Question

If \( F = F(x, y, z) = \dfrac{x^2 y^2 z^2}{x^2 + y^2 + z^2} \),  \( G = G(x, y, z) = \log\left(\dfrac{xy + yz + zx}{x^2 + y^2 + z^2}\right) \), and \( H = F + G \), then \[ x \dfrac{\partial H}{\partial x} + y \dfrac{\partial H}{\partial y} + z \dfrac{\partial H}{\partial z} = \text{...........} \]
 

• A. \( 0 \)
• B. \( 4F \)
• C. \( 2G \)
• D. \( 6H \)

Hint:

Use Euler’s theorem for quickly evaluating expressions involving homogeneous functions. Always check the degree of homogeneity by comparing the powers in numerator and denominator.

Answer 1

The Correct Option is B

Step 1: Use Euler’s theorem for homogeneous functions.
If a function \( f(x, y, z) \) is homogeneous of degree \( n \), then \[ x \dfrac{\partial f}{\partial x} + y \dfrac{\partial f}{\partial y} + z \dfrac{\partial f}{\partial z} = n f \] Step 2: Analyze the function \( F(x, y, z) = \dfrac{x^2 y^2 z^2}{x^2 + y^2 + z^2} \)
The numerator \( x^2 y^2 z^2 \) is of degree 6, and the denominator \( x^2 + y^2 + z^2 \) is of degree 2. 
So, \( F \) is homogeneous of degree \( 6 - 2 = 4 \). 
Thus, by Euler's theorem: \[ x \dfrac{\partial F}{\partial x} + y \dfrac{\partial F}{\partial y} + z \dfrac{\partial F}{\partial z} = 4F \] Step 3: Analyze the function \( G(x, y, z) = \log\left(\dfrac{xy + yz + zx}{x^2 + y^2 + z^2}\right) \)
Let \( A = \dfrac{xy + yz + zx}{x^2 + y^2 + z^2} \). Note that both the numerator and denominator are homogeneous of degree 2, so \( A \) is homogeneous of degree 0. Then \( G = \log(A) \), which is also homogeneous of degree 0. Hence: \[ x \dfrac{\partial G}{\partial x} + y \dfrac{\partial G}{\partial y} + z \dfrac{\partial G}{\partial z} = 0 \] Step 4: Add the derivatives for \( H = F + G \)
\[ x \dfrac{\partial H}{\partial x} + y \dfrac{\partial H}{\partial y} + z \dfrac{\partial H}{\partial z} = x \dfrac{\partial F}{\partial x} + y \dfrac{\partial F}{\partial y} + z \dfrac{\partial F}{\partial z} + 0 = 4F \] Final Answer: \( \boxed{4F} \)