Question

Consider the curve $x=a(\cos \theta +\theta \sin \theta )$ and $y=a(\sin \theta -\theta \cos \theta )$.

What is $\frac{d^{2}y}{d{x}^2}$ equal to?

• A. ${\sec }^2\theta$
• B. $-cose{c}^2\theta$
• C. $\frac{{\sec }^3\theta }{a\theta }$
• D. None of the above

Answer 1

Answer: (C)

$\frac{{\sec }^3\theta }{a\theta }$

The given equation is:

$x=a(\cos \theta +\theta \sin \theta )$

$y=a(\sin \theta -\theta \cos \theta )$

Differentiating x and y w.r.t. to $\theta$ once we get,

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

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

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

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

Dividing the two equations we get,

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

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

Now,

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

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left(\tan \theta \right).\frac{1}{a\theta \cos \theta }$

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

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