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

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(C) Given $x = a \sin^3 t$ and $y = a \cos^3 t$.$\frac{dx}{dt} = 3a \sin^2 t \cos t$ and $\frac{dy}{dt} = -3a \cos^2 t \sin t$.$\frac{dy}{dx} = \frac{

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$\frac{32}{27a}$

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(C) Given $x = a \sin^3 t$ and $y = a \cos^3 t$.$\frac{dx}{dt} = 3a \sin^2 t \cos t$ and $\frac{dy}{dt} = -3a \cos^2 t \sin t$.$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3a \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\cot t$.$\frac{d^2y}{dx^2} = \frac{d}{dx}(-\cot t) = \frac{d}{dt}(-\cot t) \cdot \frac{dt}{dx} = \csc^2 t \cdot \frac{1}{3a \sin^2 t \cos t}$.$\frac{d^2y}{dx^2} = \frac{1}{3a \sin^4 t \cos t} = \frac{\sec t \csc^4 t}{3a}$.At $t = \frac{\pi}{3}$, $\sin t = \frac{\sqrt{3}}{2}$ and $\cos t = \frac{1}{2}$.$\frac{d^2y}{dx^2} = \frac{1}{3a (\sqrt{3}/2)^4 (1/2)} = \frac{1}{3a (9/16) (1/2)} = \frac{1}{27a/32} = \frac{32}{27a}$.

If $x = a \sin^3 t$ and $y = a \cos^3 t$, then the value of $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{3}$ is equal to

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