AI Assistant
Transcript
00:38:820Paolo Guiotto: Okay, good morning, take your seat, please.
00:46:360Paolo Guiotto: So we start with,
00:49:840Paolo Guiotto: the exercise I left you to try.
00:53:20Paolo Guiotto: It is, the number 3.4.5.
00:59:260Paolo Guiotto: we have this vector field, F of XY, is equal to…
01:06:580Paolo Guiotto: minus AXY divided by X squared plus Y square.
01:14:280Paolo Guiotto: squared… second component, the BX squared minus Y squared.
01:21:610Paolo Guiotto: divided by X squared plus Y squared squared.
01:27:400Paolo Guiotto: This on domain D, so for point XY, that belongs to the domain, which is the natural domain.
01:34:950Paolo Guiotto: That is, the full plane are to accept the the origin.
01:40:760Paolo Guiotto: Where the two components of this field are not defined?
01:45:70Paolo Guiotto: Here, these A and B are just real parameters.
01:53:270Paolo Guiotto: So the problem asks, number one, determine… values… of A and B.
02:02:550Paolo Guiotto: such that the field F is irritational, Question 2… is,
02:14:680Paolo Guiotto: for such values, well, let's put this… determine in this way. Determine values of A and B subset.
02:22:360Paolo Guiotto: F is conservative.
02:26:560Paolo Guiotto: And in this case, compute also potentials.
02:32:590Paolo Guiotto: determine…
02:37:420Paolo Guiotto: potentials.
02:45:970Paolo Guiotto: Okay.
02:48:720Paolo Guiotto: Now, let's check the first question.
02:55:280Paolo Guiotto: So here we have a field from R2 to R2. More precisely, F is from D, which is contained in R2 to R2, so it's a two-dimensional field.
03:09:420Paolo Guiotto: it has two components, say, this one we call F1, this is…
03:15:710Paolo Guiotto: The second component, we call F2.
03:19:600Paolo Guiotto: So… F is irrotational.
03:25:830Paolo Guiotto: on D,
03:29:360Paolo Guiotto: If and only if… well, here, since we have two components, the test of cross derivatives reduces just to a unique condition, you have to take, for example, the first component and do the derivative with respect to the other variable, which is
03:42:830Paolo Guiotto: Y, the second variable, and this should be identically the same of the derivative of the second component with respect to the first variable.
03:53:200Paolo Guiotto: Ondi.
03:54:770Paolo Guiotto: So, for… this means for every point, XY in R2 minus 00.
04:04:960Paolo Guiotto: Let's see what happens. So, dy of F1 is the DY of F1 is minus AXY divided X squared plus Y squared squared.
04:21:490Paolo Guiotto: Now, the minus A can be carried outside, because it's a factor.
04:27:180Paolo Guiotto: If you want, also X can be carried outside from this, because it's a constant for the Y derivative.
04:34:530Paolo Guiotto: So, we have to differentiate this. Y divided X squared plus Y squared squared. That's a fraction, so we have minus AX times… we do the square of the denominator, so we get X squared plus Y squared to power 4.
04:55:310Paolo Guiotto: Then we have the derivative of numerator with respect to Y, and this is 1 times the denominator, which is X squared plus Y squared squared, minus numerator Y, times the derivative with respect to Y of that denominator.
05:12:470Paolo Guiotto: And that's 2X squared plus Y squared times 2Y.
05:21:630Paolo Guiotto: Okay, let's see if we can simplify.
05:26:750Paolo Guiotto: So we can take this X squared plus Y squared, we take one from here, and we simplify with this, we have cube.
05:37:510Paolo Guiotto: So we have, minus AX,
05:42:340Paolo Guiotto: The denominator is X squared plus Y squared power 3.
05:51:120Paolo Guiotto: About the numerator, we have X squared plus Y squared here, minus 224Y
06:02:490Paolo Guiotto: square. So, in total, we get X squared minus What?
06:09:450Paolo Guiotto: It's a bit of a disaster today.
06:12:740Paolo Guiotto: 3y squared.
06:20:540Paolo Guiotto: So this is the DY of F1. Let's compute now the DX of F2.
06:26:690Paolo Guiotto: This is the DX of F2 is B, X squared minus Y squared divided.
06:35:290Paolo Guiotto: X squared plus Y squared squared.
06:40:650Paolo Guiotto: Here, it is not convenient to split that fraction. We do not simplify too much. So let's proceed directly with the derivative. We have the denominator is X squared plus Y squared.
06:52:440Paolo Guiotto: Squared, squared, so it means power 4 again.
06:56:90Paolo Guiotto: Now we have the derivative of numerator with respect to X. This is… 2BX.
07:04:210Paolo Guiotto: times denominator, X squared plus Y squared.
07:08:200Paolo Guiotto: squared minus numerator, BX squared, minus Y squared times denominator, the derivative, Respect to action.
07:18:870Paolo Guiotto: So this is 2, X squared plus Y squared.
07:23:470Paolo Guiotto: Thanks to Axe.
07:26:550Paolo Guiotto: Also, here we can slightly simplify this.
07:35:940Paolo Guiotto: So, I get, huh?
07:42:500Paolo Guiotto: I got the Cuba.
07:44:340Paolo Guiotto: Right? Denominator…
07:49:880Paolo Guiotto: About the numerator… We may say we have 2BX cubed.
07:59:340Paolo Guiotto: plus 2BXY squared, Then there is a 4X.
08:07:250Paolo Guiotto: minus, 4X times 2B times B squared, so minus 4.
08:12:970Paolo Guiotto: BX cubed… And we have a blast, 4X, Y squared.
08:25:470Paolo Guiotto: We can,
08:28:160Paolo Guiotto: Slightly simplify this, you see, no? This 2BX cubed with this minus 4 is yields minus 2.
08:36:400Paolo Guiotto: Then we have, this,
08:40:730Paolo Guiotto: we can say that at the end we have X squared, plus Y square cuba, So, minus 2BX cubed.
08:53:520Paolo Guiotto: And then we have plus 2B plus 4.
08:58:30Paolo Guiotto: X, Y squared.
09:00:210Paolo Guiotto: Okay, now this is the derivative with respect to…
09:05:150Paolo Guiotto: What? To respect to acts of the second complaint.
09:08:410Paolo Guiotto: Okay, now, that condition, the cross derivative checked, holds, so… DUI.
09:18:330Paolo Guiotto: of F1 is identically equal, to DX of F2, if and only if We have to copy.
09:30:580Paolo Guiotto: Well, perhaps it is better… well, okay, let's copy. Minus AX… Times of this fraction.
09:40:470Paolo Guiotto: B, no, X squared minus 3Y square.
09:47:60Paolo Guiotto: over X squared plus Y squared power 3.
09:51:390Paolo Guiotto: This must be equal.
09:53:970Paolo Guiotto: to this one.
09:55:630Paolo Guiotto: X squared plus Y squared cuba.
09:59:980Paolo Guiotto: minus 2BX cuba.
10:04:170Paolo Guiotto: plus 2B plus 4… X, Y squared.
10:12:220Paolo Guiotto: Now, this identity must be verified for every point of the domain, so for every XY in D,
10:20:240Paolo Guiotto: That means the plane accepts one
10:24:770Paolo Guiotto: Point, so with a unique exception, 00, where this thing Does not make any sense.
10:32:150Paolo Guiotto: But you notice that the denominator is the same, so when point XY is different from 0, that's a number different from 0. We can simplify from multiplying both sides by this quantity. So this reduces to…
10:48:670Paolo Guiotto: let's say to a sort of line identity, perhaps it's better if we multiply minus AX cubed.
10:55:580Paolo Guiotto: plus 3AXY squared.
11:00:670Paolo Guiotto: Equal minus 2BX cuba.
11:06:200Paolo Guiotto: plus 2B, plus 4XY squared.
11:12:760Paolo Guiotto: So we can carry everything on a unique side. For example, we can write 2B minus A X cubed.
11:22:260Paolo Guiotto: Plus, there is this, 3A.
11:25:850Paolo Guiotto: minus 2B, minus 4XY squared, right? This should be identical equal to 0.
11:37:800Paolo Guiotto: Now, when this happens.
11:40:410Paolo Guiotto: you may imagine that this quantity, to be identical equal to zero for every point XY,
11:47:120Paolo Guiotto: different from zero-zero.
11:50:610Paolo Guiotto: There should be reasonably, or at least intuitively, a unique possibility, is that these two numbers are… this one and this one are zero.
12:00:750Paolo Guiotto: Now, it's a general principle, for a valid for polynomials, so if you have any polynomial, even in several variables, so if P of XY
12:12:290Paolo Guiotto: It is something like sum of coefficients C, I don't know, to denote CJ, X to some exponent KJ, Y to some exponent HJ, no? This is the shape of a polynomial.
12:30:340Paolo Guiotto: And this polynomial is identically equal to zero on, in this case, R2.
12:38:350Paolo Guiotto: then this is possible only in the unique trivial case that makes this true, only if the coefficients are all equal to zero. So if and only if the coefficients CJ are all equal to zero for every J.
12:53:310Paolo Guiotto: Now, this is a polynomial in X and Y, no?
