If x=3 sqrt{2} cos ^3 theta and y=4 tan ^2 th | vedclass

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(C) Given $x = 3 \sqrt{2} \cos^3 	heta$ and $y = 4 	an^2 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = 3 \sqrt{2} \cd

Vedclass · Accepted Answer

$-\frac{32}{9}$

Vedclass · Answer

(C) Given $x = 3 \sqrt{2} \cos^3 	heta$ and $y = 4 	an^2 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = 3 \sqrt{2} \cdot 3 \cos^2 	heta \cdot (-\sin 	heta) = -9 \sqrt{2} \cos^2 	heta \sin 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = 4 \cdot 2 	an 	heta \cdot \sec^2 	heta = 8 	an 	heta \sec^2 	heta$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{8 	an 	heta \sec^2 	heta}{-9 \sqrt{2} \cos^2 	heta \sin 	heta}$. At $	heta = \frac{\pi}{4}$,we have $	an \frac{\pi}{4} = 1$,$\sec \frac{\pi}{4} = \sqrt{2}$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. Substituting these values: $\left(\frac{dy}{dx}ight)_{	heta=\frac{\pi}{4}} = \frac{8(1)(\sqrt{2})^2}{-9 \sqrt{2} (\frac{1}{\sqrt{2}})^2 (\frac{1}{\sqrt{2}})} = \frac{8 \cdot 2}{-9 \sqrt{2} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}} = \frac{16}{-9 \cdot \frac{1}{2}} = \frac{16}{-4.5} = -\frac{32}{9}$.

If $x=3 \sqrt{2} \cos ^3 \theta$ and $y=4 \tan ^2 \theta$,then $\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}} = $

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