题目
注 类似地,已知函数f(x,y)满足df(x,y)=(xdy-ydx)/(x^2)+y^(2),f(1,1)=(pi)/(4),则f(sqrt(3),3)=__.
注 类似地,
已知函数f(x,y)满足$df(x,y)=\frac{xdy-ydx}{x^{2}+y^{2}}$,$f(1,1)=\frac{\pi}{4}$,则$f(\sqrt{3},3)=\_\_. $
题目解答
答案
已知全微分 $ df(x, y) = \frac{xdy - ydx}{x^2 + y^2} $,可对应函数 $ f(x, y) = \arctan\left(\frac{y}{x}\right) $,因为其全微分满足该形式。由初始条件 $ f(1, 1) = \frac{\pi}{4} $,得常数 $ C = 0 $。因此,
\[ f(x, y) = \arctan\left(\frac{y}{x}\right). \]
代入 $ x = \sqrt{3} $,$ y = 3 $,得
\[ f(\sqrt{3}, 3) = \arctan\left(\frac{3}{\sqrt{3}}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}. \]
答案:$\boxed{\frac{\pi}{3}}$