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:

Always normalize the direction vector before computing a directional derivative. The dot product of the gradient and the unit direction vector gives the desired result.

Answer 1

The Correct Option is C

Step 1: Compute the gradient of \( f(x, y, z) = xyz \):
\[ \nabla f = \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right) = (yz, xz, xy) \] At the point \( (1, 2, 3) \), we get: \[ \nabla f(1, 2, 3) = (2 \cdot 3, 1 \cdot 3, 1 \cdot 2) = (6, 3, 2) \] Step 2: Normalize the direction vector \( \vec{v} = \langle 2, 1, -2 \rangle \):
\[ |\vec{v}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] So the unit vector is \[ \hat{u} = \left\langle \dfrac{2}{3}, \dfrac{1}{3}, \dfrac{-2}{3} \right\rangle \] Step 3: Use the formula for directional derivative:
\[ D_{\hat{u}} f = \nabla f \cdot \hat{u} = (6, 3, 2) \cdot \left( \dfrac{2}{3}, \dfrac{1}{3}, \dfrac{-2}{3} \right) \] \[ = 6 \cdot \dfrac{2}{3} + 3 \cdot \dfrac{1}{3} + 2 \cdot \left( \dfrac{-2}{3} \right) = \dfrac{12}{3} + \dfrac{3}{3} - \dfrac{4}{3} = \dfrac{11}{3} \] Final Answer: \( \boxed{\dfrac{11}{3}} \)