Question

If $x=a\sin \theta$ and $y=b\cos \theta$, then $d^{2}y/d{x}^2$ is:

• A. $(a/b^2)se{c}^2\theta$
• B. $-(b/a)se{c}^2\theta$
• C. $-(b/a^2)se{c}^3\theta$
• D. $(b^2/a^2)se{c}^3\theta$

Answer 1

The correct option is C

$-(b/a^2)se{c}^3\theta$

Explanation for the correct option.

Step 1: Differentiate $x\mathrm{and}y$ w.r.t $\theta$

$\begin{array}{rcl}x& =& a\sin \theta \\ & \Rightarrow & \frac{dx}{d\theta }=a\cos \theta \end{array}$

$\begin{array}{rcl}y& =& b\cos \theta \\ & \Rightarrow & \frac{dy}{d\theta }=-b\sin \theta \end{array}$

Step 2: Find$\frac{dy}{dx}\mathrm{and}\frac{d^{2}y}{d{x}^2}$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{d\theta }\times \frac{d\theta }{dx}\\ & =& -\frac{b\sin \theta }{a\cos \theta }\\ & =& -\frac{b}{a}\left(\tan \theta \right)\end{array}$

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{d\theta }\left(\frac{dy}{dx}\right)\times \frac{d\theta }{dx}\\ & =& \frac{d}{d\theta }\left[-\frac{b}{a}\left(\tan \theta \right)\right]\left[\frac{1}{a}\left(\sec \theta \right)\right]\\ & =& \left[-\frac{b}{a}\left({\sec }^2\theta \right)\right]\left[\frac{1}{a}\left(\sec \theta \right)\right]\\ & =& -\frac{b}{a^2}\left({\sec }^3\theta \right)\end{array}$

Hence, option C is correct.