Question

The rate of increase, of a scalar field $f(x,y,z) = xyz$, in the direction $v=(2,1,2)$ at a point $(0,2,1)$ is

• A. $\dfrac{2}{3}$
• B. $\dfrac{4}{3}$
• C. $2$
• D. $4$

Hint:

The rate of increase of a scalar field in a direction $v$ is the directional derivative: $\nabla f . \dfrac{v}{|v|}$. Always normalize the direction vector before dotting.

Answer 1

The Correct Option is B

Step 1: Gradient of $f(x,y,z)$.
\[ f(x,y,z) = xyz \] \[ \nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (yz, \; xz, \; xy). \] Step 2: Evaluate at $(0,2,1)$.
At $(0,2,1)$: \[ \nabla f = (2. 1, \; 0. 1, \; 0. 2) = (2,0,0). \] Step 3: Direction vector.
Given direction $v=(2,1,2)$. Unit vector: \[ \hat{v} = \frac{(2,1,2)}{\sqrt{2^2+1^2+2^2}} = \frac{(2,1,2)}{\sqrt{9}} = \left(\tfrac{2}{3}, \tfrac{1}{3}, \tfrac{2}{3}\right). \] Step 4: Directional derivative.
\[ D_v f = \nabla f . \hat{v} = (2,0,0) . \left(\tfrac{2}{3}, \tfrac{1}{3}, \tfrac{2}{3}\right) = 2. \tfrac{2}{3} = \tfrac{4}{3}. \] \[ \boxed{\tfrac{4}{3}} \]