Question

Помогите пожалуйста!!!!! СРОЧНО!!!!!

Answer 1

$u=-2\cdot ln(x^2-5)-4xyz\; \; ,\; \; \vec{l}=\vec{i}+2\vec{j}-2\vec{k}\; \; ,\; \; M_0(1,1,1)\\\\\\\vec{l}=(1,2,-2)\; \; ,\; \; |\, \vec{l}\, |=\sqrt{1+4+4}=3\; \; ,\\\\\vec{l}^\circ =(cos\alpha ,cos\beta ,cos\gamma )=(\frac{1}{3}\, ,\, \frac{2}{3}\, ,\, -\frac{2}{3}\, )\\\\\\u'_{x}=-\frac{4x}{x^2-5}-4yz\; \; ,\; \; \; u'_{x}(M_0)=-\frac{4}{1-5}-4=1-4=-3\\\\u'_{y}=-4xz\; \; ,\; \; \; u'_{y}(M_0)=-4\\\\u'_{z}=-4xy\; \; ,\; \; \; u'_{z}(M_0)=-4$

$grad\, u(M_0)=u'_{x}(M_0)\cdot \vec{i}+u'_{y}(M_0)\cdot \vec{j}+u'_{z}(M_0)\cdot \vec{k}\\\\grad\, u(M_0)=-3\vec{i}-4\vec{j}-4\vec{k}\\\\\\\frac{\partial u}{\partial \vec{l}}=u'_{x}(M_0)\cdot cos\alpha +u'_{y}(M_0)\cdot cos\beta +u'_{z}(M_0)\cdot cos\gamma \\\\\frac{\partial u}{\partial \vec{l}}=-3\cdot \frac{1}{3}-4\cdot \frac{2}{3}+4\cdot \frac{2}{3}=\frac{-3-8+8}{3}=-1$