Path-Independent Vector Fields and the Fundamental Theorem of Calculus for Line Integrals

This section requires the idea of a line integral from .

This section is long, but important. To do all the activities, you will need multiple 50-minute class periods. One way to facilitate this would be to treat as if it were a preview activity and have students work on it before the second class meeting. You could also opt to de-emphasize finding potential functions and omit that activity.

is also rather long. We have arranged the subsection by placing after the activity so as to not spoil the discovery to which the activity builds. This activity can be skipped without adversely impacting the remainder of the chapter. However, if choosing to omit , you may wish to specifically point out the culminating theorem to students.

In , , and , we encountered situations where \(C_1\) and \(C_2\) are different oriented curves from a point \(P\) to a point \(Q\) and \(\int_{C_1}\vF\cdot d\vr = \int_{C_2}\vF\cdot d\vr\). In this section, we explore vector fields which have the property that for all points \(P\) and \(Q\), if \(C_1\) and \(C_2\) are oriented paths from \(P\) to \(Q\), then \(\int_{C_1}\vF\cdot d\vr = \int_{C_2}\vF\cdot d\vr\).

Hopefully has prompted you to wonder about the phenomenon of the value of a line integral depending only on the initial and terminal points of the oriented path (rather than the oriented path itself) and how a potential function comes into play. We say that a vector field \(\vF\) defined on a region \(D\) is path-independent if \(\int_{C_1}\vF\cdot d\vr = \int_{C_2}\vF\cdot d\vr\) whenever \(C_1\) and \(C_2\) are oriented paths in \(D\) such that both curves start at point \(P\) and end at point \(Q\).

In and , we encountered situations where we had evidence that a vector field was path-independent. However, since the definition of path-independence requires that the value of the line integral be the same for every possible path from one point to the other (regardless of choice for the initial and final points), it doesn't appear that verifying a vector field is path-independent is an easy task.

Fortunately, one familiar class of vector fields can be shown to be path-independent. Let \(f\colon \R^3\to \R\) be a function for which \(\nabla f\) is continuous on a region \(D\). Suppose that \(P\) and \(Q\) are points in \(D\) and let \(C\) be a smooth oriented path from \(P\) to \(Q\) that is also contained in our region \(D\). We consider \(\int_C\nabla f \cdot d\vr\) by fixing an arbitrary parametrization \(\vr(t)\) of \(C\), \(a\leq t \leq b\). Since we can write \(x\), \(y\), and \(z\) in terms of \(t\), along \(C\), the gradient of \(f\) is given by \[\begin{aligned}\end{aligned}\]. Hence, we can write the line integral of the vector field \(\vF = \nabla f\) over \(C\) in the following way: \[\begin{aligned}\int_C\nabla f\cdot d\vr \amp = \int_a^b \nabla f(\vr(t))\cdot \vr'(t)\, dt \\ \amp = \int_a^b \langle f_x(\vr(t)), f_y(\vr(t)), f_z(\vr(t))\rangle \cdot \vr'(t)\, dt\end{aligned}\]

If \(\vr(t) = \langle x(t), y(t), z(t)\rangle\), then the integrand above is \[\begin{aligned}\amp \phantom{=} \langle f_x(\vr(t)), f_y(\vr(t)), f_z(\vr(t))\rangle\cdot \langle x'(t),y'(t),z'(t)\rangle \\ \amp= f_x\left( x(t),y(t),z(t) \right) \, x'(t) + f_y \left( x(t),y(t),z(t) \right) \, y'(t) \\ \amp\phantom{=} + f_z \left( x(t),y(t),z(t) \right) \, z'(t)\end{aligned}\]. Or more simply, \[\begin{aligned}\end{aligned}\].

Notice that this is exactly what the chain rule tells us \(\frac{d}{dt} f(\vr(t))\) is equal to. Therefore, we may apply the Fundamental Theorem of Calculus to obtain \[\begin{aligned}\end{aligned}\].

In other words, gradient vector fields are path-independent vector fields, and we can evaluate line integrals of gradient vector fields by using a potential function. Remember that what we are calling gradient vector fields are also commonly called conservative vector fields.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Recall that an oriented curve \(C\) is closed if the curve has the same initial and terminal point. A typical example of a closed curve would be a circle (with an orientation of which way to go around), but we could also consider something like the square with vertices \((1,1)\), \((-1,1)\), \((-1,-1)\), and \((1,-1)\), oriented clockwise (or counterclockwise). Recall that we sometimes use the symbol \(\oint\) for a line integral when the curve is closed and that if \(C=C_1+C_2\), then \(\int_C\vF\cdot d\vr=\int_{C_1}\vF\cdot d\vr+ \int_{C_2}\vF\cdot d\vr\).

We summarize the result of with the theorem below. Although this theorem is not a terribly useful way to show that a vector field is path-independent, it can be a useful way to show that a vector field is not path-independent: If you can find a closed curve around which the circulation is not zero, then the vector field is not path independent.

The following activity gives you a chance to reason about path-independence based purely on a graphical representation of a vector field.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Recall that in single variable calculus, The Second Fundamental Theorem of Calculus tells us that given a constant \(c\) and a continuous function \(f\), there is a unique function \(A(x)\) for which \(A(c) = 0\) and \(\frac{dA}{dx} (x) = f(x)\). In particular, \(A(x) = \int_c^x f(t)\, dt\) is this function. We are about to investigate an analog of this result for path-independent vector fields, but first we require two additional definitions.

If \(D\) is a subset of \(\R^2\) or \(\R^3\), we say that \(D\) is open provided that for every point in \(D\), there is a disc (in \(\R^2\)) or ball (in \(\R^3\)) centered at that point such that every point of the disc/ball is contained in \(D\). For example, the set of points \((x,y)\) in \(\R^2\) for which \(x^2+y^2 \lt 1\) is open, since we can always surround any point in this set by a tiny disc contained in the set (as illustrated by point \(P\) in ). However, if we change the inequality to \(x^2+y^2\leq 1\), then the set is not open, as any point on the circle \(x^2+y^2=1\) cannot be surrounded by a disc contained in the set; any disc surrounding a point on that circle will contain points outside the set, that is with \(x^2+y^2>1\) (as illustrated by the point \(Q\) in ). We will also say that a region \(D\) is path-connected provided that for every pair of points in \(D\), there is a path from one to the other contained in \(D\).

We summarize the result of below. Much like the Second Fundamental Theorem of Calculus, which tells us that a function is an antiderivative for another function, but leaves the antiderivative in terms of a definite integral, this theorem tells us that a function is a potential function for a vector field, but the definition of the potential function is in terms of a line integral.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.