Question

The directional derivative of \( f(x, y, z) = xyz \) at the point \( (1, 2, 3) \) in the direction of the vector \( 2\hat{i} + \hat{j} - 2\hat{k} \) is ............

• A. $\dfrac{5}{3}$
• B. $\dfrac{-5}{3}$
• C. $\dfrac{11}{3}$
• D. $\dfrac{19}{3}$

Hint:

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

Answer 1

The Correct Option is C

The directional derivative of a function $f(x, y, z)$ in the direction of a vector $\mathbf{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula: \[ D_{\mathbf{v}}f = \nabla f \cdot \hat{v} \] where $\nabla f$ is the gradient of the function and $\hat{v}$ is the unit vector in the direction of $\mathbf{v}$. 1. First, compute the gradient of $f(x, y, z) = xyz$: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \] \[ \frac{\partial f}{\partial x} = yz, \frac{\partial f}{\partial y} = xz, \frac{\partial f}{\partial z} = xy \] Thus, \[ \nabla f = (yz, xz, xy) \] 2. Now, evaluate the gradient at the point $(1, 2, 3)$: \[ \nabla f(1, 2, 3) = (2 \times 3, 1 \times 3, 1 \times 2) = (6, 3, 2) \] 3. Next, find the unit vector in the direction of the given vector $\mathbf{v} = 2\hat{i} + \hat{j} - 2\hat{k}$: The magnitude of $\mathbf{v}$ is: \[ |\mathbf{v}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] Thus, the unit vector is: \[ \hat{v} = \frac{1}{3} (2\hat{i} + \hat{j} - 2\hat{k}) = \left( \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \right) \] 4. Finally, compute the directional derivative: \[ D_{\mathbf{v}}f = \nabla f \cdot \hat{v} = (6, 3, 2) \cdot \left( \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \right) \] \[ D_{\mathbf{v}}f = 6 \times \frac{2}{3} + 3 \times \frac{1}{3} + 2 \times \frac{-2}{3} = 4 + 1 - \frac{4}{3} = \frac{12}{3} + \frac{3}{3} - \frac{4}{3} = \frac{11}{3} \] Thus, the directional derivative is $\dfrac{11}{3}$.