13.7-bo'lim : Yo'nalishli hosilalar
3. \(\displaystyle f\left( {x,y} \right) = \cos \left( {\frac{x}{y}} \right)\) uchun \({D_{\vec u}}f\) ni \(\vec v = \left\langle {3, - 4} \right\rangle \) yo‘nalishida aniqlang.
Yechim
OK, biz gradient kerakligini bilamiz, shuning uchun avval uni olamiz.
\[\nabla f = \left\langle { - \frac{1}{y}\sin \left( {\frac{x}{y}} \right),\frac{x}{{{y^2}}}sin\left( {\frac{x}{y}} \right)} \right\rangle \]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.
\[\left\| {\vec v} \right\| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} = \sqrt {25} = 5\hspace{0.5in}\vec u = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{1}{5}\left\langle {3, - 4} \right\rangle = \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \]3-qadam
Yo'nalishli hosila u holda,
\[\begin{align*}{D_{\vec u}}f & = \left\langle { - \frac{1}{y}\sin \left( {\frac{x}{y}} \right),\frac{x}{{{y^2}}}\sin\left( {\frac{x}{y}} \right)} \right\rangle \centerdot \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{3}{{5y}}\sin \left( {\frac{x}{y}} \right) - \frac{{4x}}{{5{y^2}}}\sin\left( {\frac{x}{y}} \right) = - \frac{1}{5}\left( {\frac{3}{y} + \frac{{4x}}{{{y^2}}}} \right)sin\left( {\frac{x}{y}} \right)}}\end{align*}\]