Question

If $x=a\cos \theta ,y=b\sin \theta$, then $\frac{d^{3}y}{d{x}^3}$ is equal to:

• A. $\frac{-3b}{a^3}\cos e{c}^4\theta co{t}^4\theta$
• B. $\frac{-3b}{a^3}\cos e{c}^4\theta co{t}^3\theta$
• C. $\frac{-3b}{a^3}\cos e{c}^4\theta cot\theta$
• D. None of these

Answer 1

The correct option is C

$\frac{-3b}{a^3}\cos e{c}^4\theta cot\theta$

Explanation for the correct option

$$\begin{gathered} x=a\cos \theta \\ \Rightarrow \frac{dx}{d\theta }=-a\sin \theta \end{gathered}$$

$\begin{array}{rcl}y& =& b\sin \theta \\ & \Rightarrow & \frac{dy}{d\theta }=b\cos \theta \end{array}$

So,

$\begin{array}{rcl}\frac{dy}{dx}& =& -\frac{b\cos \theta }{a\sin \theta }\\ & =& -\frac{b}{a}cot\theta \end{array}$

Now,

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{d\theta }\left(\frac{dy}{dx}\right)\times \frac{d\theta }{dx}\\ & =& \frac{d}{d\theta }\left(-\frac{b}{a}cot\theta \right)\times \frac{1}{-a\sin \theta }\\ & =& -\frac{b}{a}\left(-\cos e{c}^2\theta \right)\times \frac{1}{-a\sin \theta }\\ & =& -\frac{b}{a^2}\cos e{c}^3\theta \end{array}$

So,

$\begin{array}{rcl}\frac{d^{3}y}{d{x}^3}& =& \frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)\\ & =& \frac{d}{dx}\left(-\frac{b}{a^2}\cos e{c}^3\theta \right)\\ & =& -\frac{3b}{a^3}\left(\cos e{c}^4\theta cot\theta \right)\end{array}$

Hence, option C is correct.