13.7-bo'lim : Yo'nalishli hosilalar
5. \(\vec v = \left\langle { - 1,4,2} \right\rangle \) yo‘nalishi bo‘yicha \(f\left( {x,y,z} \right) = 4x - {y^2}{{\bf{e}}^{3x\,z}}\) uchun \({D_{\vec u}}f\left( {3, - 1,0} \right)\) ni aniqlang.
Yechim
OK, biz gradient kerakligini bilamiz, shuning uchun avval uni olamiz.
\[\nabla f = \left\langle {4 - 3z{y^2}{{\bf{e}}^{3x\,z}}, - 2y{{\bf{e}}^{3x\,z}}, - 3x{y^2}{{\bf{e}}^{3x\,z}}} \right\rangle \]Chunki biz ham bilamizki, oxir-oqibat, biz buni ham olishimiz mumkin bo'lgan nuqtada baholanishi kerak.
\[\nabla f\left( {3, - 1,0} \right) = \left\langle {4,2, - 9} \right\rangle \]2-qadam
Shuningdek, yo'nalish vektori birlik vektor ekanligiga ishonch hosil qilishimiz kerakligini eslang. Bu vektor birlik vektor emasligi (umid qilamanki) juda aniq, shuning uchun uni birlik vektoriga aylantiramiz.
\[\left\| {\vec v} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt {21} \hspace{0.25in}\vec u = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{1}{{\sqrt {21} }}\left\langle { - 1,4,2} \right\rangle = \left\langle { - \frac{1}{{\sqrt {21} }},\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right\rangle \]3-qadam
Yo'nalishli hosila u holda,
\[{D_{\vec u}}f\left( {3, - 1,0} \right) = \left\langle {4,2, - 9} \right\rangle \centerdot \left\langle { - \frac{1}{{\sqrt {21} }},\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{14}}{{\sqrt {21} }}}}\]