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DERIVATIVE OF ARC LENGTHPowerPoint Presentation

DERIVATIVE OF ARC LENGTH

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### DERIVATIVE 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

Taking the limit as Q → P and using the fact that, perpendicular to QM ( see

We obtain

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

positive.

Thus we have

See Figure 1

Cor. 1 : perpendicular to QM ( seeIf the equation of the curve is x = f(y), then

Cor.2 : For parametric cartesian equations with parameter

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

Cor. 3 :

Exercises perpendicular to QM ( see

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 perpendicular to QM ( see

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

and Q(r+Δr, θ+Δ θ) be two adjacent points on the

curve AB as shown in the Figure 2.

Let arc AP = S, arc PQ= ΔS & the chord PQ = Δc.

Drop PM perpendicular to OQ

From the right angled triangle PMQ, we have

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

Also, from the right angled perpendicular to QM ( seeΔle 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 sin2Δθ⁄ 2) --------------- (iii)

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

Δc2 = (r sin Δθ)2 + (Δr + 2r sin2Δθ ⁄ 2)2

Now taking the limit as Q perpendicular to QM ( see→ P, Δθ → 0, then the

above equation reduces to

As s increases with the increase of perpendicular to QM ( seeθ ds / dθ is positive.

Hence

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

Cor. 2 : from Cor. 1. we have perpendicular to QM ( see

Also,

Example 2 : perpendicular to QM ( seeFind 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 perpendicular to QM ( seeYOU

END OF SHOW

Solution : perpendicular to QM ( seeGiven r2 = a² cos 2θ

Differentiating with respect to θ, we obtain

Previous slide

r = f( perpendicular to QM ( seeθ)

B

Q(r+Δr, θ+Δθ)

Figure 2

M

Δc

Δs

P(r, θ)

φ

s

r

A

Δθ

ψ

X

θ

O

T

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