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Chapter 17: Problem 64

A vector field in polar coordinates has the form \(\mathbf{F}(r, \theta)=f(r, \theta) \mathbf{u}_{r}+g(r, \theta) \mathbf{u}_{\theta},\) where the unit vectors are defined in Exercise 62. Sketch the following vector fields and express them in Cartesian coordinates. $$\mathbf{F}=\mathbf{u}_{r}$$

Short Answer

Expert verified
To summarize: 1. We sketched the vector field, and found that it is a radial field with vectors pointing directly outward from the origin at all points. 2. We converted the polar unit vectors to Cartesian unit vectors. 3. We expressed the given vector field in Cartesian coordinates as \(\mathbf{F} = \frac{x}{\sqrt{x^2+y^2}}\mathbf{i} + \frac{y}{\sqrt{x^2+y^2}}\mathbf{j}\).

Step by step solution

01

Sketching the vector field

To sketch the vector field, we first need to understand the behavior of the given vector field which is \(\mathbf{F}=\mathbf{u}_{r}\). The unit vector \(\mathbf{u}_r\) points radially outward, so the vector field corresponds to a radial field. Therefore, at any point in space, the vector will point directly outward (away from the origin). To sketch this, draw small arrows at various points pointing away from the origin.
02

Convert polar unit vectors to Cartesian unit vectors

Recall that the polar unit vectors can be expressed in Cartesian coordinates as: \(\mathbf{u}_r = \cos(\theta)\mathbf{i} + \sin(\theta)\mathbf{j}\) \(\mathbf{u}_\theta =-\sin(\theta)\mathbf{i}+ \cos(\theta)\mathbf{j}\) We are given: \(\mathbf{F}=\mathbf{u}_{r}\), we can now express this in Cartesian coordinates using the relationship between polar and Cartesian unit vectors.
03

Convert the given vector field to Cartesian coordinates

Substitute the Cartesian expression for \(\mathbf{u}_r\) into the given vector field: \(\mathbf{F} = \cos(\theta)\mathbf{i} + \sin(\theta)\mathbf{j}\) But polar coordinates are related to Cartesian coordinates as follows: \(x = r\cos(\theta)\) or \(\cos(\theta) = \frac{x}{\sqrt{x^2+y^2}}\) \(y = r\sin(\theta)\) or \(\sin(\theta) = \frac{y}{\sqrt{x^2+y^2}}\) Now replace \(\cos(\theta)\) and \(\sin(\theta)\) in the expression for \(\mathbf{F}\) with their Cartesian equivalents: \(\mathbf{F} = \frac{x}{\sqrt{x^2+y^2}}\mathbf{i} + \frac{y}{\sqrt{x^2+y^2}}\mathbf{j}\)

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Key Concepts

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

Polar Coordinates
Polar coordinates provide a unique way to describe the position of a point in a plane using a distance and an angle. Instead of using traditional x-y coordinates, polar coordinates specify a point by its distance from a central point (called the origin) and the angle it makes with a reference direction, typically the positive x-axis.
1. **Radial Distance (r):** This is the distance from the origin to the point. Think of it as how far you need to travel to reach the point, starting from the origin.
2. **Angular Component (胃):** This is the angle formed by drawing a line from the origin to the point and measuring the counterclockwise rotation from the positive x-axis.

An essential aspect of polar coordinates is the ability to convert them to Cartesian coordinates for different calculations. The formulas for conversion:
  • For the x-coordinate: \[ x = r \cos(\theta) \]
  • For the y-coordinate: \[ y = r \sin(\theta) \]
Cartesian Coordinates
Cartesian coordinates describe the location of a point in a plane using two perpendicular axes: the x-axis and the y-axis. These coordinates are typically straightforward as they rely on distance measurements along these two axes.
1. **x-coordinate (x):** This value indicates the horizontal distance from the y-axis. It can be positive or negative, depending on if the point is to the right or left of the y-axis.
2. **y-coordinate (y):** This value shows the vertical distance from the x-axis. It is positive above the x-axis and negative below.

In vector fields, converting from polar to Cartesian coordinates is important because many physical interpretations and operations, such as addition and differentiation, are easier to perform directly in Cartesian coordinates. The conversion formulas help translate polar expressions into Cartesian forms, easing mathematical manipulation:
Unit Vectors
Unit vectors are vectors that have a magnitude of one, which means they are of standard length but preserve direction.

1. **Polar Unit Vectors:** These are specific to polar coordinates, where the position and direction are described uniquely. The principal polar unit vectors are:
  • **\( \mathbf{u}_r \):** This vector points radially outward from the origin. It represents the direction of increase of the radial coordinate \( r \).
  • **\( \mathbf{u}_\theta \):** This vector is perpendicular to \( \mathbf{u}_r \) and points in the direction of increasing \( \theta \), the angular component.
Converting these polar unit vectors to their corresponding vectors in Cartesian coordinates allows expressions in different coordinate systems to be unified. For instance:
  • The radial unit vector, \( \mathbf{u}_r \), converts to: \[ \mathbf{u}_r = \cos(\theta)\mathbf{i} + \sin(\theta)\mathbf{j} \]
  • The angular unit vector, \( \mathbf{u}_\theta \), becomes: \[ \mathbf{u}_\theta = -\sin(\theta)\mathbf{i} + \cos(\theta)\mathbf{j} \]
Understanding these conversions is pivotal for addressing vector fields, especially when switching between polar and Cartesian forms.

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Most popular questions from this chapter

Conditions for Green's Theorem Consider the radial field \(\mathbf{F}=\langle f, g\rangle=\frac{\langle x, y\rangle}{\sqrt{x^{2}+y^{2}}}=\frac{\mathbf{r}}{|\mathbf{r}|}\) a. Explain why the conditions of Green's Theorem do not apply to F on a region that includes the origin. b. Let \(R\) be the unit disk centered at the origin and compute \(\iint_{R}\left(\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}\right) d A\) c. Evaluate the line integral in the flux form of Green's Theorem on the boundary of \(R\) d. Do the results of parts (b) and (c) agree? Explain.

Prove the following identities. Assume \(\varphi\) is a differentiable scalar- valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \cdot(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot(\nabla \times \mathbf{F})-\mathbf{F} \cdot(\nabla \times \mathbf{G})$$

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) \(T(x, y, z)=100+x^{2}+y^{2}+z^{2} ; D\) is the unit sphere centered at the origin.

The potential function for the force field due to a charge \(q\) at the origin is \(\varphi=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{|\mathbf{r}|},\) where \(\mathbf{r}=\langle x, y, z\rangle\) is the position vector of a point in the field, and \(\varepsilon_{0}\) is the permittivity of free space. a. Compute the force field \(\mathbf{F}=-\nabla \varphi\) b. Show that the field is irrotational; that is, show that \(\nabla \times \mathbf{F}=\mathbf{0}\)

Zero circulation fields. For what values of \(b\) and \(c\) does the vector field \(\mathbf{F}=\langle b y, c x\rangle\) have zero circulation on the unit circle centered at the origin and oriented counterclockwise?
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