 + by2 + 2hxy + 2gx + 2/y + c = 0 is given byn (nx - Iz) + 2h (nx - lz) (ny - mz) + b {ny - mz) + 2gn {nx - lz) + 2fn [ny - mz) + n~

X• Equation of a right circular cylinder bf radius r and axis -

and intersecting the given conicl m n

c - a.-a y-p c-y is given by/ m 11

[(x -a)2 + (y- P)2 + {z- y)2] )/2 + m2 + n2) -[/(*- a) + m (y-$)+n(z- y)]2= 2 (l2 + m2 + n2)

• Equation of the tangent plane at (a, P, y) to the cylinder ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is given by

actx + // (xP -t- ya) + byP + g (x + a) +/(y + P) + c = 0; ’


The CylinderREMARK

> If the axis of the right circular cylinder is z-axis, then the equation of right circular cylinder is obtained by putting a = 0, P = 0,Y = 0 and / = 0, m = 0, n = 1 in (2) above, is

x2 + y2 = r2.

* 25.5. EQUATION OF A TANGENT PLANE TO THE CYLINDERTo obtain the equation of a tangent plane to the cylinder whose equation is

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0at the point P(a, (3, y).

aax + k(x& +ya) + by& + g(x + a) +/(y + P) + c = 0. ...(1)Corollary. The tangent plane (1) touches the cylinder

ax + 2hxy + by2 + 2gx + 2fy + c = 0along a generator.

Proof. Since the equation of a cylinder whose axis is the axis of z is given by /(x, y) = ox2 + 2hxy + £>y2 + 2gx + 2/y + c = Q. ...(2)

The equation of a generator through the point (a, f3, y) parallel to the axis of the cylinder is

• ••(3)= '"t (say)00 1

any point on this line is (a, P, + y).Now the equation of a tangent plane at (a, p, r^ + y) to (1) using the formula (1), we get

am + 2/t(xp + ya) + 6yp + g(x + a) +/(y + p) + c = 0 Thus (1) and (4) are identical. Hence the tangent plane (1) touches the cylinder along the

- (4)

generator.SOLVED EXAMPLES

Example 1. Find the equation to the cylinder whose generators are parallel to the line* _ y _ z1 -2 3

and the guiding curve is the ellipse x + 2y = 1, z = 3.Solution. Let P(a, p, y) be any point on the cylinder. Then the equation of the generator

through P(a, P, y) and parallel to the line* _ y _z1-2 3

x-a y-P z-yis given by ^^/■(say). .-(I)-21Any point on this line is (r + a,-2r +P, +y). This point lies on the ellipse

x2 + 2y2 = 1, z = 3, then we get(r+a)2 + 2(-2r+P)2= 1.

3r + y=3.Eliminating r between (2) and (3), we get

•••(2)and ..•(3)

,2

(3a + 3 - y)2 + 2 (" 6 + 2y + 3p)2 = 9.Thus the locus of P(a, p, y) is

(3x - z + 3)2 + 2(3y + 2z - 6)2 = 99x2 + z2 + 9-6zx~6z + 18x + 18y2 + 8z2 + 72 + 24yz - 48z - 72y = 9 9x2+ I8y2 + 9z2+24yz-6zx+ I8x - 54z-72y + 72 = 0.

Example 2. Find the equation of the right circular cylinder of radius 2 and whose axis is the

3-V + 2 = 1a + 3

or

oror

lineXZ± = 1 = ZZ±.

2 3 1Solution. Let P{x, y, z) be any point on the cylinder. The eqution of its axis is

x-l y z~ 3 2 'I* 1


guiding curve.

* 25.2. RIGHT CIRCULAR CYUHDBHDefinition. A surface generated by a moving straight line (generator) which moves m

a way that it is always at a constant distance from the fixed like (axis), is called a right circular cylinder and the constant distance is called the radius of thiJ right circular cylinder. r/* 25.3. EOmriOH OF A CyUUDEB PASSIUG THROUGH A GIVEN CONIC

To obtain the equation of a cylinder whose generator are parallel to the fixed line (axis) and intersecting the given conic.

Let the equation of fixed line through the origin 0 and having the direction-ratios l. m. n be

x_2__±l m n ...0)

and that of the given conic beax2 + 2/ixy + by2 + 2gx + 2fy + c - 0, z - 0.

Let P (a, p, y) be any point on the surface of the cylinder. Then the equation of its generator is•-(2)

...(3)l m nThis generator (moving straight line) meets the plane 2-0 at the point whose co-ordinates

p - 0 . This point lies on the conic (2), we getn nare a- * «

f •0^.y + O'*!3Vij0

u

9+ t< SPi09.* PUB

t

The CylinderUNIT

THE CYLINDERSTRUCTURE

• Cylinder• Right Circular Cylinder• Equation of a Cylinder Passing through a Given Conic• Equation of a Right Circular Cylinder• Equation of a Tangent Plane to th Cylinder• Summary• Student Activity• Test Yourself


• About the cylinder and how to get the equation of the cylinder passing through the given conic

• How to find the equation of right circular cylinder• How to find the tangent plane to the given cylinder

• 25.1. CYLINDER \Definition. A surface generated by a variable straight line which moves parallel to a fixed

line and intersecting a given curve or touching a given surface, is called a cylinder.Axis of the cylinder. The fixed line parallel to which a variable straight line moves, is called

die axis of the cylinder.jve^straight line which intersect a given curve. This given curve is called the

\k

Calculus & Geometry > MULTIPLE CHOICE QUESTIONS : Choose the most appropriate one :

The vertex of the cone fyz + gzx f- hxy = 0 is : (0, 0, 0)

2. The degree of every homog (b) Two

(a) (b) (1, 1, 1) (d) (0,0, 1).(c)(0, ly0) , sneous equation of a cone is :

(c) Three(a) One (d) None of these.

ANSWERS

Fill in the Blanks :3-fyz + gzx + lixy = 0 4. / —’^-0

Z. Z-1. Two 2. (0,0,0)


Multiple Choice Questions :1. (a) 2. (b)

□□□

t


The Coni'

• STUDENT ACTIVITY1. Find the equation of the cone whose vertex is the origin and which passes through the curve

9 9given by a* +by = 2z, lx + my + nz = p-

Prove that the equation ax2 + by2 + cz2+ 2ux+ 2vy+ 2wz +d = 0 represents a cone ifu v2 w2 ,

+ — = d.2.

— + —b ca

• TEST YOURSELF-31. Show that the reciprocal cone of the cone

±^gy ± 'ihz = 0 fyz + gzx + hxy = 0.

2. Prove that perpendicular drawn from the origin to the tangent planes to the ax2 + by2 + cz2 = 0 lie on the cone

*2 v2 z2' “ H----- --- Ox

a b c \3. Find the equation of the cone reciprocal to the cone fyz + gzx + hxy = 0.

iscone

ANSWERS

f2x2 + g2y2 + h2z2 - 2ghyz - 2hfzx -fgxy = 0.1.


1. The equation of a cone with its vertex (0, 0, 0) is homogeneous of degree...........2. Every homogeneous equation of second degree represents a cone whose vertex is.......3. The equation of a cone of second degree passing through the axis is...........4. The equation of the cone whose vertex is (0, 0, 0) and base the curve z = k,f{x, y) = 0 is

> TRUE OR FALSE :Write ‘T’ for true and ‘F’ for false statement:

x y z1. If the line - = J- = - is a generator of the cone fix, y, z) = 0 (a homogeneous equationl m n

of second degree), then m, n) = 0.2. The equation of a cone with its vertex at (0, 0, 0) is not homogeneous.3. Every homogeneous equation of second degree in x, y, z represents a cone.

(T/F)(T/F)(T/F)


Calculus & Geometry and F-gli-]af=0r.O-0., G~0,H = 0.Similarly

Substitute these values in (1), we getbcx2 + cay2 + abz2 = 0.

r.s' ’ , r r l

Divide by abc, we getJC2 V2 '22 ••

a O r cThis is the required equation of a reciprocal cone.Example!. Prove that the equation'..t'..', f2x2 +g2y2+-h2z2- 2f>hyz-ah fix -2fgxy = 0 ,•i

orrepresents a cone which touches the co-ordinate planes.

Solution. The given equation is__ ___ iV(/x) ± V(gy) ± V(/iz) = 0 or i(fx) ± V(gy) = + 'f(hz) . .'-■v

Squaring of both sides, we get___ ^ . ' ___/* + gy ± 2V(/y) <f(gy) =1iz or fx' + gy~hz = + I'Jfgxy.

Again squaring of both sides we have . - •'. * .J: ■\:(fx'-gyirhzff,4fgxyi } ■ V;

A2 + gV + h2z2 + 2fgxy - Ighyz - 2Jhzx = Afgxy fx2 + g2y2 + h\2 -2fgxy- 2ghyz - 2Jhzx = 0. . '

t

or i"

drThis is a homogeneous equation .of degree 2,-hence this represents'a cone.Iftheplane.x = Ointer$ect(l),then -

^2y2 +/i2Z2 - 2g/iyz - 0 ior {jgy - hz)2 = 0. ;. It is therefore obtained a perfect sqaure'hence x = 0 touches (1): Similarly y = 0, z = 0 also

touch the cone (1)

• SUMMARY•t• Equation of a cone with its vertex at the origin-is given.by

ax2 + by2,+ cz2-+2fyz + 2gzx + 2hxy = 0.

• If y = — is the generator of a cone ax1 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0, then

♦

r.r:al2 + btn2 +' c/i2 + 2fmn *+ 2glx + 2hlm - 0. t '

• Equation of a cone passing through coordinate'axes’is • ; ^fyz + gzx + hxy,=;0. . ;

• Tangent plane at (a, P, y) to the cone‘lu2.+. /)y2 + cz2 + 2fyz + 2gzx + 2hxy = 0‘is given byx (aa + /jP + gy) + y (/ta + 6p 4-^) = z (ga +yp + cy) = 0.

/

\ • The equation of a cone reciprocal to thergiven cone ax2 +‘by2 + cz2 + 2^z + 2gzx + 2/wy =-0 is given by

Ax2 + By2 + Cz2 + IFyz + 2Czi + 2Hxy = 0. where A=bc-f, B'=ca - g2, 0=ab -12, F~ gh- af, G = hf- by, H ^fg-^ch. ^ ^

• Angle between the plane u* + uy iMvz = 0 and the cone tf(x, y, z) = ax2 + by2'+ cz2 + 2fyz ‘+ 2gzx + 2lixy = 0 is given by

V + /+V * 1

*■ r"

[2p-10 = tan . r' .£_y:J(a + b + c) (m2 + v2 + w2) v, w)sJ. ‘ ' w ' i ■ J t ’ ' - - , - . -a h g u

h b f v g f c w U V w o

• The cone or2 + &y2 + cz2 + 2^ + 2£z* + 2/izr + 2/ucy = 0 has three mutually perpendicular generators if a + + c = 0. J

• The cone a*2 + fcy2 + cz2 + 2/yz + 2gzr + 2/ixy - 0 has three mutually perpendicular tangent planes if/2 + g2 + h2 = ab + bc + ca.

2where P -r

■.:r

t


J.

1 : The Cone\2\2 • ; •' : ; 1 [ tL

A . ,Eliminating /, m, n from (l) and (3), we get

x-a.

1 p-m^2 K n•••(3)= 1.

n

211 ^P-v■■+-h

^Z^-+^lZjyt = (z-yf,

= 1— a-Yz-yz-ya

-(4)or 2 *2aThis is the required equation of a cone. Further, the section of this cone (4) by * = 0 is

(Pz-yy)2n2-2 (X 4 = (*-Y)2i- + (By putting x = 0 in (4))b2a2

Vi £b

J_ 2 ^ v-T2y + 7+\

If this equation represents a rectangular hyperbola in yz-plane then, we have

te+£_1U.0a2 b2 b2

orc

(•.• coefficient of y2 + coefficient of z2 ~ 0)

The locus.of P(a, p, y) is obtained by generating a,fi,ytox,y, z- That is

Hence proved.x2 y2 + z2 a2 b2

= 1.

Example 2. Prove that the equation2 -vby2 + cz + 2ux + 2vy + 2>vz + d = 0

2 .2 2represents a cone if ~a + ~b + ^ =

Solution. Making the equation homogeneous of degree 2 by multiplying the proper power of/, we get

f(x, y, z, t) s ax2 + by2 + cz2 + 2uxt + 2vyr + 2wzr + dt2 = 0. Differentiating (1) partially, w.r.t. x, y, z and r, we get

^ = lax + 2ut, = 2by + 2vr,

•••(I)

dx2cz + 2wt, ^ = 2ujc + 2vy + 2ivz + 2dt

atif.dz

Now putting t ~ 1 and taking ^- = 0, = 0, ^ = 0, 0, we get

rv l^ = f0 and t~ 1lax + 2u = 0 • dx« u ax + u = 0 or x = - —■or a

v wSimilarly, y = -b’z = - ux + vy + wz + d = Q.

Substitute the values of jc, y and z in (2), we get«a V2 w2

+ -7- +a b c

c...(2)and

u2 v2 IV2 A— - —------+ d = Q ora b c

= d.

. . ru• TEST YOURSELF-2Find the equation to the cone whose vertex is the point (a. b. c) and whose generating lines intersects the conic px2 + qy2 = l, z = 0.

2. Find the equation of the cone, whose vertex is (ct, p, y) and base ax2 + by2 = J, z = 0.3. Prove that the equation of a cone whose vertex is the point (a, (3, y) and base curve as

z2 = 4by, x = 0 is (y* - Oz)2 = 4b (a - x) (ay - pc).4. Find the equation to the cone whose vertex is (a, p, Y) and base y2 = 4ax, z = 0.

