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If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$, find $\frac{d^2 y}{d x^2}$
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Verified Answer
$x=a(\operatorname{cost}+t \sin t)$
$\therefore \quad \frac{d x}{d t}=a(-\sin t+\sin t+t \cos t)=a t \cos t$
and $y=a(\sin t-t \cos t)$
$\frac{d y}{d t}=a(\cos t-\cos t+t \sin t)=a t \sin t$
$\therefore \quad \frac{d y}{d x}=\tan t$
It is valid $t \neq 0, t \neq(2 n+1) \pi / 2$
$\frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{d y}{d x}\right) \times \frac{d t}{d x}=\frac{1}{a t} \sec ^3 t$
$\therefore \quad \frac{d x}{d t}=a(-\sin t+\sin t+t \cos t)=a t \cos t$
and $y=a(\sin t-t \cos t)$
$\frac{d y}{d t}=a(\cos t-\cos t+t \sin t)=a t \sin t$
$\therefore \quad \frac{d y}{d x}=\tan t$
It is valid $t \neq 0, t \neq(2 n+1) \pi / 2$
$\frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{d y}{d x}\right) \times \frac{d t}{d x}=\frac{1}{a t} \sec ^3 t$
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