Question

If $u = \sin^{-1}\left( \frac{x+2y+3z}{x^8+y^8+z^8} \right)$, then the value of $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z}$ is given by}

• A. $-7 \tan u$
• B. $7 \tan u$
• C. $7 \sin u$
• D. $-7 \sin u$

Hint:

For any function $u = \sin^{-1}(f(x,y,z))$ or $u = \tan^{-1}(f(x,y,z))$, if $f$ is homogeneous of degree $n$, the Euler expression is always equal to $n \frac{F(u)}{F'(u)}$. This saves time during exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Euler's Theorem for homogeneous functions states that if $V(x, y, z)$ is a homogeneous function of degree $n$, then: \[ x\frac{\partial V}{\partial x} + y\frac{\partial V}{\partial y} + z\frac{\partial V}{\partial z} = n V \]
Key Formula or Approach:
If we have a function $u$ such that $F(u) = V(x, y, z)$ is a homogeneous function of degree $n$, then: \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} = n \frac{F(u)}{F'(u)} \]

Step 2: Detailed Explanation:
Let us define $V(x, y, z) = \sin u$: \[ V(x, y, z) = \frac{x+2y+3z}{x^8+y^8+z^8} \]
We check the homogeneity of $V$ by substituting $x \to tx$, $y \to ty$, and $z \to tz$: \[ V(tx, ty, tz) = \frac{tx+2ty+3tz}{(tx)^8+(ty)^8+(tz)^8} = \frac{t(x+2y+3z)}{t^8(x^8+y^8+z^8)} = t^{1-8} V(x, y, z) = t^{-7} V(x, y, z) \]
Thus, $V(x, y, z)$ is a homogeneous function of degree $n = -7$.
By applying the modified Euler's formula: \[ F(u) = \sin u \implies F'(u) = \cos u \] \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} = -7 \frac{\sin u}{\cos u} = -7 \tan u \]
Therefore, the value of the expression is $-7 \tan u$.

Step 3: Final Answer:
The value of the partial differential expression is $-7 \tan u$.