# Vector fields; Line Integrals; Work (in plane)



My notes of the excellent lectures 19 and 20 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

We will learn about vector fields in the plane and determining the line element, using work as an example.

## Vector fields

A vector field is defined as:

A multi-variable function that returns a vector value for each input.

An vector field $$\vec{F}$$ with inputs $$x,y$$, and $$M,N$$ functions of $$x,y$$, is expressed as: $$\vec F(x,y) = \hat\imath M + \hat\jmath N = \left\langle M,N \right\rangle \label{eq:vectorfield}$$

Examples are

• Velocity in a fluid, $$\vec v$$
• Force fields, $$\vec F$$

Vector fields are drawn as a grid of ($$x,y)$$ input values, from where arrows represent the magnitude and direction of the output vector. An example of a vector field is shown below.

### Examples

#### One

This doesn’t depend on $$x$$ or $$y$$ $$\vec F = 2\hat\imath + \hat\jmath = \left\langle 2,1 \right\rangle \nonumber$$

#### Two

Only depends on $$x$$ $$\vec F = x\hat\imath = \left\langle x,0 \right\rangle \nonumber$$

#### Three

$$\vec F = x\hat\imath + y\hat\jmath = \left\langle x,y \right\rangle \nonumber$$

#### Four

This is $$\left\langle x,y\right\rangle$$ rotated 90° counnterclockwise $$\vec F = -y\hat\imath + x\hat\jmath = \left\langle -y,x \right\rangle \nonumber$$

This is the velocity field for uniform rotation at angular velocity (1 rad/s).

## Line integrals in a vector field (Work)

We will explore line integrals in a vector field, using work as an example.

### Linear Work

In physics, work is defined as:

Whenever a force is applied to an object, causing the object to move, work is done by the force.

Work is force $$F$$ times distance $$r$$, but only the component of the force in the direction of the displacement does any work. $$\rm{W} = F\,r\,\cos\theta \nonumber$$ where $$\theta$$ is the angle between force vector $$\vec{F}$$ and displacement vector $$\vec r$$ as shown in the figure below.

Recall that the dot-product is the product of the “length” of vectors “in the same direction”.

$$\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}|\ |\overrightarrow{b}|\ \cos\theta \nonumber$$

If you have a force exerted on a particle moving in a straight line, then the work done by the force corresponds to the dot-product of the force vector with the displacement vector (=how much you have moved the particle). $$\shaded{ \rm{Work}=\vec{F}\cdot\vec{r} } \label{eq:lineairwork}$$

### Work in a vector field and line integrals

If you move over a straight line, equation $$\eqref{eq:lineairwork}$$ works well, but if you move over a complicated trajectory or the force keeps changing, then you want to integrate that over time.

Consider a vector field $$\vec F(x,y)$$, and a tiny displacement $$\Delta\vec r$$, the change in position vector.

The work $$\Delta W$$ tells you how much energy the force provides to perform this tiny displacement. Or, if you go against the force, the energy that you have to provide to move the particle. $$\Delta \rm{W} = \vec F\cdot\Delta\vec r \nonumber$$

The particle doesn’t just move $$\Delta\vec r$$, but moves along a trajectory, then the force from the vector field $$\vec F$$ might be different at at every point.

To find the total work done along the trajectory curve $$C$$, we have to cut the trajectory in little pieces. For each of them, we have a $$\Delta\vec r$$ and force $$\vec F$$. Then we sum them together. $$\rm{W} = \lim_{\Delta r_i\to 0} \sum_i\vec F_i\cdot\Delta\vec r_i \nonumber$$

The line integral for work follows $$\shaded{ \rm{W} = \int_C\vec F\cdot d\vec r } \nonumber$$

Where

• $$\int_C$$, is an integral along curve $$C$$
• $$d\vec r$$, is an infinite small piece of the trajectory
• $$\vec F$$, the value of the force field at that location

