https://r2.erweima.ai/mathbot/image/1726278458349atzlpnlx.jpg

The image presents the following problem:

We are given the function:

\[
z(x, y) = 2(ax + by)^2 - (x^2 + y^2)
\]
with the condition \(a^2 + b^2 = 1\).

The task is to verify that:

\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0.
\]

### Step 1: First Partial Derivatives

Let's compute the first partial derivatives of \(z(x, y)\).

1. **Partial derivative with respect to \(x\):**

\[
z_x = \frac{\partial}{\partial x} \left( 2(ax + by)^2 - (x^2 + y^2) \right)
\]

The first term is \(2(ax + by)^2\), so we apply the chain rule:

\[
\frac{\partial}{\partial x} \left( 2(ax + by)^2 \right) = 4(ax + by) \cdot a.
\]

The second term is \(-(x^2 + y^2)\), so:

\[
\frac{\partial}{\partial x} \left( -(x^2 + y^2) \right) = -2x.
\]

Thus, we have:

\[
z_x = 4a(ax + by) - 2x.
\]

2. **Partial derivative with respect to \(y\):**

Similarly, for the partial derivative with respect to \(y\):

\[
z_y = \frac{\partial}{\partial y} \left( 2(ax + by)^2 - (x^2 + y^2) \right)
\]

The first term \(2(ax + by)^2\) gives:

\[
\frac{\partial}{\partial y} \left( 2(ax + by)^2 \right) = 4(ax + by) \cdot b.
\]

The second term is \(-(x^2 + y^2)\), so:

\[
\frac{\partial}{\partial y} \left( -(x^2 + y^2) \right) = -2y.
\]

Thus, we have:

\[
z_y = 4b(ax + by) - 2y.
\]

### Step 2: Second Partial Derivatives

Now, we compute the second partial derivatives.

1. **Second derivative with respect to \(x\):**

\[
z_{xx} = \frac{\partial}{\partial x} \left( 4a(ax + by) - 2x \right)
\]

The derivative of \(4a(ax + by)\) with respect to \(x\) is:

\[
4a^2.
\]

The derivative of \(-2x\) is \(-2\). Therefore:

\[
z_{xx} = 4a^2 - 2.
\]

2. **Second derivative with respect to \(y\):**

\[
z_{yy} = \frac{\partial}{\partial y} \left( 4b(ax + by) - 2y \right)
\]

The derivative of \(4b(ax + by)\) with respect to \(y\) is:

\[
4b^2.
\]

The derivative of \(-2y\) is \(-2\). Therefore:

\[
z_{yy} = 4b^2 - 2.
\]

### Step 3: Sum of Second Derivatives

We now sum the second partial derivatives:

\[
z_{xx} + z_{yy} = (4a^2 - 2) + (4b^2 - 2) = 4a^2 + 4b^2 - 4.
\]

Using the condition \(a^2 + b^2 = 1\), we substitute:

\[
4(a^2 + b^2) - 4 = 4 \cdot 1 - 4 = 0.
\]

Thus, we have verified that:

\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0.
\]

### Summary:

The given condition is verified: \(\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0\).

---

Do you need further details or have any questions? 

Here are some related questions:

1. How does the Laplace equation relate to this type of verification?
2. What is the physical interpretation of the equation \(a^2 + b^2 = 1\)?
3. Can this problem be generalized for other functions?
4. What role does the chain rule play in partial differentiation?
5. How does this relate to the concept of harmonic functions?

**Tip:** The Laplacian operator often appears in problems involving heat conduction, fluid flow, and electromagnetism.

Sea z(x, y) = 2(ax + by)^2 - (x^2 + y^2) con a^2 + b^2 = 1. Verifique que ∂^2z/∂x^2 + ∂^2z/∂y^2 = 0.

Learn how to verify the Laplace equation for the function z(x, y) = 2(ax + by)^2 - (x^2 + y^2) with a step-by-step solution. This problem involves partial derivatives and the Laplacian in 2D space, suitable for university-level calculus students.

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