Activity 13.4.7.
Let \(\vF=\langle F_1,F_2\rangle\) be a continuous, path-independent vector field on an open, path-connected region \(D\text{.}\) We will assume that \(D\) is in \(\R^2\) and \(\vF\) is a two-dimensional vector field, but the ideas below generalize completely to \(\R^3\text{.}\) We want to define a function \(f\) on \(D\) by using the vector field \(\vF\) and line integrals, much like the Second Fundamental Theorem of Calculus allows us to define an antiderivative of a continuous function using a definite integral. To that end, we assign \(f(x_0,y_0)\) an arbitrary value. (Setting \(f(x_0,y_0)=0\) is probably convenient, but we won’t explicitly tie our hands. Just assume that \(f(x_0,y_0)\) is defined to be some number.) Now for any other point \((x,y)\) in \(D\text{,}\) define
\begin{equation*}
f(x,y) = f(x_0,y_0) + \int_C\vF\cdot d\vr\text{,}
\end{equation*}
where \(C\) is any oriented path from \((x_0,y_0)\) to \((x,y)\text{.}\) Since \(D\) is path-connected, such an oriented path must exist. Since \(\vF\) is path-independent, \(f\) is well-defined. If different paths from \((x_0,y_0)\) to \((x,y)\) gave different values for the line integral, then we would not be sure what \(f(x,y)\) really is.
To better understand this mysterious function \(f\) we’ve now defined, let’s start looking at its partial derivatives.
(a)
Since \(D\) is open, there is a disc (perhaps very small) surrounding \((x,y)\) that is contained in \(D\text{,}\) so fix a point \((a,b)\) in that disc. Since \(D\) is path-connected, there is a path \(C_1\) from \((x_0,y_0)\) to \((a,b)\text{.}\) Let \(C_y\) be the line segment from \((a,b)\) to \((a,y)\) and let \(C_x\) be the line segment from \((a,y)\) to \((x,y)\text{.}\) (See Figure 13.4.6.) Rewrite \(f(x,y)\) as a sum of \(f(x_0,y_0)\) and line integrals along \(C_1\text{,}\) \(C_y\text{,}\) and \(C_x\text{.}\)
An oriented curve from \((x_0,y_0)\) to \((x,y)\text{.}\) It begins with a curved arc from \((x_0,y_0)\) to \((a,b)\) and then proceeds vertically from \((a,b)\) to \((a,y)\) and then horizontally to \((x,y)\text{.}\)
(b)
Notice that we can parametrize \(C_y\) by \(\langle a,t\rangle\) for \(b\leq t\leq y\text{.}\) Find a similar parametrization for \(C_x\text{.}\)
(c)
Use the parametrization from above to write \(\int_{C_y}\vF\cdot d\vr\) and \(\int_{C_x}\vF\cdot d\vr\) as single variable integrals in the manner of Section 13.3. Use the fact that \(\vF(x,y) = \langle F_1(x,y),F_2(x,y)\rangle\) to express your integrals in terms of \(F_1\) and \(F_2\) without any dot products.
(d)
Rewrite your expression for \(f(x,y)\) using a line integral along \(C_1\) and the single variable integrals above.
(e)
Notice that your expression for \(f(x,y)\) from the previous part only depends on \(x\) as the upper limit of an single variable integral. Use the Second Fundamental Theorem of Calculus to calculate \(f_x(x,y)\text{.}\)
(f)
To calculate \(f_y(x,y)\text{,}\) we continue to consider a path \(C_1\) from \((x_0,y_0)\) to \((a,b)\text{,}\) but now let \(L_x\) be the line segment from \((a,b)\) to \((x,b)\) and let \(L_y\) be the line segment from \((x,b)\) to \((y,b)\text{.}\) Modify the process you used to find \(f_x(x,y)\) to find \(f_y(x,y)\text{.}\)
(g)
What can you conclude about the relationship between \(\grad f\) and \(\vF\text{?}\) What does this tell you about \(\vF\) beyond that it is path-independent and continuous?
