Question:
Find the equation of the tangent and the normal to the following curves at the indicated points:
$x^{2 / 3}+y^{2 / 3}=2$ at $(1,1)$
Solution:
finding the slope of the tangent by differentiating the curve
$\frac{2}{3 x^{1 / 3}}+\frac{2}{3 y^{1 / 3}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\frac{y^{1 / 3}}{x^{1 / 3}}$
$\mathrm{m}($ tangent $)$ at $(1,1)=-1$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}$ (normal) at $(1,1)=1$
equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$
$y-1=-1(x-1)$
$x+y=2$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-1=1(x-1)$
$y=x$