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

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

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

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

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

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