12:59:310Paolo Guiotto: It's not identical equal to 0 for every XY, it is for every XY different from 0, 0. But, of course, when XY are 0, 0, that is true, no? Because when you put X and Y equals 0, that polynomial is 0.
13:15:40Paolo Guiotto: So we can say that, in fact, this is equivalent to say that 2B minus A
13:22:690Paolo Guiotto: X cubed plus 3A minus 2B minus 4, XY squared is, in fact, identical equal to 0, because the .00 does not add anything here.
13:37:170Paolo Guiotto: And so, if we apply the, this principle, we have that 2B minus A must be equal to 0, and at the same time, 3A minus 2B minus 4 must be equal to 0.
13:54:120Paolo Guiotto: So we got a system here. Let's see what this system
13:58:740Paolo Guiotto: else, so we have, for example, A is equal to 2B, we plug this into the second equation.
14:05:130Paolo Guiotto: So 3A will be 6B minus 2B.
14:09:490Paolo Guiotto: minus 4 equals 0, so we get 4B equal to 4, and this means B equal to 1, and therefore A equal to 2.
14:20:720Paolo Guiotto: So, the conclusion is that,
14:27:00Paolo Guiotto: the field F… is irritational, If and only if
14:34:590Paolo Guiotto: The value of A is 2, and the value of B is 1.
14:39:490Paolo Guiotto: So, among all these, let's say, fields.
14:43:990Paolo Guiotto: there is one field for each pair of A and B, no?
14:48:840Paolo Guiotto: Among all these possible fields, there is only one of them that can… that is irrotational. That one with A equal 2 and B equal 1, okay?
15:00:380Paolo Guiotto: Now, as you can see, apart for the calculations, what have we done?
15:05:360Paolo Guiotto: We just applied the definition. Irrotational means the… the… the… the cross… the cross derivative check, is… is true.
15:16:310Paolo Guiotto: is verified. In this case, this reduces just to a unique condition, because we have two components, two variables.
15:25:120Paolo Guiotto: And there is a unique combination of different components with different variables. So, the first component with the second variable, the second component with the first variable, and we need to check if these derivatives are the same.
15:38:660Paolo Guiotto: So what have we done? We have computed the derivatives, so there are…
15:42:590Paolo Guiotto: A certain number of calculations, but these are just ordinary derivatives.
15:47:630Paolo Guiotto: And then, at the end, we impose the identity.
15:51:140Paolo Guiotto: In this case, we obtained an identity that means a polynomial identically zero.
15:58:960Paolo Guiotto: And this happens because of general principle.
16:02:120Paolo Guiotto: Exactly, if and only if all the coefficients are zero.
16:11:390Paolo Guiotto: And, from this, we get two conditions.
16:15:740Paolo Guiotto: for A and B, and therefore we get
16:20:20Paolo Guiotto: Finally, the values of AB, for which we have the field irritation. Now, we have question two.
16:26:990Paolo Guiotto: That asks, like, I just modified a little bit the… the… how the question is posed. At the end, it's the same thing.
16:34:950Paolo Guiotto: Let's say that the question is determined values of A and B for which F is conservative.
16:42:500Paolo Guiotto: Okay? Now, what I've done in the first question is not useless for… to respond to this second question. We know that there is a relation between these two properties, irrational and conservative. They are not equivalent.
16:56:950Paolo Guiotto: So I cannot say the same A and B, no? But I can say that to be conservative, the field must be rotational. So necessarily, the unique possibility for AB to be conservative, F,
17:10:810Paolo Guiotto: is A equal 2 and B equal 1?
17:13:940Paolo Guiotto: However, this does not yet prove anything, because we know that there are irritational fields which are not conservative.
17:21:859Paolo Guiotto: Okay? So you see the… the logical, approach. So we can say that to be… conservative.
17:33:430Paolo Guiotto: F… must be… irritational.
17:43:460Paolo Guiotto: So… Necessarily by what we just,
17:48:920Paolo Guiotto: come to C, A is 2, and B is 1.
17:54:350Paolo Guiotto: No?
17:55:550Paolo Guiotto: However, I cannot just stop here, because now I know that this is the unique possibility to be rotational.
18:02:500Paolo Guiotto: And a conservative field must be rotational. So, what they can do now is to focus on this particular field.
18:10:390Paolo Guiotto: Which is, minus, 2… They go after my… But I can copy from.
18:20:550Paolo Guiotto: Minus 2XY divided the X squared plus Y squared squared.
18:28:860Paolo Guiotto: And then second component, B is 1, so we get X squared minus Y squared divided X squared plus Y squared.
18:39:570Paolo Guiotto: square. So the unique possible F that can be conservative is this one.
18:46:540Paolo Guiotto: But we don't know yet if it is conservative.
18:49:950Paolo Guiotto: At this stage, we have no other condition that ensures this, except for the definition. So, this F…
19:00:420Paolo Guiotto: is conservative.
19:03:180Paolo Guiotto: If and only if this F is the gradient.
19:07:100Paolo Guiotto: of a function f is conservative. Of course, here we are working on that domain on D, and this must be on D.
19:15:500Paolo Guiotto: That is, we want that, we want to find, we want to discuss if there exists an F, such that the partial derivative with respect to X is the first component, minus 2XY divided by X squared plus Y squared.
19:32:290Paolo Guiotto: And the second DYFXY is the second component of the field, so X squared minus Y squared divided
19:42:790Paolo Guiotto: X squared plus Y squared squared.
19:47:460Paolo Guiotto: Now, how do we solve this? Exactly as we did yesterday for simple cases. Now, what changes the computational difficulties? So let's take the first equation, for example. Normally, you should choose that one that seems to be easier, no? Easier to do what?
20:06:210Paolo Guiotto: Now, we have to look at the function f, whose derivative with respect to X, is that right-hand side, so we have to
20:14:150Paolo Guiotto: I don't like to use this word here. We have to integrate, you know, this equation. So this means that, in fact, the right word, we have to do the primitive. They are two different concepts. An integral is a number, a primitive is a function. That's why I don't like to
20:29:850Paolo Guiotto: to confuse the two. So now, this one is equivalent to say that the function fxy must be a primitive in the variable X of that quantity minus 2XY divided by X squared plus Y squared squared.
20:48:790Paolo Guiotto: plus… Possibly a constant, but the constant in X.
20:54:610Paolo Guiotto: And this means that, in general, this can be a function of Y.
21:00:530Paolo Guiotto: So let's compute this primitive. Now, we have to focus on this, and notice that this, as a function of X, is a rational function, so a ratio between polynomials. So, in principle, there exists an algorithm to compute these things.
21:15:340Paolo Guiotto: Okay? So, another remark is that I can consider factors like this as constants. Minus 2 is evident, but also Y is a constant for that calculation, because we are doing the primitive in X.
21:30:250Paolo Guiotto: So the letter Y is another variable. It's not the integration variable. So I can put the minus 2Y outside, then I have this X divided X squared plus Y squared squared EX,
21:44:730Paolo Guiotto: plus this mysterious function C of Y.
21:48:660Paolo Guiotto: Now… Here are…
21:52:730Paolo Guiotto: you don't need to be a champion of computing primitives, but you just need to remind how derivative works. So, do you see that this could be immediately a derivative, should come from
22:07:640Paolo Guiotto: Computing the derivative with respect to X of which function?
22:14:370Paolo Guiotto: You don't see, okay.
22:16:250Paolo Guiotto: You see that downstairs, if you forget of Y, you have a power of X, no? So powers comes from powers, except when the exponent is 1, the denominator, then it comes from the logger.
22:28:80Paolo Guiotto: So, probably, this comes from power, this is power X squared plus Y squared to minus 2, right?
22:38:720Paolo Guiotto: So, since when I do the derivative, the exponent decreases of one unit, doing the opposite, I have to increase the exponent of one unit. So I have to pass from this to this, X squared plus Y squared to exponent minus 1.
22:53:660Paolo Guiotto: And in fact, if you take X squared plus Y squared to the exponent minus 1, what is the derivative with respect to X? It is minus 1, that comes down, that there is X squared plus Y squared, 2 exponent minus 1, minus 1, minus 2,
23:09:30Paolo Guiotto: Then you have the derivative of the argument with respect to X, and this is a 2X. And if you rewrite this, you get minus 2X divided by X squared plus Y squared squared. So, I don't need to push out this minus 2, I return it back to the
23:27:640Paolo Guiotto: The inside of the primitive.
23:30:160Paolo Guiotto: And now you read that that expression, minus 2x divided X squared plus Y squared is a derivative. It is the derivative of this guy.
23:41:140Paolo Guiotto: So, when I do the primitive, the primitive is the inverse operation of the derivative.
23:46:520Paolo Guiotto: So, you… the primitive of the derivative is the function, because it does the opposite. So, at the end, here I get, so Y times this, X squared plus Y squared, 2 minus 1 plus C of Y,
24:03:710Paolo Guiotto: or if you prefer, Y divided X squared plus Y squared plus C of Y.
24:12:770Paolo Guiotto: So, from the first equation of this system, which is the system that says F is a potential.
24:21:630Paolo Guiotto: I get that F must be this.
24:25:660Paolo Guiotto: function.
