A) \[\frac{4\sqrt{2}}{3a}\]
B) \[2\]
C) \[\frac{1}{12a}\]
D) None of these
Correct Answer: A
Solution :
Let\[y=a{{\sin }^{3}}t\]and\[x=a{{\cos }^{3}}t\] On differentiating w.r.t. t, respectively, we get \[\frac{dy}{dt}=3a{{\sin }^{2}}t\cdot \cos t\] and \[\frac{dx}{dt}=3a{{\cos }^{2}}t(-\sin t)\] \[\therefore \] \[\frac{dy}{dt}=\frac{dy/dt}{dx/dt}\] \[=\frac{3a{{\sin }^{2}}t\cdot \cos t}{3a{{\cos }^{2}}t(-\sin t)}=-\frac{\sin t}{\cos t}\] \[\Rightarrow \] \[\frac{dy}{dx}=-\tan t\] On again differentiating w.r.t. x, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-{{\sec }^{2}}t\times \frac{dt}{dx}\] \[=-{{\sec }^{2}}t\times \frac{1}{3a{{\cos }^{2}}t(-\sin t)}\] At \[t=\frac{\pi }{4}\], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{\sec }^{2}}\frac{\pi }{4}}{3a{{\cos }^{2}}\left( \sin \frac{\pi }{4} \right)}=\frac{2}{3a\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right)}\] \[=\frac{4\sqrt{2}}{3a}\]
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