If x = sin theta, y = sin^3 theta then frac{d | vedclass

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

Vedclass · Accepted Answer

$6$

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(B) Given $x = \sin 	heta$ and $y = \sin^3 	heta$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = \cos 	heta$$\frac{dy}{d	heta} = 3 \sin^2 	heta \cos 	heta$Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3 \sin^2 	heta \cos 	heta}{\cos 	heta} = 3 \sin^2 	heta$Next,differentiate $\frac{dy}{dx}$ with respect to $x$:$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3 \sin^2 	heta) = \frac{d}{d	heta}(3 \sin^2 	heta) \cdot \frac{d	heta}{dx}$$\frac{d^2 y}{dx^2} = (6 \sin 	heta \cos 	heta) \cdot \frac{1}{\cos 	heta} = 6 \sin 	heta$Finally,evaluate at $	heta = \frac{\pi}{2}$:$\frac{d^2 y}{dx^2} = 6 \sin(\frac{\pi}{2}) = 6(1) = 6$.

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

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