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a. Let \(\mathbf{a}=\langle 0,1,0\rangle, \mathbf{r}=\langle x, y, z\rangle,\) and consider the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r} .\) Use the right-hand rule for cross products to find the direction of \(\mathbf{F}\) at the points (0,1,1),(1,1,0),(0,1,-1), and (-1,1,0). b. With \(\mathbf{a}=\langle 0,1,0\rangle,\) explain why the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r}\) circles the \(y\) -axis in the counterclockwise direction looking along a from head to tail (that is, in the negative \(y\) -direction).

Short Answer

Expert verified
Using the right-hand rule and the cross product of vector \(\mathbf{a} = \langle 0, 1, 0 \rangle\) and position vector \(\mathbf{r} = \langle x, y, z \rangle\), we found that the rotation field \(\mathbf{F} = \langle 0, 0, x \rangle\). The direction of the rotation field at each point resulted in a counterclockwise rotation around the \(y\)-axis when looking from the negative \(y\)-axis towards the origin. This is because the field moves up along the positive \(z\)-axis when the point is on the positive \(x\)-axis and down along the negative \(z\)-axis when the point is on the negative \(x\)-axis. Points on the \(y\)-axis have no rotation, as \(\mathbf{F}\) is zero.

Step by step solution

01

Find the cross product of \(\mathbf{a}\) and \(\mathbf{r}\)

Recall the formula for the cross product of two vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\): $$ \mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle. $$ Applying this formula to \(\mathbf{a}\) and \(\mathbf{r}\), we get: $$ \mathbf{F} = \langle 0 \cdot z - 0 \cdot y, 0 \cdot x - 0 \cdot z, 1 \cdot x - 0 \cdot y \rangle = \langle 0, 0, x \rangle. $$
02

Substitute the given points into \(\mathbf{r}\) and calculate the values of \(\mathbf{F}\).

Using \(\mathbf{F} = \langle 0, 0, x \rangle\), we'll find the direction of \(\mathbf{F}\) at each point: a. For point (0,1,1), we have \(\mathbf{F} = \langle 0, 0, 0 \rangle\). b. For point (1,1,0), we have \(\mathbf{F} = \langle 0, 0, 1 \rangle\). c. For point (0,1,-1), we have \(\mathbf{F} = \langle 0, 0, 0 \rangle\). d. For point (-1,1,0), we have \(\mathbf{F} = \langle 0, 0, -1 \rangle\).
03

Use the right-hand rule to determine the direction of \(\mathbf{F}\) at each point.

Applying the right-hand rule on each point: a. At point (0,1,1), \(\mathbf{F} = \langle 0, 0, 0 \rangle\), there is no direction since the rotation field is zero. b. At point (1,1,0), \(\mathbf{F} = \langle 0, 0, 1 \rangle\), the direction is up along the positive \(z\)-axis. c. At point (0,1,-1), \(\mathbf{F} = \langle 0, 0, 0 \rangle\), there is no direction since the rotation field is zero. d. At point (-1,1,0), \(\mathbf{F} = \langle 0, 0, -1 \rangle\), the direction is down along the negative \(z\)-axis.
04

Explain why the rotation field circles the \(y\)-axis counterclockwise.

By observing the direction of \(\mathbf{F}\) at each point, we realize that the field moves up along the positive z-axis when the point is on the positive x-axis, and downwards along the negative z-axis when the point is on the negative x-axis. Additionally, points on the y-axis have no rotation, as \(\mathbf{F}\) is zero. Overall, the direction of the field results in a counterclockwise rotation around the \(y\)-axis when looking from negative \(y\)-axis towards the origin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right-Hand Rule
The right-hand rule is used to determine the direction of vectors resulting from the cross product of two vectors. It’s particularly useful in physics and engineering to determine rotations and forces. Here's how it works:

  • Point your index finger in the direction of the first vector, \( \mathbf{a} \).
  • Point your middle finger in the direction of the second vector, \( \mathbf{r} \).
  • Your thumb will then point in the direction of the resulting vector, \( \mathbf{F} \).

This rule helps us visualize the direction of the vector resulting from a cross product without performing any complex calculations. It's a quick, handy visualization technique.
For example, in the exercise, points that lie along the y-axis resulted in a zero vector \( \mathbf{F} \), meaning no direction. Points along the positive x-axis result in upward movement, while those along the negative x-axis move downward.
Cross Product
The cross product, or vector product, is a binary operation on two vectors producing a third vector orthogonal to both. Mathematically, if we have vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the cross product is given by:

\[\mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle.\]

This results in a vector perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \). In the exercise, the cross product \( \mathbf{a} \times \mathbf{r} \) developed the rotation field vector \( \mathbf{F} \) as \( \langle 0, 0, x \rangle \).

