recognize conservative vector fields
compute gradients and potentials
In this section we'll return to the concept of work. Many vector fields are actually the derivative of a function. When this occurs, computing work along a curve is extremely easy. All you have to know is the endpoints of the curve, and the function \(f\) whose derivative gives you the vector field. This function is called a potential for a vector field. Once we are comfortable finding potentials, we'll show that the work done by such a vector field is the difference in the potential at the end points. This makes finding work extremely fast.
Let \(\vec F\) be a vector field. A potential for the vector field is a function \(f\) whose derivative equals \(\vec F\text{.}\) So if \(Df=\vec F\text{,}\) then we say that \(f\) is a potential for \(\vec F\text{.}\) When we want to emphasize that the derivative of \(f\) is a vector field, we call \(Df\) the gradient of \(f\) and write \(Df = \vec \nabla f\text{.}\) If \(\vec F\) has a potential, then we say that \(\vec F\) is a gradient field.
We'll quickly see that if a vector field has a potential, then the work done by the vector field is the difference in the potential. If you've ever dealt with kinetic and potential energy, then you hopefully recall that the change in kinetic energy is precisely the difference in potential energy. This is the reason we use the word “potential.”
Let's practice finding gradients and potentials.
HintFind the gradient of \(f\text{,}\) i.e. find \(Df(x,y)\text{.}\)
Next compute \(D^2f(x,y)\) (you should get a square matrix).
What are \(f_{xy}\) and \(f_{yx}\text{?}\)
Find the derivative of \(\vec F(x,y)\) (it should be a square matrix).
Now find a function \(f(x,y)\) whose gradient is \(\vec F\) (i.e. \(Df=\vec F\)).
What are \(f_{xy}\) and \(f_{yx}\text{?}\)
Find the derivative of \(\vec F\text{.}\)
Explain why is there no function \(f(x,y)\) so that \(Df(x,y)=\vec F(x,y)\text{?}\)
HintBased on your observations in the previous exercise, we have the following key theorem.
Let \(\vec F\) be a vector field that is everywhere continuously differentiable. Then \(\vec F\) has a potential if and only if the derivative \(D\vec F\) is a symmetric matrix. We say that a matrix is symmetric if interchanging the rows and columns results in the same matrix (so if you replace row 1 with column 1, and row 2 with column 2, etc., then you obtain the same matrix).
For each of the following vector fields, find a potential, or explain why none exists.
HintIf you haven't yet, please watch this YouTube video.
\(\vec F(x,y)=(2x-y, 3x+2y)\)
\(\vec F(x,y)=(2x+4y, 4x+3y)\)
\(\vec F(x,y)=(2x+4xy, 2x^2+y)\)
\(\vec F(x,y,z)=(x+2y+3z,2x+3y+4z,2x+3y+4z)\)
\(\vec F(x,y,z)=(x+2y+3z,2x+3y+4z,3x+4y+5z)\)
\(\vec F(x,y,z)=(x+yz,xz+z,xy+y)\)
If a vector field has a potential, then there is an extremely simple way to compute work. To see this, we must first review the fundamental theorem of calculus. The second half of the fundamental theorem of calculus states,
\(f\)\([a,b]\)\(F\)\(f\text{,}\)\(F(b)-F(a) = \int_a^b f(x) dx\text{.}\)
If we replace \(f\) with \(f'\text{,}\) then an anti-derivative of \(f'\) is \(f\text{,}\) and we can write,
\(f\)\([a,b]\text{,}\) \begin{equation*} f(b)-f(a)=\int_a^b f'(x) dx. \end{equation*}
This last version is the version we now generalize.
Suppose \(f\) is a continuously differentiable function, defined along some open region containing the smooth curve \(C\text{.}\) Let \(\vec r(t)\) be a parametrization of the curve \(C\) for \(t\in[a,b]\text{.}\) Then we have
\begin{equation*} f(\vec r(b))-f(\vec r(a))=\int_a^b Df(\vec r(t))D\vec r(t)\ dt. \end{equation*}Notice that if \(\vec F\) is a vector field, and has a potential \(f\text{,}\) which means \(\vec F = Df\text{,}\) then we could rephrase this theorem as follows.
