C4 revision guide

7/21/2019 C4 revision guide

1/40

Pure C4

Revision Notes

March 2013


2/40


3/40

C4 14/04/13 1

Contents

Core

4

1

Algebra ................................................................................................................. 3

Partial fractions ........................................................................................................................................................ 3

2

Coordinate Geometry .......................................................................................... 5

Parametric equations ................................................................................................................................................ 5

Conversion from parametric to Cartesian form ...................................................................................................................... 6

Area under curve given parametrically................................................................................................................................... 7

3

Sequences and series ......................................................................................... 8

Binomial series (1 +x)n for anyn.......................................................................................................................... 8

4 Differentiation .................................................................................................... 10

Relationship between and

......................................................................................................................... 10

Implicit differentiation ........................................................................................................................................... 10

Parametric differentiation ...................................................................................................................................... 11Exponential functions, a

x...................................................................................................................................... 11

Related rates of change .......................................................................................................................................... 12

Forming differential equations ............................................................................................................................... 12

5 Integration .......................................................................................................... 13

Integrals of ex and

.......................................... .................................................................................................. 13

Standard integrals ................................................................................................................................................... 13

Integration using trigonometric identities .............................................................................................................. 13

Integration by reverse chain rule ......................................................................................................................... 14

Integrals of tanxand cotx .................................................................................................................................................... 15

Integrals of secxand cosecx ................................................................................................................................................ 15

Integration using partial fractions .......................................................................................................................... 16

Integration by substitution, indefinite .................................................................................................................... 16

Integration by substitution, definite ....................................................................................................................... 18

Choosing the substitution ...................................................................................................................................................... 18

Integration by parts................................................................................................................................................. 19

Area under curve .................................................................................................................................................... 20

Volume of revolution ............................................................................................................................................. 20

Volume of revolution about thexaxis ................................................................................................................................. 21

Volume of revolution about theyaxis ................................................................................................................................. 21

Parametric integration ............................................................................................................................................ 21

Differential equations ............................................................................................................................................. 22

Separating the variables ........................................................................................................................................................ 22

Exponential growth and decay ............................................................................................................................... 23


4/40

C4 14/04/13 SDB2

6 Vectors ................................................................................................................ 24

Notation .................................................................................................................................................................. 24

Definitions, adding and subtracting, etcetera ........................................................................................................ 24

Parallel and non - parallel vectors......................................................................................................................................... 25

Modulus of a vector and unit vectors ................................................................................................................................... 26

Position vectors ..................................................................................................................................................................... 27

Ratios ..................................................................................................................................................................................... 27

Proving geometrical theorems .............................................................................................................................................. 28

Three dimensional vectors ..................................................................................................................................... 29

Length, modulus or magnitude of a vector ........................................................................................................................... 29

Distance between two points ................................................................................................................................................ 29

Scalar product ........................................................................................................................................................ 29

Perpendicular vectors ............... ............... .............. ............... ............... ............... .............. ................. .............. ................ ...... 30

Angle between vectors .......................................................................................................................................................... 31

Vector equation of a straight line .......................................................................................................................... 31

Intersection of two lines ......................................................................................................................................... 32

2 Dimensions ............. ................ ............... ............... ................ .............. ................ .............. ................ ............... ................ ... 32

3 Dimensions ............. ................ ............... ............... ................ .............. ................ .............. ................ ............... ................ ... 33

7

Appendix ............................................................................................................. 34

Binomial series (1 +x)

n

for any n proof ........................................................................................................... 34

Derivative of xq for q rational ................................................................................... ......................................... 34

for negative limits ....................................................................................... .............................................. 35Integration by substitution why it works ............................................................................................................ 36

Parametric integration ............................................................................................................................................ 36

Separating the variables why it works ................................................................................................................ 37

Index ............................................................................................................................. 38


5/40

C4 14/04/13 3

1 Algebra

Partial fractions

1) You must start with a proper fraction: i.e. the degree of the numerator must be less than

the degree of the denominator.

If this is not the case you must firstdo long division to find quotient and remainder.

2) (a) Linear factors (not repeated)

........)().....)(.(

......+

bax

A

bax

(b) Linear repeated factors (squared)

........)()().....(.)(

......22

+

+

bax

B

bax

A

bax

(c) Quadratic factors)

........)().....)(.(

......22

++

+

+ bax

BAx

bax

or ........)().....)(.(

......22

+++

+

++ cbxax

BAx

cbxax

Example: Express)1)(1(

252

2

xx

xx

+

+ in partial fractions.

Solution: The degree of the numerator, 2, is less than the degree of the denominator, 3, so

we do not need long division and can write

)1)(1(

252

2

xx

xx

+

+

211 x

CBx

x

A

+

++

multiply both sides by (1x)(1+x2)

5 x+ 2x2 A(1 +x2) + (Bx+ C)(1 x)

5 1 + 2 = 2A A= 3 clever value!, put x= 1

5 =A+ C C= 2 easy value, put x= 0

2 =AB B= 1 equate coefficients of x2

)1)(1(

252

2

xx

xx

+

+

21

2

1

3

x

x

x +

++

.

N.B. You can put in any value for x, so you can always find as many equations as youneed to solve for A, B, C, D . . . .


6/40

C4 14/04/13 SDB4

Example: Express2

2

)3)(12(

227

+

xx

xx in partial fractions.

Solution: The degree of the numerator, 2, is less than the degree of the denominator,3, so we do not need long division and can write

2

2

)3)(12(

227

+

xx

xx 3)3(12 2

+

+

x

C

x

B

x

A multiply by denominator

x2 7x+ 22 A(x 3)2 + B(2x 1) + C(2x 1)(x 3)

9 21 + 22 = 5B B= 2 clever value, put x= 3

7/2+ 22 = 2.52A A= 3 clever value, put x=

22 = 9AB + 3C C= 1 easy value, put x= 0

2

2

)3)(12(

227

+

xx

xx

3

1

)3(

2

12

32

+

xxx

Example: Express9

392

23

+

x

xxx in partial fractions.

Solution: Firstly the degree of the numerator is not lessthan the degree of thedenominator so we must divide top by bottom.

x2 9 ) x3 + x2 9x 3 ( x + 1

x3 9x

x2 3

x2 96

9

392

23

+

x

xxx = x+ 1 +

9

62

Factorise to give x2 9 = (x 3)(x+ 3) and write

9

62

33)3)(3(

6

++

+

x

B

x

A

xx multiplying by denominator

6 A(x+ 3) + B(x 3)

6 = 6A A= 1 clever value, put x= 3

6 = 6B B= 1 clever value, put x= 3

9

392

23

+

x

xxx = x+ 1 +

3

1

3

1

+

xx


7/40

C4 14/04/13 5

2 Coordinate Geometry

Parametric equations

If we definexandyin terms of a single variable (the letters tor are often used) then thisvariable is called a parameter: we then have the parametric equation of a curve.

Example: y= t2 3, x= 2 + t is the parametric equation of a curve. Find

(i) the points where the curve meets thex-axis,

(ii) the points of intersection of the curve with the line y= 2x+ 1.

Solution:

(i) The curve meets thex-axis wheny= 0 t2= 3 t= 3

curve meets thex-axis at (2 3, 0) and (2 + 3, 0).

(ii) Substitute forxandyin the equation of the line

t2

3 = 2(2 + t) + 1 t2 2t 8 = 0 (t 4)(t+ 2) = 0

t = 2 or 4

the points of intersection are (0, 1) and (6, 13).

Example: Find whether the curvesy= t2 2, x= 2t+ 3 and y= t+ 3, x= 3t 4 intersect.

If they do give the point of intersection, otherwise give reasons why they do notintersect.

Solution: If they intersect there must be a value of t which makes their x-coordinatesequal, so find this value of t

2t+ 3 = 3t 4 t= 7

But for this value of t the y-coordinates are 47 and 10.

So we cannot find a value of t to give equal x-and equal y-coordinates and thereforethe curves do not intersect.


8/40

C4 14/04/13 SDB6

Conversion from parametric to Cartesian form

Eliminate the parameter (tor or ) to form an equation betweenxandyonly.

Example: Find the Cartesian equation of the curve given by y= t2 3, x= t+ 2.

Solution: x= t+ 2 t=x 2, and y= t2 3

y= (x 2)2 3,

which is the Cartesian equation of a parabola with vertex at (2, 3)

With trigonometric parametric equations the formulae

sin2t + cos

2t = 1 and sec

2t tan

2t = 1

will often be useful.

Example: Find the Cartesian equation of the curve given by

y= 3sin t+ 2, x= 3cos t 1.

Solution: Re-arranging we have

3

1cosand,

3

2sin

+=

=

xt

yt , which together with sin

2t + cos

2t = 1

13

1

3

222

=

++

xy

(x+ 1)2 + (y 2)2 = 9

which is the Cartesian equation of a circle with centre (1, 2) and radius 3.

Example: Find the Cartesian equation of the curve given byy= 3tan t, x= 4sec t.

Hence sketch the curve.

Solution: Re-arranging we have4

secand,3

tanx

ty

t == ,

which together with sec2t tan

2t = 1

134 2

2

2

2

= yx

which is the standard equation of a hyperbolawith centre (0, 0)

andx-intercepts (4, 0), (4, 0).

6 4 2 2 4 6

4

2

2

4 y

x


9/40

C4 14/04/13 7

Area under curve given parametrical ly

We know that the area between a curve and thex-axis is given by A = ydx

ydx

dA= .

But, from the chain ruledtdx

dxdA

dtdA =

dtdxy

dtdA =

Integrating with respect to t

A = dtdtdx

y .

Example: Find the area between the curve y= t2 1, x= t

3+ t, the x-axis and the linesx= 0 and x= 2.

Solution: The area is A= .

A = we should write limits for t, notx

Firstly we need to find y anddt

dx in terms of t.

y= t2 1 and

dt

dx= 3t

2+ 1.

Secondly we are integrating with respect to t and so the limits of integration must be for

values of t.

x= 0 t= 0, and

x= 2 t3+ t = 2 t3+ t 2 = 0 (t 1)(t2+ t+ 2) = 0 t= 1.

so the limits for t are from 0 to 1

A =1

0

dxdt

dt = 1

0(t

2 1) (3t

2+ 1) dt

= 1

03t4 2t2 1 dt

=

1

0

35

32

53

ttt = 15

11

Note that in simple problems you may be able to eliminate t and find y dx in the usualmanner. However there will be some problems where this is difficult and the above techniquewill be better.

??


10/40

C4 14/04/13 SDB8

3 Sequences and series

Binomial series (1 + x)n for anyn

(1 +x)n = ....!3

)2)(1(

!2

)1(1 32 +

+

++ x

nnnx

nnnx

This converges provided that |x| < 1.

Example: Expand (1 + 3x)2

, giving the first four terms, and state the values of x for

which the series is convergent.

Solution:

(1 + 3x)2

= 1 + (2) 3x +2 3( 2) ( 3) ( 2) ( 3) ( 4)(3 ) (3 ) ...

2! 3!x x

+ +

= 1 6x + 27x2

108x3

+ ...This series is convergent when |3x| < 1 |x| < 1/3.

Example: Use the previous example to find an approximation for2

1

0 9997.

Solution: Notice that2

1

0 9997 = 09997 2 = (1 + 3x)2 when x= 00001.

So writing x= 00001 in the expansion 1 6x + 27x2 108x3

09997 2 1 + 00006 + 000000027 + 0000000000108 = 1000600270108

The correct answer to 13 decimal places is 10006002701080 not bad eh?

Example: Expand 21

)4( x , giving all terms up to and including the term inx3, and state

for what values ofxthe series is convergent.

Solution: As the formula holds for (1+x)n we first re-write

2

1

)4( x =

2

1

2

1

414

x

=

2

1

412

x

and now we can use the formula

= 2 .....4!34!242

11

3

23

21

212

21

21

+

+

+

+

xxx

= 2 512644

32xxx

.

This expansion converges for 14


11/40

C4 14/04/13 9

Example: Find the expansion of6

132

xx

x in ascending powers ofx up tox

2.

Solution: First write in partial fractions

2

1

3

2

6

132

++

=

xxxx

x, which must now be written as

1

2211

332

23

)1()1()1(2

1)1(3

2 +=

+

xx

xx

=

+

+

+

+

22

2!2

)2)(1(

2)1(1

2

1

3!2

)2)(1(

3)1(1

3

2 xxxx

=216

11

36

17

6

12

xx .


12/40

C4 14/04/13 SDB10

4 Differentiation

Relationship between and

1

1

So y= 3x2

xdy

dxx

dx

dy

6

16 == .

Implicit differentiation

This is just the chain rule when we do not know explicitlywhatyis as a function of x.

Examples: The following examples use the chain rule (or implicit differentiation)

3 sin sin cos 5 10 5 using the product rule

( ) ( ) ( ) ( )

++=++=+

dx

dyxyxyx

dx

dyxyx

dx

d32333333

2222232

Example: Find the gradient of, and the equation of, the tangent to the curve

x2+y2 3xy = 1 at the point (1, 2).

Solution: Differentiating x2+y2 3xy= 1 with respect to x gives

2x+ 2ydx

dy

+

dx

dyxy 33 = 0

xy

xy

dx

dy

32

23

= 4=

dx

dy when x= 1 andy= 2.

Equation of the tangent is y 2 = 4(x 1) y= 4x 2.


13/40

C4 14/04/13 11

Parametric d ifferentiation

Example: A curve has parametric equations x = t2+ t, y= t3 3t.

(i) Find the equation of the normal at the point where t= 2.

(ii) Find the points with zero gradient.

Solution: (i) When t = 2, x= 6 and y= 2.

33 2 = tdt

dyand 12 += t

dt

dx

when t= 2

Thus the gradient of the normal at the point (6, 2) is 5/9

and its equation is y 2 =5

/9 (x 6) 5x+ 9y= 48.

(ii) gradient = 0 when 012

33 2=

+

=t

t

dx

dy

3t2 3 = 0

t= 1

points with zero gradient are (0, 2) and (2, 2).

Exponential functions, ax

y= ax

lny = ln ax = xln a

aydx

dya

dx

dy

ylnln

1==

( ) aaadxd

xx ln=

3

32 1 95


14/40

C4 14/04/13 SDB12

Related rates of change

We can use the chain rule to relate one rate of change to another.

Example: A spherical snowball is melting at a rate of 96 cm3s

1when its radius is 12 cm.

Find the rate at which its surface area is decreasing at that moment.

Solution: We know that V= 4/3r3 and that A= 4r2.

Using the chain rule we have

dt

drr

dt

dr

dr

dV

dt

dV== 24 , since 24 r

dr

dV=

dt

dV = 4r2

dt

dr

96 = 4 122 dt

dr

dt

dr =

6

1

Using the chain rule again

dt

drr

dt

dr

dr

dA

dt

dA== 8 , since r

dr

dA8=

12166

1128 == scm

dt

dA

Forming differential equations

Example: The mass of a radio-active substance at time t is decaying at a rate which isproportional to the mass present at time t. Find a differential equation connecting themass m and the time t.