Here are key features of the cross product:
  • The magnitude of the cross product vector relates to the area of the parallelogram formed by the two original vectors.
  • The direction of the resulting vector is given by the right-hand rule.
  • It highlights properties of rotational fields, as seen in this exercise.
Rotation Field
A rotation field, like \( \mathbf{F} = \mathbf{a} \times \mathbf{r} \) from the exercise, is a field describing how objects rotate around an axis. Rotation fields are crucial in understanding fluid dynamics, electrical fields, and more.

Key characteristics include:
  • At some points, the rotation field can be zero, indicating no rotation or change (like the points on the y-axis in the exercise).
  • The rotation direction can shift from positive to negative based on the position, as seen when transitioning from positive to negative x-axis points.
  • These fields help visualize and predict the movement and behavior of rotating systems.

Understanding rotation fields offers insights into the physics of moving objects, essential in real-world applications like meteorology and mechanical engineering.
Coordinate Systems
Coordinate systems allow us to describe and analyze vectors in space. In the context of vector calculus, the standard coordinate system used is the Cartesian or rectangular coordinate system. It uses three axes—x, y, and z.
  • The x-axis typically runs horizontally.
  • The y-axis is vertical.
  • The z-axis is perpendicular to the x-y plane.

Coordinates like \((x, y, z)\) specify positions in a three-dimensional space. In the exercise, different points such as \((1, 1, 0)\) and \((-1, 1, 0)\) were used to demonstrate the change in the rotation field \( \mathbf{F} \) along with the axes.

When working with vectors and cross products, being comfortable with these coordinate concepts helps visualize and solve problems effectively. Coordinate systems form the foundation for mathematical calculations and interpretations in dimensions.

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Most popular questions from this chapter

\(\mathbb{R}^{2}\) Assume that the vector field \(\mathbf{F}\) is conservative in \(\mathbb{R}^{2}\), so that the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is independent of path. Use the following procedure to construct a potential function \(\varphi\) for the vector field \(\mathbf{F}=\langle f, g\rangle=\langle 2 x-y,-x+2 y\rangle\) a. Let \(A\) be (0,0) and let \(B\) be an arbitrary point \((x, y) .\) Define \(\varphi(x, y)\) to be the work required to move an object from \(A\) to \(B\) where \(\varphi(A)=0 .\) Let \(C_{1}\) be the path from \(A\) to \((x, 0)\) to \(B\) and let \(C_{2}\) be the path from \(A\) to \((0, y)\) to \(B .\) Draw a picture. b. Evaluate \(\int_{C_{1}} \mathbf{F} \cdot d \mathbf{r}=\int_{C_{1}} f d x+g d y\) and conclude that \(\varphi(x, y)=x^{2}-x y+y^{2}\) c. Verify that the same potential function is obtained by evaluating the line integral over \(C_{2}\)

Consider the vector field \(\mathbf{F}=\langle y, x\rangle\) shown in the figure. a. Compute the outward flux across the quarter circle \(C: \mathbf{r}(t)=\langle 2 \cos t, 2 \sin t\rangle,\) for \(0 \leq t \leq \pi / 2\) b. Compute the outward flux across the quarter circle \(C: \mathbf{r}(t)=\langle 2 \cos t, 2 \sin t\rangle,\) for \(\pi / 2 \leq t \leq \pi\) c. Explain why the flux across the quarter circle in the third quadrant equals the flux computed in part (a). d. Explain why the flux across the quarter circle in the fourth quadrant equals the flux computed in part (b). e. What is the outward flux across the full circle?

Show that \(\left|\mathbf{t}_{u} \times \mathbf{t}_{v}\right|=a^{2} \sin u\) for a sphere of radius \(a\) defined parametrically by \(\mathbf{r}(u, v)=\langle a \sin u \cos v, a \sin u \sin v, a \cos u\rangle,\) where \(0 \leq u \leq \pi\) and \(0 \leq v \leq 2 \pi\).

Consider the radial vector fields \(\mathbf{F}=\mathbf{r} /|\mathbf{r}|^{p},\) where \(p\) is a real number and \(\mathbf{r}=\langle x, y, z\rangle\) Let \(C\) be any circle in the \(x y\) -plane centered at the origin. a. Evaluate a line integral to show that the field has zero circulation on \(C\) b. For what values of \(p\) does Stokes' Theorem apply? For those values of \(p,\) use the surface integral in Stokes' Theorem to show that the field has zero circulation on \(C\).

The cone \(z^{2}=x^{2}+y^{2},\) for \(z \geq 0,\) cuts the sphere \(x^{2}+y^{2}+z^{2}=16\) along a curve \(C\) a. Find the surface area of the sphere below \(C,\) for \(z \geq 0\). b. Find the surface area of the sphere above \(C\). c. Find the surface area of the cone below \(C,\) for \(z \geq 0\).

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