Question

The maximum value of the directional derivative of \(f=2x^2+3y^2+z^2\) at \((2,1,3)\) in the direction of \(\hat i+2\hat j-2\hat k\) is

• A. \(2\sqrt{34}\)
• B. \(\sqrt{34}\)
• C. \(\sqrt{2}\)
• D. \(3\sqrt{5}\)

Hint:

The maximum directional derivative of \(f\) at a point is \(|\nabla f|\).

Answer 1

The Correct Option is A

Concept:
The directional derivative of \(f\) in the direction of a unit vector \(\hat u\) is \[ \nabla f\cdot \hat u \] The maximum directional derivative at a point is the magnitude of the gradient: \[ |\nabla f| \]

Step 1: Find the gradient of \(f\). Given, \[ f=2x^2+3y^2+z^2 \] Therefore, \[ \nabla f= \frac{\partial f}{\partial x}\hat i+ \frac{\partial f}{\partial y}\hat j+ \frac{\partial f}{\partial z}\hat k \] \[ \nabla f=4x\hat i+6y\hat j+2z\hat k \]

Step 2: Substitute the point \((2,1,3)\). \[ \nabla f(2,1,3)=4(2)\hat i+6(1)\hat j+2(3)\hat k \] \[ \nabla f(2,1,3)=8\hat i+6\hat j+6\hat k \]

Step 3: Find maximum value. \[ \text{Maximum directional derivative}=|\nabla f| \] \[ |\nabla f|=\sqrt{8^2+6^2+6^2} \] \[ =\sqrt{64+36+36} \] \[ =\sqrt{136} \] \[ =2\sqrt{34} \]

Step 4: Final answer. \[ \boxed{2\sqrt{34}} \]