13.7-bo'lim : Yo'nalishli hosilalar
4. \(f\left( {x,y,z} \right) = {x^2}{y^3} - 4xz\) uchun \({D_{\vec u}}f\) ni \(\vec v = \left\langle { - 1,2,0} \right\rangle \) yo‘nalishida aniqlang.
Yechim
OK, biz gradient kerakligini bilamiz, shuning uchun avval uni olamiz.
\[\nabla f = \left\langle {2x{y^3} - 4z,3{x^2}{y^2}, - 4x} \right\rangle \]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.
\[\left\| {\vec v} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} = \sqrt 5 \hspace{0.5in}\vec u = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{1}{{\sqrt 5 }}\left\langle { - 1,2,0} \right\rangle = \left\langle { - \frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }},0} \right\rangle \]3-qadam
Yo'nalishli hosila u holda,
\[{D_{\vec u}}f = \left\langle {2x{y^3} - 4z,3{x^2}{y^2}, - 4x} \right\rangle \centerdot \left\langle { - \frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }},0} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{\sqrt 5 }}\left( {4z - 2x{y^3} + 6{x^2}{y^2}} \right)}}\]