Question

If x = a (1 + cos θ), y = a(θ + sin θ), prove that $$\frac{d^{2}y}{d{x}^2}=\frac{-1}{a}at\theta =\frac{\pi }{2}$$

Answer 1

Here, 

$$x=a(1+\cos \theta )\text{ and }y=a(\theta +\sin \theta )$$

$$\text{ Differentiating w . r . t .}\theta ,\text{ we get }$$

$$\frac{dx}{d\theta }=-a\sin \theta \text{ and }\frac{dy}{d\theta }=a+a\cos \theta$$

$$\therefore \frac{dy}{dx}=\frac{a+a\cos \theta }{-a\sin \theta }=\frac{1+\cos \theta }{-\sin \theta }$$

$$\text{ Differentiating w . r . t . x, we get }$$

$$\frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left\{\frac{dy}{dx}\right\}\frac{d\theta }{dx}$$

$$\frac{d^{2}y}{d{x}^2}=-\left\{\frac{-{\sin }^2\theta -\cos \theta -{\cos }^2\theta }{{\sin }^2\theta }\right\}\frac{d\theta }{dx}$$

$$=\frac{1+\cos \theta }{{\sin }^2\theta }\times \frac{-1}{a\sin \theta }$$

$$=\frac{-(1+\cos \theta )}{a{\sin }^3\theta }$$

$$\text{ At }\theta =\frac{\pi }{2}:\frac{d^{2}y}{d{x}^2}=\frac{-(1+\cos \frac{\pi }{2})}{a{(\sin \frac{\pi }{2})}^3}=\frac{-1}{a}$$