1.

f|

Self-Instructional Material 175.>1

, 1 J|

Calculus- & Geometry r2 (ax^ + by2 + cz\2 + 2fyiZi +2^ciXj + 2/u:|>'|) + r{2nx\ + 2vyi + 2ivi|) + rf = 0.For all values of r (3) must be an identity. Therefore we must have

ax2 + by2 + cz2 + 2/y1Z| + Igz^i + 2hxlyl = 0 2uX] + 2vy[ + 2»vzj = 0

d = Q. |From equation (5) it has been observed that the point F(.x,i,y,tZ|) satisfies an equation of

degree one which gives that the surface is a plane if u, v and w are not all zero. But the surface is assumed to be a cone, hence u. v, >v all should be zero. i.c.. u = v~ w = 0. Now from equation (6) we obtained that ^ = 0. Therefore substituting w = v = w = 0 and rf- 0 in (1) we get (1) reduces to a homogeneous equation of degree 2 in a*, y. z.

ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy - 0.Hence the equation of a cone with its vertex is a homogeneous equation of degree two in

or ...(3)

...(4)...(5)

and ...(6)

x, y, z.Conversely, Every equation w.'iich is homogeneous of degree two in x, y, z represents a cone

with its vertex at the origin.Let the homogeneous equation of degree two in x, y, z be given by

ax2 + by2 + cjz2 + 2fyz + 2gzx + 2hxy ~ 0.

Let P(xi, yj.Zj) be any point. V^hicli obviously satisfies (1). Therefore for ail values of /■. the point (>*1, oq, rzj) also satisfies the equation (I). Thus any point (/:*», OV'm) °n the line OP. where O is the origin satisfies the equation (1). Hence the line OP is the generator of the surface (1) through the origin. Consequently the surface|(l) represents a cone with vertex at the origin O.

...(I)

• 24.3. AN IMPORTANT RESULTThe direction-cosines of the generator of the cone satisfy the equation of the cone with its

vertex at the origin.Let the equation of a line representing the generator of a cone with its vertex at the origin be

I m n ...O)

and let the equation of a cone be

ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0. ...(2)Any point on the line (1) is (Ir.mr, nr), where r is the distance of this point from the origin

O, because l,m,n are the actual direction-cosines. Since any point on the generator will satisfies the equation of a cone (2), then

r2 \al2 + bn2 + c/t2 + 2fmn + 2gnl + 2/i/m] = 0a/2 + bm1 + cn2 + 2fmn + 2gnl + 2hlm = 0 2 2 2Conversely. \i al +bm +cn + 2fmn + 2gnl + 2hhn = 0. then the line with direction-cosines

l, m, n is a generator of the cone (2) with its vertex at origin.

(v r*0)or

• 24.4. EQUATION OF A CONE PASSING THROUGH THE AXESSince we know that the equation of a cone with its vertex at the origin is satisfied by the

direction-cosines of its generator. Here the x-axis. y-axis. and z-axis are the generators of a required cone.

Let the equation of a cone with its vertex at the origin of degree two be given as oa2 + by2 + cz2 + 2Jyz + 2gzx + 2hxy = 0.

Since a*, y, z axes are the generators of (1) and d.c.'s of a, y and z are respectively, i. 0. 0; 0, 1.0; 0, 0, 1. then we have a = 0,6 = 0 and c = 0.

Thus (1) becomes

...(1)

2^z + 2gzx + 2hxy = 0 fyz+gzx + hxy = 0.

This is the required equation of a cone of degree two with its vertex at the origin passing through the co-ordinates axes.SOLVED EXAMPLES

or

X y zExample 1. The plane — + ^ + - = 1 meets the co-ordinate axes in A, B and C. Prove that,

the equation to the cone generated by lines drawn from O to meet the circle ABC is


The ConeUN I T

THE CONESTRUCTURE

The ConeEquation of a Cone with Its Vertex at the Origin An Important ResultEquation of a Cone Passing through the Axes• Test Yourself-1Equation of a Cone having a given Vertex and given base (Guiding Curve) Condition for the General Equation of Degree Two to be a Cone• Test Yourself-2Tangent Line and the Tangent Plane Condition on Tangency The Reciprocal Cone• Summary• Student Activity• Test Yourself-3


• How to find equation of a cone with given vertex and given base curve.♦ How to find the cone reciprocal of the given cone.

• 24.1. THE CONEDefinition. A surface generated by a moving straight line which passes through a fixed point

and touches a given surface or intersect a given cur\’e, is called the cone.Vertex. A fixed point through which a moving line passes, is called the vertex of the cone. Guiding curve. The moving line which interest the given curve, this curve is called the guiding

curve.Generator. The moving line in any position is called the generator.

• 24.2. EQUATION OF A CONE WITH ITS VERTEX AT THE ORIGINTo obtain the equation of a cone having its vertex at the origin is a homogeneous equation

of degree two in x, y, z.Let us assume that the general equation of degree two in x, y and z represents a cone with its

vertex at origin 0. Let its equation beax' + by2 + cz2 + 2fyz + 2gzx + 2hxy + lux + 2vy + 2wz + = 0.

Let F be any point on the cone whose co-ordinates is Z]). Then the equation of itsgenerator OF is given by

— = -2- = -^-. y\ zi

Any point Q on this line is (rxj, ryh rzj) wherex y z , .— = r (say).x\ y\ z\

Since OP is the generator of (1) and Q is any point on this line OP, then Q will satisfy the equation (1) for all values of r.

arfxff + b(ryxf + cfrzj)2 + 2/r2y1z1 + 2gr\xx + Ih^x^y^

...(2)

+ 2urx\ + 2vryx +-2wrz| + = 0

Self-Instructional Material 171-

. 1

Calculus & Geometry OBJECTIVE EVALUATION

> FILL IN THE BLANKS :In the equation of. a sphere the coefficients of x2,y2, z2 must be............

2. The equation ax' + by2 + c2 + 2ux + 2vy + 2wz + d = § represents a sphere if3. The equation of a sphere whose centre is (0, 0, 0) and radius is a is...........

> TRUE OR FALSE :Write ‘T’ for true and ‘F’ for false statement:L The centre of the sphere ;t2 + y2 + z2 + 2ux + 2vy + 2wz + J = 0 is (- u, - v, - w).2. The centre of the sphere (jc - 2) (jc - 4) + (y - 1) (y - 3) + (z - 3) (z - 5) = 0 is (3, 2, 4). (T/F)3. The centre of the great circle is the centre of the sphere.

(T/F)

(T/F)


2 2 2The centre of the sphere x + y + z -ax--£>y-cz = 0is:

(a) (0,0,0)1.

a b c(h) 2>2’2

■a b_ £ l 2’ _ 2’ _ 2

The radius of the "sphere 2x2 + 2y2 + 2z2 = 50 is :(b) 25 (d) 2.

(c) -- (d) (a, b, c).

2.(a) 5(c) 50The number of spheres that are touching the co-ordinate axis are : (a) Infinite

3.(b) 0(d) None of these.(c) 1

ANSWERS

Fill in the Blanks :1. unity

True or False :l.T 2. T 3. T

Multiple Choice Questions : 1. (b) 2. (a) 3. (a)

2. a = b = c 3. x2 + y2 + z2 = a2

□□□


The Sphere• Two spheres x2 + y2 + z2 + 2u tx + 2viy + 2wlz + = 0 andx2 + y2 -\-z2 + 2w2* + 2V2)’ + 2w2z + d2 = 0 cut orthogonally if 2m]H2 + 2v]v2 + 2w1w2 - d] +d2-

• STUDENT ACTIVITY1. A plane passes through a fixed point (p, q, r) and cuts off th axes in A, B, C. Show that the

locus of the centre of the sphere OABC is ^ -* y 2

= 2.

4

2 2 2Find the equation of the tangent plane to the sphre x +y +z -2*-4;y-6z + 9 = 0at

(1, 1, 1).

2.

• *TEST YOURSELF-4Find the equation of a sphere which touches the plane 3x + 2y-z + 2 = 0 at the point (1,-2, 1) and also cuts orthogonally the sphere x2 + y + z2 ~ 4x + 6y + 4 = 0.

1.

2. Prove that the sphere which cuts two spheres Si = 0 and S2 = 0 at right angles will cut the sphere A|Si + X2S2 = 0 at right angles.

3. Show that the spheresx2 + y2 + z2 - 4x - 2y + 2z~ 3 = 0

.t2 + y2 + z2 - - 8y - lOz + 41 = 0andtouch externally.

4. Show that the two spheresx2 + y2 + z2 + 6y + 2z + 8 = 0

x2 + y2 + z2 + 6x + 8y + 4z + 20 = 0andare orthogonal.

ANSWERS

1. x2 + y2 + z2 + lx+ I0y-5z+ 12 = 0.


Calculus & Geometry SOLVED EXAMPLESExample 1. Find the equation of the sphere that passes through the circle

spherex2 + y2 + z? - 2x + 3y - 4z + 6 = 0,;t2 + y2 + z2 + + 4y - 6z + 11 =0 orthogonally.

Solution. The equation of a sphere through the circlex2 + y2 + z2 - 2;c + 3;y - 4z + 6 = 0, 3* - 4.y + 5z - 15 = 0 is (x2 + / + z2 - 2x + 3;y - 4z + 6) + A(3x - 4^ + 5z - 15) = 0 x2 + y2 + z2 + (— 2 + 3X) x + (3 — 4X) ^ + (—4 + 5X)z + 6— 15A. = 0-...( 1)

3x - 4_y + 5z - 15 = 0 and thecuts

orSince (1) cuts the given sphere

x2 + y2 + z2 + 2x + 4;y - 6z + 11 =0 orthogonally. Then, 2uiu2 + 2vlV2 + 2wlWT = d] +

2 2 2-4 + 5X x(-3) = 6- 15A + 11x 2 + 2

22(- 2 + 3X) + 2(3 - 4X) - 3(- 4 + 5X) = 11 - 15A-5X = 1 or A = -1/5.

Putting the value of A in (1), we get2 2 2 13 x2 + / + z2 5

5(x2 + y2 + z2) -This is the required equation Example 2. Two spheres of

oror

19— y-5z + 9 = 0

13x + 19y - 25z + 45 = 0. of a sphere.radii /•[ and r2 cut orthogonally. Prove that the radius of the

- —x +

or

r\r2common circle is\/ 2 , 2 Vri +r2

Solution. Let the equation of the common circle bex2 +y2 = a2, z = 0.

Therefore, the equation of the given spheres through the circle (1) are x2 + y2 + z2 + 2Az - a2 = 0 x2 + y2 + z2 + 2jiz - «2 = 0.

z. r2 = A2 + a2, r2 = p2 + a-

...(1)

...(2)and ...(3)

Since (2) and (3) cut orthogonally, then2u\ u2 + 2v|V2 + 2w\W2 = rf] + d22A x jj. = - a2 - a2 or A2jx2 = a4 (squaring)

('■I -a)(r2 -a) = a ■ 2 1 2 , 2 2 2.2, (. ri = A + a , r2 = \i ¥ a )or

2 2 = a2(r2 + r2) or a =or ri r2yirl2 + r2

r\r2radius of the circle is' V + r22

• SUMMARY

Equation of a sphere of centre (a, (3, y) and radius r is (x - a)2 + (y - P)2 + (z - Y)2 - r2. Equation of a sphere having (xj, yi, z) and (x2, y2» z2) as end points of its diameter is

(x-X,) (x-X2) + (y -yi) (y -y2) + (z - zi) (z- z2) = 0.General quation of a sphere is x2 + y2 + z2 + 2«x + 2vy + 2wz + d = 0 with its centre (- m, — v, — w) and radius = \u -¥ v2 + w~ - d.Equation of sphere through

S = x2 + y2 + z2 + 2mx + 2vy + 2wz + d =0 P = ax +by+ cz + d = 0and

is givn by S' + AP = 0.2 0 0Tangent plane at (xt,y],zi) to th sphere x +y +z +2ux + 2vy + 2wz + d=0 is given by

xx2 + yyi +zZi + « (x + x,) + v (y +y!) + w (z.+ z^) + d = 0.


Tlu' 'Sphere6.x - By - 23 = 0 = 3z + 2.2. Find the equation of the sphere described on the line joining the points (3, 4, 1) and

(-1,0, 5) as diameter and find also the equation of the tangent plane at (- 1,0,5).3. Prove that the equation of the sphere which touches the sphere

4{x2 + y2 + z) + 10* - 25y - 2z = 0 at (1, 2, - 2) and passes through the point (- 1, 0,0) isx2 + y2 + z2 + 2x - 6y + l = 0.

4. Find the equation of the sphere touching three co-ordinate axes. How many such spheres can be drawn. 2 2 25. Find the equations of the spheres which pass through the circle * +y +z =5, x + 2y + 3z = 5 and touch the plane 4* + 3y = 15.

6. Show that the sum of the squares of the intercepts made by a given sphere on any three mutually perpendicular straight lines through the fixed point is constant.

7. A sphere touches the three co-ordinate planes and passes through the point (2, 1,5). Find its equation.

ANSWERS

2. 2*-y + 4z-5-0 and 4* - 2y - z - 16 = 0.3. x2 + y2 + z2 ~ 2* - 4y - 6z + 2 = 0; x + y - z + 6 - 0.5. x2 + y2 + z2 ± 2rx ± 2ry ± 2rz + 2r2 - 0; Eight spheres.

x2 + y2 + z2 + 2* + 4y + 6zx - 11 = 0; 5 (*2 + y2 + z2) - 4* - 8y - 12z - 3 = 0. 7. x2+y2 + z2-lOx-lOy-10z + 50 = 0.5.

• 23.8. INTERSECTION OF TWO SPHERES(a) Angle of intersection of two spheres.When two spheres intersect each other, then the angle of intersection of two spheres at the

common point of intersection is equal to the angle between their tangent planes at the common point.

(b) Condition for orthogonal intersection of two spheres.When two spheres intersect each other, the intersection is said

to be orthogonal if the angle of intersection is a right angle.Let Ci and C2 be the centres of two spheres and P be the

common point of intersection as shown in fig. 5.Since ZC1PC2 = 90°, then in AC1PC2

C[C22 = C]/*2 + C2P2-Let r{ and r2 be the radius of two spheres respectively then

r\ + r2 - C]C2 •Let the equation of two spheres be

x2 + y2 + z2 + 2«iX + 2v]y + 2wtz + di =0 x2 + y2 + z2 + 2u2x + 2v2y + 2w2z + d2 = 0.

Then centres C\ and C2 are respectively (- «], - vj, - wj) and (- u2, - v2, - w2) and2^ 2 2 2222 ri =«i” + ';i + W! - d\ and r2 =u2 + v2 +w2 - d2.

The distance between the centres Ci and C2 isCiC2- = («2 - «,)“ + (v2 - V|)2+ (w2- W])2.

-d)

...(2)

...(3)and

(1) becomes222 222 2 ^ 2 Mi + V[ + Wi — d\ u2 + v2 + w2 — d2 = (u2 — U[) + (v2 - V|) + (W2 — wi)

2tti«2 + ^vlv2 + 2h’1H'2 = di+ d2.This is the required condition for orthogonality of two spheres.

REMARKS

or

>• If 11*! + r2 | = C,C2, then the spheres touch externally.> If | ri - r2 | = CiC2, then the spheres tocuh internally.> If 0 be the angle between the tangents at the point of intersection of the spheres, then

r] + rl-cyc\2r\r2

cos 0 =


Calculus & Geometry 2 , 2 , x +y +i 2 + lux + Ivy + 2wz + d = 0 ax + by + cz + dt = 0.and let the plane be

If the plane (2) is the tangent plane of (1), then the perpendicular distance from the centre (- - v, - w) of the sphere (1) to the plane (2) will be equal to the radius of the sphere.

Since the radius of the sphere (1) is 'u2 + v2 + w2 -d.vw2 +v2+ w2-</= perpendicular distance from (-«,-v,-w) to the' plane

• •(2)

ax + by+ cz + df =0- au - bv - cw + d>U2 + v2 + w2 — ^ - 1or

Vi2 + b2 + 'c2ve get the required conditionSquaring of both the sides.

(u2 + v2 + w2 - d) (a2 + b2 + c2) = (au +bv + cw -di)2.

SOLVED EXAMPLESExample 1. Find the equation of a tangent plane to the sphere

x2 + y2 +.z2 - lx - 4y - 6z + 9 = 0 at (l, 1, 1). Solution. Equation of tangent at (*], /i, £]) to the sphere

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 is + xyi + zz| h m (x + Xi) + v (y + >>]) + w (z + ^[) + d = 0.

Here (xt,yi, zt) = (1,1,1) ard the sphere isx2 + y2 + z2 - 2x - 4y - 6z + 9 = 0.

So, the required tangent plane is- (x+xi) - 2 (y+y!) - 3 (z + z,) + 9 = 0xxj + yy [ + zzi

x + y + z-(x+ 1) - 2 (y + 1) - 3 (z + l) + 9 = 0 - y - 2z + 3 = 0

y + 2z - 3 = 0.

i.e..oror

ane 2x-2y + z+ 12 = 0 touches the sphere y2 + Z2-2x-4y + 2z-3 = 0. •>

Example 2. Show that the p> x2 +

Solution. The given sphere isx2 + y2 + z2i- 2x -4y + 2z - 3 = 0.

Centre is (1,2, - 1) and its radius is Vl + 4 + 1 + 3 = 3.Now, the perpendicular distaice from (1,2, - 1) to the plane 2x - 2y + c + .12 = 0 is

2(l)-2l(2)-l + 12

V4T4TT= 3

sphere. Hence the plane 2x-2y + <: + 12 = 0 touches the givenwhich is equal to the radius of the sphere.

Example 3. Find the equation of the two tangent planes to the sphere x2 + y2 + - 2y - 6^ + 5 = 0 which are parallel to the plane 2x + 2y - z-0.

Solution. The equation of any plane parallel to the given plane 2x + 2y - z = 0 is 2x + 2^ - z = A. ...d)

Since (1) touches the given spheres. Then the perpendicular distance from the centre (0. 1. 3) to the plane (1) is equal to the radius ^ of the sphere.

2x0+2xl-3-^ = ±VJ •V4 + 4 + 1

- 3 - X = ± 3V5* or X = -3-(±3V5). Substitute the value of A. in (I), we get

2x + 2y - z = - 3 - (± 3^5)2x +2y - z + 3 ±| 3^5 = 0.

This is the required equations of tangent planes.

or

or

TEST YOURSELF-3I 2 2 2 *1. Find the equation of tangent planes of the sphere x +y+ z + 2x -Ay + 6z - 7 -- 0 which

intersect in the line


= v(4)2 - (- 3)2 = V16-9 - V?-.Centre of the circle. The equation of a straight line through the point (0, 1, 2) and

perpendicular to the plane x + 2y + 2z= 15 is given by'<“*>■ '

Any point on this line is (r, 2r+ l,2r + 2). let this point be the centre of the circle so (r, 2r + 1, 2r + 2) lies on the plane x Jr2y + 2z = 15. Then

r + (2r + 1)2 + 2(2/-+ 2) = 15 9r = 9.

The Sphere

J

r - 1.Hence the centre is (1, 3, 4).Example 2. Find the equation of the sphere through the circle

.r2 + y2 + z2 = 9, 2x + 3;y + 4z = 5and the point (1.2, 3).

Solution. Since the equation of the circle is given by *2 + y2 -i- z2 = 9, 2x + 3y + 4z = 5 5 = x2 + y2 + z2 - 9 = 0 F = 2jc + 3y + 4z - 5 = 0.

The equation of the sphere through the circle is S + XP = 0

i.e..and

x2 + y2 + z2 - 9 + X(2x + 3y + 4z - 5) = 0.Since (1) is passing through the point (1, 2, 3) also. Then

1 +22 + 32-9 + A(2 + 6+I2-5)=0 or X = -j

Substitute the value of X in (1), we get3 (;c2 + y2 + z2) - 2* - 3y - 4z - 22 = 0.

This is the required equation of the sphere.

or

• TEST YOURSELF-21. Find the radius and centre of the circle of intersection of the sphere

x2 + y2 + z2 - 2y - 4z = 11 and the plane x + 2y + 2z = 15.2. Find the radius of the circle given by the equations 3x2 + 3y2 + 3z2+ x-5y-2 = 0,

x + y = 2.3. A variable plane is parallel to the given plane ^ ^ ^ = 0 and meets the axes in A, B, C

respectively. Prove that the circle ABC lies on the surfacec a'l fac- + - +zx - + - + xy t + — c b a c b ab-=o.yz

4. Find the equation of the sphere passing through the circles y2 + z2 = 9, x = 4 and y2 + z2 = 36, x= 1.

ANSWERS

1. Vt"; (1, 3, 4). 2. 1A/2. 4. x2 + y2 + z2 + 4x - 41 = 0.

• 23.6. EQUATION OF THE TANGENT PLANEThe equation of the tangent plane to a sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d-Q at the

point (xj, yj, Z]) on the sphere is

xx i + yy, + ZZ! + u(x + x,) + >>(y + yj) + w(z + Zi) + d = 0.

Corollary. The equation of a tangent plane at (xj.yi, zi) to the sphere x2 + y2 + z2 = r2 is givenby

XX,+yyi+zz, = r2.

• 23.7. CONDITION FOR TANGENCY OF A PLANE TO THE SPHERELet the equation of a sphere be given

Self-Instructiopal Material 165

Calculus & Geometry In the fig. 4 the dotted line represents the cross-section of a sphere cuts off by a plane {2). Let C be the centre of the sphere (1). Draw a perpendicular from C to the plane (2) which meets the plane at O', join C to P, whe-e P is on the surface of the sphere and on the dotted line. Then CP lies in the plane and CC is perpendicular to the plane. Thus CC is perpendicular to every line which is in the plane. Therefore ZCCP = 90°

In ACCT, ZCCP = 90°, thenCP2 = CP2 - CC2 or CP = ^CP2-CC2.

Since CP is the radius of the1 sphere which is constant and CC is the length of the perpendicular drawn from C to the given plane. Thus CP comes out to be a constant for all positions of P on the cross-section.Therefore, the point P moves in plane (2) in such a way that the distance of P from fixed point C is always contant. Hence P traces out a circle whose centre is C and the radius is C'P. Consequently the equation (1) and (2) simultaneously give the equation of a circle.

Great Circle. The cross section of a sphere by a plane through the centre of the sphere is called the a great circle. The centre and radius of a great circle are same as centre and radius of the sphere.

• 23.4. INTERSECTION OF TWO SPHERESTo prove that the intersectiol^ of two spheres comes out be a circle. Let the equation of two spheres be given

51 = x2 + y2 + z + 2u\X + 2v|y + 2wjz + dj =0 ...(1)52 = x2 + y2 + z‘ + 2u2X + 2v2y + 2w'2Z + d2~0. ...(2)

Subtracting (1) and (2), we getS,-5j = 0.

i.

and

i • •

2 (mj -u2)x + 2 (vx -v2)y +2 (>*T -w2)z+d1-d2 = 0. ...(3)

This equation (3) represents a plane. Thus the points of intersection of two spheres are as same as obtained by the intersection of either (1) or (2) and the plane (3). Hence these points of intersection lie on a circle.

• 23.5. SPHERES THROUGH A GIVEN CIRCLE

From above it has been observed that the intersection of a sphere and a plane give a circle or the intersection of two spheres give a circle.

Let the given circle be represented by the equations5 = a:2 + y2 + z2 + 2ux + 2vy + 2wz + d — 0 P ^ ax + by + cz + d] =0

then the equation S + XP = 0 is sa isfied by those points which are common to both 5, = 0 and P = 0 for all values of X. In fact S + XP = Q represents the equation of a sphere through the given circle 5 = 0 and P = 0 together.

Likewise the equation

and

S2X = 0Si +represents a sphere through the circSOLVED EXAMPLES

e 5] = 0 and = 0 together.\

Example 1. Find the co-ordinates of the centre and the radius of the circle x + 2y + 2zil5,x2 + y2 + z2 -2y-4z- 11=0.

Solution. The centre of the sphere*2 + y2 + z2-2y-4z- II =0

\ •

is (0, 1, 2) and its radius isr= VO +'l +4+11 =4.

Let p be the perpendicular distance from (0, 1, 2) to the plane x + 2y + 2z = 15. 0 + 2(l)|f2(2)-15

Vl +4 + 4 3Thus the radius of the circle is


,1"

The SphereSolution. Let the equation of a plane be

^-^ = 1 a b c

which cuts the axes in A{a, 0, 0), B(0, b, 0) and C(0, 0, c) and also passes through the fixed point (p, q, r), then we have

...(1)

*+?+^=i.a b c»l !...(2)

The equation of a sphere OABC isx2 + y1 + z1 - ax - by - cz = Q

a b £2’2’2 ‘

its centre is

Let (a, (3, y) be the centre of a sphere OABC, then a „ b c

a = 2’ P = 2’ T=2

a = 2a,b = 2(3, c = 2y. Eliminating a, b, c from (2) and (3), we get

JL + JL + J_i 2a + 2p + 2yZ+R+I-2’ a P y

.1:1

...(3)or

orO n -Thus the locus of (a, P, y) is

^ + ^ + - = 2. x y z

Hence proved.

• TEST YOURSELF-1Find the centre and radius of the following sphere :

lx2 + 2y2 + 2z-2x + 4y-bz-l5 = 0.Find the equation of the sphere whose centre is (- 3,4,5) and radius 7.Find the equation of a sphere whose centre is (2, - 3,4) and which passes through the point(I‘2,-1).Find the equation of the sphere whose centre is (1, 3,5) and which passes through the point (3,5,6).Find the equation of the sphere which passes through the points (1,-3,4), (1,-5, 2), (1,-3,0) and whose centre lies on the plane x + y + z = 0.Find the equation to the sphere through the points (0,0,0), (0, 1,-1), (-1,2,0) and (1,2, 3).(a) Find the equation of the sphere on the join (3, - 1,5) and (4,5, 1) as diameter.(b) Find the equation of a sphere whose extremities of a diameter are (1, 1,1) and (2, 3, 5).

1.

2.3.

4.

5.

6.

7.

ANSWERS

2 2

3. x2 + y2 + z2 - 4x+ 6y - 8z-22 = 0. 4. *2+/+ z2 - 2x - 6y - 10z + 26 = 0.5. x2 + y2 + z2 - 2x + 6y - 4z 4-10 = 0.6. 8 (x2 + y2 + z2) - 15* - 25y - 1 lz = 07. (a) *2+y2 + z2-7*-4y-6z+12 = 0. (b) *2+y2 + z2-3*-4y-6z+10 = 0.

1 2. *2 + y2 + z2 + 6* - 8y - lOz +1=0.1.

• 23.3. THE PLANE SECTION OF A SPHEREWhen a plane cuts the sphere, a cross-section of a sphere is obtained as circle.To prove that the cross section of a sphere cuts by a plane is a circle and find also its centre

and radius.Let the equation of the sphere and the plane be

x2 + y2 + z2 + 2ux + 2vy + 2wz + d-=0- ax + by + cz + di =0.

