Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a head-on elastic collision between two particles. (since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, \(v_{1}-v_{2}=-\left(v_{1}^{\prime}-v_{2}^{\prime}\right),\) where \(v_{1}\) and \(v_{2}\) are the initial velocities and \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) the corresponding final velocities.

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

By writing a \(\cdot \mathbf{b}\) in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is, \(\frac{d}{d t}(\mathbf{a} \cdot \mathbf{b})=\frac{d \mathbf{a}}{d t} \cdot \mathbf{b}+\mathbf{a} \cdot \frac{d \mathbf{b}}{d t}\).

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

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free