Question Number 204499 by DeArtist last updated on 19/Feb/24
$$\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{directional}\:\mathrm{derivative}\:\mathrm{of}\: \\ $$$${F}\left({x},{y},{z}\right)\:=\:\mathrm{2}{xy}−{z}^{\mathrm{2}} \:\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right)\:\mathrm{in} \\ $$$$\mathrm{a}\:\mathrm{direction}\:\mathrm{towards}\:\left(\mathrm{3},\mathrm{1},−\mathrm{1}\right)\: \\ $$$$\mathrm{in}\:\mathrm{what}\:\mathrm{direction}\:\mathrm{is}\:\mathrm{the}\:\mathrm{directional}\:\mathrm{derivative} \\ $$$$\mathrm{maximum}?\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this}\:\mathrm{maximum}? \\ $$
Answered by TonyCWX08 last updated on 19/Feb/24
$$ \\ $$$$\bigtriangledown{F}\: \\ $$$$=\:\left(\frac{\partial{F}}{\partial{x}},\frac{\partial{F}}{\partial{y}},\frac{\partial{F}}{\partial{z}}\right) \\ $$$$=\left(\mathrm{2}{y},\mathrm{2}{x},−\mathrm{2}{z}\right) \\ $$$$ \\ $$$$\bigtriangledown{F}\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right)\:=\:<−\mathrm{2},\mathrm{4},−\mathrm{2}> \\ $$$$ \\ $$$${let}\:{P}\:=\:\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right) \\ $$$${let}\:{Q}\:=\:\left(\mathrm{3},\mathrm{1},−\mathrm{1}\right) \\ $$$$ \\ $$$${P}\overset{\rightharpoonup} {{Q}}\:=\:<\mathrm{1},\mathrm{2},−\mathrm{2}> \\ $$$$\hat {{u}} \\ $$$$=\frac{<\mathrm{1},\mathrm{2},−\mathrm{2}>}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$=\frac{<\mathrm{1},\mathrm{2},−\mathrm{2}>}{\mathrm{3}} \\ $$$${D}_{\hat {{u}}} {F}\:=\:{Directional}\:{Derivative} \\ $$$$\:=\:<−\mathrm{2},\mathrm{4},−\mathrm{2}>×\frac{\mathrm{1}}{\mathrm{3}}<\mathrm{1},\mathrm{2},−\mathrm{2}> \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(−\mathrm{2}+\mathrm{8}+\mathrm{4}\right) \\ $$$$=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$$ \\ $$$${Direction}\:{of}\:{Directional}\:{Derivative} \\ $$$$=\frac{<−\mathrm{2},\mathrm{4},−\mathrm{2}>}{\:\sqrt{\mathrm{4}+\mathrm{16}+\mathrm{4}}} \\ $$$$=\frac{<−\mathrm{2},\mathrm{4},−\mathrm{2}>}{\:\sqrt{\mathrm{24}}} \\ $$$$=\frac{<−\mathrm{2},\mathrm{4},−\mathrm{2}>}{\mathrm{2}\sqrt{\mathrm{6}}} \\ $$$$=<−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}},\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}> \\ $$$$ \\ $$$${Maximum}\:{Value} \\ $$$$=\sqrt{\mathrm{2}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$=\sqrt{\mathrm{1}} \\ $$$$=\mathrm{1} \\ $$