Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x=a\left(\cos t+t\sin t\right)\mathrm{and}y=a\left(\sin t-t\cos t\right) \\ \mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \frac{dx}{dt}=\frac{d}{dt}\left[a\left(\cos t+t\sin t\right)\right]=-a\sin t+a\sin t+at\cos t \\ =at\cos t \\ \mathrm{and} \\ \frac{dy}{dt}=\frac{d}{dt}\left[a\left(\sin t-t\cos t\right)\right]=a\cos t-a\cos t+at\sin t \\ =at\sin t \\ \mathrm{Now},\left(\frac{dy}{dx}\right)=\frac{{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}}}{{\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}}}=\frac{at\sin t}{{\displaystyle at\cos t}}=\tan t \\ \frac{d^{2}y}{d{x}^2}=\frac{\mathrm{d}}{dx}\left(\frac{dy}{dx}\right)=\frac{\mathrm{d}}{dx}\left(\tan t\right) \\ =\frac{\mathrm{d}}{dt}\left(\tan t\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}={\sec }^{2}t\times \frac{1}{at\cos t} \\ =\frac{1}{at{\cos }^{3}t} \\ {\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\mathrm{\pi }}{4}}=\frac{1}{a\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right){\cos }^3\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}=\frac{8\sqrt{2}}{a\mathrm{\pi }} \\ \mathrm{Hence},\mathrm{at}t=\frac{\mathrm{\pi }}{4},\frac{d^{2}y}{d{x}^2}=\frac{8\sqrt{2}}{a\mathrm{\pi }}. \end{gathered}$$