Subtopic: Homogeneous function – Euler's theorem on homogeneous functions
Question Statement:
If $u(x, y) = \sin^{-1}\left(\frac{x^2 + y^2}{x + y}\right)$, the value of the expression $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ at any point where $u = \frac{\pi}{4}$ is ________ (answer in integer).
Answer Key:
1
Step-by-Step Solution & Explanation
Let $v = \sin(u) = \frac{x^2 + y^2}{x + y}$.
Observe that $v(tx, ty) = \frac{t^2(x^2 + y^2)}{t(x + y)} = t v(x, y)$, so $v$ is a homogeneous function of degree $n = 1$.
By Euler's theorem applied to $v$:
$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = 1 \cdot v$
Since $v = \sin(u)$, we have $\frac{\partial v}{\partial x} = \cos(u) \frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial y} = \cos(u) \frac{\partial u}{\partial y}$:
$\cos(u) \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right) = \sin(u)$
Dividing by $\cos(u)$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \tan(u)$
For $u = \frac{\pi}{4}$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \tan\left(\frac{\pi}{4}\right) = 1$