Solution: Remember thatdt

dm means the rate at which the mass is increasingso in this

case we must consider the rate of decay as a negative increase

dt

dm m

dt

dm = km, where k is the (positive) constant of proportionality.

1 ln m =

kt2 + lnA


15/40

C4 14/04/13 13

5 Integration

Integrals of ex and

+= cedxe xx

+= cxdxx

||ln1

for a further treatment of this result, see the appendix

Example: Find dxx

xx +

2

3 3

Solution: dxx

xx +

2

33

= + dx

xx

3 = x2 + 3 lnx + c.

Standard integrals

x must be in RADIANS when integrating trigonometric functions.

f(x) dxxf )( f(x) dxxf )(

xn

1

1

+

+

n

xn

sinx cosx

x1 ln x cosx sinx

e

x

e

x

secxtanx secxsec2x tanx

cosecxcotx cosecx

cosec2x cotx

Integration us ing trigonometric identities

Example: Find dxx2

cot .

Solution: cot2x = cosec

2x 1

= dxxdxx 1coseccot 22

= cotx x + c.


16/40

C4 14/04/13 SDB14

Example: Find .sin 2 dxx

Solution: sin2x = (1 cos 2x)

= dxxdxx )2cos1(sin 212 = x sin 2x + c.

You cannotchangex to 3xin the above result to find dxx3sin2

. see next example

Example: Find dxx3sin 2 .

Solution: sin23x = (1 cos 2 3x) = (1 cos 6x)

= dxxdxx )6cos1(3sin 212 = x

1/12sin 6x + c.

Example: Find sin 3xcos 5x dx.

Solution: Using the formula 2 sinAcosB = sin(A+B) + sin(AB)

sin 3xcos 5x dx

= 1/2sin 8x + sin (2x) dx = 1/2sin 8x sin 2x dx= 1/16cos 8x + cos 2x + c.

Integration by reverse chain rule

Some integrals which are not standard functions can be integrated by thinking of the chainrule for differentiation.

Example: Find dxxx 3cos3sin 4 .

Solution: dxxx 3cos3sin 4

If we think of u= sin 3x, then the integrand looks like dx

du

u

4

if we ignore theconstants, which would integrate to give

1/5u

5

so we differentiate u5 = sin

53x

to give ( ) ( ) xxxxxdx

d3cos3sin153cos33sin53sin 445 ==

which is 15 times what we want and so


17/40

C4 14/04/13 15

dxxx 3cos3sin 4 = cx +3sin 5151

Example: Find dxxx

)32( 2

Solution: dxxx

)32( 2

If we think of u= (2x2 3), then the integrand looks like

dx

du

u

1if we ignore the

constants, which would integrate to ln u

so we differentiate ln u = ln 2x2 3

to give ( )2 2 21 4

ln 2 3 42 3 (2 3)

d xx x

dx x x = =

which is 4 times what we want and so

dx

x

x

)32( 2 = ln 2x2 3 + c.

In general cxfdxxf

xf+= )(ln)(

)('

Integrals of tan xand cot x

tanx dx = dxxx

cos

sin =

dx

x

x

cos

sin,

and we now have cxfdxxf

xf+=

)(ln

)(

)('

tanx dx = ln cosx + c

tanx dx = ln secx + c

cotxcan be integrated by a similar method to give

cotx dx = ln sinx + c

Integrals of sec xand cosec x

secxdx = dxxx

xxxdx

xx

xxx +

+=

++

tansec

tansecsec

tansec

)tan(secsec 2

The top is now the derivative of the bottom

and we have cxfdxxf

xf+= )(ln)(

)('

secxdx = ln secx+ tanx + c


18/40

C4 14/04/13 SDB16

and similarly

cosecxdx = ln cosecx+ cotx + c

Integration using partial fractions

For use with algebraic fractions where the denominator factorises.

Example: Find

+dx

xx

x

2

6

2

Solution: First express2

62 +xx

x in partial fractions.

2

62 +xx

x

21)2)(1(

6

++

+ x

B

x

A

xx

x

6x A(x+ 2) + B(x 1).

put x = 1 A= 2,

put x = 2 B= 4

++=+ dxxxdxxx

x

2

4

1

2

2

62

= 2 ln |x 1| + 4 ln |x+ 2| + c.

Integration by substitution, indefinite

(i) Use the given substitution involving x and u, (or find a suitable substitution).

(ii) Find eitherdx

du or

du

dx, whichever is easier and re-arrange to find dx in terms

of du, i.e dx= ... du(iii) Use the substitution in (i) to make the integrand a function of u, and use your

answer to (ii) to replace dx by ...... du.

(iv) Simplify and integrate the function of u.

(v) Use the substitution in (i) to write your answer in terms of x.

Example: Find dxxx 53 2 using the substitution u= 3x2 5.

Solution: (i) u= 3x2 5

(ii) x

dudxxdx

du

66 ==

(iii) We can see that there an x will cancel, and 3 5 dxxx 53 2 = x

duux

6= du

u

6

(iv) = duu 21

6

1 = c

u+

23

2

3

6

1


19/40

C4 14/04/13 17

(v) = cx

+

9

)53( 23

2

Example: Find + dxx 211

using the substitution x = tan u.

