Math4Teens Path Integrals: UNIZOR.COM - Math4Teens - Calculus - Integrals Part 2 - Path Integrals

Duration: 27:16 3268 transcript words EN summary

Video summary

The video introduces the concept of path integrals, also known as line or curve integrals, which extend the standard definition of definite integrals from straight intervals to curves in two or three-dimensional space. While regular integrals calculate the area under a graph on a flat interval by summing infinitesimal rectangles, path integrals apply this logic to curved paths. The speaker explains that even though the surface formed along a curve is not perfectly rectangular, dividing the path into sufficiently small segments allows us to approximate these areas as nearly rectangular strips. By taking the limit as the number of segments approaches infinity and their size approaches zero, the calculation method remains fundamentally similar to standard integration, simply multiplying the function's value at each point by the infinitesimal length of the curve segment. To mathematically define a path in three-dimensional space, the video utilizes parametric equations where the coordinates $x$, $y$, and $z$ are expressed as functions of a single parameter, typically denoted as $s$. This parameter represents the progression along the curve, such as the distance traveled from a starting point to an ending point. Consequently, any scalar function defined along this path can be rewritten entirely in terms of this single parameter $s$. The integral is then computed by integrating this transformed scalar function with respect to $ds$ from the initial parameter value to the final one, effectively reducing the complex geometric problem into a standard definite integral over a real interval. The lecture further distinguishes between two primary types of path integrals: those involving scalar functions and those involving vector functions, with significant applications in physics. For scalar functions, the video provides an example of calculating average barometric pressure along a mountainous trail, where the parameter $s$ represents the distance covered. In contrast, vector path integrals are essential for calculating physical work done by variable forces, such as wind acting on a sailboat or friction opposing a car's motion. The speaker demonstrates that when force and displacement vectors are not aligned, the calculation involves the dot product of the force vector and the differential displacement vector ($d\mathbf{r}$). This approach accounts for angles between directions and allows for the precise computation of work even when forces vary in magnitude and direction along a curved trajectory. Ultimately, the video concludes that despite the apparent complexity introduced by curves and multi-dimensional spaces, path integrals are not mysterious but rather a systematic reduction to regular definite integrals. Whether dealing with scalar quantities like pressure or vector quantities like force, the process involves parameterizing the curve, expressing all variables in terms of that parameter, and performing a standard integration. The speaker emphasizes that this mathematical framework bridges calculus and physics, enabling the solution of real-world problems involving motion along non-linear paths. By mastering these concepts, students can analyze phenomena ranging from fluid dynamics to mechanical work, confirming that advanced integral calculus is built upon the same foundational principles as basic integration.

