Question

If $x=3\cos \theta -2{\cos }^3\theta$ and $y=3\sin \theta -2{\sin }^3\theta$. Then ${\left(\frac{d^{2}x}{d{y}^2}\right)}_{\theta =\pi }$ will be

• A. -3
• B. $-\frac{1}{3}$
• C. 3
• D. does not exist

Answer 1

Answer: (B)

$\frac{dx}{d\theta }=3\sin \theta \cos 2\theta ;\frac{dy}{d\theta }=3\cos \theta \cdot \cos 2\theta$
$\frac{dx}{dy}=\tan \theta$
$\Rightarrow \frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{d}{dy}\cdot (\tan \theta )\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{d}{d\theta }(\tan \theta )\cdot \frac{d\theta }{dy}$
$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{{\sec }^2\theta }{3\cos \theta \cdot \cos 2\theta }=\frac{1}{3\cdot {\cos }^3\theta \cdot \cos 2\theta }$
${\left(\frac{d^{2}x}{d{y}^2}\right)}_{\theta =\pi }=\frac{1}{3\cdot (-1)\cdot 1}=-\frac{1}{3}$