..•(1)...(2)and


Calculus <& Geometry Let P(x,y,z) be any point on the surface of the sphere. Join P to A and P to 8 such that 6APB is a right angled triangle right angled at P. Therefore the lines PA and PB are perpendicular to each other. Then

d.r.’s of PA arex-X|,y->»|,z-*Zj d.r.’s of PB are x- atj, y - yj* * ” 2:2

since PA is perpendicular to PB, then we haveand

(X -Xt) (x -ir2) + (y -yi) (y -y2) + (z- zQ U - zi) = 0-4

This is the required equation of the sphere.

• 23.2. GENERAL FORM OF A SPHERETo obtain the equation of a 'sphere in general form.The equation of a sphere of radius r having centre (a, p, y) is given by

(x - a)2 + (y -|P)2 + (z - y)2 = f2 x2 + y2 + z2- 2ttt - 2py - 2yz + a2 + p2 + y2 “ ^ = 0

From (1) we observed that the equation (1) is a second degree in x,y,z having no terms of xy, yz and zx and coefficients of x2, y2 and z2 equal to 1. Therefore the equation of the type

x2 + y2 + ^ + 2ux + 2vy + 2wz + d = 0having the same character as (J) has, is called general equation of a sphere.Now comparing (2) with (1), we get

...(1)or

...(2)

- Pj=-v, y = - u’ and d = a2 + p2+ y2 - r2. _— -----------Thus the centre and radius of (2) are given by (-u,~v,-w) and vit + v2 + w2 - d

respectively.SOLVED EXAMPLES ,

Example 1. Find the equation of the sphere whose centre is (2, - 3, 4) and radius VsT. Solution. Since we know tfie equation of a sphere whose centre is (a, p, y) and radius r as

follows :

a =

a)2 + (y|- P)2 + (l-y)2 = '-2-Here a = 2, P = -3, y = 4 and r - ■v51, then

(x - 2)2 + (y + 3)2 + (z - 4)2 = 51x2 + y2 + z2 - 4x + 6y - 8z - 22 = 0.

Example 2. Find the centre and radius of the sphere given byx2 + y2 + - 4x + 6y + 2z + 5 = 0.

(x-

or

iSolution. The given sphere isx2 + y2 + z2 - 4x + 6y + 2z + 5 = 0.

Compare this equation with the equationx2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0,2w - - 4 or m = - 2 2v = 6 or v = 3 2;v = 2 or iv-l and d = 5.

Centre is (- u, - v, - w) = (2, - 3, - 1)radius = vm2 + v2 + w2-d

= V(- 2)2 + 32 + l2- 5 = V4 + 9+ l- 5= V9=3.

Example3. Find the equation of the sphere on the point (2, - 3, 1) and (3. -1.2) as diameter. Solution. Since we have the equation of a sphere whose end points of a diameter are

(xi,yi, zi) and (x2, y2, Zi)- The equation is(x-x,) (x-xj) + (y-yi) (y -y2) + <* - zi) (z - z2) = 0 (jfi.yi.Zi)s(2,-3.1) and (x2,y2,Z2) = (3.-1.2).

we get

and

HereThen we get

(x - 2) (x- 3) + (y + 3) (y + l) + (z - 1) (z - 2) = 0 x2 + y2 + z2-5x + 4y-3z+ 11 =0.

Example 4. A plane passes through a fixed point (p, q. r) and cuts off the axes in A. B. C. Show that the locus of the centre of the sphere OABC is

or

£ + 4^ = 2.x y z

-162 Self-Instructional Material

The SphereUNIT

THE SPHERESTRUCTURE

• Equation of a Sphere• General Form of a Sphere

• Test YourseW-1• The Plane Section of a Sphere• Spheres through a Given Circle

• Test Yourself-2• Equation of the Tangent Plane• Condition for Tangency of a Plane to the Sphere

• Test Yourself-3• Intersection of Two Spheres


QttfC «•***«'


# How to calculate the radius of the circle defined as the section of the sphere of the plane

• How to find the tangent planeHow to find the condition that two spheres cut orthogonally

• 23.1. EQUATION OF A SPHERE{a) To obtain the equation of a sphere whose centre and radius are given.Let C be the centre of a sphere whose co-ordinates are assumed (a, p, y) and let r be the radius

of this sphere. Let P(x,y,z) be any variable on the surface of the sphere whose equation to be determined as shown in fig. 1.

Since we have P (x, y. z)CP = r or CP2 = r2. -.(I) r.

Further since CP is the distance between two points C(a, (3, y) and P(x, y, z). Then _________________________ C (a, p, y)

CP = V(*-a)2 + 0--P)2 + (z-Y)2. Now substitute this value of CP in (1), we get

(x - a)2 + (y - (3)2 + (Z- y)2 = r2. This is the required equation of a sphere.

••.(2)Fig. 1

Corollary. If the centre of the sphere is origin i.e., (0,0,0) and radius is r then its equation is x2 + y2 + z2 = r2.

Proof. In the formulaP(.xyz)(x-a)2 + (y- P)2 + (z - y)2 = 22

Putting a = 0, p = 0, y = 0, we get 2 2 2 2X + y + Z = r . Hence obtained. B(x2,y2,z2)Oj-Vp Z0A(b) To obtain the equation of a sphere whose

end points'of a diameter are given.Let and B(x2,y2,Z2) be the end

points of a diameter of a sphere as shown in fig. 2.

C

Fig. 2


Calculus <&. Geometry > MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one

x - i y-3 z+1The line —^---- ^---------------

(a) (1,3,1)(c) (1,3.-1)

1. passing through the point:

(- 1, 3, - 1)(- 1,^3,- 1).

The equation ^ ~ g = q represents the equation of:

3 2(b)(d)

2.

(b) v y-axis (d) None of these.

The shortest.distance between the lines

(a) x-axis (c) z-axis /-r

x+l y+2 z+3 y + 2 a- + 1 z + 3 .3. rand 4i 2 5(a) 1 (b) 6 (c) 0 (d) 2.

4. The shortest distance between two coplanarlines is : (a) 0 (b) 1 (c) 2 (d) Noneof:these

ANSWERSFiU in the Blanks :

A-a_y-P_z-V X~X\ y-yi z-z\1 " 0 " 01. 3./ m n *i-X\ yi-y\ Zi-Z\

4. 2, 3, 3 True or False :

I. T 2.F 3. T 4. TMultiple Choice Questions :

2. (a) • 3. (c)1. (c) 4. (a)□□□

160 Self-Instructional Material.

-3 ,x-2 y-4 z-5rand-j

Find the shortest distance between the lines =£

y-2 z- The Straight Line2.5 '4.3

• TEST YOURSELF-5

1. Find the shortest distance between the linesx-3_ y-8_ z — 3 x + 3_ y + 7_ z-6

3 - ~ 1 ~ 1 ' -3_2_4Find also its equations and the points in which it meets the given lines.

2. Find the length and equation of the shortest distance between the following linesx-3 y-5' z~7 x + 1 y+1 Z+l

(0 1 " - 2 ~ 1 ; 7 “ - 6 “ - r.... A"-3 y-5 z-2a-1 y+1 z+1(n)-F=^=T~;“y=V=—'

-3 _ y-5 _ z-7. x + 1 _y + 1 _ z+1 1 " -2 ~ 1 . ; 7 " -6 " 1 '(iii) ”

f

ANSWERS

1. 3V30;^ = 3’“8

2. (i) 2^,^

30V29’ :

(iii)2V29,^3 y-5-z“7

^-p;(3f 8, 3) and (- 3,-7,6)

y-2 z- 35

- 13 4

- 11a - 2y + 7z + 9 = 0 = 27a + 26y - 33z + 22(ii)

3 4OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. The equation of a straight line through the point (a, (3, y) having the d.c.’s /, m, n is...........2. The equation of A-axis in symmetrical form is...........3. The equation of the straight line passing through the two given points (Ai.yi.zi) and

(A2, y2,23) is...........X4. The direction ratios of the line —-

V TRUE OR FALSE :Write ‘T’ for true ‘F’ for false :

-0 y-1 z-2 are3 3

A-a y-P z-Y—1 is the equation of a straight line in symmetrical form.1. / (T/F)mx y z2. The equation of y-axis is — = ^ = y (T/F)

3. The line of intersection of the planes aiA + b\y + c\z + d\ =0 and aix + bjy + C2Z + ^2 = 0 gives the line in non-symmetrical form.

4. The equation w + Av = 0 represents the line of intersection through the planes u = 0 and(T/F)

v = 0. (T/F)


Calculus & Geometry • SUMMARY• Equation of a straight line in general form is axx + b{y + C]Z + = 0 = <32 + b2y + 'cxz + dz.• Equation of a straight line in symmetrical form is

x-q y-fl g-yl m n

where /, m, n are d.c’s of the line.• Let ax + by + cz +d=0b& a plane and x , a = ^^ = -—

l . m n(i) Line is parallel to the plane if al + btn + cn = 0 and aa + b$ + cy + d*0

(ii) Line is perpendicular to the plane if — = ^ .

(iii) Line lies in the plane if al + bm + cn = 0 and a(X + b$ + cy + d = 0.X — o. y — R z~y

• Angle betwen the line—-—=z—c- =----- 1 and the plane ax + + c^.+ i/-0 is given by

al + bm + cn 1 •

a line. Then

• -ism. va+ bsup2 + c2 V/2 + m? + n2 .

i y-p\ z-yi x — c&2 z-y2x-aLines and are coplanar if

. /iq2 - a! p2 - p[ y2 ~ yx

"i

mj n m2 n21

l = 0mj1

h m2 ”2x-a Lll= y - povrw -• Perpendicular distance of from P (X), yi, Zj) is given by

/ n2 2

yi-p z xi-a y, - P-y. Z\~y xi-aPNl = + +/ /m n n in

Shortest distance between t x-tt] y-pj z-y

ic linesx-O-2 y~$2 z-y2

and is given by/ hmx m2 n2i

S.D = (a2 - aO / + (fo - p,) m + (y2 ~ Yi) "where /, m, n satisfy the rea tions

llx + mmx + nnx = 0 and ll2 + mm2 + nn2 ~ 0 y Z~c x y z + c

1 tan a 0 ’ 1 tan a 0

t

x• Skew lines are — =

• STUDENT ACTIVITY1. Find the image of the point (1, 3, 4) in the plane 2x-y + z + 3 = 0.

' ! !

<!S


The Straight LineTherefore the distance between these parallel planes (3) and (4) gives the length of shortest distance between (1) and (2).

And the equation of shortest distance is obtained as the line of intersection of the two planes:(i) The plane through (1) and perpendicular to the plane (3).(ii) The plane through (2) and perpendicular to the plane (4).

SOLVED EXAMPLESExample 1. Find the shortest distance between the lines

x-1 y — 2 z- 3 x-2 y-4 z-5 4 ’ 3

Solution. Let /, m, n be the direction-cosines of shortest distance, since S.D. is perpendicular to both lines so we have

2 4 53

21 + 3m + 4/i = 0 3/ + 4m + 5n = 0.

...(1)

...(2)andSolve (1) and (2), we get

l m n15-16 12-10 8-9l l m nm nor 2 -1 °r 1- 1 -2 1

21 1m <6'n V6VT

Since ^(l, 2,3) and Q(2,4,5) are the points on the given lines respectively. Thus the projection of PQ on the shortest distance is

4=l+(5-3)4=1S.D. = (2-1)-=-+(4-2) -V6V6 V6

_1___4_ _2__Vcf Vis

1

1Hence, the length of S.D. = — numerically.Vb"

Example 2. Show that the length of the shortest distance between the linesx-2 y + 1 z

2 “ 3 “4;

2x+3y-5z-6 = 0 = 3x-2y-z + 3w 97/(13^6).Solution. Since, we have

x-2 y +1 z3 “4 '

2* + 3y - 5z - 6 = 0 1 3x - 2^ - z + 3 = 0 J ■'

The equation of a plane through (2) is given by(2x + 3y - 5z - 6) + A (3x - 2y - z + 3) = 0

(2 + 3A) x + (3 - 2A) y + (- 5 - A) z - 6 + 3A = 0.Since (3) is parallel to (1) so d.r.’s of normal to (3) is perpendicular to (1). Then

2(2 + 3A) + 3(3 - 2A) + 4(- 5 - A) = 0 -4X-7 = 0 or A = -7/4.

Substitute this value of A in (3), we get

bfVf-TV

,.(1)2

...(2)

•••(3)or

or

Sj7l A 21-5+7 z-6- — 4 4

= 0

- 13x + 26y- 13z-45 = 0 13jc - 26y + 13z + 45 = 0.

Now (1) and (4) is parallel and P(2, - 1,0) is any point on (1) so length of shortest distance

or•.•(4)or

is given by.S.D. = Perpendicular distance from P(2, - 1,0) to the plane

13.x - 26y + 13z + 45 = 013 x 2 - 26(- 1) + 13(0) + 45 97

l3>/6V(13)2 + (- 26)2 + (13)2

Self-Instructional Material. 157"

Calculus & Geometry Next, the equation of shortest distance is the plane containing first line and S.D. (shortest distance) and also containing second line and S.D. which are given by

<Xi y-P,. z-Yix -l = 0 ...(3)'*i «i/ m n

a2 y-$2 z-y2x -and ^2 = 0 -(4)/«2 «2

l m n

Thus the equations (3) and (4) simultaneously gives the equation of shortest distance, (ii) Co-ordinates Mel hod (Point Method).Let the equation of two lines be

x-Q-i y-pi *-Vi= ri (say) ...(1)h mi ni

x-a2 y-p2 z~y2and = r2 (say) ...(2)m2 «2

Any point on (1) is + ab m^ + pb + Yj) and on . (2) is(2(/2r2 + ot2, '”2^2 + P21 niri + f2)-Let hne PQ be the shortest distance. Then PQ will be perpendicular to both the lines (1) and (2) so that we have d.r.’s ratios of PQ are

hn - f 02 - «!, m2r2 - mjr, + P2 - Pj, n2r2 - + Y2 - Yi ■Since PQ is perpendicular to (1) and (2), we have

^1(^2 -*!''!+ «2 “ “i) + mi(m2f2 ~ m,ri + p2 - P,) + n,(n2r2 - n,^ + Y2 ~-Yi) ^ 0 ...(3)and /2(/2r2 - l{r{ + a2 - aO + m2(m2r2 - + p2 - pi) + n2(n2r2 - + Y2 “ Yt) = 0. .(4)

Solving (3) and (4), we get rj and r2- Substitute these value of /[ and r2 in the pointsP and Q we get the two points at which the shortest distance meets. Therefore, the distance between the points P and Q gives the length of shortest distance and the equation of shortest distance is thus obtained by the line joining the point P and Q.

