Question

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

• A. $-\frac{3b}{a^3}cose{c}^4\theta {\cot }^4\theta$
• B. $\frac{3b}{a^3}cose{c}^4\theta \cot \theta$
• C. $-\frac{3b}{a^3}cose{c}^4\theta \cot \theta$
• D. none of these

Answer 1

Answer: (C)

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

$\because x=a\cos \theta \Rightarrow \frac{dx}{d\theta }=-a\sin \theta ⟶\left(1\right)$

and, $y=b\cos \theta \Rightarrow \frac{dy}{d\theta }=b\cos \theta ⟶\left(2\right)$

also, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{b\cos \theta }{-a\sin \theta }=-\frac{b}{a}\cot \theta$

Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(-\frac{b}{a}\cot \theta )=\frac{-b}{a}(-{cosec}^2\theta )\frac{d\theta }{dx}=\frac{b}{a}{cosec}^2\theta \left(\frac{-1}{a\sin \theta }\right)\left(From\right(1))=-\frac{b}{a^2}{cosec}^3\theta (\sin x=\frac{1}{cosecx})$

again, $\frac{d^{3}y}{d{x}^3}=\frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=\frac{d}{dx}(-\frac{b}{a^2}{cosec}^3\theta )=\frac{-3b}{a^3}{cosec}^4\theta \cot \theta$