Question

Найти градиент функции z в точке м. (1/pi^2)(arcsin^2)y/x+(3y^3)+5x; m(2,корень из 3)

Answer 1

$z=\frac{1}{\pi ^2}\cdot 2arcsin^2\frac{y}{x}+3y^3+5x\; ,\; \; M(2,\sqrt3)\\\\z'_{x}=\frac{1}{\pi ^2}\cdot \2arcsin\frac{y}{x}\cdot \frac{1}{\sqrt{1-\frac{y^2}{x^2}}}\cdot \frac{-y}{x^2}+5=\frac{1}{\pi ^2}\cdot 2arcsin\frac{y}{x}\cdot \frac{-y\cdot x}{x^2\sqrt{x^2-y^2}}+5=\\\\=\frac{1}{\pi ^2}\cdot 2arcsin\frac{y}{x}\cdot \frac{-y}{x\sqrt{x^2-y^2}}+5$

$z'_{x}(2,\sqrt3)=\frac{1}{\pi ^2}\cdot 2arcsin\frac{\sqrt3}{2}\cdot \frac{-\sqrt3}{\sqrt{4-3}}+5=\frac{1}{\pi }*\frac{2\pi}{3}*\frac{-\sqrt3}{1}+5=5-\frac{2\sqrt3}{3\pi }$

$z'_{y}=\frac{1}{\pi ^2}\cdot 2arcsin\frac{y}{x}\cdot \frac{x}{\sqrt{x^2-y^2}}\cdot \frac{1}{x}+9y^2$

$z'_{y}(2,\sqrt3)=\frac{1}{\pi ^2}\cdot 2arcsin\frac{\sqrt3}{2}\cdot \frac{1}{\sqrt{4-3}}+9\cdot 3=\frac{1}{\pi^2}\cdot 2\cdot \frac{\pi}{3}+27=\frac{2}{3\pi }+27$

$grad\, z|_{M}=(5-\frac{2\sqrt3}{3\pi })\, \overline {i}+(\frac{2}{3\pi }+27)}\, \overline {j}$