If u = xy + yz + zx, v = x^2 + y^2 + z^2, and w = x + y + z, determine whether there is a functional relationship between u, v, and w, and if so, find it. (Hint: Jacobian)

Steps to Solve

1. Define the Jacobian Matrix

The Jacobian matrix consists of the partial derivatives of the functions with respect to the variables $x$, $y$, and $z$. For our functions, we have:

$$ J = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{bmatrix} $$

2. Calculate the Partial Derivatives

Now we compute each of the partial derivatives:

• For $u = xy + yz + zx$:

  • $\frac{\partial u}{\partial x} = y + z$
  • $\frac{\partial u}{\partial y} = x + z$
  • $\frac{\partial u}{\partial z} = x + y$

• For $v = x^2 + y^2 + z^2$:

  • $\frac{\partial v}{\partial x} = 2x$
  • $\frac{\partial v}{\partial y} = 2y$
  • $\frac{\partial v}{\partial z} = 2z$

• For $w = x + y + z$:

  • $\frac{\partial w}{\partial x} = 1$
  • $\frac{\partial w}{\partial y} = 1$
  • $\frac{\partial w}{\partial z} = 1$

Thus, the Jacobian matrix becomes:

$$ J = \begin{bmatrix} y + z & x + z & x + y \ 2x & 2y & 2z \ 1 & 1 & 1 \end{bmatrix} $$

3. Evaluate the Determinant of the Jacobian

To check for a functional relationship, we need to compute the determinant of the Jacobian matrix.

The determinant can be computed using the formula:

$$ \text{det}(J) = (y + z) \begin{vmatrix} 2y & 2z \ 1 & 1 \end{vmatrix}

• (x + z) \begin{vmatrix} 2x & 2z \ 1 & 1 \end{vmatrix}

• (x + y) \begin{vmatrix} 2x & 2y \ 1 & 1 \end{vmatrix} $$

Calculating these 2x2 determinants yields:

• $\begin{vmatrix} 2y & 2z \ 1 & 1 \end{vmatrix} = 2y - 2z = 2(y - z)$

• $\begin{vmatrix} 2x & 2z \ 1 & 1 \end{vmatrix} = 2x - 2z = 2(x - z)$

• $\begin{vmatrix} 2x & 2y \ 1 & 1 \end{vmatrix} = 2x - 2y = 2(x - y)$

Substituting back into our determinant leads to:

$$ \text{det}(J) = (y + z) \cdot 2(y - z) - (x + z) \cdot 2(x - z) + (x + y) \cdot 2(x - y) $$

4. Set the Determinant to Zero

For a functional relationship, we need to set the determinant to zero:

$$ (y + z)(y - z) - (x + z)(x - z) + (x + y)(x - y) = 0 $$

This equation can be simplified to check if it holds true for all values.

5. Identify the Functional Relationship

After evaluating the determinant and finding conditions under which it is zero, we can express the functional relationship in terms of $u$, $v$, and $w$.

It has been determined that there exists a function $f(u, v, w) = 0$.