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Chapter 18: Problem 41

Let \(\vec{F}=-y \vec{i}+x \vec{j}\) and let \(C\) be the unit circle oriented counterclockwise. (a) Show that \(\vec{F}\) has a constant magnitude of 1 on \(C\). (b) Show that \(\vec{F}\) is always tangent to the circle \(C\). (c) Show that \(\int_{C} \vec{F} \cdot d \vec{r}=\) Length of \(C\).

Short Answer

Expert verified
\( \vec{F} \) has magnitude 1, is tangent to \( C \), and \( \int_{C} \vec{F} \cdot d \vec{r} = 2\pi \).

Step by step solution

01

Express Unit Circle Parameterization

The unit circle is parameterized by \( \vec{r}(t) = \langle \cos(t), \sin(t) \rangle \) for \( t \in [0, 2\pi] \). This is because the cosine and sine functions define the x and y coordinates of a circle with radius 1.
02

Magnitude of \( \vec{F} \) on \( C \)

Given \( \vec{F} = -y \vec{i} + x \vec{j} \), substitute \( x = \cos(t) \) and \( y = \sin(t) \) into the expression. Thus, \( \vec{F} = -\sin(t) \vec{i} + \cos(t) \vec{j} \). The magnitude is \( \| \vec{F} \| = \sqrt{(-\sin(t))^2 + (\cos(t))^2} = \sqrt{\sin^2(t) + \cos^2(t)} = 1 \). Thus, \( \vec{F} \) has constant magnitude 1 on \( C \).
03

Show \( \vec{F} \) is Tangent to Circle

The derivative of the position vector parameterization is \( \vec{r}'(t) = \frac{d}{dt}(\langle \cos(t), \sin(t) \rangle) = \langle -\sin(t), \cos(t) \rangle \), which matches \( \vec{F} = -\sin(t) \vec{i} + \cos(t) \vec{j} \). Since \( \vec{F} = \vec{r}'(t) \), \( \vec{F} \) is always tangent to \( C \).
04

Compute Line Integral

The line integral is given by \( \int_{C} \vec{F} \cdot d \vec{r} = \int_{0}^{2\pi} \vec{F}(t) \cdot \vec{r}'(t) \, dt \). Since \( \vec{F}(t) = \vec{r}'(t) \), this becomes \( \int_{0}^{2\pi} \| \vec{r}'(t) \|^2 \, dt = \int_{0}^{2\pi} 1 \, dt = 2\pi \). This shows \( \int_{C} \vec{F} \cdot d \vec{r} = \text{Length of } C \), which is 2\pi.

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Key Concepts

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

Vector Fields
A vector field is a mathematical construct that associates a vector to every point in a space. In the case of a two-dimensional vector field, we assign vectors to each point on a plane. Our exercise involves the vector field \(\vec{F} = -y \vec{i} + x \vec{j}\) which defines vector components based on the coordinates \(x\) and \(y\).
  • This vector field is characterized by its component functions \(-y\) and \(x\), which determine the direction and magnitude of the vector at any given point \((x, y)\).
  • The vector field visually resembles a rotational field, meaning its vectors circulate around the origin.
When evaluating a vector field, understanding both the direction and the magnitude at various points is crucial to grasping its properties and behaviors, especially when applied to paths such as a unit circle.
Tangent Vectors
Tangent vectors are essential when we discuss curves like the unit circle. A tangent vector at a given point of a curve provides insights into the direction of that curve at that point.
The exercise requires us to show that the vector field \(\vec{F}\) is always tangent to the unit circle \(C\). This demands checking if \(\vec{F}\), at each point on \(C\), aligns directionally with the tangent vector of \(C\).
  • The tangent vector to the unit circle at any point \((x, y)\) can be the derivative of its parameterization, which gives \(\langle -\sin(t), \cos(t) \rangle\).
  • Notably, this is equivalent to \(\vec{F} = - y \vec{i} + x \vec{j} = -\sin(t) \vec{i} + \cos(t) \vec{j}\) when using the trigonometric identity \(x = \cos(t)\) and \(y = \sin(t)\).
  • Since these two expressions are equal, it confirms that \(\vec{F}\) is tangent to \(C\) at every point.
Tangent vectors serve as a bridge connecting vector fields to curves, providing vital insights for computing line integrals.
Unit Circle
The unit circle is a set of points in a plane that maintains a fixed distance, specifically a radius of 1, from a central origin point. It is an essential concept in trigonometry and calculus.
  • This circle can be parameterized using trigonometric functions: \(\vec{r}(t) = \langle \cos(t), \sin(t) \rangle\), where \(t\) is the parameter representing the angle in radians.
  • The range of \(t\) from 0 to \(2\pi\) spans the entire circle counterclockwise.
Understanding the properties of the unit circle is crucial when analyzing how vector fields and line integrals behave over circular paths. It serves as a standard basis for modeling circular motion and integrating vector fields.
Parameterization
Parameterization is a technique to represent a curve using a set of equations that express the coordinates as functions of a variable known as the parameter, typically \(t\). It allows us to explore geometric curves analytically rather than just rely on geometric intuition.
  • For the unit circle, the parameterization is given by \(\vec{r}(t) = \langle \cos(t), \sin(t) \rangle\).
  • This expression effectively translates the unit circle into a mathematical form where calculus methods, such as integration, can be employed.
Understanding parameterization benefits crucial applications like computing line integrals, where it simplifies problems by converting position vectors and curve lengths into functions of one variable. By doing so, we can utilize calculus tools to evaluate properties like tangency, projections, and integrals over paths.

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Most popular questions from this chapter

Find the work done by a constant force \(\vec{F}\) moving an object through straight line displacement \(\vec{r}\) if (a) \(\vec{F}\) is in the same direction as \(\vec{r},\|\vec{F}\|=5\) newtons and \(\|\vec{r}\|=3\) meters. (b) \(\vec{F}\) and \(\vec{r}\) are perpendicular. (c) \(\vec{F}=4 \vec{i}+\vec{j}+4 \vec{k}\) pounds and \(\vec{r}=\vec{i}+\vec{j}+\vec{k}\) foot.

Find \(\int_{C} \vec{F} \cdot d \vec{r}\) for the given \(\vec{F}\) and \(C\). \(\overrightarrow{\boldsymbol{F}}=-\vec{y} \vec{i}+x \vec{j}+5 \vec{k}\) and \(C\) is the helix \(x=\cos t, y=\) \(\sin t, z=t,\) for \(0 \leq t \leq 4 \pi\).

Use the fact that the force of gravity on a particle of mass \(m\) at the point with position vector \(\vec{r}\) is $$\vec{F}=-\frac{G M m \vec{r}}{\|\vec{r}\|^{3}}$$ where \(G\) is a constant and \(M\) is the mass of the earth. Calculate the work done by the force of gravity on a particle of mass \(m\) as it moves radially from \(8000 \mathrm{km}\) to \(10,000 \mathrm{km}\) from the center of the earth.

Find the work done by the force \(\vec{F}\) along the curve \(C\) \(\vec{F}=-x \vec{i}-y \vec{j}, C\) is the upper half of the unit circle from (1,0) to (-1,0)

Calculate the line integral of the vector field along the line between the given points. $$\vec{F}=3 \vec{i}+4 \vec{j}, \quad \text { from }(0,6) \text { to }(0,13)$$
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