Question

The directional derivative of \( f(x, y, z) = x^2 + 3y^2 + z^2 \), at point \( (2, 1, 0) \) along the unit vector in \( x \)-direction, \( \hat{i} \), is \(\underline{\hspace{2cm}}\).

Hint:

To compute the directional derivative, find the gradient and take the dot product with the unit vector in the given direction.

Answer 1

Correct Answer: 4

The directional derivative is given by:
\[ D_{\hat{v}} f = \nabla f \cdot \hat{v} \] where \( \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \) and \( \hat{v} \) is the unit vector in the given direction. First, compute the gradient \( \nabla f \):
\[ \frac{\partial f}{\partial x} = 2x, \frac{\partial f}{\partial y} = 6y, \frac{\partial f}{\partial z} = 2z \] Thus, \( \nabla f = (2x, 6y, 2z) \). At point \( (2, 1, 0) \):
\[ \nabla f(2, 1, 0) = (2 \times 2, 6 \times 1, 2 \times 0) = (4, 6, 0) \] The unit vector in the \( x \)-direction is \( \hat{i} = (1, 0, 0) \). Now, compute the dot product:
\[ D_{\hat{i}} f = (4, 6, 0) \cdot (1, 0, 0) = 4 \] Thus, the directional derivative is \( \boxed{4} \).