AI Assistant
Transcript
00:09:560Paolo Guiotto: Morning, take your seat.
00:15:970Paolo Guiotto: So, I want to return on this problem, because… I think it's Sport 2.
00:23:960Paolo Guiotto: To be examined a bit. So you remember yesterday we started this exercise.
00:32:180Paolo Guiotto: I'll just quickly review what we did.
00:36:30Paolo Guiotto: And where we stopped. So, the first question was to determine values of ABCD in such a way that field is irrational. We did this by controlling, by checking the cross derivative at the same
00:53:500Paolo Guiotto: And at the end of this calculation, we got these conditions, C equal minus B and D equal A. So this basically reduces the field to this,
01:06:870Paolo Guiotto: shape, huh?
01:08:300Paolo Guiotto: AX plus BY over X squared plus Y squared for the first component, and minus BX plus AY over X squared plus Y squared for the second component.
01:21:100Paolo Guiotto: Second question, is, determine the values ABCD for which, F is conservative, and in this case, determine also the potentials.
01:32:310Paolo Guiotto: Now, we say that to be conservative must be irrational.
01:36:680Paolo Guiotto: And since we know that the rotationals are only those, we focus only on those fields, because the other one cannot be definitely conservative. So, let's focus on this, and therefore we look for a function F for which gradient F is
01:53:270Paolo Guiotto: The field, we do not have for the moment any other condition to be checked than the definition.
01:59:640Paolo Guiotto: So we took the first equation, we solved, and we found this for Y different from 0, A half log X squared plus Y squared plus B are tangent X over Y plus Z of Y. For Y equals 0, we have A squared log X squared plus number.
02:18:890Paolo Guiotto: Now, what we did, is… was to verify that also the second condition is,
02:27:940Paolo Guiotto: Is if the recommendation is verified or not.
02:31:790Paolo Guiotto: And, here is where we stopped. We said, for Y different from 0, we get, for the derivative of F, we get,
02:42:250Paolo Guiotto: Visa?
02:44:00Paolo Guiotto: And it turns out that, I suspected that there was an error, but in fact, there is no error.
02:50:500Paolo Guiotto: But let's see where is the trouble later. So…
02:55:70Paolo Guiotto: We have here that the left-hand side is equal to the right-hand side, the fraction, so we get C prime y equals 0, so C equals constant. I say that we should get this, but that's not the case. Now, let's finish this.
03:11:20Paolo Guiotto: So, apparently, It works, so it means that we get FXY is this.
03:19:960Paolo Guiotto: So, the unique novelty is that C is a constant, right? So, we have A half log X squared.
03:29:550Paolo Guiotto: plus Y squared… Plus VAC tangent X over Y.
03:39:310Paolo Guiotto: Arc tangent.
03:41:330Paolo Guiotto: X or Y… plus constant, and for Y equals 0, we get a half log X squared.
03:49:820Paolo Guiotto: That's constant.
03:52:250Paolo Guiotto: This is for Y different from 0.
03:55:470Paolo Guiotto: And this is for Y plus 0.
03:58:60Paolo Guiotto: Now, is that enough to say that we found a…
04:03:690Paolo Guiotto: The potential? Apparently, yes, but in fact, no.
04:07:800Paolo Guiotto: What should be suspect here is this, Y equals 0, Y different from 0, so the Y.
04:15:530Paolo Guiotto: Because we can see that this cannot be actually a potential, because this function is not even continuous, apart for the case B equals 0. Let's see why.
04:27:40Paolo Guiotto: Now, so this is a little bit, tricky case, so I will,
04:36:560Paolo Guiotto: I, I think that you can, you can shift, no? Obviously, no result, so I will practice here the… the argument. So let's look at this.
04:45:830Paolo Guiotto: Now, we have a function, F, which is defined in a way for Y different from 0. So, here we have this expression here and here, while for Y equals 0, we are on the y-axis, we have another expression.
05:04:470Paolo Guiotto: That is not just plugging Y equals Z in the first one, by the way.
05:08:520Paolo Guiotto: Now, what happens is that if you take the point… a point on this axis, for example, here, it doesn't matter what is the point. The point will have a shape at zero.
05:19:580Paolo Guiotto: What happens to this function when you go to point X0?
05:25:590Paolo Guiotto: So, imagine that we take a… we move along this line, for example. Here we have points of type XY. Now, just to have the idea that we are not moving X, but only Y, we put an hex here.
05:41:630Paolo Guiotto: Now, what happens to the function f at point X hat y?
05:46:970Paolo Guiotto: Where the function for Y different from 0 is AF log depth.
05:53:670Paolo Guiotto: X hat square plus Y squared.
05:58:160Paolo Guiotto: Then we have plus B, our tangent.
06:02:870Paolo Guiotto: of X hat divided Y plus constant.
06:07:610Paolo Guiotto: I imagine that we move this point, to that… to the vertical line, to the axis, so this means let's send Y to 0.
06:17:890Paolo Guiotto: What happens? Well, what happens depends on how I get there, because if I go there from this side, so I take a pointer in the upper half plane, so this means that you go to zero positive. In this case, X is a positive number.
06:36:30Paolo Guiotto: So, when you send Y to 0 plus.
06:43:510Paolo Guiotto: positively, so the point is moving down to the axis. You see what you get. So the first part is basically innocent, because you have A half.
06:54:860Paolo Guiotto: It's a constant. Then the argument of log goes to log X hat squared. X hat is different from zero, so that X hat square is positive, there is no problem. Log X hat square.
07:08:500Paolo Guiotto: Then, look at the second part, B,
07:11:230Paolo Guiotto: What happens to the argument of the F tangent?
07:14:620Paolo Guiotto: The denominator here goes to zero, positive.
07:19:190Paolo Guiotto: That's positive, the et cetera, so the argument of the tangent goes to
07:26:690Paolo Guiotto: plus infinity, so this guy goes to plus infinity. The F tangent goes to…
07:36:470Paolo Guiotto: by half.
07:38:320Paolo Guiotto: So we have this pi half plus the constant C.
07:44:150Paolo Guiotto: What if we go, to the, to the point X0 from the negative part? So, in this case, we send y to 0, but negative.
07:56:930Paolo Guiotto: So, for the first term, nothing changed, because Y goes to 0, Y squared goes to 0, so the argument of log goes to X hat square. So this, again, goes to A half log X hat square plus B.
08:13:180Paolo Guiotto: Minus, yes.
08:15:80Paolo Guiotto: So, when you look now at the arctangent, why both, let's change problems?
08:20:730Paolo Guiotto: Y goes to 0 minus. The argument of the arctangent goes to…
08:26:850Paolo Guiotto: minus infinity. Yeah, tangent goes to…
08:30:740Paolo Guiotto: minus 5 half. So, what happens here is we get this.
08:36:520Paolo Guiotto: And then the constant remains the constant.
08:39:330Paolo Guiotto: So it means that, If you go to point accept zero, moving, from…
08:46:840Paolo Guiotto: from below, you go to this value down here. If you move to that point from above, you go to the value you have there. They are not the same, you see? Because the first piece is the same, the constant is the same, but this one is just the opposite.
09:06:890Paolo Guiotto: And there is one unique case when this is the same.
09:12:310Paolo Guiotto: equal.
09:13:470Paolo Guiotto: Exactly. Now, what is the problem with the not being the same? If it is not the same, it means that that function is not continuous, because you have two different ways to go to that point, and that you have two different limits. And why this is a problem?
09:28:640Paolo Guiotto: Because that function must have a gradient, and to have a gradient, it's differentiable, and differentiable is more than continuous, so that function
09:37:700Paolo Guiotto: wouldn't be differentiable if B is different from 0, so it couldn't be a potential in any case, okay?
09:45:860Paolo Guiotto: So, what we have here is that… so, actually, it's something that we should always check, even in the other exercises, but we never check because there are no
09:56:680Paolo Guiotto: suspicious cases. Like, for example, look here, this was the previous example, the previous exercise we did.
10:06:450Paolo Guiotto: Yesterday, so we add this field determined A and B, such that it is rotational, and then it is conservative. We found that,
10:16:230Paolo Guiotto: to be rotational, we had this combination, A equals 2, B equals 1, and then what we did was, let's look for the potential, so let's solve this. We solved the first equation, we plug it into the second one. At the end, we arrive to say that this is the function.
10:35:770Paolo Guiotto: Now, from the logical point of view, this is not complete argument, because this shows that if F is a potential, then F is this one.
10:46:490Paolo Guiotto: But who says that that one is a potential? You should check back that the gradient of this is your function. So we have not done, but if you take back the first examples, we have done…
10:59:440Paolo Guiotto: Mmm, in the first classes, perhaps the first example,
11:08:790Paolo Guiotto: yeah, now, since you see, we checked here, we computed the derivative, and we checked back that that is potential. So, in principle, that check should always be made, and here we didn't. But in any case, so…
11:25:490Paolo Guiotto: When you do in this problem, if you arrive here and you say, okay, this is the potential.
11:32:590Paolo Guiotto: This is not correct, because at the end, you see that that function cannot be a potential, because if B is different from 0, that function is not continuous at those points on the x-axis. So this would mean that F is…
11:52:320Paolo Guiotto: Well, that's what you read. F… is not… from Teamos.
11:58:910Paolo Guiotto: on a part of the domain. So, the problem was to determine the potential on R2 minus 0. So, in that part, the axis is in that domain. It's not continuous on…
12:12:70Paolo Guiotto: Excellent.
12:13:410Paolo Guiotto: axis.
12:15:620Paolo Guiotto: So… F is not… differentiable.
12:22:440Paolo Guiotto: on X axis.
12:27:40Paolo Guiotto: So, F cannot be a potential.
