The flow of a vector field along a curve

Suppose you are flying to Olinda Brazil for a mathematics conference. Flying out of John Glenn Columbus International Airport, you have a layover in Hartsfield-Jackson Atlanta International Airport. Because you bought your tickets as cheaply as possible, you have very little time to catch your plane in Atlanta. Hence, your flight time is of crucial interest to you. Fortunately or unfortunately, hurricane Gauss (note this name is not actually possible for a hurricane) is creating “interesting” wind patterns. Of crucial interest to you is the following question:

If the flow of the air is with the path and direction of your plane, you will pick up a tailwind and get to Atlanta in plenty of time to catch your next flight. If flow of the wind is against the path and direction of your plane, you might miss your connecting flight and be forced to spend days in the Atlanta airport reading science fiction novels.

Let $\vec {W}(x,y)$ be a vector field that represents wind currents. Let $\vec {p}(t)$ be a vector-valued function describing the path of your plane where $t$ represents the times you are in flight assuming no wind. The velocity vector of the plane is given by $\vec {p}'(t)$. To see if the wind currents are pushing with the plane or against the plane for any given time, you compute: $$\underbrace {\vec {W}(\vec {p}(t))}_{\text {wind currents}}\dotp \underbrace {\vec {p}'(t)}_{\text {velocity of the plane}}$$ Here the dot product measures “how aligned” these two vectors are. If the dot product is positive, the wind is pushing with the plane and speeding-up your flight. If the dot product is negative, the wind is pushing against the plane and is slowing your flight.

To see the net accumulation of the wind on the plane’s flight, you should hence integrate with respect to time: $$\int _{t_\text {takeoff}}^{t_\text {landing}} \vec {W}(\vec {p}(t)) \dotp \vec {p}'(t) \d t$$ This integral will measure the accumulated contribution of the wind to the flight of the plane. If the integral is positive, you will arrive in Atlanta early, if the integral is negative you will arrive late. This sort of integral is called a line integral or path integral because we are integrating along a line or a path.

Now, there is some stuff to unpack here. First of all, whenever you are computing a line integral, you have two parts:

• A vector field.
• A vector-valued function that draws a path through the field.

A line integral measures the flow of a vector field along a path. The basic idea is that there is some vector field given by $\vec {F}$:

Now we add directed path $C$ that is parameterized by $\vec {p}(t) = \vector {x(t),y(t)}$. This can be thought of as a path that an object takes through the field:

Now, if we travel the opposite direction through the field,

Consider the following vector field along with a (directed) curve $C$.

Do you expect $$\int _C \vec {F}\dotp \d \vec {p}$$ to be positive, zero, or negative?

positive zero negative

We can think about this better if we break the path into pieces: $C_1$, $C_2$, $C_3$.

The field vectors are orthogonal to the direction of the path $C_1$. So this part contributes nothing to the integral.

The field vectors are flowing against the direction of $C_2$. This contributes a negative value to our integral.

The field vectors are again orthogonal to the direction of the path $C_3$. So this part contributes nothing to the integral.

Consider the following vector field along with a (directed) curve $C$.

Do you expect $$\int _C \vec {F}\dotp \d \vec {p}$$ to be positive, zero, or negative?

positive zero negative

We can think about this better if we break the path into pieces: $C_1$, $C_2$, $C_3$.

We can see that the vectors are flowing with the direction of $C_1$. Note that the magnitude of these vectors is large, so this contributes a large positive value to our integral.

The field vectors are orthogonal to the direction of the path $C_2$. So this part contributes nothing to the integral.

The field vectors are flowing against the direction of $C_3$. However, their magnitude is much less than the vectors that flowed with $C_1$. So this contributes a small negative value to our integral.

Consider the following vector field along with a (directed) curve $C$.

Do you expect $$\int _C \vec {F}\dotp \d \vec {p}$$ to be positive, zero, or negative?

positive zero negative

Think about what the tangent vectors to the parameterized curve look like, and whether they point with the field or against the field.

Consider the following vector field along with a (directed) curve $C$.

Do you expect $$\int _C \vec {F}\dotp \d \vec {p}$$ to be positive, zero, or negative?

positive zero negative

Think about what the tangent vectors to the parameterized curve look like, and whether they point with the field or against the field.

The notation of line integrals

Since line integrals appear in many different context, there is a wide variety of notation for line integrals. In fact, if we set

where $\vec {p}(t)$ draws a curve $C$ as $t$ runs from $a$ to $b$, then each integral below represents the same line integral:

• $\int _C \vec {F}\dotp \d \vec {p}$
• $\int _C \vector {M,N}\dotp \vector {\d x,\d y}$
• $\int _C M(x,y)\d x + N(x,y)\d y$
• $\int _a^b \left (M(x(t),y(t))\cdot x'(t) + N(x(t),y(t))\cdot y'(t)\right ) \d t$
• $\int _a^b \vector {M(x(t),y(t)),N(x(t),y(t))}\dotp \vector {x'(t), y'(t)} \d t$

If the path $C$ is closed, then sometimes people write a “circle” on the integral sign: $$\oint _C \vec {F}\dotp \d \vec {p}$$ This notation is not critical, but it can sometimes help us from making silly mistakes.

