Question

Помогите, пожалуйста, срочно!
Найди угол между градиентами функции u (x, y, z) и v (x, y, z) в точке М.

Answer 1

Ответ:

$u=\dfrac{z^2}{x^2y^2}\ \ ,\ \ \ v=\dfrac{3x^2}{\sqrt2}-\dfrac{y^2}{\sqrt2}+\sqrt2\, z^2\ \ \ ,\ \ \ M\Big(\ \dfrac{2}{3}\ ;\ 2\ ;\ \sqrt{\dfrac{2}{3}}\ \Big)\\\\\\\displaystyle u'_{x}=\frac{z^2}{y^2}\cdot \frac{-2}{x^3}\ \ ,\ \ \ \ u'_{x}\Big|_{M}=\frac{2}{3\cdot 4}\cdot \frac{-2\cdot 27}{8}=-\frac{9}{8}\\\\\\u'_{y}=\dfrac{z^2}{x^2}\cdot \dfrac{-2}{y^3}\ \ ,\ \ \ y'_{y}\Big|_{M}=\frac{2\cdot 9}{3\cdot 4}\cdot \frac{-2}{8}=-\frac{3}{8}$

$\displaystyle u'_{z}=\frac{1}{x^2y^2}\cdot 2z\ \ ,\ \ \ \ u'_{z}\Big|_{M}=\frac{9}{4\cdot 4}\cdot 2\sqrt{\frac{2}{3}}=\frac{9\sqrt2}{8\sqrt3}=\frac{3\sqrt3}{4\sqrt2}\\\\\\grad\, u\Big|_{M}=\Big(-\frac{9}{8}\ ;-\frac{3}{8}\ ;\ \frac{3\sqrt3}{4\sqrt2}\ \Big)\\\\\\\Big|grad\, u\Big|_{M}\Big|=\sqrt{\frac{81}{64}+\frac{9}{64}+\frac{27}{32}}=\sqrt{\frac{81+9+54}{64}}=\sqrt{\frac{144}{64}}=\frac{12}{8}$

$\displaystyle v'_{x}=\frac{3}{\sqrt2}\cdot 2x=3\sqrt2\, x\ \ ,\ \ \ v'_{x}\Big|_{M}=3\sqrt2\cdot \frac{2}{3}=2\sqrt2\\\\\\v'_{y}=-\frac{1}{\sqrt2}\cdot 2y=-\sqrt2y\ \ ,\ \ \ \ v'_{y}\Big|_{M}=-\sqrt2\cdot 2=-2\sqrt2\\\\\\v'_{z}=\sqrt2\cdot 2z\ \ ,\ \ \ \ \ v'_{z}\Big|_{M}=2\sqrt2\cdot \sqrt{\frac{2}{3}}=\frac{4}{\sqrt3}\\\\\\grad\, v\Big|_{M}=\Big(\ 2\sqrt2\ ;-2\sqrt2\ ;\ \frac{4}{\sqrt3}\ \Big)$

$\displaystyle \Big|grad\, v\Big|_{M}\Big|=\sqrt{(2\sqrt2)^2+(2\sqrt2)^2+\frac{16}{3}}=\sqrt{\frac{64}{3}}=\frac{8}{\sqrt3}\\\\\\cos\varphi =\dfrac{-\dfrac{9}{8}\cdot 2\sqrt2+\dfrac{3}{8}\cdot 2\sqrt2+\dfrac{3\sqrt3}{4\sqrt2}\cdot \dfrac{4}{\sqrt3}}{\dfrac{12}{8}\cdot \dfrac{8}{\sqrt3}}=\frac{-\dfrac{9\sqrt2}{4}+\dfrac{3\sqrt2}{4}+\dfrac{3}{\sqrt2}}{4\sqrt3}=\\\\\\=\frac{-\dfrac{6\sqrt2}{4}+\dfrac{3\cdot 2\sqrt2}{4}}{4\sqrt3}=\frac{0}{4\sqrt3}=0\ \ \ \ \Rightarrow \ \ \ \ \varphi =90^\circ$