12:37:270Paolo Guiotto: Unless… this problem disappears when, unless B is 0, Unless…
12:47:200Paolo Guiotto: B equals 0. In that case, we get F of XY equal
12:53:510Paolo Guiotto: Now, if you look at the formula, when B is 0, it disappears that a tangent of X over Y, which is a problem for the formula, we can directly write log X squared plus Y squared, because we have for both cases. So, this formula reduces… where is it?
13:13:140Paolo Guiotto: So, A half log X squared plus Y squared, plus C, And this is differentiable.
13:24:410Paolo Guiotto: And the gradient of F, which is made by the XF, the YF,
13:31:80Paolo Guiotto: is now okay, because it is equal to A alpha first component.
13:36:690Paolo Guiotto: This is X squared plus Y squared, denominator delivered with respect to X is 2X, so this cancels this, so let's put AX here.
13:47:650Paolo Guiotto: Good.
13:49:20Paolo Guiotto: AX here, and then when you do the derivative with respect to Y, you get AY divided by X squared plus Y squared.
13:57:20Paolo Guiotto: which is for the case, B equals 0, which is exactly The field we had the… Here…
14:12:130Paolo Guiotto: You see, if you put B equals 0, you get AX over X squared plus Y squared, and AY over X squared plus Y squared. It is the field we have, so that's the gradient equal to the field, and this means it is the potential.
14:29:830Paolo Guiotto: So you should always check that what you get is actually the result, because from the logical point of view, I repeat, we are just doing this kind of argument. If F is a potential, then F must be this one. But no one says
14:45:230Paolo Guiotto: If this one is a potential issue, it's the vice versa that is missing.
14:49:890Paolo Guiotto: Okay, this was to, fix this very particular situation in most of cases.
14:56:900Paolo Guiotto: We do not have such complications.
14:59:890Paolo Guiotto: Okay, let's rewind what we have seen yesterday in the second part. So, we introduced the concept of curve.
15:11:530Paolo Guiotto: As just a function of one real variable, vector-valued.
15:17:550Paolo Guiotto: We say that it is regular if it's the derivative, so it means that each of the companies has derivative.
15:27:120Paolo Guiotto: And with this, we introduced the concept of line, or at integral, line integral, of a field along a curve gamma, which is defined by this formula.
15:42:450Paolo Guiotto: At the end of the class, we noticed this important fact.
15:46:230Paolo Guiotto: If the field is conservative, so if there is potential, then the arc in the line integral depends only on the initial and final point.
15:56:80Paolo Guiotto: And this is because we have this formula, the line integral is the difference between final value and initial value.
16:04:780Paolo Guiotto: Now, let's restart from this point today.
16:08:210Paolo Guiotto: So, I just write, we proved.
16:16:90Paolo Guiotto: That.
16:20:160Paolo Guiotto: if… F.
16:23:70Paolo Guiotto: Ease.
16:24:560Paolo Guiotto: conservative.
16:32:310Paolo Guiotto: And… F.
16:36:50Paolo Guiotto: he is, gradient of little f. This is not the second condition, means the same.
16:42:410Paolo Guiotto: And I'm saying, and F is the potential of capital F.
16:47:400Paolo Guiotto: Then, The, line integral on, curve gamma of F,
16:55:920Paolo Guiotto: is equal to the function, the potential, at final point, gamma of B,
17:02:600Paolo Guiotto: minus the potential at initial point, gamma of A.
17:07:280Paolo Guiotto: Where gamma is gamma of T.
17:12:280Paolo Guiotto: with the T defined in the interval AB, With values in, in general.
17:20:40Paolo Guiotto: So we have a particular consequence of this.
17:24:710Paolo Guiotto: If the final point, gamma B, coincides with the initial point, so we are in this situation, we have the domain B,
17:33:760Paolo Guiotto: We have a pattern, gamma, that starts from a point and then ends into the same point. So this is the game when gamma A is equal to gamma B.
17:47:100Paolo Guiotto: In this case, as you may imagine, we call this gamma a circuit.
17:51:840Paolo Guiotto: Okay?
17:53:10Paolo Guiotto: So, let's introduce this definition. Definition.
17:58:460Paolo Guiotto: If gamma of A is equal to gamma of B,
18:05:590Paolo Guiotto: for gamma equal gamma of P.
18:09:410Paolo Guiotto: defined on the interval AD to RB,
18:14:10Paolo Guiotto: So, let's say final value equals to the initial value, final point equal to the initial point. We say…
18:23:430Paolo Guiotto: that… Gamma.
18:27:860Paolo Guiotto: is, circuit.
18:34:480Paolo Guiotto: and the arc integral in this case, of a field F,
18:41:340Paolo Guiotto: the, line integral, arc-line integral are synonymous, so let's keep line… line integral.
18:55:330Paolo Guiotto: of F,
18:57:590Paolo Guiotto: On a circuit is also denoted with this particular symbol to remind us that it is an integral of a circuit.
19:08:230Paolo Guiotto: And it takes off a name, it's called,
19:14:720Paolo Guiotto: circulation.
19:20:400Paolo Guiotto: off F.
19:22:870Paolo Guiotto: Along.
19:24:990Paolo Guiotto: Gamma.
19:27:290Paolo Guiotto: Now, the consequence, these are just names, the consequence of this formula, as you can see, if gamma B is equal to gamma A, that difference becomes equal to 0.
19:37:990Paolo Guiotto: So, in particular, we have this corollary.
19:46:400Paolo Guiotto: So, if, F… is conservative, We'll serve up here.
19:53:780Paolo Guiotto: Ben… each circulation of, F… Is equal to zero.
20:02:180Paolo Guiotto: This is for every gamma contained.
20:09:610Paolo Guiotto: in the share plate.
20:17:930Paolo Guiotto: So, if you want, this sounds to be another necessary condition to be conservative. If you want to be conservative, you must have circulation equal to zero. Actually, it turns out that this is also a sufficient condition, so it is an if-and-all if. This is a quite important
20:36:460Paolo Guiotto: TRM.
20:40:170Paolo Guiotto: actually F.
20:42:180Paolo Guiotto: is conservative.
20:45:640Paolo Guiotto: on D.
20:47:830Paolo Guiotto: If, and only if all circulations zero.
20:55:600Paolo Guiotto: Or, I mean, gamma.
20:58:260Paolo Guiotto: contained in it. Why I provide this, which is not…
21:03:70Paolo Guiotto: An element, it's a subset contained.
21:06:480Paolo Guiotto: For every gamma contains the secret.
21:11:140Paolo Guiotto: So, the novelty is the vice versa.
21:15:60Paolo Guiotto: Because this, in principle, says that if we compute all possible situations, which is impossible, by the way, and we show that they are equal to zero, we could say that it is, it is conservative, so it becomes a test.
21:32:140Paolo Guiotto: Now, hopefully, but this will be a consequence of something we will see
21:38:260Paolo Guiotto: After we have done integrals of multiple integrals, and in particular here, double integrals, in the case of plane, this reduces just to compute few circulations.
21:52:670Paolo Guiotto: However, before we see the proof of this.
21:58:100Paolo Guiotto: I want to show you something.
22:01:130Paolo Guiotto: So, just an example.
22:05:790Paolo Guiotto: So, we are ready…
22:12:850Paolo Guiotto: checked.
22:17:200Paolo Guiotto: that this field… F.
22:21:410Paolo Guiotto: This was the first example of field which is irritational, but not conservative.
22:29:600Paolo Guiotto: And it was, something like minus Y divided X squared plus Y squared.
22:35:860Paolo Guiotto: And X divided X squared plus plus square.
22:41:100Paolo Guiotto: is… irrational.
22:46:750Paolo Guiotto: But… not.
22:49:580Paolo Guiotto: Don't share with you.
22:51:770Paolo Guiotto: Okay, now let's show that.
22:54:570Paolo Guiotto: suppose that we don't know that it is not conservative. Let's show that there is at least a circulation here, which is circulation, which is different from zero. So, this means that it cannot be conservative because of that, okay?
23:10:850Paolo Guiotto: We can show We can.
23:17:10Paolo Guiotto: But we now…
23:25:60Paolo Guiotto: by using…
23:31:00Paolo Guiotto: But it's not working.
23:33:880Paolo Guiotto: By using simulations.
23:46:40Paolo Guiotto: Sweet.
23:47:430Paolo Guiotto: No.
23:50:590Paolo Guiotto: Okay, enough.
23:53:910Paolo Guiotto: that… F.
23:58:220Paolo Guiotto: is not conservative. So suppose that you don't already know this, and now we will check this. Okay, let's look at this field. It's a field in the plane, XY.
24:10:520Paolo Guiotto: With a unique bed pointer where the field is not defined, it's the route.
24:16:20Paolo Guiotto: Now, this is not incidental. We will take a circuit that turns around that point.
24:22:220Paolo Guiotto: This is the… the bad points are normally sources of issues.
24:28:330Paolo Guiotto: So we take this circuit, which is just a standard circle of radius 1, so we describe it by…
24:41:80Paolo Guiotto: this parametization, gamma of t is, cos T, sine T.
24:48:370Paolo Guiotto: with the T in 0 to pi.
24:53:130Paolo Guiotto: So let's compute the… the circulation of this field F along this circuit gamma, okay?
25:02:10Paolo Guiotto: Clearly, this gamma is, siquid.
25:09:270Paolo Guiotto: If you don't trust me, you see that gamma of 0 is cos 0, sine 0, because 0 is 1, sine 0 is 0, and gamma of 2 pi.
25:20:280Paolo Guiotto: is cos of 2 pi 1 sine of 2 pi is 0. So the final point inside me the initial point.
25:28:290Paolo Guiotto: Now, let's… so we do also… we have not yet computed the any line integral, so let's do also this one to see how the line integral works. So if I want to compute the circulation along this gamma of the field F,
25:43:420Paolo Guiotto: I have to do a reminder formula, so let's write the general formula. It is integral from A to B, F, that has to be evaluated at point gamma of T,
25:54:410Paolo Guiotto: scalar product with gamma prime of t.
