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Question

(C) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First, we find the first derivative $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dt} = 3a \sin^2 t

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$\frac{4 \sqrt{2}}{3 a}$

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(C) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First, we find the first derivative $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dt} = 3a \sin^2 t \cos t$ $\frac{dx}{dt} = -3a \cos^2 t \sin t$ $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -	an t$. Now, we find the second derivative $\frac{d^2 y}{dx^2}$ with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{d}{dx}(-	an t) = \frac{d}{dt}(-	an t) \cdot \frac{dt}{dx} = (-\sec^2 t) \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{1}{3a \cos^4 t \sin t}$. Evaluating at $t = \frac{\pi}{4}$: $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. $\left. \frac{d^2 y}{dx^2} ight|_{t=\pi/4} = \frac{1}{3a (\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{1}{3a (\frac{1}{4}) (\frac{1}{\sqrt{2}})} = \frac{4\sqrt{2}}{3a}$.

The second order derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

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