Subtopic: Homogeneous function – Euler's theorem on homogeneous functions
Question Statement:
Let $u(x, y) = \tan^{-1}\left( \frac{x^3 + y^3}{x - y} \right)$ for $x \neq y$. The value of $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ when $u = \frac{\pi}{4}$ is ________ (answer in integer).
Answer Key:
1
Step-by-Step Solution & Explanation
Let $z = \tan(u) = \frac{x^3 + y^3}{x - y}$.
The function $z(x, y)$ is a homogeneous function of degree $n = 3 - 1 = 2$.
By Euler's Theorem on homogeneous functions:
$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z = 2 z$
Since $z = \tan(u)$:
$\frac{\partial z}{\partial x} = \sec^2(u) \frac{\partial u}{\partial x}, \quad \frac{\partial z}{\partial y} = \sec^2(u) \frac{\partial u}{\partial y}$
Substituting:
$\sec^2(u) \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \right) = 2 \tan(u)$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2 \tan(u)}{\sec^2(u)} = 2 \sin(u) \cos(u) = \sin(2u)$
When $u = \frac{\pi}{4}$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \sin\left( 2 \times \frac{\pi}{4} \right) = \sin\left( \frac{\pi}{2} \right) = 1$