25:58:720Paolo Guiotto: So we have to do a little bit of calculations. For us, the interval AB is the interval 02 pi.
26:08:240Paolo Guiotto: I have to evaluate F at the point gamma T. Point gamma T is cos T.
26:15:420Paolo Guiotto: Sciety…
26:18:790Paolo Guiotto: I have to do the scalar product of this with gamma prime t. So let's write down here who is gamma prime t.
26:26:780Paolo Guiotto: This is the vector made of the derivatives of this, so derivative of cos, So, cos prime T sine…
26:37:750Paolo Guiotto: prime T.
26:39:750Paolo Guiotto: And so, derivative of cosine is minus sine.
26:44:280Paolo Guiotto: Divide of sine is plus cosine.
26:46:940Paolo Guiotto: So the vector you have to right here is minus sine t, for scientific.
26:55:50Paolo Guiotto: The last thing to do is to see what is this vector here. This is the field F evaluated at point cos T sine t. We have the field at the generic point XY, we have to plug X equals cos T, Y equals sine t. Let's see what happens, let's write down here.
27:14:980Paolo Guiotto: the value of this, F, at point cos T.
27:19:760Paolo Guiotto: minus sine T, Peace.
27:23:650Paolo Guiotto: First component.
27:25:290Paolo Guiotto: The third component is minus Y divided X squared plus Y squared. So minus Y is… Y is this? This is X, this is Y. So minus Y is sine P.
27:35:770Paolo Guiotto: divided X squared plus 1 squared.
27:39:660Paolo Guiotto: But, so this is cost square.
27:43:890Paolo Guiotto: Y, there is minus sign T.
27:47:610Paolo Guiotto: It's me.
27:50:150Paolo Guiotto: So, let's go back.
27:53:670Paolo Guiotto: There is a minus Y, so I should say a minus it.
27:58:340Paolo Guiotto: Okay, now is it… it is correct. Okay, divided the cos square t plus sine squared t.
28:06:550Paolo Guiotto: Second component is X divided.
28:10:670Paolo Guiotto: So X is cos T divided cos squared t plus sine squared t.
28:19:90Paolo Guiotto: Okay, let's simplify. We can do, because this is equal to 1.
28:24:290Paolo Guiotto: this the same. So at the end, we get the vector is minus sine t, Falsy.
28:32:720Paolo Guiotto: So this is the vector we have to plug there.
28:35:820Paolo Guiotto: Price of this, so this is, we said, minus… scientist, costi…
28:43:810Paolo Guiotto: And we have to do now the scalar product between these two.
28:47:240Paolo Guiotto: Okay, so let me just copy here. So, integral 02 pi…
28:51:840Paolo Guiotto: So, the vector was minus sine P, Costi…
28:57:860Paolo Guiotto: I repeat, this is F evaluated at point comma T.
29:04:510Paolo Guiotto: scalar product with gamma prime T.
29:09:40Paolo Guiotto: Which is the vector, minus sine plus, the same vector.
29:19:740Paolo Guiotto: Now, when we do the Scalaprado, we do…
29:22:720Paolo Guiotto: sum of first times first plus second times second, so we get sine square T plus cos square T.
29:34:500Paolo Guiotto: But that's, again, equal to 1.
29:37:230Paolo Guiotto: So we are integrating 1 between 0 and 2 pi, we get 2 pi, which is not 0.
29:45:890Paolo Guiotto: Okay?
29:47:310Paolo Guiotto: So, since we found just one single circulation, which is not zero.
29:53:460Paolo Guiotto: It means that this property is not true.
29:57:380Paolo Guiotto: This says, if it is conservative, all simulations must be equal to zero, not some simulation. All.
30:05:130Paolo Guiotto: And here we found 1, which is different from 0, so it means that F cannot be conservative.
30:12:560Paolo Guiotto: And that's the conclusion.
30:23:20Paolo Guiotto: So this is, let's say, a new test.
30:26:400Paolo Guiotto: to check, for the moment, to check to exclude, you know, if it is conservative. Actually, we say that theorem, it turns out that this is an if and all if, so that's, at the same time, an necessary and sufficient condition to be conservative.
30:45:150Paolo Guiotto: Now, let's see why the theorem is true.
30:48:810Paolo Guiotto: So, let me just rewrite the statement here. So, F, conservative, If, and only if, the circulations…
31:01:730Paolo Guiotto: of F.
31:02:980Paolo Guiotto: All equal to zero for every gamma.
31:06:750Paolo Guiotto: contained in this secret.
31:14:430Paolo Guiotto: Okay, so let's see the proof. I don't know if I will do in the full details, because there is a part which is a bit technical, but let's see what is the idea behind this.
31:27:550Paolo Guiotto: So we already proved that that implication is valid, no?
31:31:920Paolo Guiotto: this one… I'll read it.
31:37:680Paolo Guiotto: Duh?
31:40:310Paolo Guiotto: Now, let's check this one.
31:42:510Paolo Guiotto: So let's write well what is the hypothesis. The hypothesis is that the circulations of F, 0.
31:53:300Paolo Guiotto: For every gamma, Containing the secret.
31:59:760Paolo Guiotto: And the thesis, which is what we need to prove, is that F is conservative, so that there is a potential.
32:07:340Paolo Guiotto: Since there is no other condition for the moment.
32:10:590Paolo Guiotto: If you think about it, the only condition is to prove that there is a potential to exhibit a function f
32:18:420Paolo Guiotto: which is function f of x, defined on D.
32:23:800Paolo Guiotto: with values in R, such that the gradient of F is exactly the function F.
32:32:690Paolo Guiotto: Okay, now where do we get the idea? The idea, we get the idea from this, again, from this formula.
32:40:940Paolo Guiotto: That we mentioned here.
32:43:140Paolo Guiotto: Because this formula says, no, if you take a gamma, which is not a secret, it just…
32:50:490Paolo Guiotto: E path, il cover, huh?
32:53:460Paolo Guiotto: the, if the… if F is conservative, we have this relation, no? Now, let's imagine to do this figure.
33:04:490Paolo Guiotto: So imagine that,
33:08:690Paolo Guiotto: I forgot, there is also an hypothesis, fundamental hypothesis, is that domain D, B, connected.
33:20:310Paolo Guiotto: So… roughly… made… off.
33:27:810Paolo Guiotto: one.
33:28:940Paolo Guiotto: Peace.
33:31:920Paolo Guiotto: So, actually, the definition, more precisely, is that
33:36:260Paolo Guiotto: If you take two points, you can always join these two points to a… A pet, a club.
33:44:70Paolo Guiotto: So the idea is that to fix a point.
33:47:490Paolo Guiotto: Pick a pointer that you call X0, wherever you want, it doesn't matter, and pick a generic point X.
33:55:800Paolo Guiotto: Now, since it is connected, there is a path that joins these two points. Let's call it gamma.
34:06:650Paolo Guiotto: So, let gamma… So let X0 be in D be… fixed.
34:18:310Paolo Guiotto: and X another point in D.
34:22:70Paolo Guiotto: Gamma, impact.
34:27:199Paolo Guiotto: Joining.
34:31:900Paolo Guiotto: X0.
34:34:690Paolo Guiotto: 2X.
34:36:530Paolo Guiotto: So this means that this gamma is a path, so it's a function, gamma of P,
34:43:100Paolo Guiotto: Where T belongs to some interval AB.
34:48:170Paolo Guiotto: It varies in at D.
34:50:590Paolo Guiotto: And we say that this is the initial point, gamma, of A,
34:54:790Paolo Guiotto: And this is the final point, gamma of B.
34:59:280Paolo Guiotto: Okay, so gamma.
35:00:990Paolo Guiotto: of A equal… X0.
35:05:930Paolo Guiotto: and gamma of B equal X.
35:11:750Paolo Guiotto: Okay?
35:14:00Paolo Guiotto: Well, this is a technical point. We accept that this gamma is a regular path, it's not… so we can do the derivative.
35:23:390Paolo Guiotto: So, let's add this, irregular.
35:29:190Paolo Guiotto: Bye.
35:31:840Paolo Guiotto: Now, we notice this fact. Suppose for a moment that the field is conservative, so there is a potential.
35:39:700Paolo Guiotto: So… If F… ease.
35:44:870Paolo Guiotto: conservative.
35:47:220Paolo Guiotto: So, F is the gradient of something.
35:50:490Paolo Guiotto: Ben,
35:51:790Paolo Guiotto: We know that when we do the line integral of F along that path gamma, this is the difference between the final F at final point, F of gamma B,
36:06:830Paolo Guiotto: minus F of gamma A. This is the result we proved yesterday, we reminded this morning here, no?
36:16:390Paolo Guiotto: But since the final point, gamma B, is X,
36:21:160Paolo Guiotto: and the initial point is X0, we would have this factor. This would be F of X, minus FX0.
36:32:450Paolo Guiotto: So we would see that… F of X,
36:36:740Paolo Guiotto: Which is the potential.
36:38:680Paolo Guiotto: would be FX0, Plus the integral on gamma of effort.
36:48:450Paolo Guiotto: Now, this f at 0.0 is just a constant.
36:53:870Paolo Guiotto: So basically, the potential F of X
36:57:70Paolo Guiotto: Would be that integral plus a constant.
37:01:70Paolo Guiotto: Well, do you see that the integral here is an integral of a fat gamma that we said joins these two points, X0 and X. So let's put also this in the notation of gamma.
37:14:710Paolo Guiotto: Let's put something like, well, maybe not in the argument, because it's going to confuse you. Let's put down here, X0 and X initial and final point.
