For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Potential Functions, Conclusion & Summary

Let C be a path on F connecting (0,1) to (3, − 2). Find ∫

_{C}F .

- By the fundamental theorem of line integrals, ∫
_{C}F = P_{F}(Q) − P_{F}(P) where P_{F}is the potential function for the vector field F with P and Q being points on C.

_{C}F = f(3, − 2) − f(0,1) = − 6 − 0 = − 6.

i) Determine ∫

_{C}F for the path on the open set S.

- By the fundamental theorem of line integrals, ∫
_{C}F = P_{F}(Q) − P_{F}(P) where P_{F}is the potential function for the vector field F with P and Q being points on C.

_{C}F = f(3,1) − f(1,1) = 10 − 2 = 8.

ii) Determine ∫

_{C−}F for the path on the open set S.

- Note that C
^{−}is just the change of orientation of our path.

_{C−}F = − ∫

_{C}F it follows that ∫

_{C−}F = − 8.

iii) Determine ∫

_{C′}F for the path on the open set S. C′

- Since a potential function f exists, ∫F is path independent. Note the orientation of our path C′.

_{C′}F = ∫

_{C}F = 8.

iv) Determine ∫

_{C0}F for the path C

_{0}= C

_{1}+ C

_{2}on the open set S.

- Since ∫F is path independent any closed path has ∫
_{C}F = 0.

_{0}= C

_{1}+ C

_{2}form a closed path, that is the endpoint and starpoint are the same. Hence ∫

_{C0}F = 0.

_{1}+ C

_{2}be described by:

If ∫

_{C}F = − 1, is there a potential function f for F?

_{C}F 0 along the closed path C, no potential function f exists for F.

_{1}+ C

_{2}+ C

_{3}+ C

_{4}be described by:

and ∫

_{C}F = 0. For C′ be described by:

Does ∫

_{C′}F = 0?

- Note necessarily, just because one closed path yielded zero, that does not guarantee that every closed path is zero.

_{C1}F = ∫

_{C2}F = ∫

_{C3}F + ∫

_{C4}F be described by:

Is there a potential function f for F?

- Yes, the fact that three different path integrals yield the same result make ∫F path independent.

If ∫

_{C}F = 0 over an open and connected set, then F has a potential function f.

_{C}F = 0.

If ∫

_{C}F = k with k > 0 over an open and connected set, then F does not have a potential function f.

If F is an open and connected set, then ∫

_{C}F = 0 for every closed path C.

If F is on an open set, then ∫

_{C}F = ∫

_{C1}F + ∫

_{C2}F + ∫

_{C3}F for C = C

_{1}+ C

_{2}+ C

_{3}

If F is on an open and connected set with ∫

_{C}F = 0, then ∫

_{C−}F = 0.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Potential Functions, Conclusion & Summary

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Potential Functions 0:16
- Theorem 1
- In Other Words
- Corollary
- Example 1
- Theorem 2
- Summary on Potential Functions 1
- Summary on Potential Functions 2
- Summary on Potential Functions 3
- Case 1
- Case 2
- Case 3
- Example 2

### Multivariable Calculus

### Transcription: Potential Functions, Conclusion & Summary

*Hello and welcome back to educator.com and multivariable calculus.*0000

*Today we are going to close out, or round out the story on potential functions, and summarize everything so we have a nice set of things to follow in order to decide how potential functions behave.*0004

*Okay. So, let us get started. So, let us start off with a theorem, here.*0017

*So, we will let f be a vector field, let f be a vector field on an open set s -- excuse me -- an open set s and let p _{f} be a potential function for f on s.*0026

*Okay... continue over here... let c be a path in s connecting the points p and q.*0065

*Then, the integral of f/c from point p to point q is equal to the potential function evaluated at q, the end point, minus the potential function evaluated at p, the beginning point.*0087

*Also, let us write this here -- also, this is the really interesting part in my opinion -- the integral is independent of the path.*0109

*Okay. I should actually have the path connecting p and q, I mean we know that it connects p and q, but let us just write it out anyways -- connecting p and q.*0131

*So, basically what this theorem says is that if you have this open set s and if you know that your particular vector field has a potential function, well instead of actually evaluating the integral itself, the way we did previously, evaluating the line integral, because in some cases it might be hard, you can just take the potential function, and evaluate the potential function at q and subtract the potential function at f.*0142

