If $u=f(\frac{x}{y},\frac{y}{z},\frac{z}{x}),$ then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=$

Q: If $u=f(\frac{x}{y},\frac{y}{z},\frac{z}{x}),$ then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=$

To solve the problem, we need to evaluate the expression $ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} $ given that $ u = f\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right) $. ### Explanation: 1. **Understanding the Function $ u $:** - The function $ u $ is expressed in terms of three variables: $ \frac{x}{y} $, $ \frac{y}{z} $, and $ \frac{z}{x} $. - These are homogeneous functions of degree 0 because if you scale $ x, y, z $ by a factor $ t $, each of these ratios remains unchanged. 2. **Applying Euler's Theorem for Homogeneous Functions:** - Euler's theorem states that for a homogeneous function $ g(x, y, z) $ of degree $ n $, the following holds: \[ x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} + z \frac{\partial g}{\partial z} = n \cdot g(x, y, z) \] - In our case, since $ u = f\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right) $ is a function of variables that are homogeneous of degree 0, the function $ u $ itself is homogeneous of degree 0. 3. **Applying the Theorem to $ u $:** - Since $ u $ is homogeneous of degree 0, applying Euler's theorem gives: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = 0 \cdot u = 0 \] 4. **Conclusion:** - The expression evaluates to 0, which corresponds to Option 2. ### Why Other Options are Incorrect: - **Option 1 (1):** This would imply that the function $ u $ is homogeneous of degree 1, which is not the case here. - **Option 3 (-1):** This would imply that the function $ u $ is homogeneous of degree -1, which is also incorrect. - **Option 4 (2):** This would imply that the function $ u $ is homogeneous of degree 2, which is not applicable here. Thus, the correct answer is indeed Option 2, as the expression evaluates to 0 due to the homogeneity of the function $ u $ of degree 0.

Question

Accepted Answer

To solve the problem, we need to evaluate the expression $ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} $ given that $ u = f\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right) $.

### Explanation:

1. **Understanding the Function $ u $:**
   - The function $ u $ is expressed in terms of three variables: $ \frac{x}{y} $, $ \frac{y}{z} $, and $ \frac{z}{x} $.
   - These are homogeneous functions of degree 0 because if you scale $ x, y, z $ by a factor $ t $, each of these ratios remains unchanged.

2. **Applying Euler's Theorem for Homogeneous Functions:**
   - Euler's theorem states that for a homogeneous function $ g(x, y, z) $ of degree $ n $, the following holds:
     $$
     x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} + z \frac{\partial g}{\partial z} = n \cdot g(x, y, z)
     $$
   - In our case, since $ u = f\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right) $ is a function of variables that are homogeneous of degree 0, the function $ u $ itself is homogeneous of degree 0.

3. **Applying the Theorem to $ u $:**
   - Since $ u $ is homogeneous of degree 0, applying Euler's theorem gives:
     $$
     x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = 0 \cdot u = 0
     $$

4. **Conclusion:**
   - The expression evaluates to 0, which corresponds to Option 2.

### Why Other Options are Incorrect:

- **Option 1 (1):** This would imply that the function $ u $ is homogeneous of degree 1, which is not the case here.
- **Option 3 (-1):** This would imply that the function $ u $ is homogeneous of degree -1, which is also incorrect.
- **Option 4 (2):** This would imply that the function $ u $ is homogeneous of degree 2, which is not applicable here.

Thus, the correct answer is indeed Option 2, as the expression evaluates to 0 due to the homogeneity of the function $ u $ of degree 0.