Question

Find the slope normal to the curve $x^{2/3}+y^{2/3}=2$ at the point (1, 1).

Answer 1

On differentiating the equation wrt $x$, we have

$\frac{2}{3}{x}^{-\frac{1}{3}}+\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=0$

$\therefore \frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^{-\frac{1}{3}}$

This is the slope of the tangent at any point on given curve. The slope of normal $=-\frac{dx}{dy}={\left(\frac{x}{y}\right)}^{\frac{1}{3}}$

Hence, slope of normal at $(1,1)$ is $1$.