Question

Let \( \varphi(x, y, z) = 3y^2 + 3xyz \) for \( (x, y, z) \in \mathbb{R}^3 \). Then the absolute value of the directional derivative of \( \varphi \) in the direction of the line \[ \frac{x-1}{2} = \frac{y-2}{-1} = \frac{z}{-2} \] at the point \( (1, -2, 1) \) is .............

Hint:

The directional derivative is the dot product of the gradient vector and the unit direction vector.

Answer 1

Correct Answer: 6.5

Step 1: Direction vector.
The direction vector is given by \( \mathbf{v} = \left( \frac{1}{2}, -1, -2 \right) \), as the parametric equations can be written as \( x = 1 + 2t, y = 2 - t, z = -2t \).

Step 2: Gradient of \( \varphi \).
The gradient of \( \varphi(x, y, z) \) is: \[ \nabla \varphi = \left( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z} \right) = \left( 3yz, 6y + 3xz, 3xy \right) \] At the point \( (1, -2, 1) \), we evaluate the gradient: \[ \nabla \varphi(1, -2, 1) = \left( 3(-2)(1), 6(-2) + 3(1)(-2), 3(1)(-2) \right) = (-6, -18, -6) \]

Step 3: Directional derivative.
The directional derivative is given by \( \nabla \varphi \cdot \mathbf{v} \). We calculate: \[ \nabla \varphi \cdot \mathbf{v} = (-6, -18, -6) \cdot \left( \frac{1}{2}, -1, -2 \right) = -6 \times \frac{1}{2} + (-18) \times (-1) + (-6) \times (-2) \] \[ = -3 + 18 + 12 = 27 \] Thus, the absolute value of the directional derivative is \( 27 \).

Step 4: Conclusion.
The absolute value of the directional derivative is 27.