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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.

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Draw PL, QM perpendiculars on the x-axis & also PN perpendicular to QM ( see Figure 1) From the right angled triangle PQN, we have PQ 2 = PN 2 + NQ 2 Δc 2 = Δx 2 + Δy 2

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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

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Cor. 1 : Cor. 1 : If the equation of the curve is x = f(y),then Cor.2: Cor.2: For parametric cartesian equations with parameter i.e. for x = f(t) and y = g(t), we have Cor. 3:

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Exercises 1.Find for the curves: (i)y = c cosh x/c(ii)y = a log[a 2 / (a 2 – x 2 )] 2.Find for the curves: (i) x = a cos 3 t, y = b sin 3 t (ii) x = a e t sin t, y = a e t 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 )

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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 PQ 2 = PM 2 + MQ 2 --------------- (i)

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Also, from the right angled Δle PMO, we have PM = r sin Δθ. --------------- (ii) From the figure 2., we havefigure 2 MQ = OQ – OM = r + Δr – r cos Δθ = Δr + r (1– cos Δθ) = Δr + 2r sin 2 Δθ 2) --------------- (iii) From (i), (ii) and (iii) we have Δc 22 Δc 2 = (r sin Δθ) 2 + (Δr + 2r sin 2 Δθ 2) 2

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

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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

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

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

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Example 2 : Find ds/dθ for r 2 = a 2 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 r 2 sin 2θ = 2 a 2 meets the initial line in T, show that (i) ds/dθ = r 3 / 2 a 2 (ii) PT = r

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THANK YOU END OF SHOW

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

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Figure 1 A B P Q K L M N y xo s ΔsΔs ΔcΔc ΔxΔx ΔyΔy y = f(x) Last SlideViewed

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

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POLAR COORDINATES. There are many curves for which cartesian or parametric equations are unsuitable. For polar equations, the position of a point is defined.

POLAR COORDINATES. There are many curves for which cartesian or parametric equations are unsuitable. For polar equations, the position of a point is defined.

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