37:27:460Paolo Guiotto: In such a way that we can rewrite this here. So, gamma…
37:33:760Paolo Guiotto: Here, depends on the initial point and on the final point.
37:38:540Paolo Guiotto: Otherwise, you wouldn't see where X is on the right-hand side. Your X, which is the variable of this function, is exactly here.
37:47:400Paolo Guiotto: So now this says, if f is conservative and the little f is the potential, then the little f must be a constant plus that integral.
37:58:140Paolo Guiotto: Now, this is just to say, okay, let's take that integral.
38:04:380Paolo Guiotto: And let's check that that integral is a potential of the field F.
38:09:960Paolo Guiotto: So, from this…
38:16:520Paolo Guiotto: Sweet.
38:17:480Paolo Guiotto: Led to… the… Following.
38:27:110Paolo Guiotto: idea.
38:28:910Paolo Guiotto: So now, we don't know.
38:31:30Paolo Guiotto: It's the goal to prove that the field has a potential, okay? So we cannot say… to prove that it has a potential, let's assume that it has a potential. You see, it's a logical call.
38:42:970Paolo Guiotto: No, you cannot do that, but you can say, if it has a potential, the potential must be like that, so why don't we try with that function? Maybe it is a potential. So we say, define…
38:55:430Paolo Guiotto: They're fine.
38:57:130Paolo Guiotto: F of X, huh?
39:00:160Paolo Guiotto: This is now a definition as the line integral along the path gamma that joins X0 to X.
39:09:630Paolo Guiotto: of the field F.
39:11:840Paolo Guiotto: You may say, but you should add a constant. Yes, I could do, but the constant will disappear in the gradient, so adding a constant is not relevant. The point is that I have to show that this function that I define here
39:26:620Paolo Guiotto: has gradient equal to F.
39:29:860Paolo Guiotto: Okay? So, and… Let's… prover.
39:39:400Paolo Guiotto: that, The gradient of this function, F, is the field F.
39:47:650Paolo Guiotto: This should remind to you something.
39:52:390Paolo Guiotto: Because you have already seen a version of this practice.
39:58:920Paolo Guiotto: So, or reminds.
40:07:630Paolo Guiotto: Do you remind of this fact? I will use the same notations without error.
40:13:700Paolo Guiotto: So, this function, integral of F between X0 and X, so let's add F of YDY.
40:25:120Paolo Guiotto: Peace?
40:28:940Paolo Guiotto: If I set this function, I call it little f of x, Ben?
40:35:710Paolo Guiotto: And the good assumptions, which are the function F, capital F, is continuous, so basically very mild assumption.
40:45:840Paolo Guiotto: Dan?
40:48:80Paolo Guiotto: Well, you know why you should know, because this… this has a very important name. This is called the fundamental
40:57:140Paolo Guiotto: TRM… of integral… calculus.
41:03:280Paolo Guiotto: So it's fundamental, it's just unique result you see in the theorem of integrals.
41:09:560Paolo Guiotto: Just this one, basically.
41:13:860Paolo Guiotto: No.
41:16:130Paolo Guiotto: That function is called an integral function, and the key property is that that function has derivative equal to…
41:29:630Paolo Guiotto: But…
41:31:180Paolo Guiotto: the function that you have in the… that you have inside the integral. So that function is a function whose derivative is the capital F.
41:39:430Paolo Guiotto: So now, if you look at these results in such a way, in some sense, we are trying to do the same thing.
41:45:880Paolo Guiotto: We take our capital F, we define the line integral on a path that joins X0 to X. It's not the integral from X0 to X. That's an integral in one real variable, that's an integral along a path, so it's something different.
42:01:920Paolo Guiotto: But the philosophy is the same. I have a function of X, and I prove that the derivative of that function, which is the gradient here, is the function that you have in the integral.
42:12:230Paolo Guiotto: Okay? So this is the strategy and the goal we want to achieve. Now, here there is the first initial problem.
42:21:500Paolo Guiotto: Which is…
42:22:870Paolo Guiotto: Okay, I say that, given two points, let's take a gamma that joins these two points, and define f of x in this way.
42:32:470Paolo Guiotto: But in general, there won't be just a unique path joining two points in the domain. There will be infinitely many paths.
42:38:960Paolo Guiotto: So, first the problem is,
42:46:730Paolo Guiotto: is the function f well-defined.
42:52:960Paolo Guiotto: why this matters. Because, you see.
42:56:540Paolo Guiotto: Let's do a figure. This is the point at zero.
43:00:210Paolo Guiotto: This is the point X. I'm saying, pick a path that joins these two points, like this one, gamma X0 to X,
43:12:340Paolo Guiotto: And compute f of x as the integral along that gamma X0.
43:21:300Paolo Guiotto: to AXA.
43:23:350Paolo Guiotto: All that.
43:25:30Paolo Guiotto: Yes, but there could be, in principle, another gamma, and you see here, I can find inferiorly many gammas that join these two points. This is another one. So let's say gamma tilde.
43:36:700Paolo Guiotto: that joins X0 to X.
43:39:810Paolo Guiotto: I could compute my F with the other path, no? With gamma tilde.
43:47:670Paolo Guiotto: from X0 to X.
43:50:860Paolo Guiotto: Who says that I get the same value?
43:54:790Paolo Guiotto: Because if I do not get the same value, I'm not defining anything, because I'm not saying how to define this thing, no? You see, if it depends on the path, it's something disturbed.
44:06:680Paolo Guiotto: But here is where the assumption comes.
44:10:430Paolo Guiotto: So, the assumption is that the circulation of F is always zero. Why there is a circulation here? Because if you take this part, you write them.
44:23:630Paolo Guiotto: from point X0 to X, and the second path in green from point X0 to X,
44:30:630Paolo Guiotto: Well, you could algebraically say that the integral… let's write a lighter notation, so just gamma on F equals to integral on gamma tilde without those X0 and X, etc, okay?
44:45:980Paolo Guiotto: This is the same of saying that the difference between the two must be zero, right?
44:53:800Paolo Guiotto: But what is this differential?
44:58:920Paolo Guiotto: Now, this difference is like, if you do this, you start from point X0, and you do the first way to point X along the red path.
45:11:20Paolo Guiotto: And then move back to 0.0 along the green part.
45:16:900Paolo Guiotto: So, but in the opposite direction.
45:20:150Paolo Guiotto: Now, you may expect that,
45:23:270Paolo Guiotto: running on the green path, but in the opposite direction, will change the sign of the, of the line integral. So, in fact, this minus gamma tilde is like having the
45:38:520Paolo Guiotto: in the line integral of same field F, but on the part that goes in the opposite direction. So let's call it minus gamma tilde, this part.
45:51:20Paolo Guiotto: And when you sum the two, contributions, it's like if you're doing a unique arc integral, line integral along the blue path. So the sum of these two is…
46:04:400Paolo Guiotto: the line integral on F, along this blue pattern, let's give a name, and let's call it, gamma hat, okay?
46:13:900Paolo Guiotto: And this gamma head is… what kind of party is this one?
46:18:900Paolo Guiotto: You see that you start from X0 and you return to X0. That's a sequel.
46:24:600Paolo Guiotto: This is a secret.
46:29:210Paolo Guiotto: And therefore, this must be zero by… hypothesis.
46:36:70Paolo Guiotto: So, by hypothesis, this one is zero, it means that that difference is zero, it means the two are the same, it means our function is well-defined.
46:45:980Paolo Guiotto: Okay?
46:47:240Paolo Guiotto: So this now solves the first problem. We don't have any, any issue with the definition of DSF. And second, we have to prove
46:58:460Paolo Guiotto: Second… We now can prove that gradient F is equal to this F within that F.
47:07:870Paolo Guiotto: Now, to compute the gradient, we should do the story a little bit formally, but I will do roughly, we can understand how it goes, okay? So, to compute the gradient, what I do is I compute a variation, F at point X plus H,
47:25:870Paolo Guiotto: minus F of X.
47:30:930Paolo Guiotto: yeah, let's finish this one. It takes 5 minutes. I will expect that this will be…
47:41:20Paolo Guiotto: If the function is differentiable, this should be equal
47:45:690Paolo Guiotto: be reminded that differentiable means this, minus this, minus gradient F at point X, scala H, is a little O of H, you know? This means differentiable.
47:58:940Paolo Guiotto: Now, I start computing this, and I hope to see that it is something linear in nature.
48:05:200Paolo Guiotto: And that something linear will be the gradient of my F plus a little correction. So let's do this F of X plus H minus F of X. We help to compute this thing with a figure.
48:20:370Paolo Guiotto: So the figure, this is the domain, this is the point x0, where we start. Here we have the point X, and let's say that
48:28:980Paolo Guiotto: close to point X, there is point X plus H.
48:33:590Paolo Guiotto: Particularly, this means that this little segment
48:37:820Paolo Guiotto: Between the two is the segment H.
48:41:530Paolo Guiotto: Now… my function f at point X plus H is computed through a path that joins these two, okay?
48:51:50Paolo Guiotto: Let's call it gamma, X0, X0 plus X plus H.
49:02:130Paolo Guiotto: I don't put arrows, otherwise it becomes very heavy, but of course, there is an arrow above everything here, and there is another… so, for the value here, I have integral along this path that goes from X0 to X plus H.
49:25:190Paolo Guiotto: of the field F.
49:27:830Paolo Guiotto: Now, for the value f of x, I have the same. I have another part that goes from here to here. This is gamma X0X, and this gives an integral along part gamma X0X of the field F.
49:45:120Paolo Guiotto: Now, as you can see, the two red parts are not
49:49:60Paolo Guiotto: Clothed, and not a silk indicator.
49:53:250Paolo Guiotto: But, of course, you see there is a little piece here that could close this circuit.
49:58:690Paolo Guiotto: Now, if I have this difference, I have… let's put also arrows, because these two parts are going from X0 to the final point, so they have arrows like this.