Solution: (i) x = tan u.

(ii) udu

dx 2sec= dx = sec2u du.

(iii) + dxx 211

= + duuu2

2sec

tan1

1

(iv) = duu

u2

2

sec

sec since 1 + tan2 u = sec2u

= += cudu

(v) = tan1

x + c.

Example: Find

dxx

x

4

3

2 using the substitution u

2=x

2 4.

Solution: (i) u2=x

2 4.

(ii) We know that ( ) dxdu

uudx

d

22

= so differentiating gives

dux

udxx

dx

duu == 22 .

(iii) We can see that an x will cancel and ux = 42 so

dxx

x

4

3

2 = dux

u

u

x3

(iv) = += cudu 33

(v) = cx + 43 2

A justification of this technique is given in the appendix.


20/40

C4 14/04/13 SDB18

Integration by substitution, definite

If the integral has limits then proceed as before but remember to change the limits from

values of x to the corresponding values of u.Add (ii) (a) Change limits from x to u, and

new (v) Put in limits for u.

Example: Find 6

223 dxxx using the substitution u= 3x 2.

Solution: (i) u= 3x 2.

(ii) 3=dx

du

3

dudx =

(ii) (a) Change limits from x to u

x= 2 u= 3 2 2 = 4, and x = 6 u= 3 6 2 = 16

(iii) 6

223 dxxx =

+164

2

1

33

2 duu

u

(iv) = +16

4

2

1

2

3

91 2 duuu

=

16

42

3

2

3

2

5

2

5

91 2

+

uu

(v) = [ ]64102434

52

91 + [ ]832

34

52

91 + = 524 to 3 S.F.

Choosing the substitution

In general put u equal to the awkward bit but there are some special cases where this willnot help.

1

put u =x

2+ 1

put u = x 2 2 5 put u = 2x+ 5 or u2 = 2x+ 5 4 put u or u2 = 4 x2 only ifnis ODD

put x = 2 sin u only ifnis EVEN (or zero)

this makes 4 = 4cos = 2 cos u


21/40

C4 14/04/13 19

There are many more possibilities use your imagination!!

Integration by parts

The product rule for differentiation is

To integrate by parts

(i) choose u anddx

dv

(ii) find v and

dx

du

(iii) substitute in formula and integrate.

Example: Find dxxx sin

Solution: (i) Choose u=x, because it disappears when differentiated

and choose xdx

dvsin=

(ii) u=x and1=dx

du

=== xdxxvxdxdv

cossinsin

(iii) = dxdxdu

vuvdxdx

dvu

dxxx sin = xcosx dxx)cos(1= xcosx + sinx + c.

Example: Find

dxxln

Solution: (i) It does not look like a product,dx

dvu , but if we take u= lnx and

1=dx

dv then

dx

dvu = lnx 1 = lnx

(ii) u= lnx xdx

du 1= and 1=

dx

dv v =x


22/40

C4 14/04/13 SDB20

(iii) = dxxxxdxx1

ln1ln

= xlnx x + c.

Area under curve

We found in Core 2 that the area under the curve is written

as the integral .We can consider the area as approximately the sum of therectangles shown.

If each rectangle has width x and if the heights of therectangles are y1,y2, ...,yn

then the area of the rectangles is approximately the areaunder the curve

bb

aa

y x y dx

and as x 0 we haveb b

aa

y x y dx

This last result is true for any function y.

Volume of revolution

If the curve of y=f(x) is rotated about the xaxis then the volume of the shape formed can

be found by considering many slices each of width x: one slice is shown.

The volume of this slice (a disc) is approximately y2x

Sum of volumes of all slices from a to b 2b

a

y x

x

x +x

a b

y = f (x)

y

x

y

a b


23/40

C4 14/04/13 21

and as x0 we have (using the result aboveb b

aa

y x y dx )

2 2b b

aa

y x y dx .

Volume of revolution about the xaxis

Volume when y=f(x), between x= a and x= b, is rotated about the xaxis

is V= b

adxy 2 .

Volume of revolution about the yaxis

Volume when y=f(x) , between y= c and y= d, is rotated about the yaxis

is V= d

cdyx

2 .

Volume of rotation about they-axis is not in the syllabus but is included for completeness.

Parametric integration

When x and y are given in parametric form we can find integrals using the techniques inintegration by substitution.

think of cancelling the dts

See the appendix for a justification of this result.

Example: If x= tan t and y= sin t, find the area under the curve from t= 0 to t =

/4.

Solution: The area = dxy for some limits on x = dtdtdx

y .

We know that y = sin t, and also that

x= tan t tdt

dx 2sec=

area = dxy =dx

y dtdt

24 4

0 0sin sec tan sect t dt t t dt

= =

= [ ] 40sec

t = 12

To find a volume of revolutionwe need dxy2 and we proceed as above writing


24/40

C4 14/04/13 SDB22

Example: The curve shown has parametric equations

x= t21, y= t3.

The section betweenx= 0 andx= 8 above thex-axis

is rotated about thex-axis through 2radians. Findthe volume generated.