*This is pretty much the same thing as the fundamental theorem of calculus in single variable, this is just a version of it.*0169

*For some function, some vector field that emits a potential function, this is how we can actually calculate the integral instead of going into the line integral process.*0175

*What is also interesting is that the integral is independent of the path, so it does not matter if I get there via straight line or a curve, or if I bounce around all over the set and eventually get there.*0184

*It does not matter. The integral actually ends up being the same. That is extraordinary, absolutely extraordinary and a profound importance in physics.*0196

*Let me actually write out what this is so that you have it here.*0206

*In other words, the integral of a vector field which admits a potential function over a domain is equal to the difference of the potential function evaluated at the end points, and is path independent.*0210

*So, if this is p and this is q, if you are moving in that direction, integrating along that path, you are going to get the same value whether you go like this or whether you go like that, whether you go like that, it does not matter. The path does not matter.*0282

*The integral stays the same, simply because on this open set, the vector field admits a potential function.*0309

*So, the existence of a potential function is a very, very special kind of property that is related to the vector field under discussion.*0315

*Okay. So, let us go ahead and write out a corollary to this.*0323

*All of the previous hypotheses, let f be a vector field on an open set s which admits a potential function... okay, I will just write if previous hypotheses, then the integral of f around every closed path, every closed curve -- I will write path instead of curve -- in s = 0.*0337

*Because a closed path, you remember, the beginning point and the end point are the same. When you evaluate them of course you are going to get a - a, b - b, it is going to be 0.*0381

*Okay. So... and if there exists a closed path in s such that the integral of f does not equal 0, then f does not have a potential function.*0393

*This is the contra-positive version of the corollary. What we wrote... this part, this is the contra-positive.*0424

*So, if a vector field admits a potential function on an open set s, then the integral over every closed path in s, every closed path is going to be 0.*0430

*If you can find a closed path in s where the integral does not equal 0, that means the vector field does not have a potential function.*0440

*Again, it just depends on the direction we are working in. That is all this is -- okay... does not have a potential function -- you know what, let me not actually does not have a potential function, let me just write does not have a p _{f}.*0448

*Okay. Let us do an example here. So, example 1.*0468

*So, in a previous lesson we had a vector field f which was equal to 4xy and 2x ^{2}.*0473

*Okay? That admitted a potential function, which ended up being 2x ^{2}y.*0493

*Well, let p be the point (-4,7) and let q be the point (3,-2) and c be any path connecting them.*0501

*Then, the integral of f over c is equal to, well, it is equal to the potential function evaluated at... let me see, what are our 2 points... q, right? it is the ending... so we are going from p to q.*0533

*So it is going to be the potential function evaluated at 3 - 2, minus the potential function evaluated at (-4,7).*0550

*There we go. Then when we actually do this evaluation, we end up with something like again, if I did my arithmetic correct... and by all means, please check my arithmetic but arithmetic is not important, it is the mathematics that is important.*0565

*I think you get something like -36 - 224, and I hope my number are right... -360... so this is the final answer, this is what is important, right here... this.*0579

*You do not actually have to go ahead and parameterize the path and solve the integral, you do not have to take the f(c(t)) · c'(t), evaluate that integral, just take... because it admits a potential function, you can just evaluate it as a potential function at the two end points and that is going to be your integral value.*0596

*Okay. So, let us see what we have. Now, as it turns out -- I will write it over here -- as it turns out, in this case, the converse of this theorem is also true.*0615

*So, remember that we said the converse if... if we have a theorem that says if a, then b, the converse is if b, then a, you just switch the places of b and a.*0646

*Different than the contra-positive, the contra-positive, if the positive is a, then b, the contra-positive is if not a, then not b.*0656

*Those two are equivalent, so, if I have a theorem, the contra-positive is automatically true.*0663

*In this particular case, the converse is also true, but it is not true just automatically. It actually has to be proven to be true.*0670

*Again, we are not going to through the proof, we are not interested in the details of the proof, we are interested in the results, but as long as you differentiate between the two.*0677

*A positive statement and its contrapositive come together. They are both, they are equivalent.*0685

