Question Number 433 by 123456 last updated on 25/Jan/15
$$\mathrm{find}\:\mathrm{tagent}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{surface} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{3}} ={z}^{\mathrm{4}} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{28},\mathrm{8},\mathrm{6}\right) \\ $$
Answered by prakash jain last updated on 03/Jan/15
$${z}=\left({x}^{\mathrm{2}} +{y}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\frac{\partial{z}}{\partial{x}}=\frac{\mathrm{1}}{\mathrm{4}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{3}} \right)^{\frac{−\mathrm{3}}{\mathrm{4}}} \centerdot\mathrm{2}{x} \\ $$$$\mathrm{At}\:\left(\mathrm{28},\mathrm{8}\right)=\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }\centerdot\mathrm{2}\centerdot\mathrm{28}=\frac{\mathrm{14}}{\mathrm{216}}=\frac{\mathrm{7}}{\mathrm{108}} \\ $$$$\frac{\partial{z}}{\partial{y}}=\frac{\mathrm{1}}{\mathrm{4}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{3}} \right)^{\frac{−\mathrm{3}}{\mathrm{4}}} \centerdot\mathrm{3}{y} \\ $$$$\mathrm{At}\:\left(\mathrm{28},\mathrm{8}\right)=\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }\centerdot\mathrm{3}\centerdot\mathrm{8}=\frac{\mathrm{1}}{\mathrm{36}} \\ $$$$\mathrm{Equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plane} \\ $$$${z}−{z}_{\mathrm{0}} ={f}_{{x}} \left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} \right)\left({x}−{x}_{\mathrm{0}} \right)+{f}_{{y}} \left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} \right)\left({y}−{y}_{\mathrm{0}} \right) \\ $$$${z}=\mathrm{6}+\frac{\mathrm{7}}{\mathrm{108}}\left({x}−\mathrm{28}\right)+\frac{\mathrm{1}}{\mathrm{36}}\left({y}−\mathrm{8}\right) \\ $$