Question

If $x=\cos \theta +\theta \sin \theta$, $y=\sin \theta -\theta \cos \theta$, then $y_2$ is

• A. $\frac{{\sec }^3\theta }{\theta }$
• B. $\frac{\theta }{{\sec }^3\theta }$
• C. $\frac{\theta }{{\cos }^3\theta }$
• D. ${\sec }^2\theta$

Answer 1

The correct option is A $\frac{{\sec }^3\theta }{\theta }$

The given information in the question is:

$x=\cos \theta +\theta \sin \theta$ and $y=\sin \theta -\theta \cos \theta$

Taking their first derivatives with respect to $\theta$ we get,

$\Rightarrow \frac{dx}{d\theta }=-\sin \theta +\sin \theta +\theta \cos \theta$

$\Rightarrow \frac{dx}{d\theta }=\theta \cos \theta$ and

$\Rightarrow \frac{dy}{d\theta }=\cos \theta -\cos \theta +\theta \sin \theta$

$\Rightarrow \frac{dy}{d\theta }=\theta \sin \theta$

so $\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{\theta \sin \theta }{\theta \cos \theta }$

$\Rightarrow \frac{dy}{dx}=\tan \theta$

Now $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\tan \theta \right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left(\tan \theta \right)\times \frac{d\theta }{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}={\sec }^2\theta \times \frac{1}{\theta \cos \theta }$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{{\sec }^3\theta }{\theta }$