24:27:400Paolo Guiotto: Now, it is not yet, over.
24:30:520Paolo Guiotto: Because we don't know what is this C of Y.
24:34:480Paolo Guiotto: We have another equation to exploit, the second one, and that equation will give
24:41:590Paolo Guiotto: Maybe the final answer to the problem.
24:44:840Paolo Guiotto: It's not necessarily going to tell that we have an F, we have seen that it could destroy all the arguments.
24:52:360Paolo Guiotto: Now, DY, let's check this second equation. DYF equal to, I should… FX squared minus Y squared
25:04:660Paolo Guiotto: divided the X squared plus Y squared.
25:09:850Paolo Guiotto: is equivalent to… you see, instead of now doing the same operation I've done above, integrating in Y, say.
25:18:140Paolo Guiotto: I don't do that, it would be a bit messy, no? Because I know already that F must be this one. So let's check that this one, if this one verifies also this second line, no? So let's do the derivative with respect to Y of that expression, Y divided
25:36:130Paolo Guiotto: X squared plus Y squared plus C of Y.
25:40:960Paolo Guiotto: This… now, the question, is it true that it is equal to my X squared minus Y squared divided X squared plus Y squared
25:49:140Paolo Guiotto: Square up. If yes.
25:51:490Paolo Guiotto: we are done. If not, it means that probably this cannot be true, so that cannot be a potential.
25:58:620Paolo Guiotto: Now, what is the derivative with respect to Y of the fraction? We have a fraction, so we do the square of denominator first.
26:06:550Paolo Guiotto: So we get this to power 2. Then we have the derivative with respect to Y of the numerator, and that's 1 times denominator, X squared plus Y squared, minus the numerator, y times the derivative with respect to Y of the denominator, which is…
26:24:40Paolo Guiotto: To what?
26:25:800Paolo Guiotto: Then we have also that C, that becomes, of course, C prime of Y, so the derivative of that function, that's a function of a unique variable, Y, and this must be equal
26:37:120Paolo Guiotto: You have to check whether it is equal to this one or not, so…
26:42:810Paolo Guiotto: X squared minus Y squared, divided X squared plus Y squared squared.
26:48:660Paolo Guiotto: Well, if you do the calculation here, you discover that you have X squared plus Y squared minus 2Y squared, so X squared minus Y squared, which is exactly that fraction.
26:59:290Paolo Guiotto: So at the end, you get this is true, provided C prime of Y is… well, since this is an identical, because identical means for every point xy, not just for one point, or for some point, but for all points of the domain.
27:14:810Paolo Guiotto: So that's what… why I put these three lines, to remind me that this means
27:19:580Paolo Guiotto: for every X, Y in the domain D. Huh?
27:24:880Paolo Guiotto: So this means that C prime must be identically equal to 0, and therefore C is…
27:31:220Paolo Guiotto: in C of Y is a function that must be constant in Y,
27:35:760Paolo Guiotto: but not a constant, a function of X, because it cannot be a function of X. We already, no, discussed this, so it's a constant. And therefore, this means that the function F, XY
27:49:120Paolo Guiotto: Equal, I do not remind, what is it?
27:53:650Paolo Guiotto: Why divided?
27:56:570Paolo Guiotto: X squared plus Y squared plus a constant, where constant is any constant.
28:03:680Paolo Guiotto: is such that this function, by construction, gradient F, is the field F.
28:12:160Paolo Guiotto: So, we can conclude that.
28:16:680Paolo Guiotto: So let's draw the conclusion. We say…
28:19:420Paolo Guiotto: The field F, the initial field with the parameters inside, this one.
28:28:50Paolo Guiotto: To be conservative, it must be irrational.
28:33:40Paolo Guiotto: So this implies that the two values for IB are
28:37:100Paolo Guiotto: A equal, 2 and B equal 1, okay? Then the field can be only this one. Now we say.
28:46:580Paolo Guiotto: To be conservative, we need to demonstrate that there exists a potential. So, a function f whose gradient is the field, this means to solve these two equations. So, a unique F that must solve two equations.
28:59:860Paolo Guiotto: we took the first one, we integrated in the first variable, and we have seen that the unique possibility for F is to be this, no? We see a function of Y.
29:13:740Paolo Guiotto: Then, plugging this into the second equation, so imposing that this verifies also the second condition, not just the first one, we get that C now must be a constant.
29:24:960Paolo Guiotto: So the UNICAF that verifies both conditions is this one. Actually, there are infinitely many, and that's not something strange, because if you add that constant to whatever, you get still a potential now, because the gradient of the constant is 0.
29:46:460Paolo Guiotto: Yes.
30:02:810Paolo Guiotto: No, I've not understood. What was the suggestion?
30:10:680Paolo Guiotto: Yeah.
30:30:740Paolo Guiotto: That's it.
30:37:830Paolo Guiotto: Not. I'm sorry.
31:05:130Paolo Guiotto: Nope, I understand.
31:07:460Paolo Guiotto: We are doing the derivative of this, there is not the squares here.
31:11:350Paolo Guiotto: So the formula is, derivative of diffraction
31:18:720Paolo Guiotto: I don't want to use letters like S here, but let's say, if I have to do the derivative of a fraction, say, phi of epsi, this is fraction.
31:29:730Paolo Guiotto: the square of the denominator, then you have phi prime of C minus phi C prime.
31:35:720Paolo Guiotto: And that's exactly what we did here. Now.
31:38:540Paolo Guiotto: That's the denominator, you'll see a square here.
31:41:920Paolo Guiotto: What I have to commit is to believe that we've got this one, not of this one.
31:46:00Paolo Guiotto: Bye.
31:48:650Paolo Guiotto: Yeah, but that's wrong.
31:51:30Paolo Guiotto: So be careful when you, use, formula.
31:56:350Paolo Guiotto: So now the problem is over.
31:58:900Paolo Guiotto: Do you have any questions?
32:04:110Paolo Guiotto: We may say that basically all the exercises of the chapter are of the same nature of this one. So let me do another one, for example, the 347.
32:19:580Paolo Guiotto: Here we have four parameters, A, B, CD, real.
32:26:770Paolo Guiotto: And the field is…
32:29:120Paolo Guiotto: pretty much like the previous one. It is AX plus BY over X squared plus Y squared.
32:39:250Paolo Guiotto: and, CX plus DY… over X squared plus Y squared, for XY, again, in R2, Different from zero.
32:56:30Paolo Guiotto: Okay?
32:57:680Paolo Guiotto: Now, let's say that, same questions. So, number one, determine,
33:05:920Paolo Guiotto: A, B, C, D, such that F is irrotational.
33:13:750Paolo Guiotto: Number two, determine ABCD such that F is conservative.
33:19:600Paolo Guiotto: And in this case, potentials, of this field.
33:26:90Paolo Guiotto: So, same question.
33:29:830Paolo Guiotto: Now, you see that another way to look at this way to proceed is that if I have many parameters in a field.
33:38:860Paolo Guiotto: And my question, the important question, is always the potential.
33:43:100Paolo Guiotto: It's almost never the rotational, but…
33:46:190Paolo Guiotto: The point is that this condition Imposed to the fielder.
33:51:570Paolo Guiotto: excludes some cases, no? The cases when the field is not irrotational cannot be considered as conservative.
33:59:630Paolo Guiotto: So that… and since irrotational is easier, because we have to differentiate, why conservative? We have to do the opposite, we have to integrate, and to integrate is never a trivial task, okay? This means that to check number one is easier, no? So here, F…
34:18:170Paolo Guiotto: Jeez.
34:19:540Paolo Guiotto: Irrotational.
34:22:179Paolo Guiotto: If and only if… We have that.
34:26:380Paolo Guiotto: The first component is AX plus BY divided X squared plus Y squared.
34:35:260Paolo Guiotto: Should I derive respect to X or respect to Y, this?
34:38:850Paolo Guiotto: This is the first component I have to do the derivative with respect to the second variable.
34:43:790Paolo Guiotto: Here we have only two variables, so there are only two possibilities.
34:48:600Paolo Guiotto: And then, the second company is CX plus DYE.
34:52:739Paolo Guiotto: divided by X squared plus Y squared. This one has to be computed with respect to X.
34:59:970Paolo Guiotto: Now, this must be an identity for every X and Y in the domain, so for every XY, basically, apart 400.
35:11:480Paolo Guiotto: Okay?
35:13:120Paolo Guiotto: Well, do not be confused, because I know that there are little details that can… Can make you…
35:22:220Paolo Guiotto: To think of wrong things.
35:25:900Paolo Guiotto: for example.
35:27:320Paolo Guiotto: Here, we… let's look at this, that we have already done the calculation. We really do the similar calculation.
35:33:880Paolo Guiotto: In this case, no, we checked the… we had to check that condition, so this means that the left-hand side, that was the derivative with respect to Y of F1 equals to the right-hand side, the X of F2, for every XY domain, which is F2 minus 0. So we may say, for every XY,
35:53:210Paolo Guiotto: different from zero. A minute later, we said that, actually, this is for every XY, no?
36:01:500Paolo Guiotto: So you may say, but I… I said that we should check for XY different from 0, and now you say for every XY, sorry, for every XY, also taking 0.
