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 $P = (2, 1, 2)$.

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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:

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

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Step 2: Evaluate $\nabla f$ at the point $P = (2, 1, 2)$

Substitute $x = 2$, $y = 1$, and $z = 2$ into $\nabla f(x, y, z)$:

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

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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 $\cos\theta$:

$\cos\theta = \frac{\mathbf{v} \cdot \nabla f}{\|\mathbf{v}\| \, \|\nabla f\|} = \frac{21}{\sqrt{21} \cdot \sqrt{21}} = \frac{21}{21} = 1.$

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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 $P$.

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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!

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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.