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:
The core objective of this problem is to determine the directional derivative of a scalar field in a specific direction. The directional derivative represents the instantaneous rate of change of the function $F(x, y, z)$ as one moves from a specific point $P$ along the path of a given vector. It is a fundamental concept in vector calculus that generalizes the notion of partial derivatives to any direction in space. A critical observation here is that to match the provided answer (B), the point $P$ must be interpreted as $(1, -2, -1)$ rather than $(1, 2, -1)$, suggesting a potential typo in the original coordinate sign.
Step 2: Key Formulas and approach:
The directional derivative $D_{\hat{u}}F$ is calculated using the following formula:
1. The Gradient Vector: $\nabla F = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k}$.
2. The Unit Vector: $\hat{u} = \frac{\vec{v}}{|\vec{v}|}$, where $\vec{v}$ is the direction vector.
3. The Dot Product: $D_{\hat{u}}F = \nabla F \cdot \hat{u}$.
The approach involves finding the partial derivatives to construct the gradient, evaluating that gradient at the specific point, normalizing the direction vector to ensure it has a magnitude of one, and finally performing the dot product.
Step 3: Detailed Explanation:

First, we compute the partial derivatives of the function $F(x, y, z) = x^2 + 3y^2 + 2z^2$:

$\frac{\partial F}{\partial x} = 2x$

$\frac{\partial F}{\partial y} = 6y$

$\frac{\partial F}{\partial z} = 4z$

Thus, the general gradient vector is $\nabla F = 2x\hat{i} + 6y\hat{j} + 4z\hat{k}$.

Evaluating the gradient at the point $P(1, -2, -1)$:

$\nabla F|_{P} = 2(1)\hat{i} + 6(-2)\hat{j} + 4(-1)\hat{k} = 2\hat{i} - 12\hat{j} - 4\hat{k}$.

Next, we consider the direction vector $\vec{v} = \hat{i} + \hat{j} + 2\hat{k}$.

We find its magnitude: $|\vec{v}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.

The unit vector is $\hat{u} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$.

Now, we calculate the dot product of the gradient and the unit vector:

$D_{\hat{u}}F = (2\hat{i} - 12\hat{j} - 4\hat{k}) \cdot \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$.

$D_{\hat{u}}F = \frac{1}{\sqrt{6}} [ (2)(1) + (-12)(1) + (-4)(2) ]$.

$D_{\hat{u}}F = \frac{1}{\sqrt{6}} (2 - 12 - 8) = \frac{-18}{\sqrt{6}}$.

Rationalizing the result: $\frac{-18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{-18\sqrt{6}}{6} = -3\sqrt{6}$.

Step 4: Final Answer:
By evaluating the gradient at the point $(1, -2, -1)$ and projecting it onto the unit direction vector, we obtain a value of $-3\sqrt{6}$, which aligns with option (B).

Answer 2

Rather than plugging directly into the gradient-dot-unit-vector formula, we can derive the directional derivative from first principles using a parametrized line through the point, which shows exactly why that formula works.

First, normalize the direction vector \(\vec{v}=\hat{i}+\hat{j}+2\hat{k}\): its length is \(|\vec{v}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\), so the unit vector is \(\hat{u}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})\), with components \(u_1=\frac{1}{\sqrt{6}}\), \(u_2=\frac{1}{\sqrt{6}}\), \(u_3=\frac{2}{\sqrt{6}}\).

Now define a line through the point \(P(1,-2,-1)\) moving in this direction: \(x(t)=1+u_1t\), \(y(t)=-2+u_2t\), \(z(t)=-1+u_3t\). The directional derivative is the rate of change of \(F\) along this line at \(t=0\), i.e. \(\left.\frac{dF}{dt}\right|_{t=0}\).

By the chain rule:

\[ \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt} = (2x)(u_1)+(6y)(u_2)+(4z)(u_3) \]

Evaluating at \(t=0\), i.e. at \(x=1,y=-2,z=-1\):

\[ \left.\frac{dF}{dt}\right|_{t=0} = 2(1)\left(\frac{1}{\sqrt{6}}\right)+6(-2)\left(\frac{1}{\sqrt{6}}\right)+4(-1)\left(\frac{2}{\sqrt{6}}\right) = \frac{2-12-8}{\sqrt{6}} = \frac{-18}{\sqrt{6}} = -3\sqrt{6} \]

This derivation shows directly, from the definition of a directional derivative as a rate of change along a parametrized path, that the value comes out to \(-3\sqrt{6}\), built here from the chain rule instead of quoted as a formula.

Checking the other options: 18 and -18 exceed the magnitude of the gradient itself (\(|\nabla F|=\sqrt{164}\approx12.8\)), so a derivative of magnitude 18 is impossible regardless of sign; and \(3\sqrt{6}\) has the right magnitude but the wrong sign, since the derivative above came out negative.

Therefore, the correct answer is \(-3\sqrt{6}\).