36:12:560Paolo Guiotto: So there is no confusion, because this thing makes sense only when XY is different from 0, 0.
36:19:850Paolo Guiotto: You cannot definitely write. So the problem with 0, 0 is that when you write 0, you have a 0 divided 0, nonsense, so it does not make any sense, this. While this one makes sense also for XY equals 0, and it is trivially true.
36:36:840Paolo Guiotto: Okay? So, it's… there is no confusion here possible.
36:42:300Paolo Guiotto: Okay, so let's check this. This is the reality of the fraction, the rule we remembered here, okay? So you have the square of the denominator downstairs.
36:55:00Paolo Guiotto: Then you have derivative with respect to… the variable is Y here, so derivative with respect to Y of that numerator
37:03:890Paolo Guiotto: B times the denominator, X squared plus Y squared, minus denominator, AX plus BY, times the derivative
37:15:740Paolo Guiotto: Which is too wide.
37:19:990Paolo Guiotto: And similarly, at right, we have X squared plus Y squared, squared, Here, then…
37:28:230Paolo Guiotto: Here we have to take the derivative with respect to X of numerator, now it is C.
37:35:400Paolo Guiotto: X squared plus Y squared, minus numerator, CX plus DY, times derivative with respect to X of denominator, which is 2X.
37:46:670Paolo Guiotto: Now, this must be for every XY different from 0.
37:52:470Paolo Guiotto: So, since the two denominators are the same and different from zero, we can multiply the identity.
37:58:900Paolo Guiotto: both sides by that quantity, X squared plus Y squared squared, and so we can simplify this, and we get this identity.
38:07:580Paolo Guiotto: X squared plus Y squared minus 2Y times AX plus B.
38:16:800Paolo Guiotto: Y equal CX squared plus Y squared minus 2X CX plus this DY.
38:28:430Paolo Guiotto: Now, this must be… true for every point XY.
38:32:980Paolo Guiotto: Different from zero.
38:35:290Paolo Guiotto: Okay?
38:36:660Paolo Guiotto: But they can also write.
38:38:780Paolo Guiotto: for XY equals 0, because when you put XY equals 0, whatever are these coefficients, A, B, C, D, this becomes a 0 equal to zero, so I can also remove this restriction for every XY.
38:51:180Paolo Guiotto: Now, an error here would be, well, this is an identity polynomial equal to zero, so let's put all the coefficients equal to zero. So, B must be zero, A must be 0, C must be zero, and D must be 0.
39:05:660Paolo Guiotto: That's not correct.
39:07:680Paolo Guiotto: Because we have to write as a polynomial. This means that, let's look at terms. We have a term which is X squared.
39:15:50Paolo Guiotto: Who is the coefficient of this? It is B minus C.
39:22:310Paolo Guiotto: minus C, and there is also here an X squared that comes, right? So it is plus 2C.
39:32:450Paolo Guiotto: Similarly, I have a coefficient for Y squared, So this would be B…
39:38:100Paolo Guiotto: Then we have a minus 2B from this, no?
39:42:960Paolo Guiotto: Then I have, minus C from the right-hand side, and mountain gas.
39:48:610Paolo Guiotto: Then, I have a term like YX, no?
39:52:420Paolo Guiotto: XY… Since I'm… I'm writing everything at left, so this is the coefficient is minus 2A, yeah?
40:02:340Paolo Guiotto: And from the right-hand side… on the right, I read minus 2D, but when I carry, on the left-hand side, it becomes a blue plus 2D.
40:13:310Paolo Guiotto: And they guessed that Nothing else.
40:17:730Paolo Guiotto: Okay?
40:18:780Paolo Guiotto: This equal to 0.
40:22:330Paolo Guiotto: Now, this is the polynomial where these are the monomials. You see that they are different.
40:28:950Paolo Guiotto: Okay, so you cannot say that if I… if I say, but this is also for one variable polynomials. Now, if you say AX squared plus BX squared,
40:42:850Paolo Guiotto: Now, let's do an example.
40:45:850Paolo Guiotto: minus 3X squared, look at this, equals 0 identically. It does not mean that A, B, and minus 3 must be 0, you see? Minus 3 cannot be 0. So, you must have that what multiplies A plus B minus 3?
41:01:940Paolo Guiotto: what multiplies X squared r? This one must be 0, so this quantity is 0, not all the coefficients.
41:10:930Paolo Guiotto: So you have to write the polynomial as sum of monomials, different monomials. Then, at this stage, you can say that since this must be identically zero, this is these conditions. B minus C plus 2C is B plus C equals 0.
41:30:820Paolo Guiotto: Then here we have B minus 2B, so minus B, so minus B plus C, again, equals 0. So this is right, doing that.
41:40:550Paolo Guiotto: And then we have another minus 2A plus 2D equals 0.
41:46:400Paolo Guiotto: So, from this, we get these conditions. So, let's say C is equal to minus B,
41:53:590Paolo Guiotto: and D is equal to A.
41:58:450Paolo Guiotto: So the coefficients, A, B, C, D, must verify these conditions, so they are not completely arbitrary, because once you have A, you have automatically D, and once you have B, you have automatically C.
42:11:190Paolo Guiotto: But still, they are not like in the previous example, where we ended to a specific pair, A equal 2, B equals 1. In this case, we can say that there is a sort of reduction of the number of cases
42:26:250Paolo Guiotto: There are, let's say, two free parameters, for example, A and B, and two other parameters, C and D, that are determined from A and B.
42:36:620Paolo Guiotto: So the field F takes this shape.
42:41:440Paolo Guiotto: So, if we keep A and B as parameters, so the field is AX plus BYE,
42:48:760Paolo Guiotto: divided X squared plus Y squared. The second component, that is CX plus DY, now C is minus B, so minus BX plus DY, D is A, AY, X squared plus Y squared.
43:05:390Paolo Guiotto: This is the fin. As you can see.
43:08:530Paolo Guiotto: We restricted a bit the class of these fields.
43:14:270Paolo Guiotto: These are those who are irritation.
43:17:500Paolo Guiotto: Now, so this responds to question one. Question 2.
43:22:610Paolo Guiotto: Well, so we may say F… Here, rotational…
43:27:950Paolo Guiotto: If and only if F passed this 4. Now, F, conservative.
43:35:920Paolo Guiotto: I could start with the four parameters, no?
43:39:860Paolo Guiotto: And I could write if and only if F is the gradient of little f, and then look at little f, that solves the system, blah blah blah.
43:48:700Paolo Guiotto: But I would carry around useless cases, because when A, B, C, D are not of this type.
43:55:750Paolo Guiotto: I'm sure that the field cannot be conservative.
43:59:270Paolo Guiotto: So I would waste time in doing calculation, probably doing B-states, because I carry around parameters, too many parameters. So, I say, F conservative implies F irrotational.
44:13:350Paolo Guiotto: So, necessarily, F is… this guy here.
44:18:570Paolo Guiotto: But this is a one-way implication. So this says, F conservative, then F is this one.
44:25:920Paolo Guiotto: But does not mean that this one is conservative. It says it must be there, but maybe not all of these are conservative, who knows? So we have now to discuss this.
44:39:200Paolo Guiotto: such as F… So it's a way to restrict, let's say, the search. Such F is conservative.
44:50:630Paolo Guiotto: If and all if there exists an F such that gradient F is the field F on the domain B,
45:01:40Paolo Guiotto: That is, and we have our system, the derivative with respect to X of the function F,
45:07:310Paolo Guiotto: is equal to the first component, AX plus BY divided X squared plus Y squared.
45:14:800Paolo Guiotto: Second, DYFXY, equal…
45:20:100Paolo Guiotto: Here, be careful, because the structure is minus BX plus AY divided X squared plus Y squared.
45:30:740Paolo Guiotto: Now we, work on these equations. There is no particular difference between the two, so…
45:36:220Paolo Guiotto: We again start from the first.
45:39:20Paolo Guiotto: Now, this one… is equivalent to say that the function fxy
45:44:880Paolo Guiotto: must be one of the primitives of AX plus PY,
45:50:210Paolo Guiotto: over X squared plus Y squared in the variable X,
45:57:600Paolo Guiotto: plus a constant in X, so in principle, a function of Y.
46:04:220Paolo Guiotto: Now, let's do this calculation, and let's see what happens.
46:08:460Paolo Guiotto: Here, the reminder that the, the, the calculation is in the variable X.
46:16:220Paolo Guiotto: So, it would be convenient to split this numerator, because if you split, you have this, X divided X squared plus Y squared. I already wrote the A coefficient outside.
46:30:150Paolo Guiotto: plus B, primitive of Y. In this case, I can carry out also the Y, because the Y for the integration variable, which is X, is a constant. So I could say BY, and this is 1 over X squared plus Y squared in the X.
46:49:370Paolo Guiotto: And then we have this,
46:51:650Paolo Guiotto: For the moment, mysterious functions see of what?
46:55:200Paolo Guiotto: Now, these two are different, if you… if you look at… so, if you… it disturb you, the fact that you see this Y for a second, for example, imagine that you take a specific value of Y, possibly not 0, because this…
47:08:980Paolo Guiotto: Change a bit the story, but take Y, for example, equals to 1, so you read X over the X squared plus 1.
47:17:410Paolo Guiotto: And here, you need 1 over X squared plus 1.
47:21:30Paolo Guiotto: So this one is the arctangent, that one is what?
47:25:500Paolo Guiotto: Yeah, it comes from the logarithm, because the numerator, more or less, we have the derivative of the denominator, so to have exactly that, I put a 2 here, and I write a 1 half outside. Now, this is the derivative with respect to X of the log of X squared plus Y squared.
47:45:540Paolo Guiotto: So we need a bit of intuition, but that's not something that you learn here. It's something that you have seen last year when you computer primitives. At the end, you have always to recognize that someone is the derivative of someone else.
47:59:130Paolo Guiotto: And in fact, and then you confirm, if you do the derivative with respect to X of this, derivative of log is 1 over the argument, which is X squared plus Y squared.
48:10:20Paolo Guiotto: And then you have to put also the derivative of the argument, because the argument is not X, but in this case it's X squared plus Y squared, so you get 2X.
48:19:710Paolo Guiotto: Therefore, the first piece is A half log of X squared plus Y squared.
48:27:240Paolo Guiotto: plus BY, and about this one.
48:31:580Paolo Guiotto: probably we have to recognize something like the R tangent. So, to do the R tangent, I have to put this in the form 1 plus a square, and to do that, I factor the Y squared, so I put out here 1 plus X over Y,
48:51:430Paolo Guiotto: square, so this is DX.
48:54:860Paolo Guiotto: So this looks like the arctangent of X over Y. And in fact, if I do the derivative with respect to X of the arctangent of X divided by Y, this is 1 over 1 plus the square of the argument, which is X over Y,
49:14:620Paolo Guiotto: times the derivative with respect to X of the argument, and this is…
49:21:540Paolo Guiotto: 1 over Y, which is just a constant, and this is exactly the 1 over Y we have here. So I put the 1 over Y here, and they have… and then there is also the C
49:33:590Paolo Guiotto: Still Y.
49:35:210Paolo Guiotto: So I have a half.
49:37:330Paolo Guiotto: log of X squared plus Y squared.
49:41:450Paolo Guiotto: plus B, the arctangent of X divided Y, plus C of Y.
49:50:830Paolo Guiotto: Now… If you are a bit rigorous, you should be disturbed a bit by the attacks of a Y,
49:57:840Paolo Guiotto: Because when Y is 0,
50:00:510Paolo Guiotto: Remind that our variable, is in the plane XY. The unique forbidden point is this one, 00, but anywhere else.
50:12:730Paolo Guiotto: So, if this is the point, the function should make sense, no?
50:18:950Paolo Guiotto: And then for, in particular, for Y equals 0, so it means I am here.
50:23:420Paolo Guiotto: I should have an S.
50:26:30Paolo Guiotto: There is no reason, apparently, but this one says, Y equals 0.
50:31:250Paolo Guiotto: You cannot write this, so we may say, okay, for… if Y is 0, what would change in this calculation?
50:39:970Paolo Guiotto: Well, I would have, for Y different from 0.
50:45:940Paolo Guiotto: For Y equals 0, I would have A half, let's do quickly, the primitive, the first primitive would be, 2X over X squared, so I would have 2X over X squared in X.
51:04:800Paolo Guiotto: Then I, I have, plus, what is it? B,
51:11:760Paolo Guiotto: Well, you see, the second would be zero, no? Because if Y is 0, I get just 0.
51:18:940Paolo Guiotto: plus C of 0.
51:21:920Paolo Guiotto: So this would be, let's cancel this, let's put this 1 over X, so I would get A,
51:29:250Paolo Guiotto: The primitive of 1 over X. What is this?
51:34:350Paolo Guiotto: Not exactly. Log of what?
51:39:880Paolo Guiotto: No. There is no… for mathematicians, there is no…
51:45:870Paolo Guiotto: Absolute value of X. Okay. So, plus C of 0.
51:50:580Paolo Guiotto: Now, it seems to be different from this one, apparently.
51:58:610Paolo Guiotto: But…
52:03:870Paolo Guiotto: Notice that when you put the Y equals 0 in the log.
52:08:380Paolo Guiotto: you get log of x squared.
52:10:670Paolo Guiotto: and log of X squared
52:14:680Paolo Guiotto: is exactly 2 log of absolute value of X, because you can always put an absolute value here.
52:21:620Paolo Guiotto: And therefore, if you put a 2 here, and multiply by 2 and carry the 2 inside, here you have a half log of X squared plus C of 0.
52:35:100Paolo Guiotto: Okay, so we may say that at the end, the function FXY
52:40:10Paolo Guiotto: we'll have this, shape, A half, Blas.
52:44:880Paolo Guiotto: B… no, sorry, A half log…
52:50:360Paolo Guiotto: Let's finish this calculation, then we'll do the break.
52:55:770Paolo Guiotto: plus arctangent.
53:00:380Paolo Guiotto: affects… of a Y plus C of Y,
53:05:670Paolo Guiotto: This for Y different from 0?
53:09:560Paolo Guiotto: While for Y equals 0, we get the same thing, basically, a half log of X squared.
53:18:170Paolo Guiotto: So this part is confirmed. The other part is just 0, so we do not write, plus the value of this C at 0. So that's the function.
53:28:780Paolo Guiotto: Now, we have to check if this verifies also the second equation. You remind.
53:34:930Paolo Guiotto: we have, for the moment, just imposed the first condition, huh? We are looking for the app that verified the first. Now, let's impose the second one, let's see what happens.
53:45:670Paolo Guiotto: Now… These… F verifies.
53:55:150Paolo Guiotto: also.
53:56:760Paolo Guiotto: The second condition, DYF equal, it was, minus BX plus, No. It's this one.
54:07:270Paolo Guiotto: minus BX plus AY.
54:10:770Paolo Guiotto: over X squared plus Y squared.
54:14:530Paolo Guiotto: If and only if, well, since we have this double shape for F, let's do the case Y different from 0, okay? So we take this formula here, and we do the derivative. Let's see if it comes equal to that.
54:30:30Paolo Guiotto: So, if you know, if the Y, for…
54:35:300Paolo Guiotto: Y different from 0. There is also Y equals 0, but let's take just this case. Y different from 0 means that we are considering all the plane except y equals 0. Y equals 0 is this.
54:50:10Paolo Guiotto: This is the set of points where the Y is 0, is the x-axis.
54:55:840Paolo Guiotto: So, Y different from 0 means whatever remains, so basically the plane without the x-axis. Now, DTY is…
55:09:240Paolo Guiotto: So we have A half, The derivative of the logarithm.
55:14:590Paolo Guiotto: And that's the fraction. Downstairs you have the argument of log, upstairs you have the derivative, now it's respect to Y of the argument, so you get 2Y.
55:28:440Paolo Guiotto: Plus.
55:30:560Paolo Guiotto: Let's do the derivative of the second term. We have plus B.
55:37:920Paolo Guiotto: The up tangent is 1 over 1 plus X over Y squared.
55:45:800Paolo Guiotto: This is the arctangent. Then I have to do the derivative of the argument of the arctangent with respect to Y.
55:55:940Paolo Guiotto: And what is this?
56:02:850Paolo Guiotto: minus… What is the derivative of this guy with respect to Y?
56:18:700Paolo Guiotto: Minus X over Y squared, exactly.
56:23:510Paolo Guiotto: Okay, so we got this, plus there is also the derivative of this, function here, okay, plus…
56:30:980Paolo Guiotto: C prime of Y. Let's see what happens. This is, so AF, simplified this two, so we have, let's, say AY divided X squared plus Y squared, this is the first. Oh, since I have the minus, I put the minus here.
56:49:500Paolo Guiotto: So let's rework a bit this. We do the common denominator. Downstairs, we have X squared plus Y squared. We put B here, and we have a Y square here. Then we have X over Y.
57:04:520Paolo Guiotto: Square.
57:06:190Paolo Guiotto: plus, C prime of Y.
57:12:20Paolo Guiotto: So, this cancels this…
57:16:30Paolo Guiotto: And therefore, this X goes here, so we have AY minus BX divided X squared plus Y squared plus C prime of Y.
57:30:280Paolo Guiotto: Let's see if it looks like the other one. The other one is this one.
57:36:840Paolo Guiotto: Okay, so I impose that it is equal to minus BX plus Y.
57:42:460Paolo Guiotto: VCAT, which is the same.
57:45:870Paolo Guiotto: Because it is minus BX.
57:48:700Paolo Guiotto: plus AY over X squared plus Why Square?
58:01:660Paolo Guiotto: There is something strange here, huh?
58:10:910Paolo Guiotto: But it seems correct.
58:13:560Paolo Guiotto: Okay.
58:14:820Paolo Guiotto: So this, these two are the same. So, this is possible if and only if C prime of y is equal to 0.
58:33:700Paolo Guiotto: That is something that I don't like, I will tell you in a second.
58:38:400Paolo Guiotto: But probably, I will think about during the break. So, this means that C is constant.
58:45:280Paolo Guiotto: is constant.
58:48:210Paolo Guiotto: And therefore, it means that,
58:56:10Paolo Guiotto: I'm… why I'm worried? Because of this, of a why.
59:07:890Paolo Guiotto: This… this thing shouldn't be there, so…
59:11:890Paolo Guiotto: Okay, since I'm not convinced of this, let's take the break now, 5 minutes. Meanwhile, I will check this calculation, okay?
59:26:970Paolo Guiotto: Okay, sorry, I don't know exactly what is wrong, because…
59:32:210Paolo Guiotto: I don't… I've not found the error. The problem is that, to me, B should be equal to 0.
59:40:420Paolo Guiotto: Because if you put A equals 0, no, in that field, so focus just on that field, BY over X squared plus Y squared minus BX, for example, B equals 1.
59:54:870Paolo Guiotto: Now.
59:55:970Paolo Guiotto: The left is the derivative with respect to X of that function, the tangent of X divided by. The right is minus the derivative with respect to Y of our tangent of the reciprocal.
00:08:730Paolo Guiotto: Now, the two are the same, because of a relation for the tangent. You can flip the…
00:16:850Paolo Guiotto: coefficient, but that field is not conservative.
00:20:920Paolo Guiotto: So there is not a function such as… you see, it's the opposite.
00:26:700Paolo Guiotto: So, that's strange. I don't… I thought that I did some mistake here, but I checked the calculation. They seem to be correct. But since I'm not particularly good today, so let's suspend this. I will write the end of the solution of this exercise. To me.
00:45:230Paolo Guiotto: It should come that,
00:47:260Paolo Guiotto: Here, we should have… instead of having a minus, minus, we should have a plus on one side and a minus on the other, in such a way that when you…
00:58:520Paolo Guiotto: when you put together these two, you should get this C prime. One part, this one, is the same for the two. So this cancels, but the other one is not the same.
01:12:920Paolo Guiotto: And so they sum, and you would get this C prime of y equal something like, querying on the unique side, minus 2B X divided X squared plus Y squared.
01:25:430Paolo Guiotto: So this is possible if and only if this number is zero, because in that case, if it is different from zero, it would mean that C depends on X, but that's not possible.
01:39:110Paolo Guiotto: So this is what I expected to obtain, but since I don't see where is it the error now, let's suspend this. I will return later on this, and I will fix the error, so you will see
01:53:180Paolo Guiotto: Maybe this afternoon, tonight, the solution of this. So, let's suspend a bit this point.
02:02:290Paolo Guiotto: Okay.
02:03:460Paolo Guiotto: In any case, do… I'd say, basically, all the exercises, 3, 4, also, do…
02:16:610Paolo Guiotto: So yesterday, I left you the first exercises, so do… 3, 4, 8… And 9…
02:31:640Paolo Guiotto: Well, the next exercises involve a concept that we are going to introduce now.
02:38:340Paolo Guiotto: So, let's talk about this, which is the concept of line integral.
02:51:100Paolo Guiotto: Well, this in physics is the concept of work, work made by a force to move a mass between two points. The idea is that, let's imagine that,
03:06:360Paolo Guiotto: we have a fielder, for example, a field in the plane, but the definition is general, so I will use the notation, the general notation for the field, but I do figures in the case of the plane.
03:20:340Paolo Guiotto: So let's imagine that this is, the space,
03:25:420Paolo Guiotto: RD, where the field acts. So, they're saying that it works on a domain of RD into RD.
03:36:890Paolo Guiotto: We take, let's say, two points of the domain, so let's assume that this is the domain, and we are in the domain. Let's say we take two points.
03:48:320Paolo Guiotto: Let's call them X and Y.
03:55:780Paolo Guiotto: So, if, if, we consider a trajectory, So, a line.
04:01:970Paolo Guiotto: along which, imagine here there is a point mass, which is displaced by the field F, along that line.
04:13:260Paolo Guiotto: Now, in physics, it is in… they use this quantity, they introduce this quantity to measure the work done by this force.
04:21:720Paolo Guiotto: So the idea is that, if you… if this is the trajectory, so you have to imagine that this
04:29:370Paolo Guiotto: point is in movement from the initial point to the final point, so it may be like a car, which is moving along this road. There will be a certain velocity of this car.
04:42:820Paolo Guiotto: Now, the velocity, what is the velocity? Now, we will formalize this concept in a second, but the velocity is basically the tangent to the curve, because it is connected to the derivative.
04:56:600Paolo Guiotto: So, at each point of this curve, you have the velocity and the force applied to, say that this is the force, and this is the velocity, V, of the point mass that… when the point mass passed to that point.
05:13:660Paolo Guiotto: Now, the work of the force is something that is, proportional to how much F is alleged with the velocity.
05:24:460Paolo Guiotto: Because, for example, if the force is perpendicular to the velocity, so the force pulls in a direction which is perpendicular to the direction of motion.
05:35:640Paolo Guiotto: the force will not have any influence on the motion, and therefore, it's like if this has a contribution equals zero. While, in the opposite case, if the force is aligned with the velocity, the force will pull, no? The particle will push the particle,
05:54:940Paolo Guiotto: more and more. So, the idea is that we have to try to quantify, find the methods to quantify this. When the force is perpendicular to velocity, we have no contribution, and the contribution is maximum when the force is aligned to the velocity.
06:11:630Paolo Guiotto: Now, there's a quantity that makes this. So, in other words, if this is the angle theta between the force and the velocity, we want to say that this quantity is the product between the strength of the force, which is the norm of the force.
06:30:900Paolo Guiotto: The strength of the velocity.
06:34:450Paolo Guiotto: And the cosine of the angle made between the two vectors, because the cosine 0 means that theta is 90 degrees, or something like that, so the vectors are perpendicular.
06:47:540Paolo Guiotto: Theta, for example, equals 0, means that the two vectors are perfect. That coefficient is 1, so you have the maximum.
06:55:950Paolo Guiotto: contribution. Now, this quantity is nothing but the scalar product between the two vectors, F and V.
07:04:10Paolo Guiotto: So, the idea is that the work in physics, the work.
07:11:600Paolo Guiotto: done by F to displace, let's say, let's use an improper notation, then we will formalize a notation, a precise notation.
07:23:810Paolo Guiotto: to displace the particle from point X to point Y is basically the sum of all these quantities along the line, along the curve, so the sum of
07:36:540Paolo Guiotto: products, Scala products, F, Scala D.
07:40:710Paolo Guiotto: Now, this is, of course, a very…
07:43:570Paolo Guiotto: informal, and we do not understand really what is the real definition here. So, to have a real definition, we have to introduce some formal
07:55:280Paolo Guiotto: entities. Like, for example, let's start describing what is a path, a curve, a line in the domain D.
08:04:200Paolo Guiotto: Now…
08:05:00Paolo Guiotto: how do you describe lines in plain? Now, if you think about the two, simple lines, they could be.
08:13:700Paolo Guiotto: things like that. Now, a line, for example, I am in a Cartesian plane, XY. This is the diagram of a function. It's something like Y equals F of X. This is a way to describe a line, and this means that points along the line, they have this form. X, F of X.
08:33:880Paolo Guiotto: So the coordinates are of this type, XF of X. Not all curves are described like that. For example, if you take a circle.
08:44:510Paolo Guiotto: That description is not convenient, because the circle is not the graph of a function.
08:50:640Paolo Guiotto: But usually, if the radius is R, we may say that this point can be represented in this way. R cosine alpha, r sine alpha.
09:04:870Paolo Guiotto: Where alpha is the angle made by the point and the positive direction of the axis.
09:11:319Paolo Guiotto: No? Now, what these two things have in common?
09:15:149Paolo Guiotto: Is that you have a point here, point.
09:19:990Paolo Guiotto: is a function of X. Here, also, point is a function of, in this case, the angle alpha.
09:29:960Paolo Guiotto: So what they have in common is that they are… they are both functions of one single variable.
09:37:69Paolo Guiotto: And this variable is a real number.
09:39:830Paolo Guiotto: So, this is what makes a line, and what makes different a line from a plane.
09:45:460Paolo Guiotto: So, we introduce a formal definition. We call this a curve, in general.
09:51:140Paolo Guiotto: So, F function.
09:54:680Paolo Guiotto: We will use letters like gamma.
09:58:350Paolo Guiotto: So, gamma is a vector-valued function of real variable. So, a function we normally use letter T, that reminds of time.
10:09:290Paolo Guiotto: Because if this is a trajectory of motion, this means that this is a function of time that goes from some interval I of the real line
10:21:430Paolo Guiotto: to our… B… So, it's just… it's called… Cover.
10:35:100Paolo Guiotto: in RD.
10:38:310Paolo Guiotto: If the function gamma is continuous, if gamma…
10:43:960Paolo Guiotto: is continuous. Then we talk about, Got my… is… Continuous.
10:52:700Paolo Guiotto: Cerva?
10:55:630Paolo Guiotto: If, here, the function gamma is continuous with first derivative continuous, we say C1, so…
11:06:660Paolo Guiotto: So this means gamma and gamma prime.
11:11:260Paolo Guiotto: are both continuous.
11:13:600Paolo Guiotto: Then we say that the curve gamma ease… Regular.
11:23:740Paolo Guiotto: Cerva.
