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

Step 1: Compute the gradient of $f(x, y, z)$

The gradient of $f$ is the vector of partial derivatives with respect to $x$, $y$, and $z$.

• $\frac{\partial f}{\partial x} = 2xy + \sqrt{1 + z}$
• $\frac{\partial f}{\partial y} = x^2$
• $\frac{\partial f}{\partial z} = \frac{x}{2\sqrt{1 + z}}$

Step 2: Evaluate the gradient at the point $(1, 2, 3)$

• $\frac{\partial f}{\partial x}(1, 2, 3) = 2(1)(2) + \sqrt{1 + 3} = 4 + 2 = 6$
• $\frac{\partial f}{\partial y}(1, 2, 3) = (1)^2 = 1$
• $\frac{\partial f}{\partial z}(1, 2, 3) = \frac{1}{2\sqrt{1 + 3}} = \frac{1}{2 \times 2} = \frac{1}{4}$

Thus, the gradient of $f$ at $(1, 2, 3)$ is: $\nabla f(1, 2, 3) = \langle 6, 1, \frac{1}{4} \rangle$

Step 3: Normalize the direction vector $\mathbf{v} = \langle 2, 1, -2 \rangle$

To normalize the direction vector, we compute its magnitude: $\|\mathbf{v}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ Thus, the unit vector in the direction of $\mathbf{v}$ is: $\mathbf{u} = \left\langle \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \right\rangle$

Step 4: Compute the directional derivative

The directional derivative is given by the dot product of the gradient and the unit vector: $D_{\mathbf{u}} f(1, 2, 3) = \nabla f(1, 2, 3) \cdot \mathbf{u}$ $D_{\mathbf{u}} f(1, 2, 3) = \left\langle 6, 1, \frac{1}{4} \right\rangle \cdot \left\langle \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \right\rangle$ $D_{\mathbf{u}} f(1, 2, 3) = 6 \times \frac{2}{3} + 1 \times \frac{1}{3} + \frac{1}{4} \times \frac{-2}{3}$ $D_{\mathbf{u}} f(1, 2, 3) = 4 + \frac{1}{3} - \frac{1}{6}$ We now compute this: $D_{\mathbf{u}} f(1, 2, 3) = 4 + \frac{2}{6} - \frac{1}{6} = 4 + \frac{1}{6} = 4.1667$

Final Answer:

The directional derivative is approximately $\boxed{4.17}$.

Would you like more details on any of these steps or further clarifications?

Here are 5 related questions you might find interesting:

1. How would the directional derivative change if the direction vector were different?
2. What is the geometrical interpretation of the gradient vector?
3. How do you find the directional derivative for functions with more variables?
4. Can we compute directional derivatives if the direction vector is not normalized?
5. How does the magnitude of the direction vector affect the directional derivative?

Tip: The directional derivative tells you how fast a function is changing in a specific direction from a given point.