Question

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

A

$2$

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B

$\frac{1}{12a}$

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C

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

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D

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

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Solution

## The correct option is C $\frac{4\sqrt{2}}{3a}$Explanation for the correct optionStep 1: Solve for the first order derivativeGiven functions are $a{\mathrm{sin}}^{3}\left(t\right)$ and $a{\mathrm{cos}}^{3}\left(t\right)$Let, $u=a{\mathrm{sin}}^{3}\left(t\right)$ and $v=a{\mathrm{cos}}^{3}\left(t\right)$We know that,$\frac{d\left[u\left(t\right)\right]}{d\left[v\left(t\right)\right]}=\frac{\frac{du}{dt}}{\frac{dv}{dt}}$We have,$\frac{du}{dt}=3a{\mathrm{sin}}^{2}\left(t\right)\mathrm{cos}\left(t\right)$ $\left[\because \frac{d}{\mathrm{dx}}f\left[g\left(x\right)\right]=f\text{'}\left[g\left(x\right)\right]·g\text{'}\left(x\right)\right]$And,$\frac{dv}{dt}=-3a{\mathrm{cos}}^{2}\left(t\right)\mathrm{sin}\left(t\right)$Thus,$\begin{array}{rcl}\frac{du}{dv}& =& \frac{\overline{)3a}{\mathrm{sin}}^{2}\left(t\right)\mathrm{cos}\left(t\right)}{-\overline{)3a}{\mathrm{cos}}^{2}\left(t\right)\mathrm{sin}\left(t\right)}\\ & =& -\mathrm{tan}\left(t\right)\end{array}$Step 2: Solve for the second order derivativeWe know that,$\frac{{d}^{2}\left[u\left(t\right)\right]}{d{\left[v\left(t\right)\right]}^{2}}=\frac{\frac{d}{dt}\left(\frac{du}{dv}\right)}{\frac{dv}{dt}}$So,$\begin{array}{rcl}\frac{{d}^{2}\left[u\left(t\right)\right]}{d{\left[v\left(t\right)\right]}^{2}}& =& \frac{\frac{d}{dt}\left[-\mathrm{tan}\left(t\right)\right]}{\frac{dv}{dt}}\\ & =& \frac{-{\mathrm{sec}}^{2}\left(t\right)}{3a{\mathrm{cos}}^{2}\left(t\right)\mathrm{sin}\left(t\right)}\\ & =& -\frac{1}{{\mathrm{cos}}^{2}\left(t\right)}·\frac{1}{3a{\mathrm{cos}}^{2}\left(t\right)\mathrm{sin}\left(t\right)}\\ & =& -\frac{1}{3a{\mathrm{cos}}^{4}\left(t\right)\mathrm{sin}\left(t\right)}\end{array}$Step 3: Solve for the required valueAt $t=\frac{\mathrm{\pi }}{4}$,$\begin{array}{rcl}-\frac{1}{3a{\mathrm{cos}}^{4}\left(t\right)\mathrm{sin}\left(t\right)}& =& -\frac{1}{3a{\mathrm{cos}}^{4}\left(\frac{\mathrm{\pi }}{4}\right)\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}\right)}\\ & =& \frac{1}{3a{\left(\frac{1}{\sqrt{2}}\right)}^{4}\left(\frac{1}{\sqrt{2}}\right)}\\ & =& \frac{4\sqrt{2}}{3a}\end{array}$Therefore, second-order derivative of $a{\mathrm{sin}}^{3}\left(t\right)$ with respect to $a{\mathrm{cos}}^{3}\left(t\right)$ at $t=\frac{\mathrm{\pi }}{4}$ is $\frac{4\sqrt{2}}{3a}$.Hence, option(C) is correct.

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