Now, let’s use a table to describe our vector field.

Let $\vec {F}:\R ^2\to \R ^2$ be roughly described by the following table of values:

Let $\vec {p}(t)$ be the vector-valued function describing the line $\l$ running from the point $(1,3)$ to $(-1,1)$ as $t$ runs from $0$ to $1$. Setting $\d t = 1/3$, estimate: $$\int _\l \vec {F}\dotp \d \vec {p}$$

Since we want the line to be drawn from point $(1,3)$ to point $(-1,1)$ as $t$ runs from $0$ to $1$, and the line passes through three rectangles, it makes sense we set $\d t = 1/3$. We parameterize our line as $$\vec {p}(t) = \vector {\answer [given]{1-2t},\answer [given]{3-2t}}$$ Now we can estimate $$\int _\l \vec {F}\dotp \d \vec {p}$$ by computing $\d \vec {p} = \vector {\answer [given]{-2/3}\answer [given]{-2/3}}$ and then computing: $$\begin{align*} \vector{1,-1}&\dotp \vector{-2/3,-2/3}\\ &+\vector{1,0}\dotp \vector{-2/3,-2/3}\\ &+\vector{1,-1}\dotp \vector{-2/3,-2/3} \end{align*}$$

Thus we see that: $$\int _\l \vec {F}\dotp \d \vec {p} \approx \answer [given]{-2/3}$$

We can also estimate line integrals given information found in a graph.

Below we have a very simple directed curve $C$ (it’s a line) along with field vectors from a vector field $\vec {F}$.

Use $\d x= -2$ to estimate: $$\int _C \vec {F}\dotp \d \vec {p}$$

If $\vec {F}(x,y) = \vector {M(x,y),N(x,y)}$, we have that $$\int _C \vec {F}\dotp \d \vec {p} = \int _C M(x,y) \d x + N(x,y)\d y.$$ We know that $\d x= \answer [given]{-2}$ and $$\d y = y'(x) \d x$$ Since $C$ is a line of slope $\answer [given]{1/2}$, $$\d y = \answer [given]{-1}.$$ Since we want to estimate $$\int _C M(x,y) \d x + N(x,y)\d y$$ we will compute $$\sum \left (M(x,y) \d x + N(x,y) \d y\right )$$ where each given field vector will contribute a term in our sum above. Write with me $$\begin{align*} \answer[given]{0}&\cdot (-2) + \answer[given]{2} \cdot (-1) && \text{(for the $1$st field vector)}\\ &+ (-1) \cdot (-2) + 2 \cdot (-1) && \text{(for the $2$nd field vector)}\\ &+ (-2)\cdot (-2) + 1 \cdot (-1) && \text{(for the $3$rd field vector)}\\ &+ (-3)\cdot (-2) + 0\cdot (-1) && \text{(for the $4$th field vector)}\\ &+ (-1)\cdot (-2) + (-1)\cdot (-1) && \text{(for the $5$th field vector)}\\ &+ (0)\cdot (-2) + (-1)\cdot (-1)&& \text{(for the $6$th field vector)} \end{align*}$$

Adding this all up, we find the net flow of the field vectors along the curve is $\answer [given]{11}$.

Computations with line integrals

Any smooth path can be approximated with a polygonal path. These can be quite easy to integrate. Check out our next example.

Let $\vec {F}(x,y) = \vector {0,x}$ and let $C$ be the polygonal path below parameterized in a counterclockwise direction:

Compute $$\oint _C \vec {F}\dotp \d \vec {p}$$

We need to parameterize our paths in a counterclockwise direction. While there are lots of ways to do this, we’ll break it into four line segments each parameterized as $t$ runs from $0$ to $1$ (though any parameterization would work).

Where: $$\begin{align*} \vecl_1(t) &= \vector{4t,\answer[given]{0}}\\ \vecl_2(t) &= \vector{\answer[given]{4-t},2t}\\ \vecl_3(t) &= \vector{3-2t,\answer[given]{2}}\\ \vecl_4(t) &= \vector{\answer[given]{1-t},2-2t} \end{align*}$$

and each draws the line as $t$ runs from $0$ to $1$. Write:

$$\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} = \int_0^1 &\vec{F}(\vecl_1(t))\dotp \vecl_1'(t) \d t \\ &+ \int_0^1 \vec{F}(\vecl_2(t))\dotp \vecl_2'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_3(t))\dotp \vecl_3'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_4(t))\dotp \vecl_4'(t) \d t \end{align*}$$

For each of the integrands above, say $\vecl (t) = \vector {x(t),y(t)}$, we will write $$\vec {F}(\vecl (t))\dotp \vecl '(t) = \vector {0,x(t)} \dotp \vector {x'(t),y'(t)}$$ and combine them into a single integral. Write with me

$$\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} &= \int_0^1 \left(\answer[given]{2(4-t) -2(1-t)} \right)\d t\\ &= \eval{\answer[given]{6t}}_0^1\\ &=\answer[given]{6}. \end{align*}$$

