Practicing Success
If $x=acos^3\theta , y = a sin^3\theta .$ The value of $\frac{d^2y}{dx^2}$ at $\theta =\frac{\pi}{6}$ is : |
$\frac{32}{27}a$ $\frac{32}{27a}$ -4 $-\frac{1}{\sqrt{3}}$ |
$\frac{32}{27a}$ |
The correct answer is Option (2) → $\frac{32}{27a}$ $x=a\cos^3\theta , y = a \sin^3\theta$ $\frac{dx}{dθ}=-3a\cos^2θ\sin θ$ $\frac{dy}{dx}=3a\sin^2θ\cos θ$ $⇒\frac{dy}{dx}=-\frac{\sin θ}{\cos θ}$ so $\frac{dy}{dx}=-\tan θ$ so $\frac{d^2y}{dx^2}=-\sec^2 θ\frac{dθ}{dx}$ $=\frac{-\sec^2θ}{-3a\cos^2θ\sin θ}$ at $θ=\frac{π}{6}$ is $\frac{2^5}{3a×{\sqrt{3}}^4}$ $=\frac{32}{27a}$ |