(b) When one line in symmetrical form and other in general form.Let the equation of two lines be given by

x-a Y-P z-y -■-(!)

I m n= a\X + byy + CjZ + = 0= a2x + 6?y + c2z + ^2= 0

through the line (2) is given by + Av = 0

uand ...(2)v

The equation of a plane...(3)u

where X is a parameter.Now find the X in such a way that (1) is parallel to (3) and substitute this value of X in (3)

we get the equation of a plane parallel to the line (1), Therefore, the perpendicular distance from any point oh the line (1) (a, p, y) (say) to the plane (3) gives the length of the shortest distance and the equation of shortest distance is obtained as the line of intersection of the two planes :

(i) the plane containing the line (1) and perpendicular to the plane (3) (ii) the plane containing the line (2) and perpendicular^ to the plane (3).

(c) When both the lines are in general form.Let wps 0, V| = 0 and «2 = 0, v2 = 0 be the two lines in general form that is,

ui^ aix + b\y + C\Z +d] =0 vl=a2X + b2y + c2z + d2 = 0 ...(1)

u2 = a^x + b^y + c2z + d2 = 0 v2 = ajx + b4y + ciz + d4 = 0and ...(2)

The equation of a plane through the line (1) is given byui + ^iwt = 0 ...(3)

and the equation of a plane through the line (2) is given byu2|+X2V2 — 0. •••(4)

Now find the Xj and X2 is such a way that the plane (3) and the plane (4) are, parallel. Andsubstitute the values of X|andX2 in (3) and (4) respectively. Thus (3) and (4) become parallel.


The Straight Line

• TEST YOURSELF-31. Find the projection of the line 3x-y + 2z-l=0 = x + 2y-z = 2 on the plane

3x + 2y + z = 0.2. Find the equations of the perpendicular from origin to the line

ax + by + cz + d = 0 = a,x + b,y + c'z+ ct ■3. Prove that the lines :

_y_ _z ^ _x__y___ L.£_X_ia P y ’ act b$ cy ' l m n

will lie in a plane if^(t-c)+|(c-a)+^(a-fc) = 0.

4. Prove that the lines :x-\ y-2 z-3 _x-2 y-3 z-4

4 ’ 3are coplanar; find their point of intersection. Also find the equation of the plane in which they

52 3 4

lie.

ANSWERS

1. 3;c + 2;y + z = 0 = 3* - Sy + 7z + 4. 2. {be' - b’c) x + {ca - ca) y + {ab' - ab) z = 0.4. (- 1, - 1, - 1); x - 2y + z = 0.

• 22.6. SHORTEST DISTANCE BETWEEN TWO LINESShortest distance between the lines can only have the meaning if, the lines do not intersect

or the lines are not lying in the same plane, those lines which have this character, are called skew lines and length between them which is perpendicular to both skew lines is called shortest distance.Length and Equation of Shortest Distance :

To obtain the length and equations of shortest distance between two non-intersecting (or skew)lines.

(a) When both lines in symmetrical form.(i) Projection Method. Let the two skew lines be given by

x-a, .y- p, z-Tx x-V-2 y~$2 V2and/•>/ "2mj m2i

Let /, m, n be the direction cosines of a shortest distance. Since shortest distance is perpendicular to both the given lines. Then we have

ll\ + mm) + nni = 0 U2 + mm2 + nn2 = 0.

Solving (1) and (2) by cross multiplication method, we have

...(1)

...(2)and

l m nmI/i2-m2ni «i^2-,I2^i /]m2-/2mi

Thus, direction ratios of shortest distance are mpij - m2ni, nxl2 - n2lx, fnii - l2m\. Therefore, the direction-cosines of shortest distance are

^1^2 — ^2^1 l\m2 ~ ^2nl\m1n2 - m2/t]/ = , n =, m =LM LM

where ____________________ __________________LM = V[(mln2 - m2ni)2 + (n\l2 - n2l{)2 + (/1m2 - ^i)2].

Further since, first line is passing through the point P(a,, p],Yi) and second line is passing through Q(<x2, p2, Y2)- Suppose the shortest distance meet the given lines in R and S respectively as shown in fig. 7.

Therefore the proejetion of PQ on the shortest distance gives the length of shortest distance which is given by

Shortest distance= («! - 02) / + (Pi - P2) ro + (Yi -Y2) n.


Calculus & Geometry \-$_z-yx-a _ y = r[ (say).ba c

Any point on this line is (ari + a, bri + p, cry, + y) let this point be Q, then it will lie on the given.plane ox + + cz + = 0, so we have >

aiari + a) + P (fer + p) + c(cr| + y) = 0 aa + ^P + cya2 + b2 + c2 ■-

Putting this value of r, in the co-ordinates of Q we get the point Q.Thus the line passing through the point P and Q !is the required line of projection.

SOLVED EXAMPLES

or ''I = -

%\

y-3 s-2‘ % — jExample 1. Find the distance of the point P(3,8, 2) from the line —-—

measured parallel to the plane 3x + 2y - 3z + 17 = 0.Solution. Any point on the! given line is A/ (2r + 1,4r + 3, 3r + 2).Let it be the foot of the perpendicular drawn from P to the line.The d.r.’s of PH are 2r - 2, [4r - 5, hr.The d.r.’s of the normal to the given plane 3x + 2y - 2z + 17 = 0 are 3, 2, - 2. Since PH is being measured parallel to the given plane.

3 (2r- 2)1+ 2 (4r + 5) - 2 (3r) = 0 6r — 6 + 8r - 10 - 6r = 0 8r- 16 = 0 or r-2.

4 3

oror

Putting this value of r in the co-ordinates of N, the co-ordinates of N are (5, 11, 8).PN = V(51- 3)2 + (11 - 8)2 + (8 - if

= V4 + 9 + 36 =V49 = 7.Example 2. Show that the lines

x+3 y>5 z-7 , x+l y+1 z+ l 3 and 5 =^r

are coplanar. Find the equation of the plane containing them.Solution. Let,

f 2 3

x+3 y+5 z-72 _ 3 “^3“

x+l_y)l-l_z+l i

Any point on the line (1) is P(2rl - 3, 3/y - 5, - 3rj + 7) and any point on the line (2) is e(4r2-l,5r2-l,~r2-l).

If the lines (1) and (2) are coplanar, they will intersect 2r1-3 =4r2-l or 2r1-4r2 = 2 3r1-5=5r2-l or 3^ - 5r2 = 4

- 3r] + 7 - - r2 - 1 or 3rj - r2 = 8.

= r{ (say) -(I)

and = r2 (say). .-(2)

...(3)

...(4)and ...(5)

Solving (3) and (4), we getn = 3, r2=l.

On putting these values in R.H.S. of (5), we get 3(3)-l = 9-l = 8 = R.H.S.

Thus both points coincides. Hence the lines are coplanar. The point of intersection' is (3,4,-2).

The equation of a plane containing (1) and (2) (i.e., a plane through the line (1) and parallel to the line (2)) is.

x+3 y+5 z-72 3 -3 = 0

5i-l

(x + 3) {-3+ 15} +(y + 5) {-12 + 2} + (z-7) {10- 12} = 0 12 (jc + 3) - Itf (y + 5) - 2 (* - 7) = 0 12x - lOy -2z = 0 or 6x - 5y - z = 0.

4

ororor

li>4 Self-Instructional Material

The-Straight Linefl(02 - a,) + b($2 - PO + c(y2 - Yi) = 0.Now eliminating a, b, c from (4), (5) and (6) we get the required condition

02-ai P2-P1 Y2-Y1

...(6)

/ = 0.Wl1

h m2 »2

Again eleminating a, by c from (3), (4) and (5), the required plane is x-cti y-Pj z - Yi

/ = 0.m.1

h fnt2 n2

(b) When one line in symmetrical form and other in general form :To obtain the condition that the lines

x-a y-fl z-y/ m n

aix + b\y + C\Z + dj = 0, a2* + b^y + c2z + = 0 • ••(2)

are coplanar.If the given lines intersect, they are coplanar.Any point on the first line is (Ir + a, mr + P, nr + y): This point will lie on the second line.

Then we haveai(/r + a) + b\(mr + P) + C\(nr + y) + d] = 0

aja + i^ + CiY + dia|/ +'b\m-+ Cj/i

a2a + i>2p + c2Y + d2 a2l + b2ni + c2n

Equating the values of r, we get the required condition ajCc + ^p + CiY + di Q2a + />2P + c2y + d2

ad + b2m +c2n

or r = -

Similarly r ~-

aj + b]m + Ci«

• 22.5. PROJECTION OF A LINE ON A GIVEN PLANE -•'Y ■

X — CL y — ft Z ~yLet —:— = ^=---- L be a line and ax + 6y + cz + d= 0 be a plane on which the projection

of the line is to be required. Therefore two ways to find the line of projection.(i) The equation of any plane through the given line is

A(x - a) + B(y - P) + C(z ~ y) = 0 Al + Bm + C« = 0.

Since the given plane will be perpendicular to the plane (1) if Aa + Bb + Cc = 0.

Solving (2) and (3), we get

/ m

...(Dwhere ...(2)

•••(3)

A CBme - nb na - Ic lb - ma

Putting these values of A, B, C in (1), we get(me - nb) (x - a) + (na - /c) (y - P) + (lb - ma) (z - y) - 0

Hence the equations ax + by-\-cz + d = 0 and (4) together are the equations of the line of...(4)

projection.(ii) Any point on the given line is P(lr + a,mr + p, nr + y). If this point lies on the given plane, then

a (Ir + a) + b (mr + $) + c (nr+ y) + d = 0 aa. + b$ + cy + d A(a, p, y)or r = - al + mb + nc

Putting this value of r in the co-ordinates of P, so we get the point of intersection of the given line and the given plane.

Since the line passes through the point A(a, p, y) so draw a perpendicular from A to .the given plane, which meets at Q. Let Q be the foot of the perpendicular.

Now, the d.r.’s of the normal to the given plane are a, b, c and hence these are the d.r.’s of the line through (a, P, Y) and perpendicular to the given plane, therefore the equations of this perpendicular line are

PQ

Fig. 6

Self-Instructional Material: 153

Calculus <£ Geometry y %tation of a plane containing the line ,+- = 1, jr = 0 and parallel to

t> c ■'Example 2. Find the eqi

the line — - -= l,_y = 0. a c

Solution. The equation of a plane containing the line

b c^ + ^ x*+£ + *b c- 1 + Xx = 0 or -1=0. ...(1)is b c

Now d.r.’s of normal to this plane are

i

Since this plane is parallel to the linex z , n .— - — - 1, y = 0 i.e., a c

x - a _ _ z - 00 ca

whose d.r.’s are a, 0, c.1 1Xa + ~.0 + — .c = 0

Xa +1=0 or X = - l/a.Putting this value of X in (1), the equation of required plane is

x

cor

z+*-i=ob ca£_E_i + l=0.or b ca

• TEST YOURSLF-2P = ax + by + cz + d 0,1. Find the equation of the plane through the line

Q = a'x + b'y + c'z + d' = l and parallel to the line - = -¥- = —■2. Find the equation of a pi me through the line of intersection of the planes -c + 2y + 3z -- 4 = 0

and 2x + y- z + 5 = 0 ana perpendicular to the plane 5x + 3y + 6z + 8 = 0.3. Find the direction cosinei of the line whose equations are x + y = 3 and x + y + z = 0 and show

that it makes an .angle 305 with the plane y - z + 2 = 0.4. Prove that the lines

3x + 2y + z-5 = 0 = A: + y-2z-3 and 2x-y-z = 0 = 7x + lOy -8z- 15 are perpendicular.

ANSWERS

1. P(a'l + b'm + c'n) =

3' V2 V2

<2(al + bm + cn) 2. 51x+ 15y - 50z + 173 = 0.1

• 22.4. COPLANAR LINES(a) When both lines are in symmetrical form :To obtain the condition that two lines may intersect and the equation of the plane in which

they lie.Let the equation of two lines be given by

*-«,! y-p! z-Yi-•-(1)/ *1m.i

X-CLl _y-$2 _z-Y2 h i m2 n2

The equation of a plane Ithrough the line (1) isa(x - a,) jf 6(y - pi) + c(z - 7i) = 0

dli + +cni = 0If the line (2) lies in the same plane, then the normal of this .plane is perpendicular to (2), we

...(2)and

...(3)...(4)where

have} ...(5)a/2 + b{n2 + cn2 - 0.

Further since the point (a2, P2, Y2) also lies on the plane; then


!

The Straight Line

• 22.2. EQUATION OF A PLANE THROUGH A GIVEN LINE(a) If the line is in symmetrical form. Let the line be

x-g y- ft z-y •••(I)l m nSince the plane is to passes through the given line. Therefore the plane will pass through the

fixed point (a, (3, y)- Then the equation of a plane through (a, (3, y) is a{x - a) + - [3) + c(z - y) = 0.

