If x=a cos ^3 theta and y=a sin ^3 theta,then | vedclass

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(A) Given $x=a \cos ^3 	heta$ and $y=a \sin ^3 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^2 	heta (

Vedclass · Accepted Answer

$\frac{4 \sqrt{2}}{3 a}$

Vedclass · Answer

(A) Given $x=a \cos ^3 	heta$ and $y=a \sin ^3 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^2 	heta (-\sin 	heta) = -3a \cos^2 	heta \sin 	heta$ $\frac{dy}{d	heta} = 3a \sin^2 	heta (\cos 	heta) = 3a \sin^2 	heta \cos 	heta$ Now,find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -	an 	heta$ Now,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-	an 	heta) = \frac{d}{d	heta}(-	an 	heta) \cdot \frac{d	heta}{dx} = -\sec^2 	heta \cdot \frac{1}{dx/d	heta}$ $\frac{d^2y}{dx^2} = -\sec^2 	heta \cdot \frac{1}{-3a \cos^2 	heta \sin 	heta} = \frac{1}{3a \cos^4 	heta \sin 	heta}$ At $	heta = \frac{\pi}{4}$: $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ $\frac{d^2y}{dx^2} = \frac{1}{3a (\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{1}{3a (\frac{1}{4}) (\frac{1}{\sqrt{2}})} = \frac{4\sqrt{2}}{3a}$

If $x=a \cos ^3 \theta$ and $y=a \sin ^3 \theta$,then $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{4}$ is

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