Find the equation


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 $(1,1)$


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 $)\left(x-x_{1}\right)$



equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$



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