*The converse may or may not be true, you have to check to see if it is. In this case it is, so let us go ahead and write it out. Let us go to the next page here.*0690

*It is always nice when converses are true, it always makes life a lot easier. It gives you two directions to work in depending on what your particular situation is.*0702

*Let f be a vector field on an open connected set -- you know, I am just going to go ahead and start presuming that the hypotheses are given -- okay, if the integral of f is path independent, then f admits a potential function.*0713

*In other words, a potential function exists. In other words, if you happen to have a vector field, and you happen to have two paths that you can check on two points connecting... two paths connecting two different points, and let us say they are reasonably easily parameterized, if you took the integrals and the integrals ended up being the same, you could conclude that there is a potential function.*0744

*You do not need anything else at that point. Again, it just depends what you have at your disposal.*0767

*All of these theorems are there as tools in your toolbox. You might be presented with a vector field in a particular kind of open set.*0773

*You can draw a conclusion. You might be presented with a vector field and a path that is easily parameterizable.*0781

*You might be presented with an integral value -- all kinds of things you can be presented with, we just want to present theorems that allow you to test when something has a potential function, when something does not, when you have to evaluate the line integral directly, when you have to use the potential function to evaluate it at different points, that is all that is going on here. That is why we are listing these theorems.*0788

*Okay, so, if f is path independent, then f admits a potential function. That is the converse of the other one. Now, let us see.*0812

*If the particular paths in this theorem are closed, we have the following version of the theorem.*0824

*Again, this is just the same theorem as this except for closed paths. We have the following version.*0845

*If, the integral of f = 0 for every closed path, then f has a potential function.*0857

*Now, the only reason that we gave this particular version of the theorem is for the sake of completeness. There is no way that you are going to test every single path.*0889

*There is an infinite number of paths, an infinite number of closed paths in an open set.*0897

*So, from a practical standpoint, this theorem has absolutely no value at all. From a theoretical standpoint, it does, of course.*0901

*But, again, for practical purposes, you cannot really use this in any reasonable way, because you cannot test every path.*0909

*Fortunately, we have other tools at our disposal to check to see whether a potential function exists.*0915

*Now, let us go ahead and summarize what we have got up to here. So, let me draw a line and let me go back to black.*0921

*So, summary on potential functions... summary on potential functions.*0932

*Okay. So, let us start off with let f = fg, I am only going to state it for 2-space, but of course it is valid for n-space, be a vector field on an open connected set.*0946

*Okay. On an open connected s -- let us actually give a name to this set.*0977

*Okay. Number 1. Our first test. If d2(f) = d1(g) -- I am sorry, if it does not equal... let me write this again here -- if d2(f) does not equal d1(g), then there does not exist a potential function.*0984

*Now, this is the same as df/dy not equaling dg/dx.*1021

*Capital D notation up here, regular standard notation, more common notation for partial derivatives down here.*1033

*So, if d2(f) does not equal d1(g), then there does not exist a potential function, we are done.*1042

*Number 2. Second possibility. If d2(f) does equal d1(g) and s is a rectangle or the entire plane, then there exists a potential function.*1049

*I will start abbreviating it that way, and we can find it and we can find the potential function by integrating one variable at a time.*1086

*Remember we did several examples of that by integrating one variable at a time and again, the only thing you have to do when you are actually doing that is just be careful because there are a lot of x's and y'x and z's floating around.*1102

*Lots of functions -- functions, vector fields -- you just have to keep track of everything. One variable at a time.*1113

*So, let us have... there is a third possibility... if d2(f) = d1(g) -- I always keep putting an i there, d1(g) -- but s is not a rectangle or the entire plane, then a potential function may or may not exist.*1124

*Okay. Let us do case 1. So, we move onto another test, another series of, you know, that is the thing. We have this toolbox, if one thing does not work, we move onto something else, and if something does not give us a conclusive answer, we move onto a finer -- you know -- a finer version. Something else that we can test.*1166

*That is what we do. We just run down the list of possibilities. So, if d2(f) = d1(g) but s is not a rectangle, then a potential function may or may not exist.*1191

*So, the first case is if there exists some closed path, some closed curve c such that the integral of f along c does not equal 0, then there does not exist a potential function.*1202

*Okay. So, this particular case says I take d2(f), it happens to equal d1(g), it is not a rectangle so I cannot conclude that there is a potential function, but if I can conclude that there is some potential path, some closed path where the integral does not equal 0, then I conclude that there is not a potential function on that domain.*1229

*Okay. Case 2. Now, if the integral of f around every closed path in s equals 0, then there exists a potential function.*1251

*I will put in parentheses... (not practical)... this one, again, strictly for theoretical purposes, you are going to have to find something else to test some other way of finding out whether a potential function exists.*1281

*Okay. Let us do case 3 here. Case 3. Okay. If s is, and this is the interesting one, is the entire plane from which the origin has been deleted... so if s is the entire plane from which the origin has been deleted, so our particular set s we have just taken out the origin -- deleted -- and c is the unit circle, and if the integral of f/c = 0, then their exists a potential function.*1297

*So, in this particular case, I happen to be dealing with a vector field that is not defined at the origin, so I actually have to remove the origin from consideration.*1374

*Now it is just the entire plane with the origin missing, and if I go ahead and integrate that vector field around the unit circle... just a basic unit circle, if I get an integral that is actually equal to 0, then I can conclude that a potential function exists.*1385

*This is actually pretty extraordinary. You are going to run across all kinds of vector fields, whether it be mathematics, engineering, or particularly in physics.*1402

*Simply by taking the origin out, that one point out, all of a sudden everything changes. That is really truly an extraordinary thing. that the existence of the origin as a point that is viable in our particular connected open set, it really just changes the nature of the function itself.*1413

*By removing the origin, everything changes. Again, absolutely fascinating, absolutely extraordinary.*1434

*Let us go ahead and do an example to round this out.*1442

*Example 2. Okay. So, we are going to let f(x,y)... we have seen this vector field before... equal -y/x ^{2} + y^{2}, and x/x^{2} + y^{2}.*1447

*Now, c(t), so this is our vector field, our curve is going to be the unit circle, and sin(t)... and we are going to take t from 0 to 2pi, that is the unit circle, and s, our open set, is the entire plane because this vector is defined over the entire plane excluding the origin.*1470

*Because the point (0,0), this vector field is not defined, excluding the origin.*1505

*Well, let us go ahead and form f(c(t)), we are going to evaluate this directly. So, f(c(t)) = -sin(t)/1, right? because if I put cos(t) for x and sin(t) for y, cos ^{2} + sin^{2} is 1, standard trigonometric identity, and this is going to be cos(t)/1.*1513

*Now c'(t) that equals -sin(t) and cos(t). Okay.*1541

*So, when we form f(c(t)) · c'(t), when I dot these two I end up getting sin ^{2}(t), right? -sin(t) - sin(t), sin^{2}(t) + cos^{2}(t) = 1.*1551

*Let us see here. Is equal to 1... so now we are going to... let me do this on the next page... so now the integral of f, the integral of our vector field over our unit circle is equal to the integral from 0 to 2pi of just 1 × dt, which is dt.*1574

*That is going to equal t from 0 to 2pi, which equals 2pi.*1597

*2pi does not equal 0. This implies that there does not exist a potential function.*1604

*Let me write "there does not exist a potential function" for this particular vector field on this particular open set, the plane with the origin deleted.*1618

*If it were equal to 0, then we could conclude that a potential function exists. In this particular case, potential function does not exist for this vector field.*1632

*Now, if our open set changes, a potential function for this vector field may actually exist.*1641

*Again, do not draw any conclusions, this is one open set. This particular open set is the entire plane with the origin deleted.*1650

*If I took another section of the plane that has nothing to do with the origin, maybe something in the first quadrant, something in the second quadrant, third, fourth, wherever.*1659

*This is specific. So do not draw any other conclusions that the problem, in other words, any value that you get, any conclusion that you come to is valid only for the particular open set that you are dealing with.*1669

*This particular field may or may not have a potential function if the open set is actually different, and I will let you think about that or work it out in your problem sets.*1683

*I am sure that this is a vector field that you will see over, and over, and over again.*1694

*So with that, thank you for joining us here at educator.com and multi-variable calculus, we will see you next time. Bye-bye.*1698

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