11:27:470Paolo Guiotto: So, for example, let's see a few simple examples.
11:33:970Paolo Guiotto: So let's start from the case of the, this circle. Now, we have this curve, this plane curve, we are in R2 here.
11:44:510Paolo Guiotto: Let's say that this is the circle of radius R.
11:48:650Paolo Guiotto: So now, above I use the letter alpha. Since I want to uniform a unique letter for all cases, I will always use letter T. So T here stands for the angle.
12:01:120Paolo Guiotto: made from this vector and the positive direction of the axis. This point is the point of coordinates r cosine T, r sine t.
12:11:800Paolo Guiotto: So that's the definition of what we call the gamma of t.
12:17:460Paolo Guiotto: As you can see, when you move T from 0 to 2 pi, this point describes the interior circumference, the circle.
12:30:470Paolo Guiotto: So, example, another example. Imagine I want to describe A line, a straight line.
12:40:610Paolo Guiotto: like this one. This is, normally represented in Cartesian coordinates in this form, Y equals MX plus Q, right?
12:50:380Paolo Guiotto: This means that points here, they have form X, and the Y is not arbitrary, but MX plus Q.
12:58:190Paolo Guiotto: Now, if you call XT, you have this function, gamma of t equal t, mt plus Q, when T is
13:10:60Paolo Guiotto: In the reals, so from minus infinity to plus infinity.
13:15:630Paolo Guiotto: The set of points described at that gamma is exactly that line.
13:22:320Paolo Guiotto: So if I have, for example, Ehhh…
13:26:950Paolo Guiotto: Something like, for example, a spiral, like this.
13:30:680Paolo Guiotto: I… I start here, and… I go down to the… But actually.
13:37:770Paolo Guiotto: Yeah, let's start here. I go down to the origin. I could say that this is a point of type, cost T.
13:46:530Paolo Guiotto: Sign T…
13:48:880Paolo Guiotto: But with NR, which is depending on t, and when t gets big, this becomes close to zero. For example, I can put this E to minus T, e to minus T, I define this thing, gamma of t.
14:04:250Paolo Guiotto: for T, for example, from 0 to plus infinity.
14:09:00Paolo Guiotto: And they have a curve that turns around the origin infinitely many times, shrink into the origin. And this is another example of draw, okay?
14:19:410Paolo Guiotto: So, in general.
14:24:230Paolo Guiotto: Ecova.
14:26:500Paolo Guiotto: Gamma, So, of t will be a function.
14:31:340Paolo Guiotto: It is vector-valued, and this vector will have components that we will call gamma 1 of T, gamma 2 of T, gamma
14:41:110Paolo Guiotto: of the…
14:43:600Paolo Guiotto: So this function is a special function, of, vector, is vector-valued, but of real variable, no? So this gamma, goes from interval I, which is in R1, into R
15:00:800Paolo Guiotto: V… So, in this case, the derivative of this thing
15:07:600Paolo Guiotto: Remind that for… in general, for functions of vector variable, vector value, the derivative is a matrix, it is the Jacobian matrix, no?
15:19:20Paolo Guiotto: In this case, we have what? We have a function of a unique variable.
15:24:660Paolo Guiotto: So, and there are D components. Now, you know that,
15:30:140Paolo Guiotto: In lines, I see the gradients of the components, but since there is a unique variable here, so there is a unique derivative, derivative with respect to the unique variable, which is here, is T, so this will be the derivative with respect to t of the first variable, so what we call gamma 1,
15:47:450Paolo Guiotto: prime of t, gamma 2 prime for the second line, etc, gamma D prime of T. So at the end, this matrix, again, is a vector.
16:00:140Paolo Guiotto: Okay? It's a vector, let's say in matrices is a D by 1 matrix. It's a vertical vector.
16:09:540Paolo Guiotto: While the gradient is an original vector.
16:13:40Paolo Guiotto: That's just for the aesthetic, let's say. Now, this quantity here, if gamma T is a position, we call this quantity velocity.
16:25:60Paolo Guiotto: So, if gamma… Of D is D.
16:32:440Paolo Guiotto: position.
16:34:220Paolo Guiotto: At, let's say, time, T… the gamma prime of T
16:42:440Paolo Guiotto: Is what we call the velocity.
16:48:970Paolo Guiotto: at time t.
16:51:410Paolo Guiotto: So…
16:53:220Paolo Guiotto: If we want to represent a trajectory of a mass moving in the three-dimensional space, so where we have X, Y, Z, we will describe this movement through a curve
17:08:390Paolo Guiotto: gamma of T, that in this case is a vector with three components, so there will be an X of t, a Y of T,
17:18:190Paolo Guiotto: a Z of T.
17:22:500Paolo Guiotto: And if the curve is regular, the velocity is the vector gamma prime of t.
17:30:220Paolo Guiotto: Given by the three derivatives of the three components, X prime T, Y prime t, z prime T.
17:39:340Paolo Guiotto: Okay?
17:40:520Paolo Guiotto: Now, we have the ingredients to define what is this quantity that we introduced a bit
17:48:910Paolo Guiotto: roughly here. What is the work? Of course, we won't call work, because this concept can be adapted to other situations. So, definition.
18:00:930Paolo Guiotto: Let's… F… Now, this F is a vector field.
18:08:490Paolo Guiotto: be defined on a domain D of RD, with values in… RV… B.
18:20:380Paolo Guiotto: vector.
18:22:820Paolo Guiotto: Fielder.
18:25:940Paolo Guiotto: We assume here, for technical reasons, that this is a continuous function only.
18:32:440Paolo Guiotto: So I do also a figure here.
18:35:140Paolo Guiotto: So this stands for RD, but the figure will be a figure in the plane, okay? We have our domain of the field here.
18:46:80Paolo Guiotto: We take a curve in, in, D, so let's take a curve.
18:52:880Paolo Guiotto: Which is always contained in D.
18:57:150Paolo Guiotto: let… Gamma.
19:00:480Paolo Guiotto: be a regular, so regular means with the derivative, and actually continuous derivative.
19:08:940Paolo Guiotto: Regular server.
19:12:500Paolo Guiotto: contained…
19:17:240Paolo Guiotto: in the… So this means that,
19:21:560Paolo Guiotto: Gamma of T belongs to D for all time t.
19:26:620Paolo Guiotto: in the interval of definition of the curve. So the point gamma t is, Indy.
19:35:530Paolo Guiotto: Now, since we assume that the curve is regular, we know also that at each point, there exists a velocity, that is the vector we called gamma prime of t.
19:49:150Paolo Guiotto: Now, we define them.
19:52:980Paolo Guiotto: we set…
19:54:670Paolo Guiotto: This, that we call a line integral, or arc integral, the integral on gamma of F, we use this notation.
20:04:850Paolo Guiotto: will be.
20:06:470Paolo Guiotto: Now, suppose that the interval IB
20:11:660Paolo Guiotto: the interval ABE to fix ideas.
20:16:350Paolo Guiotto: So what we want to do is we give a precise definition to this qualitative idea, that we do the sum of products between F
20:27:250Paolo Guiotto: and the V, where V is the velocity. Scarlet product between the two. So what we do here is leave some space. We take F,
20:36:110Paolo Guiotto: We evaluate at point gamma T, so at the position, if you want to stay in it.
20:42:130Paolo Guiotto: physical weight, at the position of the particle at time t. We do the scalar product with the velocity, which is gamma prime of t.
20:54:780Paolo Guiotto: Now, this quantity is what is a number, because this color product is a number, it's not a vector.
21:01:910Paolo Guiotto: And we sum this quantity when T varies between A and B. Since AB is a continuum, then that sum will be made through an integral. So we will do the integral from A to B of this quantity in the variable T.
21:18:180Paolo Guiotto: Now, this, by definition, is what is called the line integral.
21:24:570Paolo Guiotto: Line. Integral.
21:27:530Paolo Guiotto: of F… a longer… the arc gamma.
21:35:420Paolo Guiotto: In physics, if F is a force field, and gamma is the trajectory of motion, this quantity is called work made by F to displace the particle, a particle, along the path gamma.
21:52:440Paolo Guiotto: Okay?
21:56:90Paolo Guiotto: Now, first remarkable fact,
22:01:00Paolo Guiotto: is that if the field is conservative, this quantity does not depend on the path particularly, but depends only on the initial point and the final point. So we have this proposition.
22:16:980Paolo Guiotto: If, F… Jeez.
22:20:330Paolo Guiotto: conservative.
22:27:410Paolo Guiotto: Then… the, arc integral of F along, gamma depends… only.
22:44:290Paolo Guiotto: on the initial point of the pattern, say, gamma of A,
22:50:630Paolo Guiotto: and the final point of the path, gamma of B.
22:54:780Paolo Guiotto: So, you have this figure.
22:57:340Paolo Guiotto: Let's say that this is… this plane stands for the space of D. We have the domain of the field, D. We pick two points.
23:09:120Paolo Guiotto: Let's say X and Y.
23:13:210Paolo Guiotto: And we consider pets, two different paths.
23:17:620Paolo Guiotto: say, gamma 1, maybe not, because, we… we are using that index for the components. So, let's say that we consider two different paths joining these two points, so let's call this gamma tilde.
