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If \(x = a (\cos t + t\sin t)\) and \(y = a (\sin t - t \cos\: t)\), find \( \large\frac{d^2y}{dx^2}\)

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  • $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
$x=a(\cos t+t\sin t)$
Differentiating with respect to $x$
$\large\frac{dx}{dt}$$=a(-\sin t+\sin t+t\cos t)$
$\quad\;=at\cos t$
$y=a(\sin t-t\cos t)$
$\large\frac{dy}{dt}$$=a(\cos t-\cos t+t\sin t)$
$\quad\;=at\sin t$
Step 2:
$\large\frac{dy}{dx}=\Large\frac{\Large\frac{dy}{dt}}{\Large\frac{dt}{dx}}$$=\large\frac{at\sin t}{at\cos t}$
$\large\frac{dy}{dx}=$$\tan t$
It is valid when $t\neq 0,t\neq (2n+1)\large\frac{\pi}{2}$
Step 3:
$\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
$\qquad=\large\frac{d}{dt}\big(\frac{dy}{dx}\big)\times \frac{dt}{dx}$
$\qquad=\large\frac{d}{dt}$$(\tan t)\times \large\frac{dt}{dx}$
$\large\frac{dx}{dt}$$=at\cos t$
$\qquad=\sec^2t\times \large\frac{1}{at\cos t}$
$\qquad=\large\frac{1}{at}$$\sec^3t$
Here also $t\neq 0,t\neq (2n+1)\large\frac{\pi}{2}$$,t\in R-\begin{bmatrix}(2n+1)\large\frac{\pi}{2}\end{bmatrix}$
answered May 15, 2013 by sreemathi.v
 

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