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Chapter 18: Problem 71

Are the statements in Problems \(70-80\) true or false? Give reasons for your answer. If \(C\) is the line segment that starts at (0,0) and ends at \((a, b)\) then \(\int_{C}(x \vec{i}+y \vec{j}) \cdot d \vec{r}=\frac{1}{2}\left(a^{2}+b^{2}\right)\)

Short Answer

Expert verified
True: The integral equals \(\frac{1}{2}(a^2 + b^2)\).

Step by step solution

01

Understand the problem

We need to determine if the statement is true or false: The line integral of the vector field \(x\vec{i} + y\vec{j}\) along line segment \(C\) from \((0,0)\) to \((a,b)\) equals \(\frac{1}{2}(a^2 + b^2)\).
02

Set up the position vector and differential

The curve \(C\) is the line segment from \((0,0)\) to \((a,b)\). Therefore, the position vector \(\vec{r}(t)\) can be described as \((at, bt)\) for \(t\) in the interval \([0,1]\). The differential is thus \(d\vec{r} = (a \, dt, b \, dt)\).
03

Define the vector field and perform the dot product

The vector field is \(F = x \vec{i} + y \vec{j}\). Therefore, along the line segment, the vector field is \(F(at, bt) = at \vec{i} + bt \vec{j}\). The dot product \((at \vec{i} + bt \vec{j}) \cdot (a \, dt\, \vec{i} + b \, dt\, \vec{j}) = (a^2 t + b^2 t) \, dt\).
04

Evaluate the integral

We integrate the expression \(\int_0^1 (a^2 t + b^2 t) \, dt\). This simplifies to \((a^2 + b^2)\int_0^1 t \, dt = (a^2 + b^2) \left[\frac{1}{2}t^2\right]_0^1 = \frac{1}{2}(a^2 + b^2)\).
05

Conclusion

The calculated result matches with the statement given in the problem, so the statement is true.

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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 where each point in a space is assigned a vector. Imagine a weather map showing wind patterns; each vector represents the wind's direction and speed at a particular point. In mathematics, vector fields are utilized to represent force fields like gravitational or electric fields.
  • In our exercise, the vector field given is \( x \vec{i} + y \vec{j} \), where \( x \) and \( y \) denote the coordinates in 2D space.
  • The vector field at any point \((x, y)\) gives the vector which has \( x \) units in the \( \vec{i} \) direction and \( y \) units in the \( \vec{j} \) direction, essentially acting like a wind blowing at each point.
Understanding vector fields helps in visualizing various physical phenomena, such as how a force might be applied throughout an area.
Position Vectors
Position vectors allow us to describe a path or a curve in a space. They're particularly useful when you're dealing with line integrals along specific paths. Every position vector points from the origin (or another specified point) to a defined position in space.
  • In this problem, the position vector is described as \( \vec{r}(t) = (at, bt) \).
  • This vector represents the line segment from origin \((0,0)\) to the endpoint \((a,b)\).
  • As \(t\) varies from 0 to 1, the position vector traces the entire segment \(C\).
This mathematical representation helps in tracking the change in position as you move along the curve, making it crucial for integration over a path or line segment.
Dot Product
The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar. It's used to determine how much one vector goes in the direction of another. This concept is vital in physics and engineering to calculate work done when a force is applied over a distance.
  • In this exercise, you compute the dot product of the vector field and the differential position vector: \((at \vec{i} + bt \vec{j}) \cdot (a \, dt \vec{i} + b \, dt \vec{j}) = (a^2 t + b^2 t) \, dt\).
  • The dot product simplifies the expression by combining the components into a scalar quantity that can be integrated over a path.
Understanding the dot product is instrumental when performing line integrals, as it provides the magnitude of projection from one vector onto another, helping in calculating the integral accurately.
Integration
Integration is a fundamental concept in calculus that involves summing up infinitely small quantities to find areas, volumes, and other cumulative values. When integrating vector fields along a curve or line, you're calculating the total effect of the field along that path.
  • The integral in this exercise is \( \int_0^1 (a^2 t + b^2 t) \, dt \).
  • By integrating the expression, you find that \( (a^2 + b^2) \left[ \frac{1}{2}t^2 \right]_0^1 = \frac{1}{2}(a^2 + b^2) \), confirming the original statement.
Integration along paths (like line integrals) is an essential tool in multivariable calculus, enabling deeper analyses of fields and flow in a spatial context.
Multivariable Calculus
Multivariable calculus deals with functions of multiple variables and extends the concepts of single-variable calculus to higher dimensions. It's crucial for understanding systems and fields that change with respect to more than one variable, such as temperature distributions or electric fields.
  • In multivariable calculus, you often work with vector fields, curves, and surfaces, requiring more complex integrals like those seen with line integrals.
  • This exercise focuses on integrating a vector field along a path, a common task in multivariable calculus.
Mastery of multivariable calculus allows you to tackle sophisticated problems in physics, engineering, and any field where multidimensional analysis is essential.

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Most popular questions from this chapter

Are the statements true or false? Give reasons for your answer. If \(C_{1}\) is the curve parameterized by \(\vec{r}_{1}(t)=t \vec{i}+t^{2} \vec{j}\) with \(0 \leq t \leq 2,\) and \(C_{2}\) is the curve parameterized by \(\vec{r}_{2}(t)=(2-t) \vec{i}+(2-t)^{2} \vec{j}, 0 \leq t \leq 2,\) then for any vector field \(\vec{F}\) we have \(\int_{C_{1}} \vec{F} \cdot d \vec{r}=-\int_{C_{2}} \vec{F} \cdot d \vec{r}\).

Are the statements in Problems \(70-80\) true or false? Give reasons for your answer. If \(\vec{F}\) is path-independent, then there is a potential function for \(\overrightarrow{\boldsymbol{F}}\)

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.

Explain what is wrong with the statement. It is possible that for a certain vector field \(\vec{F}\) and oriented path \(C,\) we have \(\int_{C} \vec{F} \cdot d \vec{r}=2 \vec{i}-3 \vec{j}\)

Give an example of: A parameterized path \(C\) such that, for the vector field \(\vec{F}(x, y)=\sin y \vec{i},\) the integral \(\int_{C} \vec{F} \cdot d \vec{r}\) is nonzero and can be computed geometrically, without using the parameterization.
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