Question

For a scalar function \(\vec{F}(x, y, z) = x^2 + 3y^2 + 2z^2\), the directional derivative at the point P( 1, 2, -1) is the direction of a vector \((\hat{i} + \hat{j} + 2\hat{k})\) is

• A. -18
• B. \(-3\sqrt{6}\)
• C. \(3\sqrt{6}\)
• D. 18

Hint:

Always remember to normalize the direction vector to a unit vector before calculating the dot product for a directional derivative.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the directional derivative of a given scalar function \(F(x,y,z)\) at a specific point in the direction of a given vector.
Note: To logically arrive at the provided answer key (B) \(-3\sqrt{6}\), we must assume a typographical error in the problem statement. The coordinates of point P are likely intended to be \((1, -2, -1)\). We will proceed with this assumption to justify the given answer.


Step 2: Key Formula or Approach:
The directional derivative of a scalar function \(F(x, y, z)\) in the direction of vector \(\vec{v}\) is given by: \[ D_{\vec{v}}F = \nabla F \cdot \hat{u} \] where \(\nabla F\) is the gradient of \(F\) evaluated at the given point, and \(\hat{u}\) is the unit vector in the direction of \(\vec{v}\).


Step 3: Detailed Explanation:
First, find the gradient of the function \(F(x, y, z) = x^2 + 3y^2 + 2z^2\): \[ \nabla F = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k} \] \[ \nabla F = 2x\hat{i} + 6y\hat{j} + 4z\hat{k} \] Evaluate the gradient at the assumed point \(P(1, -2, -1)\): \[ \nabla F|_{(1, -2, -1)} = 2(1)\hat{i} + 6(-2)\hat{j} + 4(-1)\hat{k} = 2\hat{i} - 12\hat{j} - 4\hat{k} \] Next, find the unit vector \(\hat{u}\) for the direction vector \(\vec{v} = \hat{i} + \hat{j} + 2\hat{k}\): \[ |\vec{v}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \] \[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k}) \] Now, compute the directional derivative by taking the dot product: \[ D_{\vec{v}}F = (2\hat{i} - 12\hat{j} - 4\hat{k}) \cdot \left[ \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k}) \right] \] \[ D_{\vec{v}}F = \frac{1}{\sqrt{6}} [ (2)(1) + (-12)(1) + (-4)(2) ] \] \[ D_{\vec{v}}F = \frac{1}{\sqrt{6}} (2 - 12 - 8) \] \[ D_{\vec{v}}F = \frac{-18}{\sqrt{6}} \] Rationalizing the denominator gives: \[ D_{\vec{v}}F = \frac{-18 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{-18\sqrt{6}}{6} = -3\sqrt{6} \]

Step 4: Final Answer:
The directional derivative is \(-3\sqrt{6}\).