Question Number 7040 by FilupSmith last updated on 07/Aug/16
$${f}\left({x},{y}\right)={ax}^{\mathrm{2}} +{by}^{\mathrm{2}} \\ $$$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{find}\:\mathrm{where}\:\mathrm{the}\:\mathrm{gradient} \\ $$$$\mathrm{is}\:\mathrm{zero}\:\mathrm{for}\:\mathrm{multivariable}\:\mathrm{funtions}? \\ $$
Commented by Yozzii last updated on 07/Aug/16
$$\bigtriangledown{f}={grad}\:{f}=\begin{pmatrix}{\frac{\partial{f}}{\partial{x}}}\\{\frac{\partial{f}}{\partial{y}}}\end{pmatrix} \\ $$$${At}\:{stationary}\:{points},\:\bigtriangledown{f}=\underset{−} {\mathrm{0}} \\ $$$$\therefore\:\frac{\partial{f}}{\partial{x}}=\mathrm{0}\:\:{and}\:\frac{\partial{f}}{\partial{y}}=\mathrm{0}\:{simultaneously}\:{at}\:{some}\:\left({x},{y}\right) \\ $$$${For}\:{f}\left({x},{y}\right)={ax}^{\mathrm{2}} +{by}^{\mathrm{2}} ,\:\bigtriangledown{f}=\underset{−} {\mathrm{0}}\:{at}\:{x}={y}=\mathrm{0}. \\ $$$${There}\:{is}\:{also}\:{the}\:{directional}\:{derivative}\:{of}\:{f} \\ $$$$\hat {\boldsymbol{{u}}}\bullet\left(\bigtriangledown{f}\right)\:{at}\:{a}\:{point}\:\left({x},{y}\right)\:{in}\:{the}\:{direction} \\ $$$${of}\:{the}\:{unit}\:{vector}\:\hat {\boldsymbol{{u}}}\:{in}\:{the}\:{x}−{y}\:{plane}. \\ $$$$\hat {\boldsymbol{{u}}}\bullet\begin{pmatrix}{\partial{f}/\partial{x}}\\{\partial{f}/\partial{y}}\end{pmatrix}\:\:{gives}\:{the}\:{gradienti}\:{of} \\ $$$${the}\:{surface}\:{of}\:{f}\:{in}\:{the}\:{direction}\:{of}\:\hat {\boldsymbol{{u}}}. \\ $$
Commented by FilupSmith last updated on 07/Aug/16
$$\mathrm{This}\:\mathrm{is}\:\mathrm{extremely}\:\mathrm{interesting}! \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{this}\:\mathrm{area}\:\mathrm{of}\:\mathrm{mathematics}! \\ $$