Question

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

Answer 1

We have,

$x=a{\cos }^3\theta$ ……. (1)

$y=a{\sin }^3\theta$ ……… (2)

On differentiating to equation (1) w.r.t $\theta$, we get

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

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

On differentiating both sides w.r.t $\theta$, we have

$\frac{d^{2}x}{d{\theta }^2}=-3a({\cos }^2\theta \cos \theta +\sin \theta \left(2\cos \theta (-\sin \theta )\right))$

$\frac{d^{2}x}{d{\theta }^2}=-3a({\cos }^3\theta -2{\sin }^2\theta \cos \theta )$

Similarly,

On differentiating to equation (2) w.r.t $\theta$, we get

$\frac{dy}{d\theta }=a\left(3{\sin }^2\theta \right)\left(\cos \theta \right)$

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

On differentiating both sides w.r.t $\theta$, we have

$\frac{d^{2}y}{d{\theta }^2}=3a({\sin }^2\theta (-\sin \theta )+\cos \theta \left(2\sin \theta \left(\cos \theta \right)\right))$

$\frac{d^{2}y}{d{\theta }^2}=3a({-\sin }^3\theta +2{\cos }^2\theta \sin \theta )$

Therefore,

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

$\frac{d^{2}y}{d{x}^2}=\frac{(-{\sin }^3\theta +2\sin \theta {\cos }^2\theta )}{(-{\cos }^3\theta +2{\sin }^2\theta \cos \theta )}$

Put $\theta =\frac{\pi }{6}$,we get

${\frac{d^{2}y}{d{x}^2}|}_{\theta =\frac{\pi }{6}}=\frac{(-{\sin }^3\left(\frac{\pi }{6}\right)+2\sin \left(\frac{\pi }{6}\right){\cos }^2\left(\frac{\pi }{6}\right))}{(-{\cos }^3\left(\frac{\pi }{6}\right)+2{\sin }^2\left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{6}\right))}$

${\frac{d^{2}y}{d{x}^2}|}_{\theta =\frac{\pi }{6}}=\frac{(-{\left(\frac{1}{2}\right)}^3+2\times \frac{1}{2}\times {\left(\frac{\sqrt{3}}{2}\right)}^2)}{(-{\left(\frac{\sqrt{3}}{2}\right)}^3+2{\left(\frac{1}{2}\right)}^2\times \left(\frac{\sqrt{3}}{2}\right))}$

${\frac{d^{2}y}{d{x}^2}|}_{\theta =\frac{\pi }{6}}=\frac{(-\frac{1}{8}+\frac{3}{4})}{(-\frac{3\sqrt{3}}{8}+\frac{\sqrt{3}}{4})}$

${\frac{d^{2}y}{d{x}^2}|}_{\theta =\frac{\pi }{6}}=\frac{(-\frac{1+6}{8})}{\left(\frac{-3\sqrt{3}+2\sqrt{3}}{8}\right)}$

${\frac{d^{2}y}{d{x}^2}|}_{\theta =\frac{\pi }{6}}=\frac{7}{\sqrt{3}}$

Hence, the value is $\frac{7}{\sqrt{3}}$.