50:10:680Paolo Guiotto: So, when you do the first one, this is… this, is the integral along this path.
50:17:30Paolo Guiotto: Minus this one, along this path, we say that minus is, like, to revert the movement, so taking, like, this is the path
50:30:190Paolo Guiotto: Going back to F0. So, this integral here can be written as plus the line integral along the path, which is the same path, but in the opposite direction, of F.
50:45:730Paolo Guiotto: So there is now a plus here.
50:48:280Paolo Guiotto: Okay, so we have the red, which is going to point X plus H, plus the green.
50:55:950Paolo Guiotto: And now we do not have the blue, we add and subtract, so we do. Plus…
51:01:960Paolo Guiotto: The integral from, point, since we go in this order, we take from X plus H to X,
51:12:130Paolo Guiotto: of F. Since I am adding, and there is not this term, I have also subtract to make a correct identity, so this is minus gamma X plus H.
51:22:800Paolo Guiotto: Except on that.
51:26:300Paolo Guiotto: Okay, but now, look, if I take the sum of the red, plus the blue, plus the green, I'm doing a circuit.
51:36:320Paolo Guiotto: And this is a circulation. All these three together is a circulation along, let's say, red plus Blue.
51:49:830Paolo Guiotto: plus green.
51:53:150Paolo Guiotto: It's a circulation, this, of the field F, and we know that this is just zero, all this stuff, so this is zero.
52:01:720Paolo Guiotto: And then what remains is that last is minus integral on the path from X plus H to X of F.
52:12:160Paolo Guiotto: or I include the minus into the path, I can say plus integral on the path in the opposite direction. So instead of going from X plus H to X, I go from X to X plus H.
52:25:110Paolo Guiotto: So I have this.
52:27:40Paolo Guiotto: gamma from X to X plus H of my F.
52:33:70Paolo Guiotto: So, we got this, F of X plus H.
52:38:720Paolo Guiotto: minus… F of X, huh?
52:42:220Paolo Guiotto: This thing is the integral on this little piece from X to X plus H of the field F.
52:51:130Paolo Guiotto: Okay?
52:52:180Paolo Guiotto: So this is an exact formula.
52:56:270Paolo Guiotto: Okay, now…
52:57:850Paolo Guiotto: Since this path, imagine that in our intuition, H will be small, so these two points, X and X plus H, are very close each to the other, and here, in this path, arc line integral, I am evaluating the field on these blue points.
53:15:640Paolo Guiotto: Okay, so I could say that if the field is continuous.
53:19:790Paolo Guiotto: the value won't vary too much from the value at point X. It is here where I am now.
53:26:630Paolo Guiotto: a little bit informal. Up to this point, everything is 100% formal. So, I can say that this will be approximately the value of the field at point X.
53:38:680Paolo Guiotto: So if I replace that variable value with the constant value F of X, I get that this will be approximately equal to the integral on the path from x to x plus h of the constant vector f of x.
53:55:180Paolo Guiotto: This is independent on the path, it's constant. So this means that if I now model this segment from X to X plus H,
54:05:940Paolo Guiotto: Well, there is a standard way to parameterize a straight segment that goes from a point to another.
54:13:170Paolo Guiotto: The standard way is that you take this gamma of t is equal. Let's say that t equals 0 corresponds to point X, so you have X plus T,
54:26:590Paolo Guiotto: T equals 1 corresponds to point X plus H. If you just take the plus TH here, you see that when T is 0, you get X. When T is 1, you get X plus H. And when T is between 0 and 1, you get all these points. So for T in 01,
54:44:660Paolo Guiotto: This is the standard parameterization of a segment joining the point X to the point X plus H.
54:51:880Paolo Guiotto: Now, I plug this, in particular, the derivative of this vector, gamma. If this is gamma, the derivative will be…
55:03:120Paolo Guiotto: H. Now, the derivative is with respect to t. So, what I have here is that I am integrating on the range for the parameter, which is 0, 1, the constant vector F of X, scalar, the gamma prime, which is H,
55:22:410Paolo Guiotto: in DT.
55:24:270Paolo Guiotto: But that's a constant, and it gives F of X.
55:28:740Paolo Guiotto: scalar H time you can carry outside, because it does not depend on t, times the integral from 0 to 1 of dt, which is just equal to 1. So at the end, you get f of x scalar H.
55:44:630Paolo Guiotto: So we have now the conclusion.
55:47:80Paolo Guiotto: Because we have that F of X plus H.
55:52:20Paolo Guiotto: for this was to calculate this difference, minus f of x.
55:57:710Paolo Guiotto: At the end, you see that was exactly equal to that integral. We replaced the field F, which is variable, with the value at point X, because the idea is that points X and X plus H are close, because H is going to be small, so the field, if it is continuous.
56:16:720Paolo Guiotto: will be the value at a point on that segment here. If you take the value here, it will be approximately the value here, okay? The value of the field, no? F at this gamma of t will be approximately
56:33:290Paolo Guiotto: F at gamma of 0.
56:36:850Paolo Guiotto: Which is F of X.
56:39:430Paolo Guiotto: So I am, replacing, the, the…
56:43:610Paolo Guiotto: the generic… the value of the field, a generic point between these two, with the value at point X. So this, at the end, it says that this increment is approximately equal to F of X scala H.
57:01:670Paolo Guiotto: So it means that in an exact form, it will become like that. F , plus a correction, which is negligible in H.
57:13:920Paolo Guiotto: And in this exact form, you see that this is saying that that vector, f of x, is the vector that plays the role of gradient of F.
57:23:730Paolo Guiotto: Because this says that, by definition of differentiable function, there exists the derivative, the gradient of F at point x, and it is equal exactly to that f of x, which is what we wanted. We wanted to prove that there is this a function.
57:40:910Paolo Guiotto: Which is a potential for the field F, and this ends C proport.
57:47:630Paolo Guiotto: Okay, let's take the break, and then we do some final remarks on this.
57:54:690Paolo Guiotto: 10 minutes.
58:15:900Paolo Guiotto: Okay, so… Let's now refocus a second on the… statement.
58:24:570Paolo Guiotto: desires that… If we have a domain, connected domain made of one single piece.
58:33:860Paolo Guiotto: F is conservative independently if all the circulations are equal to zero.
58:38:330Paolo Guiotto: In principle, this is an if-and-no-if test, but in practice, it's not feasible to compute all the circulations, because there are simply many, and these are integrals, and in general, when we write an integral, we are not able to compute it.
58:55:980Paolo Guiotto: However, it is possible to prove A corollary of this.
59:03:470Paolo Guiotto: Which is something we will prove later, so I will give just a statement now, but we will return later, because we have not yet the tool to prove this.
59:14:30Paolo Guiotto: So, in general, let's say this, in general, It is impossible.
59:27:60Paolo Guiotto: to check.
59:31:190Paolo Guiotto: that, all the circulations.
59:37:360Paolo Guiotto: as, you know… For every gamma.
59:42:340Paolo Guiotto: Sickweed.
59:44:330Paolo Guiotto: contained in… D.
59:48:260Paolo Guiotto: However, in certain cases, this check can be reduced to just a finite number of circulations, And precise
59:59:120Paolo Guiotto: However…
00:04:50Paolo Guiotto: There are.
00:09:640Paolo Guiotto: of them.
00:14:830Paolo Guiotto: Jesus.
00:18:650Paolo Guiotto: where… this… Check.
00:27:350Paolo Guiotto: Cam.
00:28:410Paolo Guiotto: B.
00:29:340Paolo Guiotto: reduced the… to find… Number…
00:42:990Paolo Guiotto: Off.
00:47:350Paolo Guiotto: Circulations…
00:52:490Paolo Guiotto: So, I just give one, which is, let's say, the most used factor.
00:58:460Paolo Guiotto: this proposition.
01:00:230Paolo Guiotto: This applies to fields in plane, so let… F…
01:07:660Paolo Guiotto: be a vector field. Plain vector field means that it would be a
01:12:260Paolo Guiotto: function of two variables, add 2 to add 2.
01:15:950Paolo Guiotto: Defined on up to… Except, at most, a finite number of points, okay?
01:24:770Paolo Guiotto: If you think to the examples we have seen so far, it was always the case, because we had everything except 00, for example, okay? A finite number of points, let's call them P1, P2,
01:39:10Paolo Guiotto: piano.
01:41:880Paolo Guiotto: With values in R2.
01:45:850Paolo Guiotto: Where… So, PJ… J equal 1 to N.
01:55:00Paolo Guiotto: are points.
01:57:920Paolo Guiotto: off.
01:59:270Paolo Guiotto: R2. So let's make also a figure of what is going on to understand that.
02:04:930Paolo Guiotto: So this is the plane, huh?
02:07:340Paolo Guiotto: Where the field is defined.
02:10:00Paolo Guiotto: Except at a certain number of points, so there is a bad point, I do not have one.
02:16:240Paolo Guiotto: I put a cross to understand these are points where the field is not defined, okay? P2, P3, down here…
02:24:560Paolo Guiotto: And so on.
02:26:610Paolo Guiotto: Now, what this… this says is the following.
02:30:590Paolo Guiotto: if, first, F is… irritational.
02:39:340Paolo Guiotto: on… let's say the domain B, let's give a name to this. This is the domain B.
02:47:360Paolo Guiotto: So, if F is irrotational on the domain D, and this is not sufficient, we know, no? Irrotational is just necessary, but not sufficient condition to have that it is conservative.
03:00:210Paolo Guiotto: And two.
03:01:570Paolo Guiotto: instead of checking all circulations, which is practically impossible, it is enough to check only the circulations around these bad points. So, choosing the circles.
03:17:250Paolo Guiotto: Century at those points.
03:20:930Paolo Guiotto: So, for example, so we say that the circulation around, let's give name, say, gamma J to this pattern.
