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Find the divergence and curl of the given vector field. \(\vec{F}=\nabla f,\) where \(f(x, y, z)=\frac{1}{x^{2}+y^{2}+z^{2}}\)

Short Answer

Expert verified
The divergence is 0 and the curl is 0.

Step by step solution

01

Compute the Gradient

The vector field \( \vec{F} \) is given as \( \vec{F} = abla f \). First, compute the gradient of the scalar field \( f(x, y, z) = \frac{1}{x^{2}+y^{2}+z^{2}} \). The gradient is calculated as follows: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
02

Partial Derivatives

Calculate the partial derivatives required for the gradient:- \( \frac{\partial f}{\partial x} = -\frac{2x}{(x^{2}+y^{2}+z^{2})^2}\)- \( \frac{\partial f}{\partial y} = -\frac{2y}{(x^{2}+y^{2}+z^{2})^2}\)- \( \frac{\partial f}{\partial z} = -\frac{2z}{(x^{2}+y^{2}+z^{2})^2}\).
03

Gradient of f

Substitute the partial derivatives back into the gradient formula to get \( abla f = \left( -\frac{2x}{(x^{2}+y^{2}+z^{2})^2}, -\frac{2y}{(x^{2}+y^{2}+z^{2})^2}, -\frac{2z}{(x^{2}+y^{2}+z^{2})^2} \right) \).
04

Compute the Divergence

The divergence of a vector field \( \vec{F} = (F_1, F_2, F_3) \) is \( abla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). Calculation shows that all these terms simplify to zero for radial symmetric potentials expressed in this form: \( abla \cdot abla f = 0 \).
05

Explanation of Zero Divergence

Since \( f(x, y, z) \) is a potential field, its gradient represents a conservative field. The divergence of a conservative vector field created by the gradient of \( 1/r \) is typically zero everywhere except at infinity or undefined points.
06

Compute the Curl

The curl of a vector field \( \vec{F} \) is \( abla \times \vec{F} \). Since \( \vec{F} \) is a gradient, and the curl of a gradient is always zero, we have \( abla \times abla f = 0 \).
07

Explanation of Zero Curl

\( \vec{F} \) being a gradient field implies it's irrotational. Thus, its curl, \( abla \times \vec{F} \), is zero by default for all well-behaved scalar fields.

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Key Concepts

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

Gradient
The gradient is an essential concept in vector calculus, representing how a function changes at any given point. Imagine standing on a hill; the gradient points you in the direction of the steepest ascent and tells you how steep that climb will be. Formally, for a scalar field \( f(x, y, z) \), the gradient \( abla f \) is a vector. It consists of partial derivatives and shows how \( f \) changes along each axis.

For instance, given \( f(x, y, z) = \frac{1}{x^2 + y^2 + z^2} \), the gradient is derived as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). Each partial derivative represents how the function \( f \) changes when you tweak just one variable, keeping the others constant. This results in the vector:
  • \( \frac{\partial f}{\partial x} = -\frac{2x}{(x^2 + y^2 + z^2)^2} \)
  • \( \frac{\partial f}{\partial y} = -\frac{2y}{(x^2 + y^2 + z^2)^2} \)
  • \( \frac{\partial f}{\partial z} = -\frac{2z}{(x^2 + y^2 + z^2)^2} \)
Each component of this vector tells the rate of change of \( f \) in one dimensional direction while the others remain fixed.
Divergence
Divergence is a scalar measure of a vector field's tendency to originate from or converge at a point. Visualize it as how much a fluid spreads out from a point; higher divergence means more 'source-like' behavior.

Mathematically, for a vector field \( \vec{F} = (F_1, F_2, F_3) \), the divergence is expressed as \( abla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). When you compute the divergence of the gradient of a scalar field like \( f(x, y, z) \), especially when it is a potential field like \( f = \frac{1}{r} \), you often get zero.

This happens because conservative fields, which are formed from the gradient of potential functions, have no "net flow" out of or into any volume that isn't at infinity or undefined points. The zero divergence \( abla \cdot abla f = 0 \) shows the field is balanced without any net output or input at most points.
Curl
Understanding curl helps you grasp rotational motion within a vector field. If you imagine a paddle wheel placed in a fluid flow, the curl tells you how much that wheel would spin.

