Question

The second-order derivative of $a{\sin }^3\left(t\right)$ with respect to $a{\cos }^3\left(t\right)$ at $t=\frac{\mathrm{\pi }}{4}$ is

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

Answer 1

The correct option is C

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

Explanation for the correct option

Step 1: Solve for the first order derivative

Given functions are $a{\sin }^3\left(t\right)$ and $a{\cos }^3\left(t\right)$

Let, $u=a{\sin }^3\left(t\right)$ and $v=a{\cos }^3\left(t\right)$

We know that,
$\frac{d\left[u\left(t\right)\right]}{d\left[v\left(t\right)\right]}=\frac{{\displaystyle \frac{du}{dt}}}{{\displaystyle \frac{dv}{dt}}}$

We have,
$\frac{du}{dt}=3a{\sin }^2\left(t\right)\cos \left(t\right)$ $\left[\because \frac{d}{\mathrm{dx}}f\left[g\left(x\right)\right]=f\text{'}\left[g\left(x\right)\right]·g\text{'}\left(x\right)\right]$
And,
$\frac{dv}{dt}=-3a{\cos }^2\left(t\right)\sin \left(t\right)$

Thus,
$\begin{array}{rcl}\frac{du}{dv}& =& \frac{\cancel{3a}{\sin }^2\left(t\right)\cos \left(t\right)}{-\cancel{3a}{\cos }^2\left(t\right)\sin \left(t\right)}\\ & =& -\tan \left(t\right)\end{array}$

Step 2: Solve for the second order derivative

We know that,
$\frac{d^2\left[u\left(t\right)\right]}{d{\left[v\left(t\right)\right]}^2}=\frac{{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{du}{dv}}\right)}{{\displaystyle \frac{dv}{dt}}}$

So,
$\begin{array}{rcl}\frac{d^2\left[u\left(t\right)\right]}{d{\left[v\left(t\right)\right]}^2}& =& \frac{{\displaystyle \frac{d}{dt}\left[-\tan \left(t\right)\right]}}{{\displaystyle \frac{dv}{dt}}}\\ & =& \frac{-{\sec }^2\left(t\right)}{3a{\cos }^2\left(t\right)\sin \left(t\right)}\\ & =& -\frac{1}{{\cos }^2\left(t\right)}·\frac{1}{3a{\cos }^2\left(t\right)\sin \left(t\right)}\\ & =& -\frac{1}{3a{\cos }^4\left(t\right)\sin \left(t\right)}\end{array}$

Step 3: Solve for the required value

At $t=\frac{\mathrm{\pi }}{4}$,
$\begin{array}{rcl}-\frac{1}{3a{\cos }^4\left(t\right)\sin \left(t\right)}& =& -\frac{1}{3a{\cos }^4\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}\\ & =& \frac{1}{3a{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^4\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}\\ & =& \frac{4\sqrt{2}}{3a}\end{array}$

Therefore, second-order derivative of $a{\sin }^3\left(t\right)$ with respect to $a{\cos }^3\left(t\right)$ at $t=\frac{\mathrm{\pi }}{4}$ is $\frac{4\sqrt{2}}{3a}$.

Hence, option(C) is correct.