A) \[\frac{4\sqrt{2}}{3a}\]
B) \[2\]
C) \[\frac{1}{12a}\]
D) \[0\]
E) none of these
Correct Answer: A
Solution :
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, \[\frac{{{d}^{2}}u}{d{{v}^{2}}}=-{{\sec }^{2}}t\frac{dt}{dv}\] \[=-\frac{{{\sec }^{2}}t}{-3a{{\cos }^{2}}t\sin t}\] \[{{\left( \frac{{{d}^{2}}u}{d{{v}^{2}}} \right)}_{x=\frac{\pi }{4}}}=\frac{{{(\sqrt{2})}^{2}}}{3a{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\left( \frac{1}{\sqrt{2}} \right)}=\frac{4\sqrt{2}}{3a}\]
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