Question

If $x = a Cos ^ 3 \theta$ and $y= a Sin^3 \theta$ then $\frac{dy}{dx}$=_________

• A. $\sqrt [-3]{\frac{x}{y}}$
• B. $\sqrt [-3]{\frac{y}{x}}$
• C. $\sqrt [3]{\frac{y}{x}}$
• D. $\sqrt [3]{\frac{x}{y}}$

Answer 1

Answer: (B)

If $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$
$\Rightarrow \frac{d x}{d \theta}=3 a \cos ^{2} \theta(-\sin \theta)$
$\Rightarrow \frac{d y}{d \theta} =3 a \sin ^{2} \theta(\cos \theta)$
$\Rightarrow \frac{d y}{d x} =\frac{d y}{d \theta} \times \frac{d \theta}{d x}$
$=\left(3 a \sin ^{2} \theta \cdot \cos \theta\right) \cdot \frac{1}{-\left(3 a \sin \theta \cdot \cos ^{2} \theta\right)}$
$\frac{d y}{d x}=-\frac{\sin \theta}{\cos \theta}, \frac{d y}{d x}=-\tan \theta$
$\Rightarrow \frac{d y}{d x}=-\frac{(y / a)^{-1 / 3}}{(x / a)^{1 / 3}}$
$=-\left[\frac{y}{a} \times \frac{a}{x}\right]^{1 / 3}$
$\frac{d y}{d x} =-\left(\frac{y}{x}\right)^{-1 / 3}=-\sqrt[3]{\frac{y}{x}}$