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, $\frac{d^{2}u}{d{v}^2}=-{\sec }^{2}t\frac{dt}{dv}$ $=-\frac{{\sec }^{2}t}{-3a{\cos }^{2}t\sin t}$ ${\left(\frac{d^{2}u}{d{v}^2}\right)}_{x=\frac{\pi }{4}}=\frac{{(\sqrt{2})}^2}{3a{\left(\frac{1}{\sqrt{2}}\right)}^2\left(\frac{1}{\sqrt{2}}\right)}=\frac{4\sqrt{2}}{3a}$