Question Number 210078 by Spillover last updated on 29/Jul/24
$${Find}\:{the}\:{directional}\:{derivative}\:{of} \\ $$$${f}\left({x},{y}\right)=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:\:\:{in}\:{the}\:{direction}\:{given} \\ $$$${by}\:{the}\:{angle}\:\theta=\frac{\pi}{\mathrm{3}}\: \\ $$$${and}\:{also}\:{Evaluate}\:{directional}\:{derivatives} \\ $$$${at}\:{the}\:{point}\:\left(\mathrm{1},\mathrm{2}\right) \\ $$
Answered by Spillover last updated on 30/Jul/24
$$ \\ $$$${D}_{{v}} {f}\left({x},{y}\right)={f}_{{x}} \left({x},{y}\right){a}+{f}_{{y}} \left({x},{y}\right){b} \\ $$$${f}_{{x}} =\mathrm{12}{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \\ $$$${f}_{{y}} =−\mathrm{6}{xy} \\ $$$${v}=\left({a},{b}\right)=\left(\mathrm{cos}\:\theta,\mathrm{sin}\:\theta\right) \\ $$$${v}=\left({a},{b}\right)=\left(\mathrm{cos}\:\frac{\pi}{\mathrm{3}},\mathrm{sin}\frac{\pi}{\mathrm{3}}\:\:\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${D}_{{v}} {f}\left({x},{y}\right)={f}_{{x}} \left({x},{y}\right){a}+{f}_{{y}} \left({x},{y}\right){b} \\ $$$$=\left(\mathrm{12}{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(−\mathrm{6}{xy}\right)\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${D}_{{v}} {f}\left({x},{y}\right)=\mathrm{6}{x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{y}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{3}}\:{xy} \\ $$$${D}_{{v}} {f}\left(\mathrm{1},\mathrm{2}\right)=\mathrm{6}\left(\mathrm{1}^{\mathrm{2}} \right)−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} \right)−\mathrm{3}\sqrt{\mathrm{3}}\:\:\mathrm{2}×\mathrm{1}=−\mathrm{6}\sqrt{\mathrm{3}\:}\: \\ $$$$ \\ $$