Solution: V = 8

0

2 dxy .

x= 0 t= 1 and x= 8 t= 3,

but the curve is above thex-axis y= t3> 0 t> 0, t = +1, or 3

also y = t3, x= t

21 t

dt

dx2=

V = ==3

1

73

1

62 22 dttdtttdtdt

dxy

=

38

1

28

t

= ( )134

8

Differential equations

Separating the variables

Example: Solve the differential equation xyydx

dy+= 3 .

Solution: xyydx

dy+= 3 = y(3 +x)

We first cheat by separating the xs and ys onto different sides of the equation.

dxxdyy

)3(1 += and then put in the integral signs

dxxdyy += 31

lny = 3x+ 1/2x2 + c.

See the appendix for a justification of this technique.

5 10

10

20

x

y


25/40

C4 14/04/13 23

Exponential growth and decay

Example: A radio-active substance decays at a rate which is proportional to the mass of thesubstance present. Initially 25 grams are present and after 8 hours the mass hasdecreased to 20 grams. Find the mass after 1 day.

Solution: Let m grams be the mass of the substance at time t.

dt

dmis the rate of increaseof m so, since the mass is decreasing,

dt

dm

=

km

kdt

dm

m=

1

= dtkdmm1

ln m = kt+ ln A

ln = kt ktm Ae= .

When t= 0, m= 25 A= 25

m = 25ekt.

When t = 8, m = 20

20 = 25e8k e8k = 08

8k = ln 08 k = 0027892943

So when t= 24, m = 25e24 0027892943

= 128.

Answer 128 grams after 1 day.


26/40

C4 14/04/13 SDB24

6 Vectors

Notation

The book and exam papers like writing vectors in the form

a = 3i 4j + 7k.

It is allowed, and sensible, to re-write vectors in column form

i.e. a = 3i 4j + 7k =

7

4

3

.

Definit ions, adding and subtracting, etceteraA vector has both magnitude (length) and direction. If you always think of a vector as atranslation you will not go far wrong.

Directed line segments

The vector is the vector fromA toB,(or the translation which takes A to B).

This is sometimes called the

displacement vector from A to B.

Vectors in co-ordinate form

Vectors can also be thought of as column vectors,

thus in the diagram

=

3

7AB .

Negative vectors

is the 'opposite' of and so =

3

7.

Adding and subtracting vectors

(i) Using a diagram

Geometrically this can be done using a triangle (or a parallelogram):

Adding:

A

B

7

3

A

B

a

b

a+bb

a

a+b


27/40

C4 14/04/13 25

The sum of two vectors is called the resultant of those vectors.

Subtracting:

(ii) Using coordinates

+

+=

+

db

ca

d

c

b

a and

=

db

ca

d

c

b

a.

Parallel and non - parallel vectors

Parallel vectors

Two vectors are parallel if they have the same direction

one is a multiple of the other.

Example: Which two of the following vectors are parallel?

.1

2,

2

4,

3

6

Solution: Notice that

=

2

4

2

3

3

6 and so

3

6 is parallel to

2

4

but

1

2 is not a multiple of

2

4 and so cannot be parallel to the other two vectors.

Example: Find a vector of length 15 in the direction of

34

.

Solution:a =

3

4 has length a= a = 534 22 =+

and so the required vector of length 15 = 3 5 is 3a= 3

3

4 =

9

12.

b

aab


28/40

C4 14/04/13 SDB26

Non-parallel vectors

If a and b are not parallel and if a + b = a + b, then

a a = b b ( )a = ( )b

but a and b are not parallel and one cannot be a multiple of the other

( ) = 0 = ( )

= and = .

Example: If a and b are not parallel and if

b+ 2a + b = a + 3b 5a, find the values of and .

Solution: Since a and b are not parallel, the coefficients of a and b must balance out

2 = 5 = 7 and 1 + = 3 = 2.

Modulus of a vector and unit vectors

Modulus

The modulusof a vector is its magnitude or length.

If 73 then the modulus of 7 3 58Or, if c = 35 then the modulus of c = c = c= 3 5 34

Unit vectors

A unit vectoris one with length 1.

Example: Find a unit vector in the direction of

5

12.

Solution: a =

5

12 has length a= a = 13512 22 =+ ,

and so the required unitvector is 131 a =

=

135

1312

131

5

12.


29/40

C4 14/04/13 27

Position vectors

If A is the point (1, 4) then the position vector of A is the vector from the origin to A,

usually written as = a=

4

1.

For two points A and B the position vectors are

= a and = bTo find the vector go from AO Bgiving

= a+b = b a

Ratios

Example: A,B are the points (2, 1) and (4, 7). Mlies onAB in the ratio 1 : 3. Find thecoordinates of M.

Solution : 26 = 26 0 51 5 21 0 51 5 2 52 5 Mis (25, 25)

x

B

A

O

b

a

ba

O

B

A

M

1

3


30/40

C4 14/04/13 SDB28

Proving geometrical theorems

Example: In a triangle OBClet M and N be the midpoints of OB and OC.

Prove that BC= 2MN and that BC is parallel to MN.