Since the plane (2) is passing throguh the line (1), then its normal is perpendicular to (1)..\ alJrbm + cn = 0.

...(2)

Hence the required plane isa(x - a) + 6(y - p) + c{z - y) = 0, al + bm + cn = 0.where

(b) If the line is in general form so let the equation of a line is+ + + =0, !

a2* + b-ry + c2z + d2 = 0 [

Hence, the equation of a plane through (3)-is given by{a\X + b\y + C\z + di) + X{a2x + bjy + c2z + ^2) = 0.

• •.(3)

REMARKIf « = 0 and v = 0 are equations of two planes, then the equation of a plane through the intersection of these planes is u + \v = 0 where X is a parameter.

• 22.3. ANGLE BETWEEN A LINE AND A PLANE

x - (x y — B Z”Y------ = = -—1 and the plane ax + by + cz + d=0.To find the angle between the linel m

Let 0 be the angle between the line AB and the plane, then the angle between the line AB and normal AN to the plane will be 90° - 9. \

Direction ratios of AN are a, b, c and direction ratios of line AB are /, m, n.

al + bin + cn >9O°-0.cos (90° - 0) = 0V(q2 + b2 + c2) (l2 + m2 + n2)

al + bm + cn Asin 0 =i.e..

y(a2 + b2 + c2) (l2 + m2 + n2)This gives the angle between the line and the plane. Fig. 5

SOLVED EXAMPLESx — o. y — (3 z ~~ yExample 1. Find the equation of the plane through the line —-----^--------L and parallel

l m n-g'_ y-B' z-y'to the line n

Solution. The equation of a plane through the linem'

l m na{x - a) + 6(y - P) + c(z - y) = 0

where a, b, c are the d.r.’s of the normal to the plane (1).Since the plane (1) containing the first line whose d.r.’s are /, m, n and it is parallel to the

second line whose d.r.’s are l', m', n', then we haveal + bm + cn = 0

al' + bm' + cn' = 0.Solving (2) and (3) by cross multiplication, we get

...(1)is

-...(2).-(3)and

ba cmn'-m'n l'n-ln Im'-l'm

Putting the proportionate values of a, b and c in (1), the equation of the required plane is (mn' - m'-ti) (x -a) + (I'n - In') (y - P) + (lm' - I'm) (z-y) = 0.

Self-Instructional'Material' 151

Calculus & Geometry This is the required line in symmetrical form passing through two points (*i> y\,Z\) and

SOLVED EXAMPLES3 2=------is parallel to the plane 3x - y + 4z = 7, then findExample 1. If the line ^~~~ =

1 cc.

Solution. Dr’s of the given line are 2, 1, c.Dr’s of the normal to the plane 3* - y + 4z = 7 are 3. - 1,4.The line is parallel to the plaie if

2x3+ 1 x- 1 +cx4 = 0 c — — 5/4.

x-2 y +1 j — 2Example 2. Find the point in which the line —-— = ^ meets the plane

x -2y z = 20.Solution. Since the equations of the lines are

*-2 z-2-jj- = r (say)-

Any point on this line is P(3r + 2,4r- i. 12r + 2).The given line meets the plane x-2y + z = 20soP lies on this plane.

(3r + 2) - 2 (4r- 1) + (12r + 2) = 20 7r= 14 or r = 2.

Putting this value of r m the co-ordinates of P, the required co-ordinates of the point are (3 (2) + 2,4 (2) - 1,12 (2) + 2) i.e , (8,7,26).

Example 3. Show that the lit e joining the points A(2, -3,-1) and 5(8, -1,2) has equations x-2 y + 2 ;+ 1

1+3 4

or

36 3Find two points on the line whose distance from A is 14.Solution. The d.r.’s of the line AS are

8-2,- 1 -j(-3),'2-(- 1) or 6,2,3.Thus the equation of a line AB through A whose d.r.’s are 6, 2, 3 are

x- 2 _[ y + 3 z+ 16 3 3

These are the required equations of the line joining the points A and B.+ 3 z +1 . .- —= r (say)

so any point on this line is P(6r + 2, 2r- 3,3r- 1).Since

x-2Let 6 3

PA = 14.V(6r + 2 - 2)2 + (2r\- 3 + 3)2 + (3r - 1 + l)2 = 14

V36r2 + 4r2 + 9r2 = 14 or ^49r2 = 14 ±7r=14 or r=±2.

Putting r = 2 and -2 successively in the co-ordinates of P. we get the required points as (14, 1,5) and (-10,-7,-7).

oror

• TEST YOURSELF-1jc+1 y+3 z-2 ... .------= £-------- with the planeFind the co-ordinates of the point of intersection of the lineX. 31

3x + 4y + 5z = 20.Find the co-ordinates of the point where the line joining the points (2,-3. 1) and (3, -4,-5) meets the plane 2x + y + z = 7.

2.i x-2 y +1 z-2

Show that the distance of the point of intersection of the line ■ ^ - 4'' ~ ^planex-y+ z = 5 from the point (- 1, - 5, - 10) is 13.Find the equations of the line through the points (a, b, c) and (a\ b\ c') and prove that it passesthrough the origin if aa' + bb\ + cc ~ rr where r and / are the distances of the given points

and the3. 12

4.

from origin.

ANSWERS

y-b z-c b' -b c' - c

. x-a4. —----a -a2. (1,-2, 7).1. (0,0,4).


The Straight UneUNITj-'

THE STRAIGHT LINESTRUCTURE

• Equations of a Une in Different Forms• Test Yourself-1

• Equation of a Plane through a Given Line• Angle between a Line and a Plane

• Test Yourself-2• Coplanar Lines• Projection of a Line on a given Plane

• Test Yourself-3• Shortest Distance between Two Lines


LEARNING OBJECTIVES

iAfter going through this unit you will learn :• About different forms of a straight line• How to find the angle between the plane and the line• How to find the plane through the given line• How to find the projection of a given line on a given plane

• 22.1. EQUATIONS OF A LINE IN DIFFERENT FORMS(i) General form (non-symmetrical form).

Since every equation of first degree in x,y and z represents a plane, when two such planes intersect a line of intersection is formed, therefore the equations of two planes simultaneously represent a straight line.

ci\X + y + Cjz -f = 0a^x + £2 y + c2z + ^2 = 0 betw0 planes >n general form. Then the general form a straight line is given by

Geometry and Vectors

Z>«

P{x,y,z)&

If and

Q

a^x + b\y + C\Z + = 0 ICI2X + 62 .y + c2^ + ^2 = 0 I r

(ii) Symmetrical form.Symmetrical form of line is

Fig. 1

*-a y-|3 z-Y = APL_ m nCorollary 1. The equation of a line passing through a point (a, (3, y) having direction ratios

a, b, c. Then its symmetrical form isx-a y - (3 z-y

ba cCorollary 2. To obtain the equation of a straight line passing through two points.

y-yi _ z-zi *2-*l Z2-Z1

X -X]


ANSWERS .....Calculus <&. Geometry

25(ii)1. (i) 5

bVI?' ■2. 13x2 + 45/ + 40z2 + 24xy - 36zx + I2yz + 5(k -82y - 220z + 278 - 0.

OBJECTIVE EVALUATION > FILL IN THE BLANKS :1. The general form of a plane is...........2. The plane/a: + my+ nz= p is a...........3. A plane cuts intercepts of length 2, - 3,5 on co-ordinate axes, then the equation of plane

is4. The direction-ratios of the normal to the plane ax + by + cz = 0 are> TRUE OR FALSE :Write ‘T* for true and ‘F’ for false statement:

The intercepts form of a plane is —

.x y 22. The plane — + , + - = 1 meets the z-axis at (0, 0, a).a b c

3. The normal form of the plane is Zx + my + nz = p-> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :

X V * z1. The intercept on x-axis of the ph ne — + ^ + “

+f+i=i.b c1.

(VF)

(T/F)(T/F)

= 1 is :

(b) a (d) c.

2. The perpendicular distance from' the origin to the plane lx + my + nz = p is :(b) /

(a) b(c) 1

(a) p(d) 1.(c) n

ANSWERS

Fill in the Blanks :3 z=i* 2 3 5 11. ax + by + cz + d = 0 2. normal form

4. a, b, cTrue or False :

1. T 2. F 3. T 4. T 5. F Multiple Choice Questions :

1. (b) 2. (a)

□□□


- The Plum'

• STUDENT ACTIVITY1. A plane meets the coordinate axes in A, B, C such that the centroid of the triangle ABC is the

x y zpoint (p, q, r). Show that the equation of the plane is — + 'L + - = 3p q r

2. A variable plane at a constant distance 3p from the origin meets the axes in A, B, C. Show thatthe locus of the centroid of th triangle ABC is — + -^ + -r = -

* y z P2 •

.i

I- •

• TEST YOURSELF-21. Find the distance between the parallel planes :

(i) 2x ~ 3y ~ 6z =0 and 2x - 3_y - 6z + 14 = 0(ii) 2x-y + 3z-4-0 and 6*- 3>t + 9z + 13 = 0.Find the equation of the locus of a point P whose distance from the plane

- 2y + 3^ + 4 = 0 is equal to its distance from the point (-1,1,2).A variable plane passes through a fixed point (a, P, y) and meets the axes in A, B, and C. Show that the locus of the point of intersection of the planes through A, B and C parallel to the co-ordinate planes is

2.

3.

a p y—+ il + -L x y z

Two systems of rectangular axes have the same origin. If a plane cuts them at a distances a, b, c and a\ b', c respectively from origin, show that

_L _L L = A- J_ _L2 1^2 2 /2 /2*

A variable plane is at a constant distance p from the origin and meets the axes in A, B and C. Show that the locus of the centrqid-of the triangle ABC is aT 2 + y~ 2 + 2 = 9p~ 2.

= 1.

4.

5.

Self-Instructional Material 147 •

Calculus & Geometry Taking + ve sign, we get1 1-(* + 2y + 2z--9)= —(4*-3>>+12z+ 13)

or 13* + 26jy + 26* - 117 = 12* - + 36* + 39* + 35^-10*-156 = 0. ...(1)or

Taking - ve sign, we get1 T— (* + 2;y + 2* - 9) = - — (4* - 3_y + 12*+ 13)

13* + 26y + 26* - 117 = - (12* - 9^ + 36* + 39) 13* + 26>’ + 26* - 117 = - 12* + 9y - 36* - 39 25*+17}+ 62*-78 = 0.

Distinction of Acute and Obtuse Bisector.

ororor

Let 0 be an angle between * + 2y + 2*-9 = 0 and* +35}-J 10*-156 = 0.

1x1 + 35x2-10x2 1 + 70-20Then cos 0 =Vl +4 + 4 .Vl+'1225 + 100 3V1326

51 17cos 0 =3VI326 VI326 V1037tan 0 = > 1.

170 > 45°.

Hence the plane * + SSy - i 0* - 156 = 0 is obtuse bisector and 25*+ \ly + 62* - 78 = 0 is an acute bisector.

Example 2. Show that the origin lies in the acute angle between the planes * + 2)'+ 2*- 9 = 0 and 4* - + 12* +13=0.

Solution. First adjust the constant term in each plane to be of same sign. Thus we obtain,- * - 2>' - 2* + 9 = 0 and 4*-Sy + 12* + 13 = 0.

Now calculate aia2 + ^1^2cic2-aia2 + ^1^2 + cic2= (-1) 4 + (-2) (-3) + (-2) 12

= -4 + 6-24 = -22which is obtained negative. Hence the origin lies in acute angle between the planes.

SUMMARY

\s ax + by + cz +d = 0 where a2 + b2 + c’ * 0.General equation of a plane Euation of yz plane is *= 0.Equation of **-plane is y= 0 Equation of *y-plane is * = 0.

x y zIntercepted form of plane is — + , + —a b c

Normal form of a plane is /* + m>> + **= p.Equation of a plane passing :hrough three points (*L, y\, z\), (*2, y2, *2} and (*3, *3) is

y-y\ z~z\^2-^1 V2-y\ Z2~Z\ =0*3~*i k-yi Z3“Zi ■ : ■

• Length of perpendicular drai/n from (*[, ylt z\) to the plane ax '+by + c* + ^ = 0 is given by

= 1.

*-*1

axj + byi + cz2 + dVi2"+ b2 + c2

• Equations of bisector of angle between given planes rq* + fr]y + +•<■/(= 0 anda2X + b-^y + cyz + *(2 = 0 are given by

a\X + b\y + C\Z +'^2 aix + biy + c2z + ^2Vaf + b] + c] ~ Va2 + 62 + c2

• Equation of a plane through the planesP = a\x + biy - ciz + di=0 and Q = 02* + ^ + c2* + ^2 = 0 is given by P + XQ = 0.


1The PlaneJ_ 1 J_ = _L

a2 ^2 + c2 p2

Since (1) meets the co-ordinate axes in A, B and C, so that the coordinates of A, B and C are (a, 0,0), (0, b, 0) arid (0,0, c) respectively. Now the centroid of the tetrahedron OABC is

•••(2)or

a b c or 7,7, t • 4 4 4n + 0 + 0 + 0 0 + & + 0 + 0 O + O + c + O

4 44Let (a, (3, y) be the centroid of OABC, then

b a c7 = a'4 = P-4a = 4a, b = 4$,c = 4y.

Substitute these values of a, b and c in (2), we get1.1

a = y.4

1 1zz —-

16a2 16p2 ley2 p2a"2 + p-2 + y-2= 16p~2.or

Thus the locus of (a, p, y) isx 2 + y 2 + z 2 = I6p 2. Hence proved.

• 21.8. BISECTORS OF THE ANGLE BETWEEN THE PLANES

Let + biy + ciz + 'di =0 and ^ + ^ + CzZ + d2:=0be two planes and let P(x, y, z) be any point on the bisector plane as shown in fig. 2.

