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

Step 1: Set up the gradient.
For \(f=4e^{2x-y+z}\), each partial derivative just multiplies by the coefficient of that variable in the exponent: \(f_x=8e^{2x-y+z}\), \(f_y=-4e^{2x-y+z}\), \(f_z=4e^{2x-y+z}\).

Step 2: Evaluate the exponent at the point.
At \((1,1,-1)\): \(2(1)-1+(-1)=0\), so \(e^{0}=1\).

Step 3: Write the gradient vector.
\(\nabla f=(8,-4,4)\).

Step 4: Make the direction a unit vector.
For \(\vec a=(-4,4,7)\), its length is \(\sqrt{16+16+49}=\sqrt{81}=9\). So \(\hat a=\tfrac19(-4,4,7)\).

Step 5: Take the dot product.
\[\nabla f\cdot\hat a=\frac19\big(8(-4)+(-4)(4)+4(7)\big)=\frac19(-32-16+28).\]
Step 6: Simplify.
\[=\frac{-20}{9}.\] This is option 3.
\[ \boxed{-\dfrac{20}{9}} \]