03:34:10Paolo Guiotto: circulation around gamma J, on gamma J of F,
03:39:820Paolo Guiotto: equal 0 for every J, equal 1 to N.
03:45:660Paolo Guiotto: Where?
03:49:680Paolo Guiotto: Well, what we really need is that this gamma J our circuit.
04:01:800Paolo Guiotto: that contains, in the interior, only one singularity, only one of these points. So I secretly such that
04:12:40Paolo Guiotto: PJ, now, the language is informal here, you understand. We could give a formal shape to this, but it would be…
04:23:930Paolo Guiotto: too much, to… just to say this simple thing, such that PJ, lies… inside the…
04:37:850Paolo Guiotto: gamma J.
04:39:920Paolo Guiotto: And all the others, BK…
04:48:140Paolo Guiotto: outside… of gamma J, For all K, different.
04:57:510Paolo Guiotto: from J, as you see in the figure here. You see that gamma J turns around, let's call it PJ, this point.
05:06:140Paolo Guiotto: the point PJ, no? So…
05:09:580Paolo Guiotto: we will have… it will be always possible to do that, because we have only a finite number of points. You can take, for example, sufficiently small circles that surround each of this point. They will contain only one point, and all the others will be outside, okay?
05:27:720Paolo Guiotto: Now, if it happens that the field is rotational, and that is also easy to check, no? Because you have to do the test of the cross derivative. Here, by the way, since the field is just a field in two variables, it's a unit check, no? The DY of F1 equals the X of F2.
05:47:730Paolo Guiotto: So, if this is checked, the plus, the circulations along these secrets that turns around each of these singular points, but only on each, only on just one singular point each.
06:02:980Paolo Guiotto: are zero, so only a finite number of circulations. You don't need to check all the circulations, but only a few circulations. Then, in this case, we can say that F is conservative.
06:22:250Paolo Guiotto: At the end, this is not such a big result, because it says it is conservative. It does not compute anything.
06:31:160Paolo Guiotto: So it's just, it tells you there is a potential. But at the end, if you want to compute the potential, you have still to do the hard work of computing the potential.
06:41:270Paolo Guiotto: So let me, well, I want to do first, to show you how it works, this, let's redo the exercise we have, let's say, reviewed this morning. The exercise we did yesterday.
06:57:670Paolo Guiotto: this one with the ABCD, no? But we do with this new, result, and let's see what changed in this discussion, okay? So I… now I will repeat this exercise with this new, knowledge. So…
07:15:920Paolo Guiotto: This, by the way, is the… Let's redo the exercise.
07:22:830Paolo Guiotto: Quite easy.
07:25:530Paolo Guiotto: 3, 4, 7… So it's not just a repetition.
07:32:40Paolo Guiotto: Well…
07:32:890Paolo Guiotto: at least I… I will use part of what I've done, or I… I don't redo everything. So here I have the field FXY, which is equal to AX plus BY
07:46:290Paolo Guiotto: divided X squared plus Y squared, second component, CX plus DY, divided X squared plus Y squared.
07:57:760Paolo Guiotto: for XY.
07:59:960Paolo Guiotto: which is in R2…
08:02:790Paolo Guiotto: minus 00. You see that there is one single bad point here, okay? So the domain is the full plane, except .00.
08:15:270Paolo Guiotto: So we are exactly in this,
08:18:90Paolo Guiotto: framework, no? A field in R2 with a finite number of bed points where it is not defined. Here, it is just one, okay?
08:28:910Paolo Guiotto: So, the questions are,
08:32:760Paolo Guiotto: Question 1, A, B, C, D…
08:38:750Paolo Guiotto: such that, F is irrotational.
08:47:710Paolo Guiotto: And question two, A, B, C, D, such that F… is conservative.
08:57:20Paolo Guiotto: And then, plus potentials in this case.
09:01:840Paolo Guiotto: Now, I want to repeat question one, because it's the same, no? We do the test of the cross derivative, so I want just to show how it changed the solution with this new tool. So, question one, same thing.
09:19:630Paolo Guiotto: off… Yesterday, say?
09:26:800Paolo Guiotto: And you see, this part is more or less straightforward. You combine derivatives, you impose the identity, and we got, if you remember, this condition. B, no, C equal minus B,
09:40:979Paolo Guiotto: Reminder.
09:44:560Paolo Guiotto: B and D equally.
09:47:319Paolo Guiotto: So we get this, after this first question, FXY becomes…
09:54:980Paolo Guiotto: AX plus BY over X squared plus Y squared, and then we have minus BX plus AY divided X squared plus Y squared. Okay, this is what we get after
10:11:840Paolo Guiotto: we check, the field is rotation. And let's come to question 2, which is.
10:16:950Paolo Guiotto: a bit different here, because now, I want to find ABCD such that the field is conservative. So, I should first start saying the same thing. F conservative, conservative means that necessarily F must be irrotational.
10:33:750Paolo Guiotto: And therefore, F is this guy here, okay? So I look only at this field.
10:41:710Paolo Guiotto: Now… This field is rotational, so it is not necessarily conservative.
10:47:870Paolo Guiotto: Let's apply this tool that says, if it is irrotational, and we already checked, and the circulations around these singular points are zero, here we have one singular point, so what we have to check? So, since…
11:06:850Paolo Guiotto: F… He's irrational, on domain B is R2 minus DRG.
11:19:150Paolo Guiotto: to be.
11:21:710Paolo Guiotto: conservative.
11:24:210Paolo Guiotto: It is sufficient that… the circulation around the gamma, let's say, of F equals zero.
11:38:880Paolo Guiotto: where gamma is any circuit, you can choose just one circuit, you don't need to prove any circulation, just one circulation. The important point is that this gamma must turn around at that point.
11:52:840Paolo Guiotto: You cannot take a gamma like that, okay? You must take a gamma that turns around at that point.
12:00:940Paolo Guiotto: So… This is our gamma.
12:07:340Paolo Guiotto: web… Gamma is a circuit.
12:17:380Paolo Guiotto: turning…
12:23:90Paolo Guiotto: around.
12:26:570Paolo Guiotto: the origin. Now, you can choose whatever you want, but of course, we have to compute, so…
12:31:560Paolo Guiotto: Now, what we choose here? Well, since the field contains this nice X squared plus Y squared, which are exactly constant when X squared plus Y squared is constant, this happens along a circle, I will take a circle.
12:46:650Paolo Guiotto: It doesn't matter what is the radius. If you want, we do the calculation with the generic radius R. So let's take this gamma of T.
12:54:730Paolo Guiotto: the point that belongs to a circle centered in the origin radius R. This is R cosine T.
13:02:100Paolo Guiotto: R sine T… And t is between 0 and 2 pi.
13:08:220Paolo Guiotto: Let's use this.
13:10:150Paolo Guiotto: So, the circulation of this field along this gamma By definition, it's the integral.
13:19:440Paolo Guiotto: the integration, now I'm writing the formula now. So the general formula is A to B, F of gamma of T,
13:27:20Paolo Guiotto: scalar with gamma prime of t. Now, let's put the ingredients into this integral. AB is the integral 0 to pi, is the range for the parameters of the… of the curve. So this is the integral 0 to pi. Then I have to evaluate the field at point gamma T, so it will be R cos T.
13:46:890Paolo Guiotto: R sine T.
13:49:870Paolo Guiotto: scalar product with gamma prime. We have gamma, gamma prime is just the derivative.
13:56:660Paolo Guiotto: for each of the components, so here we have the derivative of R cos t, derivative with respect to t, which is the variable R is just a parameter.
14:05:750Paolo Guiotto: minus R sine t.
14:08:90Paolo Guiotto: And the second one is R. Koski.
14:11:390Paolo Guiotto: So here we put minus R sine T, are costly.
14:18:310Paolo Guiotto: DT. Okay, now we have to compute F at that point. F evaluated at R cos T.
14:28:180Paolo Guiotto: are scientists.
14:31:930Paolo Guiotto: Now, F is what? Is this one?
14:35:250Paolo Guiotto: So, we have fractions. For the first component, we have AXA R cos D.
14:42:370Paolo Guiotto: plus BYBR sine T. You see?
14:47:30Paolo Guiotto: This is AX plus BY.
14:51:20Paolo Guiotto: I just plugged X and Y, respectively, R cos T are single. Downstairs, I have R, cos T,
14:59:640Paolo Guiotto: square.
15:00:950Paolo Guiotto: plus r sine t squared.
15:04:290Paolo Guiotto: I use this part because, for this.
15:07:650Paolo Guiotto: This simplifies, considering the denominator. And the same here. About the numerator, numerator is, minus BX plus AY. So minus BX is, cos T,
15:23:300Paolo Guiotto: plus AYR… society.
15:28:370Paolo Guiotto: So that's F, let's simplify a bit. So this is R squared times cos squared plus sine squared, which is equal to 1, so the denominator is just R squared.
15:41:110Paolo Guiotto: we can, since it is a constant for the two components, so this is, at the end, it is equal to R squared, and also this one is equal to R squared, we can put a common factor 1 over R-squared outside.
15:56:820Paolo Guiotto: And inside, we have, you see that there is another, also an arc here that can be…
16:02:620Paolo Guiotto: that multiplies both components, so I can take out also that R. So I have A cos T plus B sine T in the first component, minus B cos T plus A sine T.
16:19:780Paolo Guiotto: E.
16:21:820Paolo Guiotto: So, at the end, I have a little cancellation here, 1 over R. Now, this is the value of the field at point R cos T. It would have been better to take R equal 1.
16:34:860Paolo Guiotto: But since we take our generic, we carry around this R. Okay, now let's plug here, in place of this. So we have integral.
16:45:920Paolo Guiotto: 02 pi, so I'm continuing this calculation. Let's put a star here.
16:56:330Paolo Guiotto: So we have this, 1 over R,
17:00:500Paolo Guiotto: A cos T plus B sine.
17:06:240Paolo Guiotto: D minus B cos… glass. A.
17:10:900Paolo Guiotto: sine T. This is color product with that one, minus R sine T, that was already there.
17:19:950Paolo Guiotto: So, minus R sine T, And the second component is our costing.
17:31:60Paolo Guiotto: So, for the moment, I'm just replacing… I'm not doing yet the calculations of integers, I'm just computing what… what is the integral I have to compute, okay?
17:41:360Paolo Guiotto: Now, here you see that there is this R that factorizes this vector.
17:46:50Paolo Guiotto: Be careful, it's not R squared, because R is a factor for all the components, so it comes out, because it's the product scalar times vector. It's not R squared. This is wrong. It's just R. So this R, with this R, they cancel, so at the end, they would have been the same, I'm taking R equal 1.
18:05:310Paolo Guiotto: Now, let's do the product, integral 02 pi. We have to do this times this, and then this plus this times this.
18:14:360Paolo Guiotto: So, with a bit of patience, so we have a minus A cos T sine T, plus… no, minus…
18:26:10Paolo Guiotto: B sine square.
18:29:390Paolo Guiotto: E.
18:30:480Paolo Guiotto: And that's the first. Then we have the second, so minus B cos square.
18:38:670Paolo Guiotto: And then we have a plus A sine T cos T.
18:45:830Paolo Guiotto: Now, you see that we can simplify directly these two.
18:52:920Paolo Guiotto: then this one is minus B that multiplies sine squared plus cos square, which is 1, so it remains minus D. So I get integral 0, 2 pi.
19:02:610Paolo Guiotto: of minus B. I've not yet computed the integral. It's now where I compute the integral, and as you can see, in this case, we are particularly lucky, because it's the integral of a constant. So it is minus B times 2 pi.
19:15:600Paolo Guiotto: So that's the value of that circulation.
19:18:840Paolo Guiotto: So when this circulation is zero.
19:23:220Paolo Guiotto: The only possibility is B equals 0.
19:26:180Paolo Guiotto: So at the end, you get the same information we got in the other way, but…
19:31:170Paolo Guiotto: here, let's say… in the other way, it would have… I myself got confused yesterday, but it's a little bit more difficult to understand, because
19:43:600Paolo Guiotto: Most of people would say, okay, I found, and this is the potential, and wouldn't realize that there is a problem with that arch tangent. While here.
19:53:580Paolo Guiotto: the… the fact is that the check is straight, so either it's 0 or not, and this comes 0 if and only if B equals zero, so this means that, so now this is an if and only if test.
20:07:470Paolo Guiotto: Okay? Because it is irritational, plus the circulations around the singular points are zero. So we can say that F is conservative.
20:21:750Paolo Guiotto: Now, conservative, if and only if F is… so, with B equals 0 becomes, what is AX divided X squared plus Y squared.
20:34:960Paolo Guiotto: And AY, if I'm not wrong, divided X squared plus Y squared.
20:40:240Paolo Guiotto: Now, this says that these are the conservative fields, but it does not tell you anything about the potential. For the potential, you have still to solve the system. There is no other way. Well, actually, there would be another way that is using this formula we learned here.
20:58:430Paolo Guiotto: We have seen that there is a formula in this theorem for the potential.
21:03:340Paolo Guiotto: Which is computing this line integrals.
21:06:280Paolo Guiotto: But this is generally a little bit tricky, so it is better to proceed directly.
21:12:90Paolo Guiotto: Okay? So, if now you want to finish and determine the potential, you have to do the same job to determine
21:20:540Paolo Guiotto: Potentials.
21:22:500Paolo Guiotto: F, we… tab.
21:26:470Paolo Guiotto: to solve… DXF equal AX divided X squared plus…
21:36:540Paolo Guiotto: Y squared and DYF equals AY
21:41:350Paolo Guiotto: divided X squared plus Y square, which is easy here. This will be the F.
21:47:710Paolo Guiotto: equal… Half.
21:50:760Paolo Guiotto: log X squared plus Y squared.
21:54:940Paolo Guiotto: Last constant.
21:56:580Paolo Guiotto: I want to repeat this part of the story.
22:00:280Paolo Guiotto: Okay.
22:01:860Paolo Guiotto: Let's say that, basically, this is the… that's all for this part on vector fields. So, in the remaining 15 minutes, let's do another
22:11:670Paolo Guiotto: exerciser.
22:13:860Paolo Guiotto: And then,
22:16:310Paolo Guiotto: You basically have to do all the exercises on the chapter. I will publish solutions, maybe I will tell you to do this and that in such a way that you can check the solution, because
22:27:450Paolo Guiotto: I don't know if I can write the solutions of all these problems, but…
22:32:950Paolo Guiotto: they are pretty much always the same problem. So let's do something like, I don't know, 3… 0.4.12.
22:43:680Paolo Guiotto: This is still on plane, huh?
22:46:790Paolo Guiotto: So here we have… also here we have four parameters.
22:50:850Paolo Guiotto: A, B, and alpha and beta are different from zero.
22:58:930Paolo Guiotto: we define this field, F, this is still a plane.
23:03:750Paolo Guiotto: Tilda?
23:06:270Paolo Guiotto: I'm doing a lot of examples in case of playing fields, because, otherwise,
23:14:230Paolo Guiotto: You have to add one more part, but it's a bit more complex, but nothing…
23:19:780Paolo Guiotto: Different, but be careful, because, for example, a test like this one.
23:24:960Paolo Guiotto: Holds only for playing fields, okay?
23:28:170Paolo Guiotto: This one with the circulation. While the circulation property holds in any dimension, so F conservative if and only if, circulation is equal to zero, this one is saying that you just need to reduce the number of circulation to a finite number.
23:44:890Paolo Guiotto: Okay? But this is a very particular situation.
23:48:810Paolo Guiotto: It holds for two-dimensional fields, okay? It is combined with the fact that you check it is rotational, so you have to understand that this is a limited case, but in any case, it works in certain situations, like probably this one.
24:08:940Paolo Guiotto: So this is for XY, of course, apart.
24:12:880Paolo Guiotto: 0, so the usual
24:15:900Paolo Guiotto: problem. Now, what if the point, the bad point is anywhere else? You translate everything, you can always suppose that it is 0, 0. Otherwise, you change variables, you do a translation, you move your bad point to 0, 0.
24:30:210Paolo Guiotto: And it's the same. So, number one, so…
24:34:530Paolo Guiotto: determine values of A, B, alpha, beta, such that F… Is, irrotational.
24:55:240Paolo Guiotto: Mama.
25:01:130Paolo Guiotto: This is a question, too, that is… is a bit bothering, however. It says, question 2, computer this line integral of the field F, huh?
25:12:350Paolo Guiotto: Where, Gamma?
25:15:00Paolo Guiotto: is, the polygonal.
25:22:10Paolo Guiotto: Joining.
25:24:860Paolo Guiotto: These points, 2-0.
25:29:510Paolo Guiotto: 0, 1, and minus 2, 0. We will see later in a figure what is it.
25:35:660Paolo Guiotto: And, number 3, determine values, determine values A, B, alpha, and B, such that F
25:45:890Paolo Guiotto: is conservative.
25:49:710Paolo Guiotto: And then, the potentials.
25:52:210Paolo Guiotto: plus preventions.
25:55:760Paolo Guiotto: Okay, so, solution.
25:58:670Paolo Guiotto: Now, we have enough, experience.
26:02:430Paolo Guiotto: Okay, so… Irrotational means that test of the cross derivatives, so, we… Have… to check…
26:19:10Paolo Guiotto: Now, the case is the case of playing field, so we have the derivative with respect to X of the second component must be equal to derivative with respect to X of the first component. So this translates into DX of F2. F2 is BY,
26:36:580Paolo Guiotto: of X squared plus Y squared to exponent beta.
26:42:740Paolo Guiotto: This must be equal to DX, you know, DY.
26:47:00Paolo Guiotto: of AEX divided, X squared plus Y squared to exponent alpha.
26:54:650Paolo Guiotto: So now, this respect to this derivative is a constant, so you carry outside. So BY…
27:01:100Paolo Guiotto: Then you have the derivative of 1 over something, you know? So you can use the formula for the derivative of some… 1 over, I don't know, let's say phi. This is minus phi prime over phi squared. So you do the square of that denominator, X squared plus
27:18:980Paolo Guiotto: Y squared to power beta, so this squared becomes to 2 beta.
27:23:220Paolo Guiotto: Then you have to put minus here, and do the derivative of that denominator with respect to X. So we will have… first of all is a power, so beta X squared plus Y squared to exponent beta minus 1.
27:40:520Paolo Guiotto: times… Then we have to derive the argument with respect to X, so we have got two X.
27:47:760Paolo Guiotto: This is the same on this side. We have AX that goes outside, then we have the same formula, so a minus denominator is X squared plus Y squared to power 2 alpha.
28:00:60Paolo Guiotto: And the numerator is alpha, X squared plus Y squared 2 alpha minus 1, then the derivative with respect to Y, so 2Y.
28:13:40Paolo Guiotto: Fail?
28:14:230Paolo Guiotto: Now, we simplify as much as we can, so this means…
28:19:560Paolo Guiotto: That, okay, well, for example, we can cancel the keyboard side, the minus,
28:26:240Paolo Guiotto: We cannot cancel variables, or we cannot cancel the X or the Y, because they are variables.
28:34:490Paolo Guiotto: But let's see what we obtain.
28:38:750Paolo Guiotto: So, at left, I would say we have, B.
28:42:300Paolo Guiotto: beta XY, then this guy, X squared plus Y squared.
28:49:300Paolo Guiotto: to which exponent? It is beta minus 2 beta minus 1, so minus beta minus 1. At right, we have similarly A alpha XYX squared plus Y squared to exponent minus alpha minus 1.
29:10:230Paolo Guiotto: Now, this thing must be true, for every point, X, Y,
29:16:30Paolo Guiotto: in the domain D, which is, I remind you, up to minus the odds, right?
29:22:490Paolo Guiotto: Actually, since we, no, well, yeah, I cannot say also for 00 because of that negative exponent.
29:31:360Paolo Guiotto: Okay, so…
29:33:970Paolo Guiotto: I cannot divide by XY, but apart for points which are on the axis, where one of the two coordinates is zero, so for all points here in the four quarters, except the axis, so, if, in particular.
29:53:310Paolo Guiotto: if XY is not in X axis.
30:01:280Paolo Guiotto: or Y axis.
30:04:750Paolo Guiotto: I can divide by XY, This implies that, B… beta.
30:12:890Paolo Guiotto: And I can also divide by that quantity, because that quantity is never zero.
30:18:620Paolo Guiotto: Okay,
30:21:500Paolo Guiotto: So I get a unique exponent. If you want to put on one side, you can cancel the minus 1 here, and you get X squared plus Y squared to,
30:37:450Paolo Guiotto: 2… Well, let's take on the left, 2 alpha minus beta. This is equal to A alpha.
30:46:750Paolo Guiotto: I multiply by X squared plus Y squared to power
30:53:470Paolo Guiotto: alpha plus 1, I cancel it, I get this.
30:56:890Paolo Guiotto: Now, look at this. At left, you have that quantity. It doesn't matter what is the X, but what can we guess from this identity? That must be, for every XY,
31:09:940Paolo Guiotto: Apart those who are in the absence.
31:16:200Paolo Guiotto: Well, it is clear that you see this quantity is variable.
31:20:440Paolo Guiotto: times a constant.
31:22:210Paolo Guiotto: Equal to a constant.
31:25:340Paolo Guiotto: Do you think, is it possible?
31:35:890Paolo Guiotto: Yes, because if this is non-zero, this is not a constant, cannot be constant. It's something in some exponent, and this is variable.
31:46:290Paolo Guiotto: So it won't be constant, unless this exponent keeps this. The only case is when this exponent is 0, so this is 0 is 1. So the first factor we gain is that
31:58:710Paolo Guiotto: Necessarily alpha equal to beta.
32:02:400Paolo Guiotto: Okay, so this means that our identity collapses to B beta equal A alpha, Right?
32:12:290Paolo Guiotto: Now, since in the assumptions we had also, these numbers are different from zero, so we can also divide now, since salsa is equal to beta, we can simplify this with this, and we get A equal to alpha.
32:29:420Paolo Guiotto: So, finally, we have alpha equal to beta, and A equal to alpha, and that's the configuration for which we have
32:38:710Paolo Guiotto: any rotational field. So, the field is rotational, F, your rotational.
32:47:20Paolo Guiotto: If and only if F has this shape.
32:52:920Paolo Guiotto: So, let's say that we keep, A and alpha as parameters. We have AX divided.
32:59:720Paolo Guiotto: X squared plus Y squared to the power alpha.
33:04:910Paolo Guiotto: Now, for the second component, I have B, B is equal to A, AY divided X squared plus Y squared to power beta, which is equal, again, to alpha. Okay, so that's the field.
33:17:580Paolo Guiotto: Question 2, it says, For the alpha you computed above.
33:26:880Paolo Guiotto: Yeah, for the values found in one, I have not written here, but suppose, compute this arc integral, this line integral. So, let's see what is it. On a polygolar, this means that we have straight lines.
33:43:410Paolo Guiotto: Now, the points, say, are 2-0, so the point to zero is this one, say?
33:51:50Paolo Guiotto: Then we have 0.01, which is about this.
33:56:630Paolo Guiotto: And then point minus… 2-0, which is this one.
34:03:560Paolo Guiotto: So, the polygonal is this one, from this to this.
34:08:450Paolo Guiotto: Okay?
34:15:650Paolo Guiotto: I think that, it… it would be better if we… if we are, A little bit,
34:25:430Paolo Guiotto: to compute the desicculation after we check if there is a potential. Because if we have a potential, it's final value by decision value, otherwise we have to do a mass of integral. So, let's,
34:37:170Paolo Guiotto: Let's do… to after… So, 3 is, this field F, X, Y,
34:50:560Paolo Guiotto: equal. You see that that factor A is,
34:54:470Paolo Guiotto: a common factor for the two components, so I can just write X divided X squared plus
34:59:780Paolo Guiotto: Y squared to power alpha, Y divided the X squared plus Y squared to power alpha.
35:09:00Paolo Guiotto: is conservative.
35:12:320Paolo Guiotto: Now, in principle, I could do the test we have seen this morning, so check the circulation. But in this case, I would proceed directly, because I already see what is the…
35:24:950Paolo Guiotto: You see what have the parcel narratives here?
35:30:730Paolo Guiotto: Suppose that you have a constant here, so X square plus 1, he's,
35:36:620Paolo Guiotto: One of this type of situations you should think about. This is… comes from a power
35:42:570Paolo Guiotto: Well, this is the power to exponent minus alpha, so it should come to exponent minus alpha plus 1. So I already see the potential, so let's proceed directly without checking the circulation. So let's compute directly the
35:56:730Paolo Guiotto: Let's bet on the fact that this is conservative, in other words. It's conservative if and only if
36:03:330Paolo Guiotto: And at the end, if we want to determine the potential, we have to do this. So, at the end, we will have to do, so…
36:10:150Paolo Guiotto: F is the gradient of F, so this means that
36:14:420Paolo Guiotto: The derivative with respect to X of F is X divided
36:19:970Paolo Guiotto: X squared plus Y squared to power alpha.
36:24:620Paolo Guiotto: And derivative with respect to Y is… Well, apart for A…
36:30:210Paolo Guiotto: AY divided the X squared plus Y squared to alpha.
36:35:760Paolo Guiotto: Now, in the first… from the first equation, we get F of XY,
36:42:140Paolo Guiotto: Is a primitive inexor of this.
36:51:130Paolo Guiotto: plus a function that might bend on Y.
36:55:430Paolo Guiotto: Now, do you see or not this, as a derivative?
37:00:190Paolo Guiotto: If you do the derivative with respect to X of X squared plus Y squared.
37:06:840Paolo Guiotto: That one would be to exponent minus alpha.
37:09:860Paolo Guiotto: So, since this should be the derivative.
37:13:10Paolo Guiotto: I should start with an exponent higher of 1 unit, so minus alpha plus 1. Let's look at what happens. It is minus alpha plus 1.
37:21:830Paolo Guiotto: Then you have the derivative of this. It is X squared plus Y squared, you decrease the exponent of 1 to minus alpha, and then you have the derivative of the argument, which is 2X.
37:33:750Paolo Guiotto: So at the end, the derivative of this is 2X,
37:38:170Paolo Guiotto: Divided, well, you have this factor, Minus alpha plus 1.
37:47:00Paolo Guiotto: divided X squared plus Y squared to exponent alpha.
37:51:560Paolo Guiotto: So, I needed to have this 2 in front, but this is a coefficient. I can put… so the A is useless, that's right outside. We put a 2 times minus alpha plus 1, and we have to divide by 2 times minus alpha plus 1.
38:08:810Paolo Guiotto: Now, what we have inside is the derivative of that, so we have A over 2 times minus alpha plus 1,
38:23:250Paolo Guiotto: This is X squared plus Y squared to minus alpha plus 1 plus a constant in X, but not in Y. So this is the candidate, FXY,
38:38:600Paolo Guiotto: we have now to plug into the second equation and verify if it is equal to the second… to the second condition. So, DYF here.
38:49:00Paolo Guiotto: is we have this A over 2 minus alpha plus 1, doing the derivative with respect to Y now, we get down this exponent, minus alpha plus 1.
39:01:440Paolo Guiotto: Then we have X squared plus Y squared 2 exponent diminished by 1, so minus alpha, times the derivative of the argument, which is 2Y now, plus C prime of Y.
39:16:300Paolo Guiotto: You see that the 2 kills the 2,
39:19:590Paolo Guiotto: T minus alpha plus 1 is T minus alpha plus 1, and we get exactly what it expected. So, AY divided X squared plus Y squared
39:31:630Paolo Guiotto: to exponent alpha plus C prime of Y.
39:35:320Paolo Guiotto: And we wonder if this is the second component of the feeder, so AY divided X squared plus Y squared.
39:43:360Paolo Guiotto: to Alpha.
39:45:180Paolo Guiotto: And this is, of course, if and all if, since these two are the same.
39:50:570Paolo Guiotto: C prime of y equals 0, so…
39:55:120Paolo Guiotto: C prime of y equals 0.
39:57:960Paolo Guiotto: And this means C is quite constant.
40:00:900Paolo Guiotto: C of Y is constant.
40:03:340Paolo Guiotto: So, yes, sir?
40:05:220Paolo Guiotto: this is with a constant, and we found the potential. Now, since we found the potential, how do we compute the
40:13:150Paolo Guiotto: line integral, you do the final value, so the value of the potential here, minus the value of the potential here. You don't need to compute the integral.
40:21:230Paolo Guiotto: So, going back to the line integral of F along that gamma.
40:26:550Paolo Guiotto: is equal to F of… the final point is minus 2, 0, minus F, the initial point was 2, 0.
40:35:360Paolo Guiotto: So, you see, if you know that there is a potential, it's much better.
40:39:600Paolo Guiotto: Because you don't need to compute any integral.
40:43:150Paolo Guiotto: Okay… I will, I will write down here the exercises to do, okay?
40:52:680Paolo Guiotto: Have a nice weekend.