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 Q17  If   x = a (\cos t + t \sin t) and   y = a (\sin t - t \cos t), find \frac{d^2 y }{dx^2 }
 

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Given functions are
 x = a (\cos t + t \sin t) and   y = a (\sin t - t \cos t)
Now, differentiate both the functions w.r.t. t  independently
We get
\frac{dx}{dt} = \frac{d(a(\cos t +t\sin t))}{dt}= a(-\sin t)+a(\sin t+t\cos t)
                                                      =-a\sin t+a\sin t+at\cos t = at\cos t
Similarly,
\frac{dy}{dt} = \frac{d(a(\sin t - t\cos t))}{dt}= a\cos t -a(\cos t+t(-\sin t))
                                                       = a\cos t -a\cos t+at\sin t =at\sin t
Now,
\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at \cos t} = \tan t
Now, the second derivative
\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \sec^2 t.\frac{dt}{dx}=\frac{\sec^2t.\sec t }{at}=\frac{\sec^3t}{at}
                                                                                                  (\because \frac{dx}{dt} = at\cos t \Rightarrow \frac{dt}{dx}= \frac{1}{at\cos t}=\frac{\sec t}{at})
Therefore,  \frac{d^2y}{dx^2}=\frac{\sec^3t}{at}

Posted by

Gautam harsolia

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