https://file.aiquickdraw.com/mathbot/file/1734147584202lblkh9gm.jpg

Let's analyze **Part (b)** of the problem step by step. The goal is to find the angle between the given velocity vector **\(\mathbf{v} = \langle 4, 2, 1 \rangle\)** and the gradient vector **\(\nabla f\)** at the point **\(P = (2, 1, 2)\)**.

---

### **Step 1: Compute the gradient of \(f(x, y, z)\)**
The level surface is given by \(f(x, y, z) = x^2 + yz - 6 = 0\). To find the gradient, compute the partial derivatives:

\[
\nabla f(x, y, z) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle.
\]

- \(\frac{\partial f}{\partial x} = 2x\),
- \(\frac{\partial f}{\partial y} = z\),
- \(\frac{\partial f}{\partial z} = y\).

Thus:

\[
\nabla f(x, y, z) = \langle 2x, z, y \rangle.
\]

---

### **Step 2: Evaluate \(\nabla f\) at the point \(P = (2, 1, 2)\)**
Substitute \(x = 2\), \(y = 1\), and \(z = 2\) into \(\nabla f(x, y, z)\):

\[
\nabla f(2, 1, 2) = \langle 2(2), 2, 1 \rangle = \langle 4, 2, 1 \rangle.
\]

---

### **Step 3: Find the angle between \(\mathbf{v}\) and \(\nabla f\)**
The angle \(\theta\) between two vectors \(\mathbf{v}\) and \(\nabla f\) is given by the formula:

\[
\cos\theta = \frac{\mathbf{v} \cdot \nabla f}{\|\mathbf{v}\| \, \|\nabla f\|}.
\]

#### Dot product:
\[
\mathbf{v} \cdot \nabla f = \langle 4, 2, 1 \rangle \cdot \langle 4, 2, 1 \rangle = (4)(4) + (2)(2) + (1)(1) = 16 + 4 + 1 = 21.
\]

#### Magnitudes:
\[
\|\mathbf{v}\| = \sqrt{4^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21},
\]
\[
\|\nabla f\| = \sqrt{4^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21}.
\]

#### Substitute into \(\cos\theta\):
\[
\cos\theta = \frac{\mathbf{v} \cdot \nabla f}{\|\mathbf{v}\| \, \|\nabla f\|} = \frac{21}{\sqrt{21} \cdot \sqrt{21}} = \frac{21}{21} = 1.
\]

---

### **Step 4: Interpret the result**
If \(\cos\theta = 1\), then \(\theta = 0^\circ\). This means the velocity vector \(\mathbf{v}\) and the gradient vector \(\nabla f\) are **parallel** at the point \(P\).

---

### **Final Answer**
The angle between \(\mathbf{v}\) and \(\nabla f\) is:

\[
\boxed{0^\circ}.
\]

Let me know if you need further clarification or want to explore related problems! 

---

#### **Five Related Questions:**
1. How do you find the gradient of a multivariable function in general?
2. What is the significance of the gradient being parallel to a direction vector on a surface?
3. Can you explain why \(\cos\theta\) is bounded between \(-1\) and \(1\)?
4. How would the calculation change if the velocity vector was not parallel to the gradient vector?
5. How can you use the gradient to determine whether a path is increasing or decreasing on a surface?

#### **Tip:**  
The gradient vector at a point is always **perpendicular** to the tangent plane of the surface at that point.

Learn how to compute the angle between a velocity vector and the gradient vector at a specific point on a level surface. This solution includes gradient computation, dot product application, and vector magnitudes. Perfect for multivariable 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