Question

If $x=acos\theta ,y=bsin\theta$, then $\frac{d^{3}y}{d{x}^3}$ is equal to

• A. $\left(\frac{-3b}{a^3}\right)cose{c}^4\theta co{t}^4\theta$
• B. $\left(\frac{3b}{a^3}\right)cose{c}^4\theta cot\theta$
• C. $\left(\frac{-3b}{a^3}\right)cose{c}^4\theta cot\theta$
• D. $noneoftheabove$

Answer 1

The correct option is C $\left(\frac{-3b}{a^3}\right)cose{c}^4\theta cot\theta$

$\because x=acos\theta \Rightarrow \frac{dx}{d\theta }=-asin\theta$ and $y=bsin\theta \Rightarrow \frac{dy}{d\theta }=bcos\theta$
$\therefore \frac{dy}{dx}=-\frac{b}{a}cot\theta \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{b}{a}cose{c}^2\theta \frac{d\theta }{dx}=-\frac{b}{a^2}cose{c}^3\theta \therefore \frac{d^{3}y}{d{x}^3}=\frac{3b}{a^2}cose{c}^2\theta (-cosec\theta cot\theta )\frac{d\theta }{dx}=\frac{3b}{a^2}cose{c}^3\theta cot\theta (-\frac{1}{sin\theta })=-\frac{3b}{a^3}cose{c}^4\theta cot\theta$