Question

150 баллов! Градиент.
дано:
функция z=arcsin(x/(x+y)) ,
точка A(5;5) ,
вектор a= -12i + 5j
Найти в точке А: grad(z) и производную по направлению вектора а

Answer 1

$grad(z)=\frac{\partial z}{\partial x}\bar{i}+\frac{\partial z}{\partial y}\bar{j}\\\frac{\partial z}{\partial x}=\frac1{\sqrt{1-\frac{x^2}{(x+y)^2}}}\cdot\left(\frac x{x+y}\right)_{\partial x}=\frac1{\sqrt{1-\frac{x^2}{(x+y)^2}}}\cdot\frac{x+y-x}{(x+y)^2}=\frac y{(x+y)^2\sqrt{1-\frac{x^2}{(x+y)^2}}}=\\=\frac y{(x+y)^2\sqrt{\frac{x^2+2xy+y^2-x^2}{(x+y)^2}}}=\frac y{(x+y)^2\sqrt{\frac{y(2x+y)}{(x+y)^2}}}$

$\frac{\partial z}{\partial y}=\frac1{\sqrt{1-\frac{x^2}{(x+y)^2}}}\cdot\left(\frac x{x+y}\right)_{\partial y}=\frac1{\sqrt{1-\frac{x^2}{(x+y)^2}}}\cdot\left(-\frac x{(x+y)^2}\right)=\\=-\frac x{(x+y)^2\sqrt{1-\frac{x^2}{(x+y)^2}}}=-\frac x{(x+y)^2\sqrt{\frac{y(2x+y)}{(x+y)^2}}}$

$grad z=\frac y{(x+y)^2\sqrt{\frac{y(2x+y)}{(x+y)^2}}}\bar{i}-\frac x{(x+y)^2\sqrt{\frac{y(2x+y)}{(x+y)^2}}}\bar{j}$
$grad z_A=\frac 5{(5+5)^2\sqrt{\frac{5(2\cdot5+5)}{(5+5)^2}}}\bar{i}-\frac 5{(5+5)^2\sqrt{\frac{5(2\cdot5+5)}{(5+5)^2}}}\bar{j}=\frac5{100\sqrt{\frac{75}{100}}}\bar{i}-\frac5{100\sqrt{\frac{75}{100}}}\bar{j}=\\=\frac1{10\sqrt3}\bar{i}-\frac1{10\sqrt3}\bar{j}=\frac{\sqrt3}{30}\bar{i}-\frac{\sqrt3}{30}\bar{j}$

$|grad(z)_A|=\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}=\sqrt{\left(\frac{\sqrt3}{30}\right)^2+\left(-\frac{\sqrt3}{30}^2\right)^2}=\frac{\sqrt6}{30}\\\cos\alpha=\frac{\frac{\partial z}{\partial x}}{|grad{z}|}=\frac{\frac{\sqrt3}{30}}{\frac{\sqrt6}{30}}=\frac1{\sqrt2}$,
$\cos\beta=\frac{\frac{\partial z}{\partial y}}{|grad(z)|}=\frac{-\frac{\sqrt3}{30}}{\frac{\sqrt6}{30}}=-\frac1{\sqrt2}\\\\\frac{\partial z}{\partial a}=\frac{\partial z}{\partial x}\cdot\cos\alpha+\frac{\partial z}{\partial y}\cos\beta\\|a|=\sqrt{x^2+y^2}=\sqrt{144+25}=13\\\cos\alpha=\frac x{|a|}=-\frac{12}{13}\\\cos\beta=\frac y{|a|}=\frac5{13}\\\frac{\partial z}{\partial a}=\frac{\sqrt3}{30}\cdot\left(-\frac{12}{13}\right)+\left(-\frac{\sqrt3}{30}\right)\cdot\frac5{13}=-\frac{17\sqrt3}{390}$