Draw PM and PN two perpendicular from P to the given planes respectivley. Therefore the plane PQ will be the angle bisector of ZMQN between two given planes if PM = ± PN.

Thus the perpendicular from P to the plane a^x + b\y + C\Z + d[=0 is given by

a^x + b^ + C\Z + d-[-0)PM =

Va,2 + bi2 + c7and the perpendicular distance from P to the other plane a^x + btf + C2Z + di^Q is given by

aix + b^y + c2z + dz ...(2)PN =Vfl22 + bi + ci

Therefore by (1) and (2) and using the result PM -±PN the equation of angle bisectors or bisector planes are

PM -±PN

aix + biy + cjz + di diX + by + c# + d2or= ±

Vq22 + fr22 + C22Vfli2 + b2 + Cl2

Here positive and negative signs indicate that there are two angle bisectors, one of them is acute and other is obtuse bisector.Distinction between Acute and Obtuse Bisector :

Let 0 be an angle between one of the bisector planes and one of the given planes. If 0 is obtained greater than ti/4 (or 45°), then the chosen bisector plane is obtuse bisector otherwise acute bisector plane.SOLVED EXAMPLES

Example 1. Find the equation of the planes bisecting the angle between the planes x + 2y + 2z = 9 and 4x - 3y + 12z + 13 = 0 and distinguish them.

Solution. The equation of angle bisectors are given by x + 2y + 2z - 9 [ 4* - 3y + 12z + 13

~ V16 + 9 + 144

!(x + 2y + 2z-9) = ±y^(4*-3y+12z + 13)

Vl +4 + 41or

Self-Instructional. Material 145 v f

Calculus & Geometry 9. Find the equation of the plane through the points (1, - 2,2), (-3, 1,-2) and perpendicular to the plane x + 2>> - 3z = 5 J

ANSWERS

3- ' 3>-p = 3.P = 3.1 21. 9, - 3, 9/2.

3. d.r.’s are 2, 1, 2; perpendicular distance = 1.4. ax + by + cz = a2 + + c2.

x y z i 27 + -r-- = 1; Normal form : — o 3 4 i7. 2x + 3;y - 3z - 5 = 0.8. 2x - y = 0.

2. — x +

5. 2* + 3y - 5z = 38.4 3 12 126. x + V29Z >/29,/7 '/29>/29 ” V29

9. a: + 16y + llz + 9 = 0.

• 21.6. LENGTH OF PERPENDICULAR FROM A GIVEN POINT TO A GIVEN PLANE

Let P(x\, y\, zj) be any point and let the general equation of a plane be ax b by + cz + d = 0.

Then the perpendicular distance from z\) to (1) is equal to 4*i + by\ + czj + rf

-.(1)

• 21.7. DISTANCE BETWEEN TWO PARALLEL PLANESTo find the distance between two parallel planes.The distance between two parallel planes is obtained by two methods.Method I. Choose any point on one of the given planes and then find its perpendicular

distance from other plane.Method II. First of all fii\d the perpendicular distance of each plane from origin and retain

their signs and find the algebraic difference of these two perpendiculars which gives the required distance between the parallel planes. But while applying this method we should be careful that the coefficient of x in both planes are of the same sign.SOLVED EXAMPLES

Example 1. Find the distance between parallel planes2j:-2y|fz + 3 = 0 and 4x-4y'+2z + 5= 0.

Solution. Let p\ and pi be the perpendicular distances from (0, 0, 0) to each planes.3 d

Pi =V(2)2+(- 2)2 + (l)2 'a2 + b2 + c23. 13 v-. *

5. • 5and P2 = V(4)2 + (-4)2 + (2)2 6 /

Thus distance between the planes =p1 - /?2 = 1 _ ^ ^'

Example 2. A variable plane is at a constant distance p from the origin.and meets the axes in A, B and C. Show that the locus of the centroid of the tetrahedron OABC is

x~2 +y-2 + z~2 = 16p~2.Solution. Let the variable plane be

a b c -(1)- +

- 1 (perpendicular distance from (0, 0, 0) to (1))V 21 1 • 1+ +ba c


The PlaneThe plane (1) meets the co-ordinate axes in A(a, 0,0), 5(0, b, 0) and C(0, 0, c). Therefore, the coordinates of the centroid of the AABC are

fa + 0 + 0 0 + b + 0 0 + 0 + / , \ a b c0r 3’3’i ’3 3 3

Since centroid of AA5C is given as ip, q,r).a 0p = - or a = 3p.

Similarly,Now substitute the values of a, b, c in (1), we get

f + ^ + f=l or ^ + -^ = 3.3p 3«? 3r

Example 2. Find the intercepts of the plane 4x+ 3y - \2z +6 = 0 on the axes.Solution. First change the given plane into intercepted form,

4x + 3y- 12z + 6.-0.4x + Sy - 12z = - 6.

Divide by - 6 throughout (1), we get—-— + -2— + —£—= L -3/2 -2 1/2

3 1Thus intercepts on x-axis, y-axis, z-axis are respectively - —, - 2, —•

Example 3. Find the equation of the plane passing through the points (1,-1,2) and (2, - 2,2) and which is perpendicular to the plane 6x-2y + 2z = 9.

Solution. The equation of any plane through the given point (1,-1, 2) isa (x - 1) + h (y + 1) + c (z - 2) = 0

b = 3q and c = 3r.

P, Q r

...(1)

•••(I)If the plane (1) passes through the second given point (2, - 2,2), then

a(2-I) + 6(-2+I)+c(2-2) = 0 . a-b+0.c=0 ...(2)or

If plane (1) is perpendicular to the given plane 6x-2y + 2z = 9 i.e., their normals are perpendicular to each other, then

•••(3)6a-2bJr2c = 0 or 3a-b + c = 0Now solving (2) and (3) by cross-multiplication, we have

b bc a ca-1-0 0-1 -1+3 °r -1 -1 2

a _ __ c_1 ~ 1 ~-2 .••(4)or

Eliminating a, b, c from (1) and (4), we get 1 (x - 1) + 1 (y + 1) - 2 (3 - 2) = 0

x + y-2z + 4 = 0,orwhich is required equation of the plane.

• TEST YOURSELF-11. Find the intercepts made on the co-ordinate axes by the plane x - 3y + 2z - 9 = 0.2. Reduce the equation of the plane x + 2y - 2z - 9 - 0 to the normal form and hence find the

length of the perpendicular drawn from the origin to the given plane.3. Find the dimension-cosines of the normal to the plane

2x + y + 2z = 3.Find also the perpendicular distance from the origin to the plane.

4. Find the equation of a plane passing through the point P{a, b, c) and perpendicular to OP, O being the origin.

5. The co-ordinates of a point A are (2, 3, - 5). Determine the equation to the plane through A at right angles to the line OA, where O is the origin.

6. Find the equation of the plane which cuts off intercepts 6,3, - 4 from the axes of co-ordinates. Reduce it to normal form and find the perpendicular distance of the plane from the origin.

7. Find the equation of a plane passing through the point (1,1,0), (1,2,1) and (- 2, 2, - 1).S. Find the equation of a plane through the point (1,2,3), perpendicular to the plane

x + 2y + 3z = 1 and parallel to z-axis.


Calculus & Geometry (i) The equation of a plane perpendicular to .x-axis is obviously parallel to y^-plane, so it equation is a: = a.

Similarly,(ii) The equation of a plane pirpendicular to y-axis isy = b.(iii) The equation of a plane jlierpendicular to z-axis is z = c.

• 21.2. INTERCEPT FORM! OF A PLANEDefinition. The equation of d plane which cuts off intercepts on the coordinate axes, is called

Intercept form of a plane. j 'The equation of a plane in intercepted form is given by

* + £ + *=i. a b cwhere a = intercept on x-axis, b = intercept on y-axis, c = intercept on z-axis

• 21.3. NORMAL FORM OF A PLANEDefinition. The equation of a plane in terms of a perpendicular distance from origin to the

plane and direction-cosines of this\perpendicular, is called Normal form.The equation of a plane in normal form is given by

or lx + my +nz=p (v l2 + m2 + n2= 1)

• 21.4. EQUATION OF A PLANE PASSING THROUGH THREE POINTSTo obtain the equation of a filane passing

though three points.P(x\>ybZi)> Q(x2, yz> zz)

R(x3, y3, z3) be three points through these the equation of a plane is to be required.

The equation of required plare is given by

Let and

1x y z *i yi zi *2 V2 Z2 *3 y3 Z3

1= 0. •••(A)1

1

• 21.5. ANGLE BETWEEN Wo PLANES r

Definition. An angle betweeh two planes is equal to the angle between the normals to the planes.

Let a]X + b\y + CiZ + d\ ~0 and a2x + b2y + c2z + d2 = 0 be two planes and let 0 be the angle between them. Then we have

Qidz + bibi +cos 0 ='Jd12 + b2 + c12-'ia22 + b22 + c22

where a],b], cj; a2, b2, c2 are the diSrection ratios of the normal to the planes.Corollary 1. The planes are perpendicular if a\a2 + b\b2 + clc2 = 0.

1 ai biCorollary 2. The planes are parallel if— = —

• • d2 b2 c2Corollary 3. The equation parallel to a given plane ax^ by+ cz + d= 0 is

ax + by + cz + X = 0, where X is to be determined by additional condition.SOLVED EXAMPLES I

Example 1. A plane meets theco-ordinate axes in A, B, C such thatthe centroid of the triangle ABC is the point (p, q, r). Show that the equation of the plane is

— + ■^■ + - = 3.P 9 r

Solution. Let the equation of the plane bex y z ,- + f + -= 1. a b c

/•

...(1)


The PlaneUNIT

THE PLANESTRUCTURE

• General forms of a Plane• Intercept form of a Plane• Normal form of a Plane• Equation of a Plane passing through Three Points• Angle between Two Planes

• Test Yourself-1• Length of perpendicular from a given point to a given plane

• Distance between Tow Parallel Planes• Bisectors of the Angle between the Planes

• Summary .• Student Activity• Test Yourself-2

Learning objectivesAfter going through this unit you will learn :

# About the different forms of the plane• How to find the plane passing through 3 points.# How to calculate the angle between 2 planes and distance between planes.• How to determine the planes bisecting the angle between the two given planes.

• 21.1. GENERAL FORMS OF A PLANE(a) General equation of first degree in x, y, z represents a plane.The general equation of a plane is given by

ax +by-i-cz+d = 0(b) General equation of a plane passing through a given point.Let y,, zj) be a given point and let the equation of a plane be ax + by + cz + d = 0 that

passes through A(xi, yj, zi), then we haveax\ + by\ + czj + if = 0

substracting these two equations, we geta(x-x1) + b(y~y1) + c(z-z1) = 0.

This is the required general equation of a plane passing through fo, y\, Zi).(c) Equation of a plane that passes through the origin is given by

ax + by + cz = 0.(d) Equations of the plane parallel to the co-ordinate axes are given by(i) If a = 0, then 6y+ cz+ d = 0 is a plane parallel to x-axis.(ii) If £ = 0, then ax + cz + d = 0 is & plane parallel to y-axis.(iii) If c = 0, then ax +by+ d = 0 is a plane parallel to z-axis.(e) Equations of the co-ordinates planes :(i) The equation to yz-plane is x = 0(ii) The equation to zx-plane is y = 0(iii) The equation of xy-plane is z = 0.(f) Equation to the planes parallel to the co-ordinate planes(i) The equation of a plane parallel to yz-plane isx = a (constant)(ii) The equation of a plane parallel to zx-plane is y = ft (constant)(iii) The equation of a plane parallel to xy-plane is z = c (constant).(g) Equations of the planes perpendicular to the co-ordinate axes :

Self-Instructional Material 141'

Calculus & Geometry • TEST YOURSELF1. Find the direction cosines ofj a line whose direction ratios are 2, 3, 6.

To show that the direction cosines of a line whose direction ratios are a, b, c are2.ba c

W + S + C2 Vq2 + &2 + C2 V + i>2 + C23. Find the direction cosines of the line segment joining the points P(2, 3, - 6) and <2(3, - 4, 5).4. If P(2, 3, - 6) and Q(3, - 4, i) are two points, find the d.c.’s of OP, PO, OQ and PQ where

O is the origin.5. Find the length of a segment of a line whose projections on the axes are 2,3,6.6. If the points P and Q are (2, 3, - 6) and (3, - 4, 5), then find the angle between OP and OQ

. where O is the origin.w 2 ? ?7. If/, m, n are the direction cosines of a line, then prove that: / +m +n =1.

8. Find the direction cosines of two lines which are given by the relationsl - 5m - 3n = 0 and ll2 + 5m2 - 3n2 = 0.

ANSWERS

- 2 3 6 1 -7 11VTtT VTtTViTT

2 3 6 3 62d.c.’s of PO are - — * -4. d.c.’s of OP are — ’ t ’7 7 7 7 77

d.c.’s of are —» »—^nd.c.’s of Pg are 5V2 5V2

r-i8>/ri

1 -? aVnT VItT VTtT

J___ 1_,_2_____________ 3_Vb" V6" ^/6” VIT VI? VIT

5\/2-15. 7, 6. 0 = cos . 8.

35

OBJECTIVE EVALUATION> FILL IN THE BLANKS :1. If a line makes the angle a, p, y with the co-ordinate axes respectively, then a + P + y *2. The numbers which are proportional to the direction-cosines are called...........3. The direction-cosines of z-axis are...........

