Question

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

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

Answer 1

Answer: (A)

Given that, $x=a(1+\cos \theta )$ ...(i) and $y=a(\theta +\sin \theta )$ ...(ii) Differentiating Eqs. (i) and (ii) w.r.t.
$\theta ,$ we get $\frac{dx}{d\theta }=-a\sin \theta$ ...(iii) and
$\frac{dy}{d\theta }=a+a\cos \theta$ ...(iv)
Dividing Eq. (iii) by Eq. (iv), we get $\frac{dy}{d\theta }/\frac{dx}{d\theta }=\frac{a(1+\cos \theta )}{-a\sin \theta }$
$\Rightarrow$ $\frac{dy}{dx}=\frac{1+\cos \theta }{-\sin \theta }$
$=-\frac{1+2{\cos }^2\frac{\theta }{2}-1}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=-\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}$
$=-\cot \frac{\theta }{2}$
$\therefore$ $\frac{d^{2}y}{d{x}^2}=\cos e{c}^2\frac{\theta }{2}.\frac{1}{2}\frac{d\theta }{dx}$
$=-\frac{1}{2}\frac{1}{a}\frac{\cos e{c}^2\frac{\theta }{2}}{\sin \theta }$
$\therefore$ ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\frac{\pi }{2}}=-\frac{1}{a}$