## Compute work

### Using $$\text{ }\vec v$$

Note that $$d\vec r$$ is not a scalar but a vector. We can not integrate over a vector. We replace $$d\vec r$$ with the velocity vector $$\vec v$$ $$\times$$ time. $$\Delta\vec r = \underbrace{\frac{d\vec r}{d t}}_{\vec v}\,dt \nonumber$$

Then compute the work as $$\shaded{ \rm W = \int_{t_1}^{t_2}\vec F\cdot\frac{d\vec r}{dt}\,dt } \nonumber$$ Here

• $$\vec F$$ at the point $$(x,y)$$ on the trajectory at time $$t$$. $$\vec F$$ depends on $$(x,y)$$, and $$(x,y)$$ depend on $$t$$.
• $$\vec r$$ at the point $$(x,y)$$ on the trajectory at time $$t$$. $$\vec F$$ depends on $$(x,y)$$, and $$(x,y)$$ depend on $$t$$.

Notes

• We assume that the force doesn’t affect the trajectory.
• The result depends on the curve, but not on the parameterization of the curve. In the 2nd example, we could have expressed the curve as $$\vec{r}=\left\langle \sin\theta, \sin^2\theta \right\rangle$$ for $$0\leq\theta\leq\frac{\pi}{2}$$.

#### Example

Let a particle move over parametric curve $$C$$, in force field $$\vec{F}$$. What work does the force field exert on the particle? $$\begin{array}{l} C: \left\{ \begin{array}{l} x = t \\ y = t^2 \\ 0 \lt t \lt 1 \end{array} \right. \\ \vec F = \left\langle -y, x \right\rangle \end{array} \nonumber$$

Picture

The formula for work $$\rm W = \int_C\vec{F}\cdot d\vec{r} = \int_0^1\underline{\vec F} \cdot \underline{\frac{d\vec r}{dt}}\,dt \\ \nonumber$$

The vector field $$\vec F$$ depends on $$(x,y)$$, that in turn depend on $$t$$ \left. \begin{align*} \underline{\vec F} &= \left\langle -y,\,x \right\rangle \\ x &= t \\ y &= t^2 \end{align*} \right\} \Rightarrow \vec F = \underline{\left\langle -t^2,\,t\right\rangle}

The derivative of $$\vec r$$ to $$t$$ $$\underline{\frac{d\vec{r}}{dt}} = \frac{d}{dt} \left\langle t,\,t^2 \right\rangle = \underline{\left\langle 1,\,2t \right\rangle} \nonumber$$

Putting them back in the integral \begin{align*} \rm W &= \int_0^1 \underline{\left\langle -t^2,t \right\rangle} \cdot \underline{\left\langle 1,2t\right\rangle}\,dt \\ &= \int_0^1 (-t^2+2t^2)\,dt \\ &= \int_0^1 t^2\,dt = \left[ \frac{1}{3}t^3\right]_0^{t=1} =\frac{1}{3} \end{align*}

### Using $$\left\langle dx,dy\right\rangle$$

$$d\vec r$$ can be expressed as $$d\vec r = \left\langle dx,dy \right\rangle \nonumber$$

For the vector field $$\vec F=\left\langle M,N\right\rangle$$ from equation $$\eqref{eq:vectorfield}$$ $$W = \int_C\vec F\cdot d\vec r = \int_C \left\langle M,N\right\rangle \cdot \left\langle dx,dy \right\rangle \nonumber$$

The work integral can then be written as the differential $$\shaded{ \rm W = \int_C\left( M\,dx + N\,dy \right) } \label{eq:workdiff}$$

Note that $$M$$ and $$N$$ depend on $$(x,y)$$ $$\Longrightarrow$$ you can’t integrate of $$dx$$ or $$dy$$, because you will end up with an expression with $$y$$ or $$x$$ instead of a number.

Instead, use the fact that along the curve, $$x$$ and $$y$$ are related to each other $$\Longrightarrow$$ you can express $$x,y$$ in terms of a single variable, $$x$$, $$y$$, or a parametric parameter $$t$$ and substitute that variable to get an usual single variable integral.