\(\vec F\)\(C\text{.}\)\(\vec F\)\(f\text{.}\)\(A\)\(B\)\(C\text{.}\)\(\vec F\)\(C\) \begin{equation*} f(B)-f(A)=\int_C \vec F\cdot d\vec r = \int_C Mdx+Ndy. \end{equation*} \(\vec F\)
If you are familiar with kinetic energy, then you should notice a key idea here. Work is a transfer of energy. As an object falls, energy is transferred from potential energy to kinetic energy. The total kinetic energy at the end of a fall is precisely equal to the difference between the potential energy at the top of the fall and the potential energy at the bottom of the fall (neglecting air resistance). So work (the transfer of energy) is exactly the difference in potential energy.
The proof of the fundamental theorem of line integrals is quite short. All you need is the fundamental theorem of calculus, together with the chain rule (chain_rule_def).
Suppose \(f(x,y)\) is continuously differentiable, and suppose that \(\vec r(t)\) for \(t\in[a,b]\) is a parametrization of a smooth curve \(C\text{.}\) Prove that \(f(\vec r(b))-f(\vec r(a)) = \int_a^b Df(\vec r(t))D\vec r(t)\ dt\text{.}\) [If you do this exercise, skip Exercise 4 ]
Suppose \(f(x,y)\) is continuously differentiable, and suppose that \(\vec r(t)\) for \(t\in[a,b]\) is a parametrization of a smooth curve \(C\text{.}\)
We want to prove that:
\begin{equation*} f(\vec r(b))-f(\vec r(a)) = \int_a^b Df(\vec r(t))D\vec r(t)\ dt \end{equation*}Let \(g(t) = f(\vec r(t))\text{.}\) Use substitution on the equation. Remember \(t \in [a,b]\text{.}\)
Explain why the equation you get after the substitution is true.
Compute \(g`(t)\) using the matrix form of the chain rule.
Substitute back in to remove \(g(\ )\) from your equation.
For each vector field and curve below, find the work done by \(\vec F\) along \(C\text{.}\) In other words, compute the integral \(\int_C Mdx+Ndy\) or \(\int_C Mdx+Ndy+Pdz\text{.}\)
HintSee Sage for a picture.
Let \(\vec F(x,y) = (2x+y,x+4y)\) and \(C\) be the parabolic path \(y=9-x^2\) for \(x\) from \(-3\) to \(2\text{.}\)
See Sage for a picture.
Let \(\vec F(x,y,z) = (2x+yz,2z+xz,2y+xy)\) and \(C\) be the straight segment from \((2,-5,0)\) to \((1,2,3)\text{.}\)
Let \(\vec F = (x,z,y)\text{.}\) Let \(C_1\) be the curve which starts at \((1,0,0)\) and follows a helical path \((\cos t, \sin t, t)\) to \((1,0,2\pi)\text{.}\) Let \(C_2\) be the curve which starts at \((1,0, 2\pi)\) and follows a straight line path to \((2,4,3)\text{.}\) Let \(C_3\) be any smooth curve that starts at \((2,4,3)\) and ends at \((0,1,2)\text{.}\)
See Sage—\(C_1\) and \(C_2\) are in blue, and several possible \(C_3\) are shown in red.
If you are parameterizing the curves, you're doing this the really hard way. Are you using the potential of the vector field?
Find the work done by \(\vec F\) along each path \(C_1\text{,}\) \(C_2\text{,}\) \(C_3\text{.}\)
Find the work done by \(\vec F\) along the path \(C\) which follows \(C_1\text{,}\) then \(C_2\text{,}\) then \(C_3\text{.}\)
If \(C\) is any path that can be broken up into finitely many smooth sub-paths, and \(C\) starts at \((1,0,0)\) and ends at \((0,1,2)\text{,}\) what is the work done by \(\vec F\) along \(C\text{?}\)
In the exercise above, the path we took to get from one point to another did not matter. The vector field had a potential, which meant that the work done did not depend on the path traveled.
We say that a vector field is conservative if the integral \(\int_C \vec F\cdot d\vec r\) does not depend on the path \(C\text{.}\) We say that a curve \(C\) is piecewise smooth if it can be broken up into finitely many smooth curves.
Compute \(\ds \int \frac{x}{\sqrt{x^2+4}}dx\text{.}\) See 1 .
Suppose \(\vec F\) is a gradient field. Let \(C\) be a piecewise smooth closed curve. Compute \(\int_C \vec F\cdot d\vec r\) (you should get a number). Explain how you know your answer is correct.