Question

$$\text{ If x }=\cos t+\log \tan \frac{t}{2},y=\sin t,\text{ then find the value of }\frac{d^{2}y}{d{t}^2}\text{ and }\frac{d^{2}y}{d{x}^2}\text{ at }t=\frac{\pi }{4}$$ ?

Answer 1

$$\text{ We have},$$

$$x=\cos t+\log \tan \frac{t}{2}\text{ and y }=\sin t$$

$$\text{ On differentiating with respect to t, we get }$$

$$\frac{dx}{dt}=\frac{d}{dt}(\cos t+\log \tan \frac{t}{2})=-\sin t+\frac{1}{\tan \frac{t}{2}}\times {\sec }^2\frac{t}{2}\times \frac{1}{2}$$

$$=-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}}=-\sin t+\frac{1}{\sin t}$$

$$=\frac{-{\sin }^{2}t+1}{\sin t}=\frac{-{\sin }^{2}t+1}{\sin t}$$

$$=\frac{{\cos }^{2}t}{\sin t}$$

$$\text{ and }$$

$$\frac{dy}{dt}=\frac{d}{dt}(\sin t)=\cos t$$

$$\text{ Now},\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left(\frac{dy}{dt}\right)=\frac{d}{dt}(\cos t)=-\sin t$$

$${\left(\frac{d^{2}y}{d{t}^2}\right)}_{t=\frac{\pi }{4}}=-\sin \left(\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}...(1)$$

$$\text{ Also },\left(\frac{dy}{dx}\right)=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos t}{\frac{{\cos }^{2}t}{\sin t}}=\frac{\sin t}{\cos t}=\tan t$$

$$\text{ Now,}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\tan t)$$

$$=\frac{d}{dt}(\tan t)\times \frac{dt}{dx}={\sec }^{2}t\times \frac{\sin t}{{\cos }^{2}t}$$

$$=\frac{\sin t}{{\cos }^{4}t}$$

$${\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\pi }{4}}=\frac{\sin \left(\frac{\pi }{4}\right)}{{\cos }^4\left(\frac{\pi }{4}\right)}=2\sqrt{2}...(2)$$

$$\text{ Hence, at }t=\frac{\pi }{4},\frac{d^{2}y}{d{t}^2}=-\frac{1}{\sqrt{2}}\text{ and }\frac{d^{2}y}{d{x}^2}=2\sqrt{2}.$$