Find the directional derivative of f(x,y,z)=z3−x2y at the point (-2, 5, 2) in the direction of the vector v=⟨−3,−1,2⟩.

To find the directional derivative of the function \( f(x, y, z) = z^3 - x^2 y \) at the point \((-2, 5, 2)\) in the direction of the vector \( \mathbf{v} = \langle -3, -1, 2 \rangle \), follow these steps:

### Step 1: Find the gradient of the function \( f(x, y, z) \)
The gradient of \( f(x, y, z) \), denoted by \( \nabla f(x, y, z) \), is the vector of partial derivatives:

\[
\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
\]

1. **Partial derivative with respect to \(x\):**
   \[
   \frac{\partial f}{\partial x} = -2xy
   \]
2. **Partial derivative with respect to \(y\):**
   \[
   \frac{\partial f}{\partial y} = -x^2
   \]
3. **Partial derivative with respect to \(z\):**
   \[
   \frac{\partial f}{\partial z} = 3z^2
   \]

Thus, the gradient is:

\[
\nabla f(x, y, z) = \langle -2xy, -x^2, 3z^2 \rangle
\]

### Step 2: Evaluate the gradient at the point \((-2, 5, 2)\)
Substitute \(x = -2\), \(y = 5\), and \(z = 2\) into the gradient:

\[
\nabla f(-2, 5, 2) = \langle -2(-2)(5), -(-2)^2, 3(2)^2 \rangle
\]
\[
\nabla f(-2, 5, 2) = \langle 20, -4, 12 \rangle
\]

### Step 3: Normalize the direction vector \( \mathbf{v} \)
The direction vector \( \mathbf{v} = \langle -3, -1, 2 \rangle \) needs to be normalized:

\[
|\mathbf{v}| = \sqrt{(-3)^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}
\]

The normalized vector is:

\[
\mathbf{u} = \frac{1}{\sqrt{14}} \langle -3, -1, 2 \rangle = \langle \frac{-3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \rangle
\]

### Step 4: Compute the directional derivative
The directional derivative of \( f(x, y, z) \) at the point \((-2, 5, 2)\) in the direction of \( \mathbf{v} \) is given by the dot product of the gradient and the normalized vector:

\[
D_{\mathbf{v}} f(-2, 5, 2) = \nabla f(-2, 5, 2) \cdot \mathbf{u}
\]
\[
D_{\mathbf{v}} f(-2, 5, 2) = \langle 20, -4, 12 \rangle \cdot \langle \frac{-3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \rangle
\]
\[
D_{\mathbf{v}} f(-2, 5, 2) = \frac{1}{\sqrt{14}} (20(-3) + (-4)(-1) + 12(2))
\]
\[
D_{\mathbf{v}} f(-2, 5, 2) = \frac{1}{\sqrt{14}} (-60 + 4 + 24)
\]
\[
D_{\mathbf{v}} f(-2, 5, 2) = \frac{1}{\sqrt{14}} (-32)
\]
\[
D_{\mathbf{v}} f(-2, 5, 2) = \frac{-32}{\sqrt{14}}
\]

Thus, the directional derivative is:

\[
D_{\mathbf{v}} f(-2, 5, 2) = \frac{-32}{\sqrt{14}}
\]

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

Here are 5 related questions:
1. How do you interpret the magnitude of the directional derivative?
2. What does it mean if the directional derivative is positive or negative?
3. How would the directional derivative change if the direction vector were different?
4. Can you find the maximum rate of change of the function at a given point?
5. How do gradient vectors relate to level curves or surfaces?

**Tip:** The magnitude of the gradient indicates the steepest ascent direction and rate of change at a point.

Find the directional derivative of f(x,y,z)=z^3−x^2y at the point (-2, 5, 2) in the direction of the vector v=⟨−3,−1,2⟩.

Find the directional derivative of f(x, y, z) = z^3 − x^2 y at the point (-2, 5, 2) in the direction of vector ⟨−3, −1, 2⟩. Learn how to compute the gradient, normalize the vector, and apply the dot product to get the solution. Suitable for Grades 11-12 and early college-level students.

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