In mathematical terms, a vector field \( \vec{F} \) has its curl defined as \( abla \times \vec{F} \). It's a vector operator that describes the infinitesimal rotation at a point. For a gradient vector field, like the one you find from a scalar function \( f \), the curl is usually zero. This is because a gradient field is irrotational by nature.

The formula \( abla \times abla f = 0 \) simplifies the understanding of conservative fields as they don't create circulations or rotations. Thus, the zero curl confirms the absence of any rotational component in such a field. This makes gradient fields an important example of irrotational fields.
Scalar Field
A scalar field assigns a single value to each point in space, and it can model a variety of phenomena, like temperature or electric potential. Imagine a map of elevations; every point on the map has a specific height value, which you can think of as a scalar.

When you're dealing with vector calculus, the role of a scalar field is crucial. It helps define vector fields through operations like the gradient. For instance, in the exercise with \( f(x, y, z) = \frac{1}{x^2 + y^2 + z^2} \), this is a scalar field describing how potential changes over space.

Scalar fields serve as the foundation for understanding gradients, as they provide the initial values from which vector fields are derived. They are key in transitioning from simple one-dimensional values to complex multi-dimensional analyses.
Conservative Vector Fields
Conservative vector fields are an essential concept where the path taken doesn’t influence the work done between two points. This makes them path-independent. A typical example is gravity, where it doesn't matter how you climb a hill; you only need to know the change in height to determine the work done.

For a vector field to be conservative, it must be the gradient of some scalar field. This is because such fields have zero curl, signifying no circulation at any point, and zero divergence outside "singular" points. In the given exercise, the vector field \( \vec{F} = abla f \) is conservative.

The ability of a field to be expressed as a gradient makes it easier to solve complex problems, as many powerful theorems apply directly to conservative fields. These properties simplify analysis and calculations in physics and engineering, where energy conservation is key.

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Most popular questions from this chapter

In Exercises \(11-14,\) a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Set up the line integral \(\int_{C} f(s) d s,\) then approximate its value using technology. $$ \begin{aligned} &C \text { is the ellipse given by } \vec{r}(t)=\langle 2 \cos t, \sin t\rangle \text { on }[0,2 \pi] ; \text { the }\\\ &\text { surface is } f(x, y)=10-x^{2}-y^{2} \text { . } \end{aligned} $$

A closed curve \(C\) enclosing a region \(R\) is given. Find the area of \(R\) by computing \(\oint_{C} \vec{F} \cdot d \vec{r}\) for an appropriate choice of vector field \(\vec{F}\). \(C\) is the curve parametrized by \(\vec{r}(t)=\langle\cos t, \sin (2 t)\rangle\) on \(-\pi / 2 \leq t \leq \pi / 2\).

In Exercises \(13-16,\) find the work performed by the force field \(\vec{F}\) moving a particle along the path \(C\). \(\vec{F}=\left\langle 2 x y, x^{2}, 1\right\rangle\) lbs; \(C\) is the path from (0,0,0) to (2,4,8) via \(\vec{r}(t)=\langle t, 2 t, 4 t\rangle\) on \(0 \leq t \leq 2,\) where distance are measured in feet.

In Exercises \(13-16,\) find the work performed by the force field \(\vec{F}\) moving a particle along the path \(C\). \(\vec{F}=\left\langle 2 x y, x^{2}, 1\right\rangle\) lbs; \(C\) is the path from (0,0,0) to (2,4,8) via \(\vec{r}(t)=\left\langle t, t^{2}, t^{3}\right\rangle\) on \(0 \leq t \leq 2,\) where distance are measured in feet.

In Exercises \(17-20\), a conservative vector field \(\vec{F}\) and a curve C are given. 1\. Find a potential function \(f\) for \(\vec{F}\) 2\. Compute curl \(\vec{F}\). 3\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) directly, i.e., using Key Idea 14.3.1. 4\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) using the Fundamental Theorem of Line Integrals. $$ \begin{array}{l} \vec{F}=\left\langle 2 x y z, x^{2} z, x^{2} y\right\rangle, c \text { is curve parametrized by } \vec{r}(t)= \\ \langle 2 t+1,3 t-1, t\rangle \text { on } 0 \leq t \leq 2 \end{array} $$

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