DERIVATIVE OF ARC LENGTH

1. 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.

2. 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

3. Taking the limit as Q → P and using the fact that, We obtain If s increases with x as shown in Figure, then is positive. Thus we have See Figure 1

4. Cor. 1 : If 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 :

5. Exercises 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 )

6. DERIVATIVE OF ARC IN POLAR FORM 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)

7. 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 sin2Δθ⁄ 2) --------------- (iii) From (i), (ii) and (iii) we have Δc2 = (r sin Δθ)2 + (Δr + 2r sin2Δθ ⁄ 2)2

8. Now taking the limit as Q → P, Δθ → 0, then the above equation reduces to

9. 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

10. Cor. 2 : from Cor. 1. we have Also,

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

12. 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

13. THANKYOU END OF SHOW

14. Solution : Given r2 = a² cos 2θ Differentiating with respect to θ, we obtain Previous slide

15. y = f(x) y B Figure 1 P Δc Δy Δs s N A Δx Q Last SlideViewed o x K M L

16. r = f(θ) B Q(r+Δr, θ+Δθ) Figure 2 M Δc Δs P(r, θ) φ s r A Δθ ψ X θ O T Last Slide Viewed