Question Number 34912 by abdo imad last updated on 12/May/18
$${let}\:{f}\left({x},{y},{z}\right)\:=\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\alpha} \:\:\:\:\:{with}\:\alpha\in{R} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\Delta{f} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\alpha\:{in}\:{order}\:{to}\:{have}\:\Delta{f}=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18
$$\bigtriangleup{f}={df}\left({x},{y}.{z}\right) \\ $$$$=\left(\frac{\partial\int}{\partial{x}}\right)_{{y},{z}} {dx}+\left(\frac{\partial{f}}{\partial{y}}\right)_{{x},{z}} {dy}+\left(\frac{\partial{f}}{\partial{z}}\right)_{{x},{y}} {dz} \\ $$$$=\alpha.\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\alpha−\mathrm{1}} .\mathrm{2}{x}.{dx}+ \\ $$$$\alpha.\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\alpha−\mathrm{1}} .\mathrm{2}{y}+\alpha.\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\alpha−\mathrm{1}} \mathrm{2}{z} \\ $$