Question Number 192398 by Mastermind last updated on 16/May/23
$$\mathrm{Let}\:\mathrm{f}:\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{define}\:\mathrm{by}\: \\ $$$$\mathrm{f}\left(\mathrm{x},\:\mathrm{y},\:\mathrm{z}\right)\:=\:\mathrm{2x}^{\mathrm{2}} −\mathrm{y}+\mathrm{6xy}−\mathrm{z}^{\mathrm{3}} +\mathrm{3z}. \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{directional}\:\mathrm{deriva}− \\ $$$$\mathrm{tive}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{u}=\left(\mathrm{2},\:\mathrm{1},\:−\mathrm{3}\right) \\ $$$$ \\ $$$$\mathrm{help}! \\ $$
Answered by aleks041103 last updated on 21/May/23
$$\left({u}\centerdot\bigtriangledown\right){f}=\mathrm{2}\partial_{{x}} {f}+\partial_{{y}} {f}−\mathrm{3}\partial_{{z}} {f}= \\ $$$$=\mathrm{2}\left(\mathrm{4}{x}+\mathrm{6}{y}\right)+\left(−\mathrm{1}+\mathrm{6}{x}\right)−\mathrm{3}\left(−\mathrm{3}{z}^{\mathrm{2}} +\mathrm{3}\right)= \\ $$$$=\mathrm{8}{x}+\mathrm{12}{y}−\mathrm{1}+\mathrm{6}{x}+\mathrm{9}{z}^{\mathrm{2}} −\mathrm{9}= \\ $$$$=\mathrm{14}{x}+\mathrm{12}{y}+\mathrm{9}{z}^{\mathrm{2}} −\mathrm{10} \\ $$