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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

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

Prove that if \(f(\mathbf{r})\) and \(g(\mathbf{r})\) are any two scalar functions of \(\mathbf{r},\) then \(\nabla(f g)=f \nabla g+g \nabla f\)

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

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 \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

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