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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

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]

Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

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 \(T=n U / 2\).

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