Find the directional derivative of 
f(x, y, z) = xy + yz + zx
 at 
P(3, −3, 3)
 in the direction from P to 
Q(2, 4, 5).

To find the **directional derivative** of the function \( f(x, y, z) = xy + yz + zx \) at the point \( P(3, -3, 3) \) in the direction from \( P \) to \( Q(2, 4, 5) \), we follow these steps:

### Step 1: Compute the Gradient of \( f \)
The gradient vector \( \nabla f(x, y, z) \) is the vector of partial derivatives of \( f \) with respect to each variable \( x \), \( y \), and \( z \).

\[
f(x, y, z) = xy + yz + zx
\]
Now, calculate the partial derivatives:

- \( \frac{\partial f}{\partial x} = y + z \)
- \( \frac{\partial f}{\partial y} = x + z \)
- \( \frac{\partial f}{\partial z} = x + y \)

So, the gradient is:
\[
\nabla f(x, y, z) = \left( y + z, x + z, x + y \right)
\]

### Step 2: Evaluate the Gradient at Point \( P(3, -3, 3) \)
Substitute \( x = 3 \), \( y = -3 \), and \( z = 3 \) into the gradient:

\[
\nabla f(3, -3, 3) = \left( (-3) + 3, 3 + 3, 3 + (-3) \right) = (0, 6, 0)
\]

So, the gradient at \( P \) is:
\[
\nabla f(3, -3, 3) = (0, 6, 0)
\]

### Step 3: Find the Unit Vector in the Direction from \( P \) to \( Q \)
The vector from \( P(3, -3, 3) \) to \( Q(2, 4, 5) \) is:
\[
\vec{PQ} = Q - P = (2 - 3, 4 - (-3), 5 - 3) = (-1, 7, 2)
\]

Now, calculate the magnitude of \( \vec{PQ} \):
\[
|\vec{PQ}| = \sqrt{(-1)^2 + 7^2 + 2^2} = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6}
\]

The unit vector in the direction from \( P \) to \( Q \) is:
\[
\hat{u} = \frac{\vec{PQ}}{|\vec{PQ}|} = \left( \frac{-1}{3\sqrt{6}}, \frac{7}{3\sqrt{6}}, \frac{2}{3\sqrt{6}} \right)
\]

### Step 4: Compute the Directional Derivative
The directional derivative \( D_{\hat{u}} f \) is given by the dot product of the gradient vector and the unit direction vector:
\[
D_{\hat{u}} f = \nabla f(3, -3, 3) \cdot \hat{u}
\]

Substitute the values:
\[
D_{\hat{u}} f = (0, 6, 0) \cdot \left( \frac{-1}{3\sqrt{6}}, \frac{7}{3\sqrt{6}}, \frac{2}{3\sqrt{6}} \right)
\]
\[
D_{\hat{u}} f = 0 \cdot \frac{-1}{3\sqrt{6}} + 6 \cdot \frac{7}{3\sqrt{6}} + 0 \cdot \frac{2}{3\sqrt{6}} = 6 \cdot \frac{7}{3\sqrt{6}} = \frac{42}{3\sqrt{6}} = \frac{14}{\sqrt{6}}
\]

### Step 5: Simplify the Result
We can rationalize the denominator:
\[
D_{\hat{u}} f = \frac{14}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{14\sqrt{6}}{6} = \frac{7\sqrt{6}}{3}
\]

Thus, the directional derivative of \( f \) at \( P(3, -3, 3) \) in the direction from \( P \) to \( Q(2, 4, 5) \) is:

\[
D_{\hat{u}} f = \frac{7\sqrt{6}}{3}
\]

---

Would you like a more detailed explanation or further clarifications?

Here are some related questions you can try:
1. How do you find the gradient of a scalar field?
2. What is the physical interpretation of a directional derivative?
3. How do you compute the unit vector in a given direction?
4. How does the gradient relate to the rate of change of a function?
5. Can you find the maximum rate of change of \( f \) at point \( P(3, -3, 3) \)?

**Tip:** The gradient vector points in the direction of the steepest increase of a function at a given point.

Find the directional derivative of f(x, y, z) = xy + yz + zx at P(3, −3, 3) in the direction from P to Q(2, 4, 5).

Learn how to compute the directional derivative of the multivariable function f(x, y, z) = xy + yz + zx at the point P(3, -3, 3) in the direction towards Q(2, 4, 5). This problem covers gradients, vector calculations, and the directional derivative formula. Suitable for undergraduate Calculus 3 students.

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