Question

The directional derivative of \(f(x,y,z)=4e^{2x-y+z}\) at the point \((1,1,-1)\) in the direction of the vector \(\vec{a}=-4\hat{i}+4\hat{j}+7\hat{k}\) is

• A. \(\dfrac{20}{9}\)
• B. \(\dfrac{48}{9}\)
• C. \(-\dfrac{20}{9}\)
• D. \(-\dfrac{48}{9}\)

Hint:

The directional derivative tells how fast a function changes if you move in a chosen direction.

Answer 1

The Correct Option is C

Concept:
The directional derivative tells how fast a function changes if you move in a chosen direction. You take the gradient (vector of partial derivatives), evaluate it at the point, and then dot it with the unit vector in the chosen direction.

Step 1:
Find the gradient of \(f=4e^{2x-y+z}\). Each partial derivative brings down the coefficient of that variable from the exponent: \[f_x = 8e^{2x-y+z},\quad f_y = -4e^{2x-y+z},\quad f_z = 4e^{2x-y+z}.\]

Step 2:
Evaluate the exponent at \((1,1,-1)\): \(2(1)-1+(-1)=0\), so \(e^0=1\). The gradient becomes \[\nabla f = (8,\,-4,\,4).\]

Step 3:
Make the direction a unit vector. The length of \(\vec{a}=(-4,4,7)\) is \(\sqrt{(-4)^2+4^2+7^2}=\sqrt{16+16+49}=\sqrt{81}=9\). So the unit vector is \(\dfrac{1}{9}(-4,4,7)\).

Step 4:
Dot the gradient with the unit direction: \[\nabla f \cdot \hat{a} = \frac{1}{9}\big[(8)(-4)+(-4)(4)+(4)(7)\big] = \frac{1}{9}\big[-32-16+28\big] = \frac{-20}{9}.\]

Answer: Option (3) — \(-\dfrac{20}{9}\).