Предмет: Алгебра, автор: CoBeH0K

Помогите решить!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Срочно!!!

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Автор ответа: NNNLLL54

1

$u=x\sqrt{y}-y\, z^2\; \; ,\; \; M_1(2,1,-1)\; \; ,\; \; M_2(0,2,0)\\\\\vec {l}=\overline {M_1M_2}=(-2,1,1)\; \; ,\; \; |\vec{l}\, |=|\overline {M_1M_2}|=\sqrt{4+1+1}=\sqrt6\\\\cos\alpha =-\dfrac{2}{\sqrt6}\; ,\; \; cos\beta =\dfrac{1}{\sqrt6}\; ,\; \; cos\gamma =\dfrac{1}{\sqrt6}\\\\u'_{x}=\sqrt{y}\; \; ,\; \; u'_{y}=x\cdot \dfrac{1}{2\sqrt{y}}-z^2\; \; ,\; \; u'_z}=-2yz\\\\u'_{x}(M_1)=1\; \; ,\; \; u'_{y}(M_1)=1-1=0\; \; ,\; \; u'_{z}(M_1)=2\\\\\overline {grad\, u(x,y)}=(u'_{x}\, ;\, u'_{y}\, ;\, u'_{z})$

Направление наибольшего возрастания функции в точке М1 совпадает с направлением градиента функции в этой точке. А величина наибольшего возрастания совпадает с модулем градиента.

$\overline {grad\, u(x,y)}\Big|_{M_1}=\Big(\, 1;0;2\Big)\; \; ,\; \; \Big|\overline {grad\, u(x,y)}\Big|_{M_1}=\sqrt{1+0+4}=\sqrt5$

Производная по направлению вектора $\vec {l}\; :$

$\dfrac{u(x,y)}{\partial\vec {l}}=u'_{x}\cdot cos\alpha +u'_{y}\cdot cos\beta +u'_{z}\cdot cos\gamma$

$\dfrac{\partial\, u}{\partial \vec{l}}\Big|_{M_1}=1\cdot (-\frac{2}{\sqrt6})+0\cdot \dfrac{1}{6}+2\cdot \dfrac{1}{\sqrt6}=0$

CoBeH0K: Функция ведь задана в неявном виде. Значит частные производные должны находиться по другому?

NNNLLL54: чётко написано: u=... это явное задание функции.............

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