Question

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

• A. $ \frac{4\sqrt{2}}{3a} $
• B. $ 2 $
• C. $ \frac{1}{12a} $
• D. $ 0 $
• E. none of these

Answer 1

Answer: (A)

Let $ u=a{{\sin }^{3}}t $ and $ v=a{{\cos }^{3}}t $ On differentiating w.r.t. t respectively $ \frac{du}{dt}=3a{{\sin }^{2}}t\cos t $ and $ \frac{dv}{dt}=-3a{{\cos }^{2}}t\sin t $ $ \therefore $ $ \frac{du}{dv}=-\frac{3a{{\sin }^{2}}t\cos t}{3a{{\cos }^{2}}t\sin t}=-\tan t $ Now, $ \frac{{{d}^{2}}u}{d{{v}^{2}}}=-{{\sec }^{2}}t\frac{dt}{dv} $ $ =-\frac{{{\sec }^{2}}t}{-3a{{\cos }^{2}}t\sin t} $ $ {{\left( \frac{{{d}^{2}}u}{d{{v}^{2}}} \right)}_{x=\frac{\pi }{4}}}=\frac{{{(\sqrt{2})}^{2}}}{3a{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\left( \frac{1}{\sqrt{2}} \right)}=\frac{4\sqrt{2}}{3a} $