Question

$x=t\cos t$, $y=t+\sin t$. Then $\frac{d^{2}x}{d{y}^2}$ at $t=\frac{\pi }{2}$ is

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

Answer 1

Answer: (B)

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

We have
$x=t\cos t$, $y=t+\sin t$
By differentiating w.r. to $t$, we get
$\frac{dx}{dt}=-t\sin t+\cos t$ and $\frac{dy}{dt}=1+\cos t$
$\therefore \frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{\cos t-t\sin t}{1+\cos t}$
$\therefore \frac{d^{2}x}{d{y}^2}=\frac{\frac{d}{dt}\left(\frac{dx}{dy}\right)}{\frac{dy}{dt}}$
$=\frac{\frac{(-2\sin t-t\cos t)(1+\cos t)-(\cos t-t\sin t)(-\sin t)}{{(1+\cos t)}^2}}{1+\cos t}$
Now, put $t=\frac{\pi }{2}$.
${\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{2}}=\frac{(-2\sin \frac{\pi }{2}-\frac{\pi }{2}\cos \frac{\pi }{2})(1+\cos \frac{\pi }{2})-\left(\cos \frac{\pi }{2}-\frac{\pi }{2}\sin \frac{\pi }{2}\right)(-\sin \frac{\pi }{2})}{\frac{{(1+\cos \frac{\pi }{2})}^2}{1+\cos \frac{\pi }{2}}}$
$=-\left(\frac{\pi +4}{2}\right)$