The 2^{nd} derivative of a sin^3 t with respe | vedclass

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Question

(A) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 3a \sin^2 t \cos t$ $\frac{dx}{dt} = 3

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $y = a \sin^3 t$ and $x = a \cos^3 t$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 3a \sin^2 t \cos t$ $\frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t$ Now,find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -	an t$ Next,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-	an t) = -\sec^2 t \cdot \frac{dt}{dx}$ Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{-3a \cos^2 t \sin t}$,we have: $\frac{d^2y}{dx^2} = -\sec^2 t \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{\sec^2 t}{3a \cos^2 t \sin t} = \frac{1}{3a} \cdot \frac{\sec^4 t}{\sin t}$ At $t = \frac{\pi}{4}$,$\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$: $\left(\frac{d^2y}{dx^2}ight)_{t = \pi/4} = \frac{1}{3a} \cdot \frac{(\sqrt{2})^4}{1/\sqrt{2}} = \frac{1}{3a} \cdot \frac{4}{1/\sqrt{2}} = \frac{4\sqrt{2}}{3a}$

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

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