Question

If $y=a{\sin }^3\theta$ and $x=a{\cos }^3\theta$, then at $\theta =\frac{\mathrm{\pi }}{3}$, $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{\sqrt{3}}$
• B. $-\sqrt{3}$
• C. $-\frac{1}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

The correct option is B

$-\sqrt{3}$

Explanation for the correct option.

Step 1: Find the value of $\frac{dy}{d\theta }$ and $\frac{dx}{d\theta }$.

Differentiate $y=a{\sin }^3\theta$ with respect to $\theta$.

$\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta \left[\because \frac{d}{\mathrm{dx}}\left(\mathrm{sinx}\right)=\mathrm{cosx}\right]$

Differentiate $x=a{\cos }^3\theta$ with respect to $\theta$.

$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta \mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\left(\mathrm{cosx}\right)\mathbf{=}\mathbf{-}\mathbf{sinx}\mathbf{]}$

Step 2. Find the value of $\frac{dy}{dx}$ at $\theta =\frac{\mathrm{\pi }}{3}$.

The term $\frac{dy}{dx}$is given as: $\frac{dy}{dx}=\frac{dy}{d\theta }·\frac{d\theta }{dx}$. So substitute the values:

$\begin{array}{rcl}\frac{dy}{dx}& =& 3a{\sin }^2\theta \cos \theta ·\frac{1}{-3a{\cos }^2\theta \sin \theta }\\ & =& -\frac{\sin \theta }{\cos \theta }\\ & =& -\tan \theta \end{array}$

So, at $\theta =\frac{\mathrm{\pi }}{3}$ its value is

$\begin{array}{rcl}{\overline{)\frac{dy}{dx}}}_{x=\frac{\mathrm{\pi }}{3}}& =& -\tan \frac{\mathrm{\pi }}{3}\\ & =& -\sqrt{3}\end{array}$

Hence, the correct option is B.