Question

prove that the whole length of the evolute of the astroid x=acos³@ and y=asin³@ is 12a​

Answer 1

we have to prove that the whole length of the evolute of the astroid x = acos³θ and y = asin³θ is 6a.

solution : here x = acos³θ and y = asin³θ

eliminating θ we get, x⅔ + y⅔ = a⅔

arc length of astroid is given by, S = $\int\limits^a_0\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$

so differentiating equation x⅔ + y⅔ = a⅔ with respect to x,

2/3 x¯⅓ + 2/3 y¯⅓ dy/dx = 0

⇒1/x⅓ + 1/y⅓ dy/dx = 0

⇒dy/dx = -y⅓/x⅓ = -(y/x)⅓

= -(asin³θ/acos³θ)⅓

= -tanθ

now x = a cos³θ

differentiating both sides we get,

⇒dx = a 3cos²θ (-sinθ) dθ

so,S = $\int\limits^a_0\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\int\limits^{2\pi}_0\sqrt{1+tan^2\theta}a3cos^2\theta(-sin\theta)\,d\theta$

= $-3a\int\limits^{2\pi}_{\pi/2}sin\theta cos\theta\,d\theta$

= $-\frac{3a}{2}\int\limits^{2\pi}_{\pi/2}(2sin\theta cos\theta)\,d\theta$

= $-\frac{3a}{2}\int\limits^{\pi/2}_{2\pi}sin2\theta\,d\theta$

= $-\frac{3a}{2}\left[\frac{-cos2\theta}{2}\right]^{2\pi}_{\pi/2}$

= 1.5a

so total length of astroid = 4S = 4 × 1.5a = 6a

Therefore total length of astroid is 6a.

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