#### Example

##### One

Redo the previous example $$\begin{array}{l} C: \left\{ \begin{array}{l} x = t \\ y = t^2 \\ 0 \lt t \lt 1 \end{array} \right. \\ \vec F = \left\langle -y, x \right\rangle \end{array} \nonumber$$

Using this method \begin{align*} \rm W = \int_C\vec F\cdot d\vec r &= \int_C\left( M\,dx + N\,dy \right) \\ &=\int_C \left\langle -y,x \right\rangle \cdot \left\langle dx, dy\right\rangle \\ &= \int_C -y\,\underline{dx} + x\,\underline{dy} \end{align*}

Use curve $$C$$ to express $$-y, dx, x, dy$$ in terms of a single variable, $$t$$ $$\left\{ \begin{array}{lll} x = \underline{t} & \Rightarrow \frac{d}{dt}x=\frac{d}{dt}t &\Rightarrow dx = \underline{dt} \\ y = \underline{t^2} & \Rightarrow \frac{d}{dt}y = \frac{d}{dt}t^2& \Rightarrow dy=\underline{2t\,dt} \end{array} \right. \nonumber$$

Substitute the above expressions for $$-y, dx, x, dy$$, and add the limits for $$t$$ to the integral \begin{align*} \rm W = \int_C \vec{F}\cdot d\vec{r} &= \int_0^1 \left(-\underline{t^2}\,\underline{dt} + \underline{t}\,\underline{2t\,dt}\right) \\ &= \int_0^1 t^2\,dt = \frac{1}{3} \end{align*} \nonumber

Note the result depends on the curve, but not on the parameterization of the curve. We could have expressed the curve as shown below, and still have gotten the same result $$C: \left\{ \begin{array}{l} x = \sin\theta \\ y = \sin^2\theta \\ 0 \lt \theta \lt \frac{\pi}{2} \end{array} \right. \\ \nonumber$$

##### Two

Let a particle move over curve $$C$$, enclosing the unit disc for $$0\leq\theta\leq\frac{\pi}{4}$$, in force field $$\vec{F}$$. What work does the force exert on the particle? $$\begin{array}{l} \vec F = \left\langle y,x \right\rangle \\ C_1: (0,0)\ \mathrm{to}\ (1,0) \\ C_2: \mathrm{unit\ circle\ from}\ (1,0)\ \mathrm{to\ the\ diagonal} \\ C_3: \mathrm{from\ the\ diagonal\ to}\ (0,0) \end{array} \nonumber$$

Picture

Substituting $$\vec F$$ in the work integral $$\eqref{eq:workdiff}$$ \begin{align*} W_i &= \int_{C_i} M\,dx + N\,dy \\ &= \int_{C_i} y\,dx + x\,dy \end{align*}

Along the curve segments

1. For $$C_1$$, along the $$x$$-axis, we can just use the $$x$$ variable, and $$y=0, \Longrightarrow dy=0$$ $$W_1 = \int_{C_1} \bcancel{y}_0\,dx + x\,\bcancel{dy}_0 = 0 \nonumber$$
2. For $$C_2$$, part of unit circle \begin{align*} x = \cos\theta &\Longrightarrow dx =-\sin\theta\ d\theta \\ y = \sin\theta &\Longrightarrow dy = \cos\theta\ d\theta \\ 0 \leq &\theta \leq \tfrac{\pi}{4} \end{align*} Substituting in the work integral for $$C_2$$ \begin{align*} W_2 &= \int_{C_2} \left(\sin\theta(-\sin\theta\ d\theta) + \cos\theta\cos\theta\ d\theta \right) \\ &= \int_0^\tfrac{\pi}{4} \underbrace{\cos^2\theta – \sin^2\theta}_{\cos(2\theta)}\ d\theta =\Big[ \tfrac{1}{2}\sin(2\theta) \Big]_0^\frac{\pi}{4} = \tfrac{1}{2} \end{align*}
3. For $$C_3$$, the work from $$\left(\tfrac{1}{\sqrt 2},\tfrac{1}{\sqrt 2}\right)$$ to $$(0,0)$$ $$x=t \Rightarrow dx =dt \\ y=t \Rightarrow dy =dt \\ 0 \leq t \leq \tfrac{1}{\sqrt 2} \nonumber$$ Substituting in the work integral for $$C_3$$ \begin{align*} W_3 &= \int_{C_3}t\,dt+t\,dt = \int_{\frac{1}{\sqrt 2}}^0 2t\,dt \\ &= \Big[t^2 \Big]^0_{\frac{1}{\sqrt 2}} = -\frac{1}{2} \end{align*}

