Question

If $x=a(1+\cos \theta ),y=a(\theta +\sin \theta )$, then prove that $\frac{d^{2}y}{d{x}^2}=1$ at $\theta =\frac{\pi }{4}$.

Answer 1

we have

$x=a(1+\cos \theta )$.....(1)

$y=a(\theta +\sin \theta )$....(2)

differentiate both equation with respect to $x$, we get

$\frac{dx}{d\theta }=a(0-\sin \theta )$

$=-a\sin \theta$.....(3)

$\frac{dy}{d\theta }=a[1+\cos \theta ]$.....(4)

Differentiate both equation with respect to $\theta$, we get

$\frac{d^{2}x}{d{\theta }^2}=-a\cos \theta$ $\frac{d^{2}y}{d{\theta }^2}=a[0-\sin \theta ]$

$=-a\sin \theta$.....(6)

on dividing eqn (6) by eqn (5), we get

$\frac{\frac{d^{2}y}{d{\theta }^2}}{\frac{d^{2}x}{d{\theta }^2}}=\frac{-a\sin \theta }{-a\cos \theta }$

$\frac{d^{2}y}{d{x}^2}=\tan \theta$

At $\frac{\pi }{4}$

$\frac{d^{2}y}{d{x}^2}=\tan \frac{\pi }{4}=1$