DERIVATIVE OF ARC LENGTHDERIVATIVE OF ARC LENGTHDERIVATIVE OF ARC LENGTHDERIVATIVE OF ARC LENGTH

Let y = f(x) be the equation of a given

curve. Let A be some fixed point on the

curve and P(x,y) and Q(x+Δx, y+ Δy) be two

neighbouring points on the curve as shown in

the Figure 1.

Let arc AP = S, arc PQ = ΔS and the chord PQ

= Δc.

Draw PL, QM perpendiculars on the x-axis & also PN

perpendicular to QM ( see Figure 1)

From the right angled triangle PQN, we have

PQ2 = PN2 + NQ2

Δc2 = Δx2 + Δy2

22

Δx

Δy 1

Δx

Δs

Δs

Δc

22

Δx

Δy 1

Δx

Δc

Taking the limit as Q → P and using the fact that, Taking the limit as Q → P and using the fact that,

We obtain We obtain

If s increases with x as shown in Figure, then is If s increases with x as shown in Figure, then is

positive. positive.

Thus we have Thus we have

2

dx

ds 1

dx

ds2

See Figure 1

Cor. 1 : Cor. 1 : If the equation of the curve is x = f(y), then

Cor.2Cor.2 : : For parametric cartesian equations with

parameter

i.e. for x = f(t) and y = g(t), we have

Cor. 3 :

ExercisesExercises

1. Find for the curves :

(i) y = c cosh x/c (ii) y = a log[a2 / (a2 – x2)]

2. Find for the curves :

(i) x = a cos3 t, y = b sin3 t

(ii) x = a et sin t, y = a et cos t

(iii) x = a( cos t + t sin t), y = a ( sin t – t cos t)

(iv) x = a ( t – sin t ), y = a ( 1 – cos t )

DERIVATIVE OF ARC IN POLAR FORM

DERIVATIVE OF ARC IN POLAR FORM

Let r = f(Let r = f(θθ) be the equation of a given curve. Let P(r, ) be the equation of a given curve. Let P(r, θθ) )

and Q(r+and Q(r+ΔΔr, r, θθ ++Δ Δ θθ) be two adjacent points on the ) be two adjacent points on the

curve AB as shown in the curve AB as shown in the Figure 2Figure 2. .

Let arc AP = S, arc PQ= Let arc AP = S, arc PQ= ΔΔS & the chord PQ = S & the chord PQ = ΔΔc. c.

Drop PM perpendicular to OQ

From the right angled triangle PMQ, we have

PQ2 = PM2 + MQ2 --------------- (i)

Also, from the right angled Δle PMO, we have

PM = r sin Δθ. --------------- (ii)

From the figure 2., we have

MQ = OQ – OM

= r + Δr – r cos Δθ

= Δr + r (1– cos Δθ)

= Δr + 2r sin2 Δθ ⁄ 2) --------------- (iii)

From (i), (ii) and (iii) we have

ΔΔcc22 = (r sin Δθ)22 + (Δr + 2r sin2 Δθ ⁄ 2)2

Now taking the limit as Q Now taking the limit as Q → P, → P, Δθ → 0, then the → 0, then the

above equation reduces to above equation reduces to

As s increases with the increase of θ ds / dθ is positive.

Hence

Cor. 1 : If the equation of the curve is θ = f(r), then

Cor. 2Cor. 2 :: from Cor. 1. we have from Cor. 1. we have

Also, Also,

Exercises : Example 1 : Find ds/dθ for r = a ( 1 + cos θ )

Solution :

Example 2 : Find ds/dθ for r2 = a2 cos 2θ

Solution :

Example 3 : Show that r (ds/dr) is constant for the curve

Example 4 : If the tangent at a point P(r, θ) on the curve

r2 sin 2θ = 2 a2 meets the initial line in T,

show that

(i) ds/dθ = r3/ 2 a2 (ii) PT = r

THANK YOU

END OF SHOW

Solution : Given r2 = a² cos 2θ

Differentiating with respect to θ, we obtain

Previous slide

Figure 1Figure 1

A

B

P

Q

K L M

N

y

xo

s

ΔsΔc

Δx

Δy

y = f(x)

Last SlideViewed

Figure 2Figure 2

OT

X

A

B

Q(r+Δr, θ+Δθ)

P(r, θ)

Δθ

θψψ

φ

M

r s

Δc

Δs

r = f(r = f(θθ))

Last Slide Viewed