Question

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

• A. $- \frac{3b}{a^{2}} \, cosec^{4} \theta \, \cot^{2} \theta$
• B. $- \frac{3b}{a^{3}} \, cosec^{4} \theta \, \cot^2 \theta$
• C. $- \frac{3b}{a^{3}} \, cosec^{4} \theta \, \cot^{4} \theta$
• D. None of these

Answer 1

Answer: (C)

$- \frac{3b}{a^{3}} \, cosec^{4} \theta \, \cot^{4} \theta$

Answer 2

We have $y = b \sin \, \theta, x = a \, cos \, \theta$. Therefore $\frac{dy}{dx}=- \frac{b}{a} \cot \theta \Rightarrow \frac{d^{2}y}{dx^{2}} =\frac{b}{a} cosec^{2 }\theta \frac{d\theta}{dx}$ $= -\frac{b}{a^{2}} cosec^{3}$ $\Rightarrow \frac{d^{3}y}{dx^{3}} =- \frac{b}{a^{2}} 3 cosec^{2} \theta\left(-cosec \theta \cot \theta \right) \frac{d\theta}{dx}$ $= \frac{3b}{a^{2}} cosec^{3} \theta \cot \theta \times\frac{-1}{a \sin \theta}$ $= \frac{-3b}{a^{3}}cosec^{4} \theta \cot \theta$