The Terrain of a certain area follows the function f(x,y,z)= x^2+3xy +2z .If you are a the point (2,0,-1) , what is the change in the terrain if you begin to walk in the direction specified by the vector 2i + j - 2k

To determine the change in terrain in the direction specified by the vector \(\vec{d} = 2\vec{i} + \vec{j} - 2\vec{k}\), we need to calculate the directional derivative of \(f(x, y, z) = x^2 + 3xy + 2z\) at the point \((2, 0, -1)\) in the direction of \(\vec{d}\).

### Step 1: Compute the Gradient of \(f\)
The gradient of \(f(x, y, z)\), denoted \(\nabla f\), is:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
\]
Calculating each partial derivative:
- \(\frac{\partial f}{\partial x} = 2x + 3y\)
- \(\frac{\partial f}{\partial y} = 3x\)
- \(\frac{\partial f}{\partial z} = 2\)

### Step 2: Evaluate the Gradient at the Point \((2, 0, -1)\)
Substitute \(x = 2\), \(y = 0\), and \(z = -1\) into each partial derivative:
- \(\frac{\partial f}{\partial x} = 2(2) + 3(0) = 4\)
- \(\frac{\partial f}{\partial y} = 3(2) = 6\)
- \(\frac{\partial f}{\partial z} = 2\)

Thus, the gradient at \((2, 0, -1)\) is:
\[
\nabla f(2, 0, -1) = 4\vec{i} + 6\vec{j} + 2\vec{k}
\]

### Step 3: Normalize the Direction Vector \(\vec{d}\)
The direction vector is \(\vec{d} = 2\vec{i} + \vec{j} - 2\vec{k}\). To normalize it, find its magnitude:
\[
|\vec{d}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3
\]
The unit vector in the direction of \(\vec{d}\) is:
\[
\hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{2\vec{i} + \vec{j} - 2\vec{k}}{3} = \frac{2}{3}\vec{i} + \frac{1}{3}\vec{j} - \frac{2}{3}\vec{k}
\]

### Step 4: Compute the Directional Derivative
The directional derivative of \(f\) at \((2, 0, -1)\) in the direction of \(\hat{d}\) is given by:
\[
D_{\hat{d}} f(2, 0, -1) = \nabla f(2, 0, -1) \cdot \hat{d}
\]
Substitute \(\nabla f(2, 0, -1) = 4\vec{i} + 6\vec{j} + 2\vec{k}\) and \(\hat{d} = \frac{2}{3}\vec{i} + \frac{1}{3}\vec{j} - \frac{2}{3}\vec{k}\):
\[
D_{\hat{d}} f(2, 0, -1) = (4\vec{i} + 6\vec{j} + 2\vec{k}) \cdot \left( \frac{2}{3}\vec{i} + \frac{1}{3}\vec{j} - \frac{2}{3}\vec{k} \right)
\]
Now, calculate the dot product:
\[
D_{\hat{d}} f(2, 0, -1) = 4 \cdot \frac{2}{3} + 6 \cdot \frac{1}{3} + 2 \cdot \left( -\frac{2}{3} \right)
\]
\[
= \frac{8}{3} + 2 - \frac{4}{3}
\]
\[
= \frac{8 + 6 - 4}{3} = \frac{10}{3}
\]

### Final Answer
The change in the terrain as you begin to walk in the direction of \(\vec{d}\) from the point \((2, 0, -1)\) is:
\[
\frac{10}{3}
\]

Would you like further clarification on any part of this solution?

---

1. What is the physical significance of the gradient vector in a function representing terrain?
2. How does normalizing a vector affect the directional derivative?
3. What is the geometric interpretation of the directional derivative?
4. How would the result change if the direction vector were different?
5. Why is it necessary to use a unit vector in the directional derivative calculation?

**Tip:** Always check if the direction vector is a unit vector before calculating the directional derivative to avoid errors.

The Terrain of a certain area follows the function f(x,y,z)= x^2+3xy +2z .If you are at the point (2,0,-1), what is the change in the terrain if you begin to walk in the direction specified by the vector 2i + j - 2k?

Learn how to calculate the directional derivative of the terrain function f(x, y, z) = x² + 3xy + 2z at the point (2, 0, -1) in the direction of vector 2i + j - 2k. This solution explores key concepts in multivariable calculus, such as gradient vectors and directional derivatives, suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation