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I am having a bit of an intuition problem here. As I am doing work problems on the work done by using line integrals it is said the work done for some but not all line integrals depends on the path. Let us focus on those. I used different parameterizations of the path.

In one case the path was nothing more that a straight line. I let x = t and y = t, being my parameterization. In another case the path was a parabola and I let x = t and y = t^2 , that being the parameterization. Now the professor said how I parameterized made no difference. So I used x = sin t and y = sin t in the line case and x = sin t and y = sin^2 t in the second case. The results being the same.

My question is when I use the sin function to parameterize how can the results possibly be guaranteed to be the same? The work function does not change , I can see that and the bound are kept the same I can see that, and the path is still the same in that but in one case I used a variable and the other case I used a sin function...how do I know the sin function does not affect the path in some way...struggling here with intuition. Thank you

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    $\begingroup$ Do you have a course textbook covering the topic of line integrals? Usually the independence of parametrization is a theorem which is proved immediately after the definition of line integrals. $\endgroup$ – Hans Lundmark Aug 9 '17 at 17:06
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    $\begingroup$ Or see this answer: math.stackexchange.com/a/1820372/1242 $\endgroup$ – Hans Lundmark Aug 9 '17 at 17:08
  • $\begingroup$ Hans...what are you talking about? I see no theorem of "independence" of parameterization. This is my problem ...how do we know this. Is there a definition of parameterization that insures this will be the outcome. ? It seems to me the paramaterization is always contrived to match the curve but in some cases it is not obvious so how do you know as long as the path is the same the work integral will yield the same result? $\endgroup$ – Sedumjoy Aug 11 '17 at 21:41
  • $\begingroup$ Well, I don't know exactly what's in your book, but usually this is explained in textbooks... $\endgroup$ – Hans Lundmark Aug 12 '17 at 9:23
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Let's make some definitions more clear before we move on to the intuition, so that we can be sure we are talking about the same things. In this context, a path is some possibly curved line between two endpoints. A parameterization of a path are the formulas $x = x(t)$ and $y = y(t)$. You can think of a path as some geometric object and a parameterization as some equations describing the path. The key here is that a path can have several parameterizations; for example, the straight line from $(0, 0)$ to $(1, 1)$ can be parameterized as $x = t, y = t$ or $x=t^2, y=t^2$, for $t \in [0, 1]$.

Now what about line integrals? A line integral is an integral over a path; ordinarily we compute a line integral using a parameterization of a path, but the original line integral won't make mention to a specific parameterization of said path. This is because any parameterization of a path will yield the same line integral. You can try this out with any line integral you will come across right now by using different parameterizations. Indeed, some parameterizations will be easier to calculate the integral than others.

Let's contrast this with what your question is asking. There are some line integrals where only the endpoints of the path of integration actually matter; the path itself can be anything, so long as the endpoints are as specified. This is much different than have different parameterizations (we have whole different paths!). These line integral for which the path of integration "doesn't matter" are called conservative, and I'm sure you will cover these types of fields in your course.

As a classic example, think of gravity (say, at the surface of the earth). We know an object falls from point $A$ to point $B$, and the question is how much work gravity has done to this object. Since gravity is a constant unidirectional force, you can convince yourself that no matter what meandering path is taken, gravity will do the same amount of work. For example, if the object goes up, the object must come back down, and so that initial negative work done by gravity is counteracted by the positive work when the object comes back down. Alternatively, if you have taken a physics course, you know that we can calculate the work done by gravity in this example using potential energy. This concept of a potential function is critical to conservative fields.

As a classic non-example, take friction. The longer and more winding the path you take, the more frictive forces act on the object, and the more work is done. Thus path matters here when calculating work (but again, parameterization does not).

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  • $\begingroup$ well said near the end...I would like to focus on your first paragraph a moment...how can the straight line from 0,0 to 1,1 be parameterized as x = t^2 and y + r^2......for t between 0 and 1? No matter what I pick it will not matter?....but there must be some criteria no ? $\endgroup$ – Sedumjoy Aug 9 '17 at 20:28
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    $\begingroup$ @Sedumjoy, in fact, any parameterization of the form $x = f(t), y = f(t)$ will describe some portion of the line $x=y$ (precisely because $x$ and $y$ are the same thing). What portion of the line $x=y$ is described depends on the range $t$ is allowed to have. Plot several points for the parameterizations I gave to get some intuition for this. $\endgroup$ – Bob Krueger Aug 10 '17 at 12:42
  • $\begingroup$ I will try...but I am struggling with this. :( . Intuitively if I plug in the parameterized part into the equation appearing in the integral it seems to me that reflects the height above the x,y plain to a surface so the distance must be longer through the volume than on the x,y plane and more work would be done. ??I will look at your explanation again. $\endgroup$ – Sedumjoy Aug 11 '17 at 15:06
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    $\begingroup$ The important part is that it is a path integral. Nothing is mentioned of parameterization because parameterization does not affect the value of the line integral. Really, choosing a different parameterization is just like $u$-substitution from ordinary calculus. $\endgroup$ – Bob Krueger Aug 11 '17 at 16:40
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    $\begingroup$ got it ...Your hint at 12:42 did the trick!..thank you it is true the path is the same as far as calculating work is concerned but different parameterizations can lead to different rates through the curve and different orientations but this will not affect work so the changes when parameterizing a path integral are mute. I found some examples that demonstrate this effect. $\endgroup$ – Sedumjoy Aug 13 '17 at 20:13

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