Question

If $x=a(1+\cos \theta ),y=a(\theta +\sin \theta )$ , then $\frac{d^{2}y}{d{x}^2}at\theta =\pi /2$ is equal to:

• A. $-\frac{1}{a}$
• B. $\frac{1}{a}$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A

$-\frac{1}{a}$

Explanation for the correct option.

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

$\begin{array}{rcl}x& =& a(1+\cos \theta )\\ & =& a+a\cos \theta \\ & \Rightarrow & \frac{dx}{d\theta }=0-a\sin \theta \\ & =& -a\sin \theta \end{array}$

$\begin{array}{rcl}y& =& a(\theta +\sin \theta )\\ & =& a\theta +a\sin \theta \\ & \Rightarrow & \frac{dy}{d\theta }=a\left(1\right)+a\cos \theta \\ & =& a\left(1+\cos \theta \right)\end{array}$

Step 2: Find first derivative.

$\begin{array}{rcl}& \Rightarrow & \frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}\\ & =& \frac{\left(1+\cos \theta \right)}{-\sin \theta }\end{array}$

Step 3: Find second derivative.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{d\theta }\left(\frac{dy}{dx}\right)\times \frac{d\theta }{dx}\\ & =& \frac{d}{d\theta }\left(-\frac{1+\cos \theta }{\sin \theta }\right)\times \frac{1}{-a\sin \theta }\\ & =& -\frac{0-\sin \theta \left(\sin \theta \right)-\left(1+\cos \theta \right)\cos \theta }{{\sin }^2\theta }\times \frac{1}{-a\sin \theta }\left[\mathrm{by}\mathrm{quotient}\mathrm{rule}\right]\\ & =& -\left(\frac{-{\sin }^2\theta -\cos \theta -{\cos }^2\theta }{{\sin }^2\theta }\right)\times \frac{1}{-a\sin \theta }\\ & =& \left(\frac{{\sin }^2\theta +{\cos }^2\theta +\cos \theta }{{\sin }^2\theta }\right)\times \frac{1}{-a\sin \theta }\\ & =& \left(\frac{1+\cos \theta }{{\sin }^2\theta }\right)\times \frac{1}{-a\sin \theta }\\ & =& -\frac{1+\cos \theta }{a{\sin }^3\theta }\end{array}$

Step 4: Find $\frac{d^{2}y}{d{x}^2}at\theta =\pi /2$

$\begin{array}{rcl}{\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\frac{\mathrm{\pi }}{2}}& =& -\frac{1+\cos {\displaystyle \frac{\mathrm{\pi }}{2}}}{a{\sin }^3{\displaystyle \frac{\mathrm{\pi }}{2}}}\\ & =& -\frac{1+0}{a\times \left(1\right)}\\ & =& -\frac{1}{a}\end{array}$

Hence, option A is correct.