The Fundamental Theorems of Calculus

We will soon see that there are many “Fundamental Theorems of Calculus.” What makes them similar is that they all share the following rather vague description:

Each version of the Fundamental Theorem of Calculus makes the “vague” statement above precise. For example, when working with a single variable, the Fundamental Theorem concludes: $$\int _a^b f'(x) \d x = f(b) - f(a)$$ In this case we are doing an integral over the “region” $[a,b]$, and the “computation” that allows “one fewer integrations” is antidifferentiation. This is the only Fundamental Theorem you have known so far in your studies. However with additional dimensions, there come additional derivatives. When working with functions $F:\R ^n\to \R$ we have the gradient as a “derivative.” This brings us to our first of several new fundamental theorems.

Fundamental Theorem for Line Integrals If $C$ is a curve that starts at $\vec {a}$ and ends at $\vec {b}$,

then: $$\int _C \grad F\dotp \d \vec {p} = F(\vec {b}) - F(\vec {a})$$

Like the first fundamental theorem we met in our very first calculus class, the fundamental theorem for line integrals says that if we can find a potential function for a gradient field, we can evaluate a line integral over this gradient field by evaluating the potential function at the end-points. In essence the potential function is like the antiderivative from your first calculus course.

The up-shot is that whenever you are dealing with a line integral, you should always start by checking to see if you are working with a gradient field.

Path independent fields

When trying to compute a line integral over a gradient field, the path one takes is not important. This is because the Fundamental Theorem for Line Integrals says: $$\int _C \grad F\dotp \d \vec {p} = F(\vec {b}) - F(\vec {a})$$ Meaning, you can just use the values of the potential function at the end-points to compute the line integral. Because of this, some people call gradient fields path independent fields. Let’s see an example.

Let $F(x,y) = x\cos (xy)$. Compute: $\grad F$ $$\grad F(x,y) = \vector {\answer {\cos (xy) -xy\sin (xy)},\answer { -x^2\sin (xy)}}$$

Now let: $$\vec {G}(x,y) = \vector {\cos (xy) -xy\sin (xy), -x^2\sin (xy)}$$ Compute $$\int _C \vec G \dotp \d \vec {p}$$ where $C$ is shown below:

$$\begin{align*} \int_C \vec{G}\dotp \d \vec{p} &= F(4,0) - F(0,2)\\ &= \answer{4}. \end{align*}$$

Conservative fields

When dealing with gradient fields and closed curves something very nice happens.

This leads us to our definition:

But don’t be bedazzled by grandiose nomenclature! Calling a field a conservative field is just another name for a gradient field.

If something is special enough to be named thrice (gradient field, path independent field, conservative field), we ought to do some more examples.

Let $\vec {F}(x,y) = \vector {-y,-x}$. Let $C$ be the polygonal path below parameterized in a counterclockwise direction:

Compute $$\oint _C \vec {F}\dotp \d \vec {p}$$

In light of the Fundamental Theorem for line integrals, we should check to see if our field is a gradient field using the Clairaut gradient test: $$\pp {x}(-x)-\pp {y}(-y) = \answer [given]{0}.$$ Since our field is a gradient field, and our curve is closed, we see that $$\oint _C \vec {F}\dotp \d \vec {p} = \answer [given]{0}.$$

Below we see a directed curve with some field vectors attached.

Assuming that the field vectors are constant along each “side” of our polygonal path, compute: $$\int _C\vec {F}\dotp \d \vec {p}$$

This problem would be easier if our field was a gradient field, and while it surely isn’t, the field is constant along two paths that together make up all of $C$:

So now we see that $$\int _C \vec {F}\dotp \d \vec {p} = \int _{C_1} \vec {F}\dotp \d \vec {p}+\int _{C_2} \vec {F}\dotp \d \vec {p}.$$ so we have that $$\vec {F}(x,y) = \begin {cases} \vector {-1,1} &\text {on $C_1$,}\\ \vector {1,-1} &\text {on $C_2$.} \end {cases}$$ Using the Clairaut gradient test we see that $\vec {F}$ isis not a gradient field when restricted to either part of $C$. Let $F$ be a potential function for $\vec {F}$. One candidate for the the potential function is $$F(x,y) = \begin {cases} -x+y &\text {on $C_1$,}\\ x-y &\text {on $C_2$.} \end {cases}$$ By the Fundamental Theorem for Line Integrals, $$\begin{align*} \int_{C_1} \vec{F}\dotp\d\vec{p} &= \eval{-x+y}_{\vector{3,3}}^{\vector{-5,3}}\\ &= \answer[given]{8}, \end{align*}$$

and

$$\begin{align*} \int_{C_2} \vec{F}\dotp\d\vec{p} &= \eval{x-y}_{\vector{-5,3}}^{\vector{-9,-5}}\\ &= \answer[given]{4}. \end{align*}$$

So $$\int _{C} \vec {F}\dotp \d \vec {p} = \answer [given]{12}.$$