Solution: Write the vectors as b, and as c.Then = = band = = c.To find , go fromMto Ousing b andthen from O toNusing c

= b + c = c bAlso, to find , go fromB to O using b andthen from O to C using c

= b +c=cb.But = b + c = (cb) = BC

BC is parallel to MN

andBC is twice as long as MN.

Example: P lies on OAin the ratio 2 : 1, and Qlies on OBin the ratio 2 : 1. Prove thatPQis parallel toABand that PQ=

2/3

AB.

Solution: Let a= , and b= ba a, band = 2/3b2/3a

=2/3 (b a)

= 2/3 PQis parallel toABand

PQ= 2/3AB.

B

C

O

c

b

O

P

A

Q

B

1

2

2

1


31/40

C4 14/04/13 29

Three dimensional vectors

Length, modulus or magnitude of a vector

The length, modulus or magnitude of the vector 1

2

3

a

a

a

=

is

= a 2 2 2`1 2 3

a a a a= = + + ,

a sort of three dimensional Pythagoras.

Distance between two points

To find the distance between A, (a1, a2, a3) and B, (b1, b2, b3) we need to find the length of

the vector . = b a 1 1 1 12 2 2 23 3 3 3

b a b a

b a b a

b a b a

= =

= AB = ( ) ( ) ( )2 2 21 1 2 2 3 3b a b a b a + + Scalar product

a.b = ab cos

where a and b are the lengths of a and b

and is the angle measured from a to b.

Note that (i) a . a = aa cos 0o = a

2

(ii) a.(b+c) = a.b + a.c.(iii) a . b = b . a since cos = cos ()

In co-ordinate form

a.b = 22112

1

2

1 . babab

b

a

a+=

= ab cos

or a.b = 332211

3

2

1

3

2

1

. bababab

b

b

a

a

a

++=

= ab cos .

a

b


32/40

C4 14/04/13 SDB30

Perpendicular vectors

If a and b are perpendicular then = 90o and cos = 0

thus a perpendicular tob a.b = 0

and a.b = 0 eithera is perpendicular to b or a or b = 0.

Example: Find the values of so that a= 3i 2j+ 2k and

b= 2i+ j+ 6k are perpendicular.

Solution: Since a and b are perpendicular a.b = 0

23 2

2 0 6 2 12 0

2 6

. = + =

2= 9 = 3.

Example: Find a vector which is perpendicular to a,

2

1

1

, and b,

1

1

3

.

Solution: Let the vector c,

r

q

p

, be perpendicular to both a and b.

r

q

p

.

2

1

1

= 0 and

r

q

p

.

1

1

3

= 0

pq+ 2r = 0 and 3p+ q+ r = 0.

Adding these equations gives 4p+ 3r = 0.

Notice that there will never be a unique solution to these problems, so havingeliminated one variable, q, we find p in terms of r, and then find q in terms of r.

4

5

4

3 rq

rp =

=

c is any vector of the form

r

r

r

4

5

4

3

,


33/40

C4 14/04/13 31

and we choose a sensible value of r = 4 to give

4

5

3

.

Angle between vectors

Example: Find the angle between the vectors

= 4i 5j+ 2k and = i+ 2j 3k, to the nearest degree.Solution: First re-write as column vectors (if you want)

a =

2

5

4

and b=

3

2

1

a= a= 5345254 222 ==++ , b = b= 14321 222 =++

and a.b =

2

5

4

.

3

2

1

= 4 10 6 = 20

a.b = ab cos 20 = 3 5 14 cos

cos =703

20 = 0.796819

= 143o to the nearest degree.

Vector equation of a straight line

r=

z

y

x

is usually used as the position vector of a

general point,R.

In the diagram the line l passes through the point A

and is parallel to the vector b.

To go from OtoRfirst go toA, usinga, and thenfromAtoR using some multiple ofb.

l

O

a

R

b

A

r


34/40

C4 14/04/13 SDB32

The equation of a straight line through the pointAand parallel to the vector b is

r = a+ b.

Example: Find the vector equation of the line through the pointsM, (2, 1, 4),

andN, (5, 3, 7).

Solution: We are looking for the line throughM (orN) which is parallel to the vector . = nm =

=

3

4

7

4

1

2

7

3

5

equation is r =

+

3

4

7

4

1

2

.

Example: Show that the pointP, (1, 7, 10), lies on the line

r =

+

3

2

1

4

3

1

.

Solution: Thexco-ord of P is 1 and of the line is 1

1 = 1 = 2.

In the equation of the line this gives y= 7 and z= 10

P, (1, 7, 10) does lie on the line.

Intersection of two lines

2 Dimensions

Example: Find the intersection of the lines

1l , r = 22 1 1 1

, and , 3 2 3 1

+ = + l r .

Solution: We are looking for values of and which give the same x and yco-ordinates on each line.

Equatingxco-ords 2 = 1 +

equatingyco-ords 3 + 2= 3


35/40

C4 14/04/13 33

Adding 5 + = 4 = 1 = 2

lines intersect at (3, 1).

3 Dimensions

This is similar to the method for 2 dimensions with one important difference you can notbecertain whether the lines intersect without checking.

You will always (or nearly always) be able to find values of and by equatingxcoordinates andycoordinates but the zcoordinates might or might not be equal andmust be checked.

