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Find the curl, \(\nabla \times \mathbf{F},\) for the following forces: \((\mathbf{a}) \mathbf{F}=k \mathbf{r} ;(\mathbf{b}) \mathbf{F}=\left(A x, B y^{2}, C z^{3}\right) ;(\mathbf{c}) \mathbf{F}=\) \(\left(A y^{2}, B x, C z\right),\) where \(A, B, C\) and \(k\) are constants.

Short Answer

Expert verified
A: \(0 ;\) B: \(0 ;\) C: \((B-C, 0, -2Ay).\)

Step by step solution

01

Understanding the Curl

The curl of a vector field \( \mathbf{F} \) is calculated using the operator \( abla \times \mathbf{F} \). For a vector field \( \mathbf{F} = (F_x, F_y, F_z) \), the curl is given by the determinant: \[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_x & F_y & F_z \end{vmatrix} \] where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors in the \( x, y, \) and \( z \) directions.
02

Evaluating Curl for \( \mathbf{F} = k \mathbf{r} \) (Part A)

For part (a), \( \mathbf{F} = k \mathbf{r} = (kx, ky, kz) \). Calculate the determinant: \[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ kx & ky & kz \end{vmatrix} \]. Solve this, and you'll find that each element of the resultant vector becomes zero: \( 0 \).
03

Evaluating Curl for \( \mathbf{F} = (Ax, By^2, Cz^3) \) (Part B)

For part (b), \( \mathbf{F} = (Ax, By^2, Cz^3) \). Calculate: \[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ Ax & By^2 & Cz^3 \end{vmatrix} \]. After computing the partial derivatives, the result is \( (0, 0, 0) \), showing the components cancel or produce zero.
04

Evaluating Curl for \( \mathbf{F} = (Ay^2, Bx, Cz) \) (Part C)

For part (c), \( \mathbf{F} = (Ay^2, Bx, Cz) \). Compute: \[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ Ay^2 & Bx & Cz \end{vmatrix} \]. Upon solving, the resulting vector is \( (B-C, 0, -2Ay) \), indicating non-zero curl components.

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Key Concepts

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

Understanding Vector Fields
Vector fields are vital in physics and mathematics, especially when studying fluid dynamics and electromagnetic fields. A vector field assigns a vector to every point in space. This vector can represent anything from velocity in a fluid to the force at different points within an electric field.
For example:
  • In a fluid, each small volume element might have a vector describing its velocity, which collectively form a velocity field.
  • In electromagnetism, an electric field at a point can be described by a vector pointing in the direction of the force experienced by a positive test charge at that point.
Visualizing vector fields can help us understand rotational patterns, flows, and other motion types. These fields often have representations in 2D or 3D, depicted using arrows to show direction and magnitude. In calculations involving vector fields, tools like the gradient, divergence, and particularly the curl (as in our exercise) become essential to analyze properties like flow vorticity and rotation.
Introduction to Partial Derivatives
Partial derivatives play a significant role in multivariable calculus, especially in fields like physics and engineering where functions depend on multiple variables. When dealing with a function of two or more variables, the partial derivative measures how the function changes as one variable changes, keeping others constant.
For instance, consider a temperature field in a room, which might depend on three spatial variables: temperature changes can be assessed by differentiating with respect to one spatial dimension while keeping the others fixed.
In formulas:
  • The partial derivative of a function \( f(x, y) \) with respect to \( x \) is denoted \( \frac{\partial f}{\partial x} \).
  • It shows the rate of change in the \( x \)-direction.
  • To find the partial derivatives for \( \mathbf{F} = (Ax, By^2, Cz^3) \), we look at each component. For example, \( \frac{\partial (By^2)}{\partial x} = 0 \) since \( By^2 \) does not change with \( x \).
Understanding partial derivatives is crucial for evaluating expressions like the curl of a vector field, in which each component is derived through a sequence of these calculations, as seen in our step-by-step solutions.
Determinant of a Matrix and its Use in Curl Calculation
The determinant of a matrix is a unique number associated with a square matrix. It's a fundamental concept used in various applications, including solving systems of linear equations and transforming vector spaces. In the context of vector calculus, determinants play a crucial role in operations like finding the curl of a vector field.
To calculate the curl of a vector field \( \mathbf{F} = (F_x, F_y, F_z) \), we use a special type of determinant called the Jacobian determinant. It involves a vector component matrix and a variable's partial derivatives:
  • The determinant for curl: \( abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_x & F_y & F_z \end{vmatrix} \)
  • This 3x3 matrix involves unit vectors and partial derivatives, combined to give the curl's components.
Using determinants in this way provides an elegant method to compute vector field curl, revealing rotational components as shown in parts (a), (b), and (c) of our exercise. Determinants thus not only simplify but also underpin much of the computation needed in advanced calculus and physics.

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Most popular questions from this chapter

A mass \(m\) moves in a circular orbit (centered ón the origin) in the field of an attractive central force with potential energy \(U=k r^{n} .\) Prove the virial theorem that \(T=n U / 2\).

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

For a system of \(N\) particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, \(U=\sum_{\alpha} U_{\alpha}=M g Y\) where \(M=\sum m_{\alpha}\) is the total mass and \(\mathbf{R}=(X, Y, Z)\) is the position of the \(\mathrm{CM},\) with the \(y\) coordinate measured vertically up. [Hint: We know from Problem 4.5 that \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

A charge \(q\) in a uniform electric field \(\mathbf{E}_{0}\) experiences a constant force \(\mathbf{F}=q \mathbf{E}_{0}\). (a) Show that this force is conservative and verify that the potential energy of the charge at position \(\mathbf{r}\) is \(U(\mathbf{r})=-q \mathbf{E}_{0} \cdot \mathbf{r}\). (b) By doing the necessary derivatives, check that \(\mathbf{F}=-\nabla U\).

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