Question

If $x=a{\cos }^3\theta$, $y=b{\sin }^3\theta$, then $\frac{d^{3}y}{d{x}^3}$ at $\theta =0$

• A. Does not exist at $\theta =0$
• B. $\frac{a}{b}$ at $\theta =0$
• C. $\frac{-a}{b}$ at $\theta =0$
• D. None of these

Answer 1

The correct option is A Does not exist at $\theta =0$

$x=a{\cos }^3\theta$, $y=b{\sin }^3\theta$
Differentiating w.r. to $\theta$
$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta$ and $\frac{dy}{d\theta }=3b{\sin }^2\theta \cos \theta$
$\therefore y_1=\frac{dy}{dx}=\frac{3b{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }$
$=-\frac{b}{a}\tan \theta$, if $\sin \theta \ne 0$, $\cos \theta \ne 0$
Therefore, $y_1$ does not exist at $\theta =0$
Hence, $y_2$ and $y_3$ do not exist at $\theta =0$