Question

If $x=a{\cos }^3\theta ;y={\sin }^3\theta$, then find the value of $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{6}$

Answer 1

$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta$

$\frac{dy}{d\theta }=3{\sin }^2\theta \cos \theta$

$\frac{dy}{dx}=\frac{3{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }$ = $\frac{-1}{a}\tan \theta$

Diffrentiate w.r.t x$\Rightarrow$

$\frac{d^{2}y}{d{x}^2}=\frac{-1}{a}{\sec }^2\theta \frac{d\theta }{dx}$

$=\frac{-1}{a}{\sec }^2\theta \left(\frac{-1}{3a{\cos }^2\theta \sin \theta }\right)$

$\frac{d^{2}y}{d{x}^2}=\frac{1}{3a^2{\cos }^4\theta \sin \theta }$

at $\theta =\frac{\pi }{6}=30$

$=\frac{1}{3a^2.\frac{3.3}{\mathrm{2.2.2.2}}\frac{1}{2}}$

$\frac{d^{2}y}{d{x}^2}=\frac{32}{27a^2}$