The second derivative of a sin^3 t with respe | vedclass

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(B) Let $x = a \cos^3 t$ and $y = a \sin^3 t$. We need to find $\frac{d^2 y}{dx^2}$.First,find $\frac{dy}{dt} = 3a \sin^2 t \cos t$ and $\frac{dx}{dt}

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

Vedclass · Answer

(B) Let $x = a \cos^3 t$ and $y = a \sin^3 t$. We need to find $\frac{d^2 y}{dx^2}$.First,find $\frac{dy}{dt} = 3a \sin^2 t \cos t$ and $\frac{dx}{dt} = -3a \cos^2 t \sin t$.Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -	an t$.Now,find the second derivative $\frac{d^2 y}{dx^2} = \frac{d}{dx}(-	an t) = \frac{d}{dt}(-	an t) \cdot \frac{dt}{dx}$.Since $\frac{dx}{dt} = -3a \cos^2 t \sin t$,then $\frac{dt}{dx} = \frac{1}{-3a \cos^2 t \sin t}$.So,$\frac{d^2 y}{dx^2} = (-\sec^2 t) \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{1}{3a \cos^4 t \sin t}$.At $t = \frac{\pi}{4}$,$\cos t = \frac{1}{\sqrt{2}}$ and $\sin t = \frac{1}{\sqrt{2}}$.$\frac{d^2 y}{dx^2} = \frac{1}{3a (\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{1}{3a (\frac{1}{4}) (\frac{1}{\sqrt{2}})} = \frac{4 \sqrt{2}}{3a}$.

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

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