Question

If $x=a(\cos t+t\sin t)andy=a(\sin t-t\cos t)$, then find the value of $\frac{d^{2}y}{d{x}^2}$ at t = $\frac{\pi }{4}$ is ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\pi }{4}}=\frac{\sqrt{k}}{a\pi }$. Find k

Answer 1

$x=a(\cos t+t\sin t)$ and $y=a(\sin t-t\cos t)$
$\frac{dx}{dt}=a(-\sin t+t\cos t+\sin t)$
$\Rightarrow \frac{dx}{dt}=at\cos t$
$y=a(\sin t-t\cos t)$
$\frac{dy}{dt}=a(\cos t+t\sin t-\cos t)$
$\frac{dy}{dt}=at\sin t\frac{d^{2}y}{d{t}^2}=a(t\cos t+\sin t)$
$\Rightarrow \frac{dy}{dx}=\tan t$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\times \frac{dt}{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\tan t\right)\times \frac{1}{at\cos t}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{{\sec }^{3}t}{at}$
$\Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\pi }{4}}=\frac{\sqrt{2}}{a\pi }$