2 2 24. If l, m, n are actual direction-cosines of a line, then l + m +n =.....> TRUE OR FALSE :Write ‘T* for true and ‘F’ for false statement:1. The direction ratios of a line massing through the points (2, 3,5) and (7,6, 8) are 5, 3, ?>.(T/F)

1 22. If 1, - 2, - 1 are the direction-ratios of a line, then its direction cosines are — > :N6 V6 ^6

-.3

1

(T/F)3. The diretion-cosines of x-axis are 0,0,1. (T/F)

> MULTIPLE CHOICE QUESTIONS :Choose the most appropriate one :1. The direction-cosines of y-axts are :

(a) 0, 1,02. The direction-cosines of a link OP are /, m, n and OP = r, then the co-ordinates of P are :

(d) (Ir, mr, nr).

(b) 0,0,1 (c) 1,0,0 (d) 1,1.1.

(a) (0,0,0) (b) (/, m, n) (c) (r, r, r)

ANSWERS

Fill in the Blanks :1. 27t 2. Direction-ratios

True or False :l.T 2.F 3. F

Multiple Choice Questions :1. (a) 2. (d) 3. (a) 4. (b)

3. 0, 0, 1 4. 1

□□□140 Self-Instructional Material

Direction Cosines and Direction Ratios• SUMMARYcos a,cos (3,cosy are the direction cosines of a straight line with a + |3 + Y*27t. We take / = cos a, m = cos (3, x = cosy. Then l2 + m2 + n2 = l.(*, y, z) -> {lr, mr, nr).If a, b, c are the direction ratios of a line, then its direction cosines l, m, n are obtained by

± b± a ± c1 = Va2 + b2 + c2' m Va2 + b2 + c2 ’ ” Va2 + c2 Direction cosines of a line segment PQ with and Q(x2,y2,Z2) is given by

X2~X\ yi-y\PQ ,m_ PQ

The projection of PQ, P{x^yl,z\),Q(x2,y2,Z2) on a line of d.c’s l, nun is given by (x2 + xl)l + (y2-y\) m + (z2- zi) n.If and l2,m2,n2 are the d.c’s of this lines and 0 is the angle between them, thencos 0 = 1^2 + mim2 + nin2 if the direction ratios of the lines are given then

Z2 ~ Z\/ = , n ~PQ

fllfl2 + ^1^2 + clc2COS 0 -

''Jet2 + bf + cf '^^2 + ^2where alt bj, Ci and a2, b2, c2 are the direction ratios of two lines.Lines are perpendicular if

lil'2 + mjnii + n]n2 = 0cii(i2 + bjb2 + C\C2 — 0.or

Lines are parallel to each other ifIi = h’ = m2, n i =/j2

o2 b2 c2or

• STUDENT ACTIVITY2 2 21. Show that sin a + sin P + sin y = 2, where cos a, cos P and cos y are the direction cosines

of a straight line.

Show that the lines whose d.c’s are given by / + w + n = 0 and 2/nn + 31n - 5lm = 0 are at right angles.

2.

Self-Instructional Material ,139

Calculus <4 Geometrycos2 a + cos2 (3 + cos2 y = 1

(1 - sin2 a) +• (i - sin2 (i) + (1 - sin2 y) = 1 3 - sin2 a - sin2 p - sin2 y = 1

sin2 a + sin2 f3 + sin2 y = 2.Example 2. Find the direction cosines of a line which is equally inclined to the axes. Solution. Let l, m, n be the direction cosines of a line which is equally inclined to the axes,

(v / = cos a, w = cos |3,/? = cos y)ororor Hence proved.

thena = P = y.

n l m n= COS P = cos y or y = y = y

+ m2 + n2

cos a

m _ n _ + 2J2lor1 1 1 ^3^1 + 1 + 1

1 1 1/ = ± ’ m =±—— * n = ±—•^3

Example 3. If P (6, 3, 2), Q(5, 1,4), R (3, - 4, 7) and S (0, 2,5) are four points. Find the projection of PQ on RS.

Solution. Firstly find d.c.’s of RS.d.r.’s of are 0 - 3, 2 + 4, 5 - 7 i.e., - 3,6, - 2.d.c.’s of KS are jy >-y-

Thus the projection of PQ on RS= I (*2“X]) l + (y2-y\)m + (z2 - zi) n |

73 (-0y +(1-3) f +(4-2) -f(5- 6) -\ /

3 12 4 - 13 13■111

Example 4. Find the direction cosines /, m, n of the two lines, which are connected by the relation l + m + n = 0 and mn - 2nl - 2/m = 0. Also, show that angle between lines is 2n/2>.

Solution. Since we have1

7 7

;/ + m + n = 0 mn -\2nl ~ 2lm = 0.

Eliminating n from (1) arid (2), we have- m (/ + m) + 21 (/ + m) - 2/m = 0 - ml - m2 + 2/2 + 2/m - 2/m = 0 or 2/2 - bn - m2 = 0

2/2 - 2/m + /m - m2 = 0

...(1)

...(2)

or

/ m l m 1 ; 1"-2(2/ + m) (/ - m) = 0 or 7 = ...(3)

1Now eliminating l from (1) and (2), we have

mn + 2/i (m jt- n) + 2m (m + «) - 0mn + 2mn +\2n2 + 2m2 + 2mn = 0 or 2m2 + 5mn + 2n2 - 0

■? i i2m + 4mn + mn + 2n = 0

(2m + n) (m + 2/i) = 0 or J = ~2 ’ - 2 " 1

or

m n.••(4)

From (3) and (4) we obtained / m______n_ l1 " 1 ’T-2 ; 1 "-2_ 1

m n

Thus the direction cosines of two lines are respectively given by 1 1! - 2 . 1 - 2 1

VtT Vtf V<f VcT Vb" Vb"Again, if 0 is the angle between these lines, then

1Y1 ' \1 1 12 2.71

cos 0 = —= + + VeJW


Direction Cosines and Direction,RatiosSince 0 = 90°, this implies cos 0 = 0, hence the condition is l\li + m\mi + n|«2 = 0.

In terms of direction ratios this condition becomes aifl2 + bibi + CjC2 = 0.

(b) Condition for parallelism. If the two lines whose d.c.'s are respectively l\, Ai],-/2, m2, n2 are parallel, then we have

sin 0 = >[(/iWi2 - tnilf}2 + ('«i«2 ~ n\mi? + (n\h ~ n2h)2] •

Since we know 0 = Otthis implies sin 0 = 0(/1m2 - m^)2 + (min2 - n^)2 + (n^j - n2^i)2 = 0.,v

This gives thatl\W2 - /nj/2 = o, Wj/12 - /liw2 = nlh - ^1^2 = 0-

«i V/12 + m12 + n12 j__/i=!hl> m2 'll22 + m^n22

l\ —12, tfi\ = m2, /tj = n2-This is the required condition.In terms of direction ratios this condition becomes

ai bi Ci

1

a2 b2 c2PERPENDICULAR DISTANCE OF A POINT FROM A LINE

To find the perpendicular distance of a point P(x\, y\, 2|) from a line passing through a point A(a, p, y) whose direction cosines are l, w, n.

Let AB be a line through /4(cc, p, y) and whose d.c.’s are /, m, n.Let PN be the perpendicular from P to AB as shown in the adjoint figure 7.

POi.jk'], z0

Therefore AN = Projection of AP on AB= (xl-a)l + (y]- p) m + (n - y) n.

AP = 'l(xl-a)2 + (yl-V)2 + (zl-S)2.

BA N...(1)Fig. 7Now

We havePN2 = AP2 - AN2

= [(x]-a.j2 + (yl-V)2 + (zi-y)2]-[(xl-a)l + (yl-&)m + (zl-y)n]2 = (xi-aj2{l-l2) + (y1-P)2(l-m2) + (z[-y)2(l-n2) .

- 2 [hn (x, - a) (y, - p) -f mn Cyt - P) (zi -y) + ln (x, - a) {z\ - y)]

= (xj - a)2 (m2 + n2) + (y, - P)2 (l2 + n2) + (z, - y)2 (l2 + m2)

- 2 [lm (xi - a) (y! - p) + mn (yl - p) (zj -y) + ln (x[ - a) (z, - y)]

[■'/ l2 + m2 + n2=l]

= {« (yi - p) - ffl (zi - Y)}2 -u* (zx-y)-n(x{- a)}2 + {m (x, - a) - ^ - p)}2PN = [n (yj - P) - m (z^ y)}2.

HOW TO SHOW THAT THREE GIVEN POINTS ARE COLLINEARIn order to show that three points A, B and C are collinear, we proceed as follows :(i) First we find the direction ratios of AB and AC.(ii) If these direction ratios are proportional, then A, B and C are collinear.

SOLVED EXAMPLESExample 1. If cos a, cos P and cos y are the direction cosines of a straight line, then prove

thatsin2 a + sin2 p + sin2 y = 2.

Solution. Since we know thatl2 + m2 + n2 = l.

Self-Instructional Material .137

Calculus & Geometri \

k\\\\

eno. >x

y,

Fig. 6 (a)i

*1 = hfl, yi = mi^, Si = «l^l; x2 - hr2i y2 = m2rh Z-2 = n2r2 projection of OQ on OP = (x2 -0)li+(y2- 0) mi + (z2 - 0) «i .

= x2h +y2mi +Z2n\ = r2 (/i/2 + w]m2+'nin2)

Fig. 6 (b)

...(1)

...(2)

(• .• x2 = l2r2, y2 = m2r2, z2 = n2r2)But we have

Projection of OQ on OP = OQ cos 0 = r2 cos 0. ...(3)From (2) and (3), we have

r2 cos 0 = r2 (lil2 + m1m2 + cos 0 = l2l2 + m\m2 + «1n2.

Aliter. In txOPQA A A

+ y\j + Zi kOP = Xi iA A A

and OQ = x2ih-y2j + z2 k.

Since | OP \ = rj and j OQ \ = r2, then

OP . 0& - OP OQ cos 0A A A | A A A

(*i 1 +y\j + Si k) ■ (x2i + y2j + z2k) = rxr2 cos 0 X\X2.+ yiy2 + ZiZ2 = nr2 cos 0 rir2 (hh + m\m2 + n!«2) = rxr2 cos 0cos 0 =/i/2 + m1m2 + «jW2

(•.• ~A.~B = ABco$G)

or[From (i)]or

or

• 20.8. ANGLE BETWEEN VHE LINES WHOSE DIRECTION RATIOS ARE GIVEN

Let aj, b\, C\ and a2, b2, c2 be these two lines will be

the direction ratios of two lines, then the direction cosines of

«i C|/i = ’ mi = ’ =

va,2 + 6i2 + ci2 Va,2 + 6j2 + Ci2b2fz= a2

Va22 + b2 + c2Cland * m2 — ’ n2 =

'la22 + b22 + C22 ' Va22 + b2 + c2Thus we get

cos 0 = l\l2 + mim2 + n\n2.Q |Zf2 + b]b2 + CiC2

cos 0 =Vfl^ + Ajf + d2 • Va22 + i22 + c22

• 20.9. CONDITION FOR PERPENDICULARITY AND PARALLELISMi

(a) Condition for perpendicularity. If the two lines whose direction cosines are l\ , ni {, ti\ and l2, m2, n2 are perpendicular, then we have

cos 0 =/|/2 + mi/n2 + rt^.


z Direction Cosines and Direction RatiosThe direction cosines of a line through*2-*i

4 «

P(.X]>y\>Zi) and Q (x2, y2, z2) arePQ »0 h)

‘22-21’ where PQ is distance between

P and Q. Moreover it has been observed that X2-X1, yi-y\ and Zi-Z\ are proportional to /, w and «, hence x2 -x{, y2-y\, z2 ~Zi are the direction ratios of a line passing through P(xhyi,zl)andQ(x2,y2,z2).

1 MPQ PQ

O —~ P' Q'

Y,

Fig. 5

• 20.6. PROJECTION OF A LINE THROUGH TWO POINTS ON ANOTHER LINE WHOSE D.C/S ARE l, m, n

Let P and Q be two points with coordinates (x,, yu zj) and (x2, y2, z2) respectively, through P and Q a line is passing whose projection on another line of direction cosines is to bedetermined. Let OX, OY and OZ be the rectangular axes. Therefore, .

------» AAA

OP = xx t + y\j + z\ k------> AAA

OQ=x2iJry2j + z2k.

PQ = OQ -OP = {x2 i + y2j + z2 k) - (xj i +y{j + z1 k)

and

PQ = fa - xi) i + (y2- y])j + {z2 - Z\) k.Now the unit vector along another line whose d.c.’s are /, m, n is

A A A A.

a = l i + mj + n k.Thus the projection of PQ on another line is.

PQ.a Projection of ffon ^isA 1«l Ml

A A

[(x2-Xi)i + Cv2-yi); + (22-2i) k] .(li + mj + nk)AAA

\ l i + mj + n k$x2 - xQ / + (y2 ~ y$ m + (z2- zt) n

'll2 + m2 + n2(v l2 + m2 + nz=l)= (xi -xi)l + iy2- y\) m + (z2-i- Zi) n

Corollary. The■ projection of a line PQ on another line whose d.c.’s are l,m,n is, ifP (0, 0, 0) and G(x)t j»lt zj)

= X\l + y\m + Z\n.

• 20.7. ANGLE BETWEEN TWO LINESTo prove that ifl\,m\,ni and l2, m2, n2 are the direction cosines of two lines and 0 is the angle

between them, thencos 0 = lil2 + m1m2 + nin2.

Let AB and CD be two lines whose direction cosines are l\, ml5 «i; l2, m2, n2 respectively and let 0 be an acute angle between AB and CD as shown in fig. 2.6 (a). Now draw two lines OP and OQ through O and parallel to AB and CD respectively. Let the co-ordinates of P and Q be (xj.y,, Z\) and (x2,y2, z^ such that OP = r1 and OQ = r2. Thus the angle between OP and OQ will be equal to 0 and also the direction cosines of OP and OQ are equal to and-/2,m2,n2respectively. Since we know that


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