23:36:950Paolo Guiotto: Maybe.
23:38:600Paolo Guiotto: Now, the arc integral along these two is the same.
23:43:600Paolo Guiotto: If the field is conservative. So, here we will have an arc integral along this first path, here we will have another arc integral along this second path. But actually, what happens is that these two guys are the same.
24:03:390Paolo Guiotto: No? So, this is remarkable because it says that it doesn't matter how you displace the mass, what is important is the initial point and the final point. How you… what you do in the middle is irrelevant.
24:16:650Paolo Guiotto: Now, let's see why.
24:19:50Paolo Guiotto: This is just a little group.
24:25:80Paolo Guiotto: But, hmm…
24:26:500Paolo Guiotto: To understand the proof, we need to do some remark, because, so let's compute the arc integral along a gamma. By definition, this is the integral from A to B of F of gamma of T
24:46:210Paolo Guiotto: Scala product with gamma prime of t. This is the definition of what is that quantity.
24:52:760Paolo Guiotto: Now, since F is supposed to be conservative, that's the main
24:58:620Paolo Guiotto: important assumption we have here. It means that this F is actually the gradient of some potential little f, so where little f is the potential
25:09:540Paolo Guiotto: of, capital F.
25:12:90Paolo Guiotto: So this formula is this, integral from A to B of gradient F evaluated at gamma of T.
25:22:330Paolo Guiotto: Scala product with gamma prime of t.
25:27:420Paolo Guiotto: Now, maybe you will not recognize what is this, but remind that the gradient of F stands for the derivative.
25:36:440Paolo Guiotto: So it's like if I have here something like F prime of gamma T,
25:43:60Paolo Guiotto: scalar product with gamma prime of t. Now, if that product would be an algebraic product, imagine that there are no arrows.
25:52:260Paolo Guiotto: And you read, F prime of gamma T times gamma prime t, this is…
25:58:330Paolo Guiotto: If there are no arrows, F of gamma… F prime of gamma T times gamma prime,
26:08:170Paolo Guiotto: Exactly. So this is the derivative of decomposition.
26:13:40Paolo Guiotto: No? Because if you do the derivative of this, how do you do this calculations? You do the derivative of this, evaluated on gamma, times the derivative of the argument. It's what we have done many times in the examples above, right?
26:26:560Paolo Guiotto: Now, what happens in this case?
26:29:730Paolo Guiotto: It turns out that it is the same, that here we have to justify this.
26:35:340Paolo Guiotto: So let's do this remark.
26:38:490Paolo Guiotto: Remarker… D quantity, gradient F, at gamma of D, scholar, the gamma Prime, of D…
26:53:10Paolo Guiotto: is exactly the same of doing. You take F of gamma T,
26:58:240Paolo Guiotto: Now, if you look at this quantity.
27:01:180Paolo Guiotto: There is some space here, because they're divided.
27:04:380Paolo Guiotto: This quantity here takes the… computes the vector gamma of t, then you do F of T, sir.
27:12:340Paolo Guiotto: But definitely is a numerical function.
27:14:580Paolo Guiotto: So, if you forget that you passed through gamma, you start from T and you end to a real. So, this is a function of a real variable, a real value.
27:24:550Paolo Guiotto: This thing, tea.
27:26:760Paolo Guiotto: goes to… F of gamma of T,
27:32:30Paolo Guiotto: is a function defined for T, real, and it is real-valued, because F, it is, yes, a function defined on the domain D, which is a vector domain, but it is real-valued. So once you evaluate F, you get real numbers.
27:50:420Paolo Guiotto: Now, this is an ordinary function.
27:53:210Paolo Guiotto: And you may wonder, what is the derivative with respect to t of this function? This is a first-year calculus function.
28:01:270Paolo Guiotto: Now, I repeat, if there are no arrows, I would have that formula above. The derivative of F of gamma T is f prime, gamma T times gamma prime t. That's the chain rule, it's called the chain rule.
28:13:850Paolo Guiotto: Now, the same happens here, but since now F is a function of vector variable, and gamma is an array.
28:21:660Paolo Guiotto: Even if it is a function of D, we get exactly that formula.
28:26:290Paolo Guiotto: Let's understand why, because imagine that, indeed.
28:33:790Paolo Guiotto: we have to do, F of gamma of T,
28:39:400Paolo Guiotto: And we want to do the derivative with respect to t.
28:43:00Paolo Guiotto: I will give you an intuitive, explanation.
28:49:870Paolo Guiotto: We have already done something similar when we proved the Lagrange form… Lagrange multiplier theorem.
28:56:60Paolo Guiotto: So I have to do F of gamma 1T, gamma 2T, etc.
29:02:280Paolo Guiotto: Until gamma DT, okay?
29:05:400Paolo Guiotto: I have to do the derivative with respect to T of this thing.
29:09:140Paolo Guiotto: Now, what should I do? If I forget of these valuables, I see I have F of gamma 1T,
29:17:230Paolo Guiotto: No? So how would I derive? I derive this deriving F with respect to this Bible times gamma 1 prime, because that's the key rule for ordinary functions we have above there.
29:31:200Paolo Guiotto: But what is the derivative of F respecting its first value?
29:35:870Paolo Guiotto: And that's the partial derivative of this variable. So, I will have…
29:42:800Paolo Guiotto: D1F, evaluated on all this point that I will write for short, again, gamma of T.
29:51:770Paolo Guiotto: Times the first derivative of gamma 1.
29:58:340Paolo Guiotto: this is the chain rule. If the function f is function only of the first variable, I have this, but F is function of this bunch of variables. So now.
30:07:680Paolo Guiotto: Once you have done this, consider the second bar. Now, look at the F as only function of this.
30:13:890Paolo Guiotto: So you apply the H-Evolution to the derivative of M with respect to this variable times the derivative of the argument with respect to t. And this guy is called the…
30:23:870Paolo Guiotto: The partial derivative with respect to the second variable of F, evaluated again at gamma T, times gamma 2 prime of t.
30:33:810Paolo Guiotto: And then you look at the function as function of the third variable, and so on, until you finish, and you arrive at the last variable, you will have the derivative with respect to the last variable of F at gamma T,
30:44:940Paolo Guiotto: times gamma D prime T.
30:48:690Paolo Guiotto: Now, the formula says that you have to sum up all these contributions.
30:53:780Paolo Guiotto: Together, to get the exact formula.
30:57:740Paolo Guiotto: And if you look at this, you recognize exactly this color product between the gradient and the derivative of gamma, because these are the components of the gradient, you see? The partial derivative with respect to X1, the partial derivative with respect to X2,
31:12:690Paolo Guiotto: Respect to X3, respect to XD, these are the components, components.
31:21:920Paolo Guiotto: off…
31:23:200Paolo Guiotto: the gradient of F. While these ones are the components of gamma prime, because gamma prime, the velocity, is exactly the vector made of the derivatives of the component.
31:37:110Paolo Guiotto: It is this one.
31:38:570Paolo Guiotto: Okay, so this is the… Components.
31:45:990Paolo Guiotto: off.
31:47:170Paolo Guiotto: Gamma prime.
31:48:980Paolo Guiotto: And so, since you have a sum of products, this is the scalar product, it is the product between the gradient F at gamma T,
31:58:340Paolo Guiotto: Scala with Gamma Prime.
32:01:750Paolo Guiotto: So, this closes the remark, and we can finish this argument.
32:07:80Paolo Guiotto: So, going back to the lily integral.
32:10:790Paolo Guiotto: So, the line integral of F along gamma is, we say, that the integral on AB of gradient F
32:20:970Paolo Guiotto: at gamma T, Which is, nothing but the field evaluated at the Kama TE.
32:28:680Paolo Guiotto: Scala product with gamma prime T.
32:34:30Paolo Guiotto: We just said that this is the derivative with respect to T of the composition. So this is… all this is the derivative with respect to T of F of gamma of t.
32:49:640Paolo Guiotto: So, we have here… integral from A to B, derivative with respect to T, of F of gamma P.
33:00:270Paolo Guiotto: DT.
33:02:920Paolo Guiotto: Now, what is integral from A to B of phi prime of TDT?
33:11:670Paolo Guiotto: What is the integral of the derivative? The integral, not the primitive now, the integral.
33:19:230Paolo Guiotto: the… it's not Q, it's a phi, but however, it's exact. Phi of B minus phi of A.
33:25:320Paolo Guiotto: final value minus initial value. This is the fundamental TRM, of integral…
33:34:920Paolo Guiotto: calculus, that's the rule you use to compute integrals in practice. And then you have the derivative, so this means that this is equal to the value of F at the point gamma B minus the value of F at point gamma A.
33:53:70Paolo Guiotto: So, as you can see, the arc integral, which is here, It's just the difference.
34:01:140Paolo Guiotto: of the potential between the final point and the initial point, so whatever happens in the middle, it does not change this value. So if you have two paths joining the same two points, but differently, you will have the same war.
34:16:330Paolo Guiotto: Okay? So this is one important feature of conservative fields that is very important, particularly in physics.
34:24:910Paolo Guiotto: Good, let's stop here, and
34:28:870Paolo Guiotto: See you on, on, Friday.