Example: Investigate whether the lines

l1, r =

+

1

2

1

3

1

2

and l2, r =

+

1

3

1

5

1

3

intersect

and if they do find their point of intersection.

Solution: Ifthe lines intersect we can find values of and to give the same x,y and zcoordinates in each equation.

Equating xcoords 2 = 3 + , I

equating ycoords 1 + 2 = 1 + 3, II

equating zcoords 3 + = 5 + . III

2 I+ II 5 = 5 + 5 = 2, in I = 3.

We must now check to see if we get the same point for the values of and

In l1, = 3 gives the point (1, 7, 6);

in l2, = 2 gives the point (1, 7, 7).

Thexandyco-ords are equal (as expected!), but thezco-ordinates are different and sothe lines do notintersect.


36/40

C4 14/04/13 SDB34

7 Appendix

Binomial series (1 + x)n for any n proof

Suppose that

f(x) = (1 +x)n = a+ bx+ cx2+ dx3+ ex4+

putx= 0, 1 = a

f(x) = n(1 +x)n 1= b+ 2cx+ 3dx2+ 4ex3+

put x= 0, n= b

f(x) = n(n1)(1 +x)n 2= 2c+ 3 2dx+ 4 3ex2+

put x= 0,

n(n

1) = 2c

! f(x) = n(n1)(n2)(1 +x)n3= 3 2d+ 4 3 2ex+ put x= 0, n(n1)(n2) = 3 2d

!

Continuing this process, we have

! and ! , etc.

giving f(x) = (1 +x)n

= 1 + nx+

! + ! + ! + ! + + ! +

Showing that this is convergent for x< 1, is more difficult!

Derivative of xq for q rational

Suppose that q is any rational number, , where r and s are integers, s 0.Then y= x

q = ys=xr

Differentiating with respect to x s = r xr1

= since q=

= qx q1 sinceys=xrand y= xq


37/40

C4 14/04/13 35

which follows the rule found for xn, where n is an integer.

for negative limits

We know that the area under any curve, fromx= atox= bis approximately

If the curve is above thex-axis, all theyvalues are positive, and if a< bthen all values of xare positive, and so the integral is positive.

1

ln|x|

ln ln

Example: Find

3

1

1dx

x.

Solution: The integral wanted is shown as A

in the diagram.

By symmetry A=A (Apositive)

and we need to decide whether the integral

is +A or A.

From x=

1 to x=

3, we are going inthe direction ofxdecreasing

all xare negative.

And the graph is below thex-axis,

theyvalues are negative, y x is positive

0 0 the integral is positive and equal toA.

The integral, A =A = 3

1

1dx

x= [ ] 31lnx = ln 3 ln 1 = ln 3

A=3

11 dxx

= ln 3

Notice that this is what we get if we write ln x in place of lnx

3

1

1dx

x= [ ] 31ln | |x

= ln 3 ln 1 = ln 3

As it will always be possible to use symmetry in this way, since we can never have onepositive and one negative limit (because there is a discontinuity atx= 0), it is correct to write

ln xfor the integral of 1/x.

x

, 0 a b

y

3 2 1 1 2 3

1

1

A

A'

y=1/x


38/40


39/40

C4 14/04/13 37

Separating the variables why it works

We show this with an example.

If y= 6y3

then

= 18y2

and so 18 18 6 Notice that we cancel the dx.

Example: Solve = x2secy

Solution: = x2secy

cosy

= x2

cos cos cancelling the dx siny = + c


40/40

Index

Binomial expansion

convergence, 8

for any n, 8

proof - for any n, 34

Differential equations

exponential growth and decay, 23

forming, 12

separating the variables, 22

Differentiation

derivative of xq for q rational, 34

exponential functions, ax, 11

related rates of change, 12

Implicit differentiation, 10

Integration1/xfor negative limits, 35

by parts, 19

by substitution why it works, 36choosing the substitution, 18

ex, 13

lnx, 13

parametric, 21

parametric why it works, 36

reverse chain rule, 14

secxand cosecx, 15

separating variables why it works, 37

standard integrals, 13

substitution, definite, 18

substitution, indefinite, 16

tanxand cotx, 15

using partial fractions, 16

using trigonometric identities, 13

volume of revolution, 20

Parameters

area under curve, 7

equation of circle, 6

equation of hyperbola, 6

intersection of two curves, 5

parametric equations of curves, 5

parametric to Cartesian form, 6

trig functions, 6

Parametric differentiation, 11

Parametric integration, 36

areas, 21

volumes, 21

Partial fractions, 3

integration, 16

linear repeated factors, 4

linear unrepeated factors, 3

quadratic factors, 3

top heavy fractions, 4

Scalar product, 29

perpendicular vectors, 30

properties, 29

Vectors, 24

adding and subtracting, 24

angle between vectors, 31

displacement vector, 24distance between 2 points, 29

equation of a line, 31

intersection of two lines, 32

length of 3-D vector, 29

modulus, 26

non-parallel vectors, 26

parallel vectors, 25

position vector, 27

proving geometrical theorems, 28

ratios, 27

unit vector, 26

C4 revision guide

Documents

Transcript of C4 revision guide