If x = a cos theta and y = b sin theta,then f | vedclass

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

Vedclass · Accepted Answer

$-2 \sqrt{2}\left(\frac{b}{a^2}\right)$

Vedclass · Answer

(C) Given $x = a \cos 	heta$ and $y = b \sin 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d 	heta} = -a \sin 	heta$ and $\frac{dy}{d 	heta} = b \cos 	heta$. Then,$\frac{dy}{dx} = \frac{dy/d 	heta}{dx/d 	heta} = \frac{b \cos 	heta}{-a \sin 	heta} = -\frac{b}{a} \cot 	heta$. Now,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{d}{d 	heta} \left( -\frac{b}{a} \cot 	heta ight) \cdot \frac{d 	heta}{dx} = \left( -\frac{b}{a} ight) (-\csc^2 	heta) \cdot \frac{1}{-a \sin 	heta} = -\frac{b}{a^2} \cdot \frac{1}{\sin^3 	heta}$. At $	heta = \frac{\pi}{4}$,$\sin 	heta = \frac{1}{\sqrt{2}}$,so $\sin^3 	heta = \left( \frac{1}{\sqrt{2}} ight)^3 = \frac{1}{2 \sqrt{2}}$. Therefore,$\left[\frac{d^2 y}{dx^2}ight]_{	heta = \frac{\pi}{4}} = -\frac{b}{a^2} \cdot (2 \sqrt{2}) = -2 \sqrt{2} \left( \frac{b}{a^2} ight)$.

If $x = a \cos \theta$ and $y = b \sin \theta$,then find the value of $\left[\frac{d^2 y}{d x^2}\right]_{\theta = \frac{\pi}{4}}$.

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