Question

$x=t\cos t$, $y=t+\sin t$ then $\frac{d^{2}x}{d{y}^2}$ at $t=\frac{\pi }{2}$ is equal to

• A. $\frac{\pi +4}{2}$
• B. $-\frac{\pi +4}{2}$
• C. $2$
• D. none of these

Answer 1

Answer: (B)

$-\frac{\pi +4}{2}$

$x=t\cos t$
$\frac{dx}{dt}=-t\sin t+\cos t$
Also given, $y=t+\sin t$
$\frac{dy}{dt}=1+\cos t$
$\Rightarrow \frac{dx}{dy}=\frac{-t\sin t+\cos t}{1+\cos t}$
$\Rightarrow \left(\frac{d^{2}x}{d{y}^2}\right)=\frac{d}{dt}\left(\frac{dx}{dy}\right)\times \frac{dt}{dy}$
$\left(\frac{d^{2}x}{d{y}^2}\right)=\left[\frac{d}{dt}\right(\frac{-t\sin t+\cos t}{1+\cos t}\left)\right]\frac{1}{1+\cos t}$
$\left(\frac{d^{2}x}{d{y}^2}\right)=\left[\frac{-2\sin t-t\cos t-\sin t\cos t-t}{(1+\cos t{)}^2}\right]\frac{1}{1+\cos t}$
Now, ${\left(\frac{d^{2}x}{d{y}^2}\right)}_{t=\frac{\pi }{2}}=-\frac{\pi }{2}-2=-\frac{(\pi +4)}{2}$