Find the directional derivative of the function $f(x, y, z) = x^2 + y^2 + z^2$ at the point $p = (1, -1, 2)$ in the direction of the vector $a = \sqrt{2}i - j - k$.

1. Problem Description

Find the directional derivative of the function $f(x, y, z) = x^2 + y^2 + z^2$ at the point $p = (1, -1, 2)$ in the direction of the vector $a = \sqrt{2}i - j - k$.

2. Solution Steps

First, we need to find the gradient of $f(x, y, z)$:

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

$$\frac{\partial f}{\partial x} = 2x$$

$$\frac{\partial f}{\partial y} = 2y$$

$$\frac{\partial f}{\partial z} = 2z$$

Thus, $\nabla f(x, y, z) = (2x, 2y, 2z)$.

Now, evaluate the gradient at the point $p = (1, -1, 2)$:

$$\nabla f(1, -1, 2) = (2(1), 2(-1), 2(2)) = (2, -2, 4)$$

Next, we need to find the unit vector in the direction of $a = \sqrt{2}i - j - k$. First, find the magnitude of $a$:

$$||a|| = \sqrt{(\sqrt{2})^2 + (-1)^2 + (-1)^2} = \sqrt{2 + 1 + 1} = \sqrt{4} = 2$$

Now, find the unit vector $\hat{u}$ in the direction of $a$:

$$\hat{u} = \frac{a}{||a||} = \frac{(\sqrt{2}, -1, -1)}{2} = \left(\frac{\sqrt{2}}{2}, -\frac{1}{2}, -\frac{1}{2}\right)$$

Finally, the directional derivative of $f$ at $p$ in the direction of $a$ is given by:

$$D_u f(p) = \nabla f(p) \cdot \hat{u}$$

$$D_u f(1, -1, 2) = (2, -2, 4) \cdot \left(\frac{\sqrt{2}}{2}, -\frac{1}{2}, -\frac{1}{2}\right) = 2\left(\frac{\sqrt{2}}{2}\right) - 2\left(-\frac{1}{2}\right) + 4\left(-\frac{1}{2}\right) = \sqrt{2} + 1 - 2 = \sqrt{2} - 1$$

3. Final Answer

The directional derivative of $f$ at the point $p$ in the direction of $a$ is $\sqrt{2} - 1$.