Question

Let \( \phi(x, y, z) = 3y^2 + 3yz \) for \( (x, y, z) \in \mathbb{R}^3 \). Then the absolute value of the directional derivative of \( \phi \) 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 - 7.5

To find the absolute value of the directional derivative of \( \phi(x, y, z) = 3y^2 + 3yz \) in the direction of a given line at the point \( (1, -2, 1) \), we start by identifying the direction vector from the line's parametric form:

The line is given by \( \frac{x-1}{2} = \frac{y-2}{-1} = \frac{z}{-2} = t \). So, we can write the direction vector as \( \mathbf{v} = (2, -1, -2) \).

First, normalize the direction vector \( \mathbf{v} \):

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

The unit direction vector \( \mathbf{u} \) is then:

\( \mathbf{u} = \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \)

Now, calculate the gradient of \( \phi \), \( \nabla\phi(x, y, z) \):

\( \nabla\phi = \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) = (0, 6y+3z, 3y) \)

At the point \( (1, -2, 1) \), the gradient is:

\( \nabla\phi(1, -2, 1) = (0, 6(-2)+3(1), 3(-2)) = (0, -9, -6) \)

The directional derivative \( D_{\mathbf{u}}\phi \) is given by:

\( D_{\mathbf{u}}\phi = \nabla\phi \cdot \mathbf{u} = (0, -9, -6) \cdot \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \)

\( = 0 \cdot \frac{2}{3} + (-9) \cdot \left(-\frac{1}{3}\right) + (-6) \cdot \left(-\frac{2}{3}\right) \)

\( = 0 + 3 + 4 = 7 \)

Thus, the absolute value of the directional derivative is \( |7| = 7 \)