If x=4 cos ^3 theta and y=3 sin ^2 theta,then | vedclass

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(B) Given,$x=4 \cos ^3 	heta$ ... $(i)$ and $y=3 \sin ^2 	heta$ ... (ii). Differentiating $(i)$ and (ii) with respect to $	heta$: $\frac{dx}{d	het

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$\frac{1}{6}$

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(B) Given,$x=4 \cos ^3 	heta$ ... $(i)$ and $y=3 \sin ^2 	heta$ ... (ii). Differentiating $(i)$ and (ii) with respect to $	heta$: $\frac{dx}{d	heta} = 4 	imes 3 \cos^2 	heta (-\sin 	heta) = -12 \cos^2 	heta \sin 	heta$ ... (iii). $\frac{dy}{d	heta} = 3 	imes 2 \sin 	heta \cos 	heta = 6 \sin 	heta \cos 	heta$ ... (iv). Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{6 \sin 	heta \cos 	heta}{-12 \cos^2 	heta \sin 	heta} = -\frac{1}{2} \sec 	heta$. Differentiating $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{1}{2} \sec 	heta ight) = -\frac{1}{2} \sec 	heta 	an 	heta \cdot \frac{d	heta}{dx}$. Substituting $\frac{d	heta}{dx} = \frac{1}{-12 \cos^2 	heta \sin 	heta}$: $\frac{d^2y}{dx^2} = -\frac{1}{2} \sec 	heta 	an 	heta \cdot \left( \frac{1}{-12 \cos^2 	heta \sin 	heta} ight) = \frac{1}{24} \cdot \frac{1}{\cos 	heta} \cdot \frac{\sin 	heta}{\cos 	heta} \cdot \frac{1}{\cos^2 	heta \sin 	heta} = \frac{1}{24 \cos^4 	heta}$. At $	heta = \frac{\pi}{4}$,$\cos 	heta = \frac{1}{\sqrt{2}}$,so $\cos^4 	heta = \left( \frac{1}{\sqrt{2}} ight)^4 = \frac{1}{4}$. Thus,$\frac{d^2y}{dx^2} = \frac{1}{24 	imes (1/4)} = \frac{1}{6}$.

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

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