Total work \begin{align*} W &= W_1+W_2+W_3 \\ &= 0+\frac{1}{2}-\frac{1}{2} = 0 \end{align*}

### Using geometric approach

A geometric approach can be easier when the vector field and the curve are relatively simple and have a geometric relation to each other.

If we take a very small piece of trajectory, then $$d\vec r$$ will be tangent to the trajectory. It will be in the same direction as the unit tangent vector $$\hat T$$. Its length will be the arc length along the trajectory, $$\Delta s$$

So, we can also write $$d\vec r$$ as $$\shaded{ d\vec r=\left\langle dx,dy\right\rangle = \hat T\,ds } \nonumber$$

It is a vector, whose

• direction is tangent to the curve,
• length is the arc length

Using this notation, the line integral for work can be expressed as $$\shaded{ \int_C \vec F\cdot d\vec r = \int_C \underbrace{\vec F\cdot\hat T}_{scalar}\,ds } \nonumber$$ $$\vec F\cdot\hat T$$ is a scalar quantity. The force projected to the tangent direction of a trajectory and then integrated along the curve.

#### Examples

Let $$C$$ be a a circle with radius $$a$$, at the origin and counterclockwise. What work does the force exert on the particle in two different force fields?

##### Radiant force field

Let a particle move over curve $$C$$, in force field $$\vec F = \left\langle x, y \right\rangle$$. What work does the force exert on the particle?

Picture

The vector field $$\vec F$$ is perpendicular to the motion $$\hat T$$ \begin{align*} \vec F \perp \hat T &\Rightarrow \vec F\cdot \hat T=0 \\ &\Rightarrow \rm W = \int_C\vec F\cdot\hat T\,ds=0 \end{align*}

##### Uniform rotation force field

Let a particle move over curve $$C$$, in force field $$\vec F = \left\langle -y,x \right\rangle$$. What work does the force exert on the particle?

Picture

The vector field $$\vec F$$ is parallel to the motion $$\hat T$$ \renewcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \vec F \parallelsum \hat T &\Rightarrow \vec F\cdot \hat T = |\vec F|=a \\ &\Rightarrow \rm W = \int_C a\ ds = a\underbrace{\int_C ds}_{\mathrm{length\ of\ }C} = a.2\pi a = 2\pi a^2 \end{align*}

If we would have calculated this using $$\left\langle dx,dy\right\rangle$$, we would have set the circle as $$\begin{array}{l} C: \left\{ \begin{array}{l} x = a\cos\theta \Rightarrow dx = -a\sin\theta\,d\theta \\ y = a\sin\theta \Rightarrow dy = a\cos\theta\,d\theta \\ 0 \lt \theta \lt 2\pi \end{array} \right. \\ \vec F = \left\langle -y, x \right\rangle \end{array} \nonumber$$

This would eventually give the same answer \begin{align*} \rm W &= \int_C M\,dx + N\,dy = \int_C -y\,dx + x\,dy \\ &= -(a\sin\theta)(-a\sin\theta\,d\theta) + (a\cos\theta)(a\cos\theta\,d\theta) \\ &= \int_0^{2\pi} a^2\bcancel{\left(\sin^2\theta + \cos^2\theta\right)}_1\,d\theta = a^2 \int_0^{2\pi}d\theta \\ &= 2\pi a^2 \end{align*}

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