if w = r ^m prove that w xx +w yy +w zz = m (m+1)r ^m-2 where r = (x^2+y^2+z^2)^1/2

We are tasked with proving that if \( w = r^m \), then the second derivatives of \( w \) satisfy the equation:

\[
w_{xx} + w_{yy} + w_{zz} = m(m+1) r^{m-2}
\]
where \( r = \sqrt{x^2 + y^2 + z^2} \).

### Step 1: Expression for \( w \)
Given \( w = r^m \), and that \( r = \sqrt{x^2 + y^2 + z^2} = (x^2 + y^2 + z^2)^{1/2} \), we will find the second derivatives of \( w \) with respect to \( x \), \( y \), and \( z \).

### Step 2: First Derivative of \( w \)
To find \( w_{x} \), we first calculate the derivative of \( r \) with respect to \( x \). Recall that:

\[
r = (x^2 + y^2 + z^2)^{1/2}
\]

Thus, the derivative of \( r \) with respect to \( x \) is:

\[
r_x = \frac{\partial r}{\partial x} = \frac{x}{r}
\]

Now, using the chain rule for \( w = r^m \):

\[
w_x = \frac{\partial w}{\partial x} = \frac{d}{dr}(r^m) \cdot \frac{\partial r}{\partial x} = m r^{m-1} \cdot \frac{x}{r} = m r^{m-2} x
\]

Similarly, the derivatives with respect to \( y \) and \( z \) are:

\[
w_y = m r^{m-2} y
\]
\[
w_z = m r^{m-2} z
\]

### Step 3: Second Derivatives of \( w \)
Now, we compute the second derivatives.

#### Second derivative with respect to \( x \):

\[
w_{xx} = \frac{\partial}{\partial x} (m r^{m-2} x)
\]

This requires applying the product rule:

\[
w_{xx} = m \left[ \frac{\partial}{\partial x}(r^{m-2}) \cdot x + r^{m-2} \cdot \frac{\partial}{\partial x} x \right]
\]

The derivative of \( r^{m-2} \) with respect to \( x \) is:

\[
\frac{\partial}{\partial x} (r^{m-2}) = (m-2) r^{m-4} \cdot \frac{x}{r} = (m-2) r^{m-4} x
\]

Thus:

\[
w_{xx} = m \left[ (m-2) r^{m-4} x^2 + r^{m-2} \right]
\]

#### Second derivatives with respect to \( y \) and \( z \):

Following the same logic, we obtain:

\[
w_{yy} = m \left[ (m-2) r^{m-4} y^2 + r^{m-2} \right]
\]
\[
w_{zz} = m \left[ (m-2) r^{m-4} z^2 + r^{m-2} \right]
\]

### Step 4: Sum of Second Derivatives
Now, sum the second derivatives:

\[
w_{xx} + w_{yy} + w_{zz} = m \left[ (m-2) r^{m-4} (x^2 + y^2 + z^2) + 3r^{m-2} \right]
\]

Since \( x^2 + y^2 + z^2 = r^2 \), this becomes:

\[
w_{xx} + w_{yy} + w_{zz} = m \left[ (m-2) r^{m-4} r^2 + 3r^{m-2} \right]
\]

Simplifying:

\[
w_{xx} + w_{yy} + w_{zz} = m \left[ (m-2) r^{m-2} + 3r^{m-2} \right]
\]

\[
w_{xx} + w_{yy} + w_{zz} = m (m+1) r^{m-2}
\]

### Conclusion:
We have shown that:

\[
w_{xx} + w_{yy} + w_{zz} = m(m+1) r^{m-2}
\]

as required.

Would you like more details on any part of this proof?

### Related Questions:
1. What happens if \( m = 2 \) in this expression?
2. How would this proof change if \( r = (x^2 + y^2 + z^2)^n \) instead of \( (x^2 + y^2 + z^2)^{1/2} \)?
3. Can you derive the third derivative of \( w \) with respect to \( x \), \( y \), and \( z \)?
4. How does this proof generalize to higher-dimensional spaces?
5. What happens to this equation if we apply it in cylindrical coordinates?

### Tip:
When solving problems involving gradients and second derivatives, organizing your work step-by-step, especially when applying the chain and product rules, helps ensure clarity and accuracy.

If w = r^m prove that w_xx + w_yy + w_zz = m(m+1)r^(m-2) where r = (x^2 + y^2 + z^2)^1/2

Explore how to prove that w_xx + w_yy + w_zz = m(m+1)r^(m-2) where r = (x^2 + y^2 + z^2)^{1/2} using advanced calculus concepts like partial derivatives, the chain rule, and the Laplacian. This step-by-step guide is ideal for university-level students studying multivariable calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation