Question

If $x^{2/3}+y^{2/3}=a^{2/3}$, then $\frac{dy}{dx}$ is equal to

• A. $-\sqrt[3]{\frac{y}{x}}$
• B. $\sqrt[3]{y/x}$
• C. $\frac{y}{x}$
• D. None of these

Answer 1

Answer: (A)

Let $x=a{\cos }^3\theta ,y=a{\sin }^3\theta$. Then,
$x^{\frac{2}{3}}+y^{\frac{2}{3}}={\left(a{\cos }^3\theta \right)}^{\frac{2}{3}}+{\left({\mathrm{asin}}^3\theta \right)}^{\frac{2}{3}}$
$=a^{\frac{2}{3}}[\left({\cos }^2\theta +\left({\sin }^2\theta \right)\right]=a^{\frac{2}{3}}$
Hence, $x={\mathrm{acos}}^3\theta ,y={\mathrm{cosin}}^3\theta$ is parametric equation of $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Now, $\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta$ and $\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$
$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta =-\sqrt[3]{\frac{y}{x}}$