Question

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

• A. $2$
• B. $\frac{1}{12a}$
• C. $\frac{4\sqrt{2}}{3a}$
• D. $\frac{3a}{4\sqrt{2}}$

Answer 1

Answer: (C)

$\frac{4\sqrt{2}}{3a}$

Let $y=a{\sin }^{3}t,x=a{\cos }^{3}t$
On differentiating with respect to $t$, we get
$\frac{dy}{dx}=3a{\sin }^{2}t\cos t,\frac{dx}{dt}=-3a{\cos }^{2}t\sin t$
$\therefore \frac{dy}{dx}=\frac{3a{\sin }^{2}t\cos t}{-3a{\cos }^{2}t\sin t}$
$\Rightarrow \frac{dy}{dx}=-\frac{\sin t}{\cos t}=-\tan t$
Again, differentiating with respect to $x$, we get
$\frac{d^{2}y}{d{x}^2}=-{\sec }^{2}t\cdot \frac{dt}{dx}$
$=\frac{-{\sec }^{2}t}{-3a{\cos }^{2}t\sin t}=\frac{1}{3a{\cos }^{4}t\sin t}$
$\Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\pi }{4}}=\frac{1}{3a{\left(\frac{1}{\sqrt{2}}\right)}^4\cdot \left(\frac{1}{\sqrt{2}}\right)}$
$=\frac{{\left(\sqrt{2}\right)}^5}{3a}=\frac{4\sqrt{2}}{3a}$