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Hi, I'm Zor. Welcome to Inor Education. Um, I would like to talk about different kind of integrals. Um, different from whatever we were using before. So, let me just remind you what was before with regular um definite integrals. You have a function and uh we defined it on some uh interval from a to b and this is the graph of the function for example and we needed to calculate basically the area under this graph. So we divide it into small pieces assumed that within this piece the value of the function is basically the same thing and the area was a rectangle we multiplied uh if this is delta x i this little segment a small one and the value of the function was let's say on the left uh side of this interval was f at x i and delta x i was actually x i - x i - 1. So then we have to multiply by delta x i. That's the area of this little thing. Then summarize it and then have the limit when all the delta x i goes to zero and the number of intervals goes to infinity. So that's basically the definition of this um function integral of a regular function and regular um real interval. Now what I would like to do is to introduce an integral of functions defined not on a straight interval from a to b but on some kind of a curve in space. Well, obviously if it's a curve, it cannot be one-dimensional. So, it's at least you need at least two dimension to have a real curve. So, um I will explain it on two or three maybe maybe three dimensions would be even better. And the curve is a curve in space. Now if it's a two-dimensional curve and I can really put it on the plane and then uh I will have a zcoordinate. So this is xy plane and on the zcoordinate I have value of the functions and value of the functions go along some curve. We also can really consider the area of it's basically a cylindrical surface. Um and uh we can also calculate the area exactly the same way. divide this little curve into small pieces and uh put the each area would would be not exactly a rectangle but a little bit curve rectangle because the u this this line is a is a curved line not a straight line but still in when when these little um uh intervals which we can call delta s let's say uh are really small uh we can consider it as a almost a rectangle. And uh obviously if the curve is smooth enough and the function is smooth enough we can actually um when the number of uh these intervals go to infinity and each one of them is infinite decimally small then we can consider it in exactly the same fashion as um we did with regular integrals. we just multiply the value of the function of si times the length of this little interval and uh summarize it and go to the limit. So basically it's the same thing. Now um in three dimension I if if the curve is not in a is not flat if it's in a space let's say in a three-dimensional space I can't even imagine how I can draw it because there is no force dimension for the value of the function but we still consider it algebraically and that's exactly what I'm going to do to define a pass or a line integral is called pass because the curve is in physics is in most cases it's a pass of movement of something. So the whole thing actually derived from from physics I mean physics initiated this type of a integral um well in as much as the regular integrals were really influenced by physical phys physics problems. So in any case, so we consider uh just as an example that you have a curve in let's say three-dimensional space and I would like to introduce a concept of a pass integral basically to define it. How can I define it? Well, it all I mean on the two-dimensional uh curve and using the value of the function at the third dimension you can actually say something about the area of the this little very narrow almost rectangle. Uh in multi-dimensional case like in three-dimensional case uh it's not really easy to to imagine what exactly happening. So basically we resort to algebra in this particular case and the first thing that we should do we should really find how can we define a curve in um and dimensional space. What is a curve? Again from the geometrical standpoint we can imagine what curve actually is. But since we would like actually to use some instruments of algebra or or calculus, we have to really define it like more mathematically and less intuitively. And this is actually a very easy thing to do. Any curve can be very easily defined using some kind of a parameter. So let's assume that we are talking about three-dimensional curve. What does it mean that we have a threedimensional curve? It means that we have three functions. Oh, sorry. S which each one depends on some parameter parameter s. So now s actually is a real number in some interval from zero to s. Now if the if we imagine that this curve goes from point A to point B, it means that X of 0, Y of 0, Z of 0, this point is A and point B would be when our um parameter has the maximum So these are two points. Now these functions of this parameter must be smooth which means we don't really have jumps in values. So as s is uh moving from zero to to capital s uh our points are not jumping all over. they really go a along a smooth curve. If s is changing a little bit, my x y and z also change a little bit. It means they are differentiable basically. So if this is um f if this is true then this particular thing can be a description of the curve. Okay. Now if in a regular integral we had an interval from from a to b and for each point we had a value of the function. Here we also can have that for each point from a to b and each point is basically calculated as some kind of a y s z. This is a point when uh this s is somewhere in between we have some value of the function which means that's what it is. So function of a point becomes actually a function of one single parameter. Now if this is a curve let's call it gamma. This is a curve. So I basically define this is a definition of integral of function f or well I don't know from some kind of a point if you wish um but I can put x of just x y z gamma I use gamma without actually this is all kind of a symbolic things I'm not defining gamma uh and here I will put a b which means I'm integrating this scalar function scalar because it's just a single value um on each point xyz along the curve gamma uh which is from point a to point b I'm just defining it as a regular integral from 0 to s function f of x of s y of s z of s d s. This is a definition. I mean nothing to it. And now this is a regular integral from uh from s is equal to 0 to s is equal to capital s. Function f it's really function of a point but each point depends on single parameter s which makes function f depending on single parameter s. So this is some kind of a function f of s if you wish this thing and d s and this is a regular integral. So this is how we have basically reduced a concept of integral uh along the path path integral or line integral or curve linear integral sometimes is is used. So this pass integral is basically defined as u reduce the definition to to definition of the regular interval and there is no mystery in it. So as long as we have defined a curve using these three functions for three-dimensional case or n functions for n dimensional case doesn't really matter. We can define integral pass integral along this particular curve along this particular path we can define it as parameterized thing nothing to it. Now um what can be an example? Well, example is some some simple for example you are would like to measure uh average barometric pressure as you are going along some mountainous area. Mountainous area means you go up and down left right etc. So you are going along certain path and you would like to average the barometric pressure. How to do it? Well, you should you should really measure your barometric pressure on at each point. Now, what can be a parameter? Parameter can be for example the length from the beginning of this pass. So, in the beginning it's zero. At the end, it's basically the total length of the pass in meters or whatever measure you would like to use. And on each point which means for each value of uh s you measure barometric pressure. So basically you are going on each point along your path and each point has three coordinates because it's somewhere in space in three-dimensional space. However, it's all reduced to a single integral. So you can say that basically then you can take this integral since at each point you have a parameter s which is the length you have covered from the beginning. So in each point you have basically this function x y and z and you can say that the barometric pressure at point um xyz is such and such but all x y and z are functions of s. So that's how you can integrate it and then divide it by the total length of the uh of the pass and you will get an average. So that's basically an example. Now this is one type of pass integrals. This is the past path integrals of a scalar function. Now scalar because we have only one single value uh of the function for each point along the pass. So the pass is multi-dimensional. Let's say let's say it's a maintenance pass somewhere and the barometric pressure is your single value function. Now in physics there is another very important concept which needs vectors. We are talking about work. So let me go back to definition of uh pass integrals but in this case for vectors not for scalar functions but for a vector function. This is a little bit more complex. Let me start from the very very simple thing. If you have um let's say um a straight road and you have some kind of an object and it moves and let's say there is a uh friction or something else but you have a fixed function f which goes along this direction and you co and you you cover the length let's say d. So what is the work which is done by this force F or vector F along this particular road uh during the distance D? Well, we know this is just a product of these guys. Okay, now this is a simple thing. Now what if F is variable? So it's still directed along the path. Let's say it's car engine. car goes straight but the road is different and therefore the engine has to move has to exort different forces to move. There are some more u uh resistance, less resistance, whatever it is. But anyway, engine goes straight, the um the car goes straight, but the engine um exorts force depending on certain conditions. So let's say x is um the length covered from the beginning and for each uh x based on let's say friction of the road f is different but still you have to cover uh the distance let's say from zero to to to to z for example. Now how to calculate um the work in this particular case? Well, again you divide it the whole area by small pieces. Now this is delta x i as usual. This is a regular integration and you have to integral from zero to maximum. Let's say it's d f ofx dx. So you if you know f of x which means you know the force which is supposed to be to to um overcome let's say the the resistance the friction or whatever it is. That's your regular integral. Okay. Now let's complicate the picture. Now what happens if um f is not the let's say force which car actually moves you. Let's say we are talking about uh the the sailboat then the sailboat goes this way but the wind goes this way. Now if you have a an angle between the um this is direction and this is the force how to calculate the work in this particular case. Well, if you remember physics, you just have to project F vector of the force onto vector of direction and multiply this by corresponding amount of uh road which you have which you have covered. So what is this? This is absolute value of the vector times let's say you have covered a very small um uh let's call it delta r vector of um of of the pass. So you calculate it times this time cosine of angle five right this is the cosine of this which is actually A scalar product of vectors dot product do that product is actually calculated as length of one length another and the cosine of angle between them. So this is the definition of the work in case force F is not directed along this straight road. Okay, let's complicate it even more. Let's say our road is not straight but a curve. And this this is how I would like to approach integral of vector function uh along a pass. So pass integral of vector function. So before we were talking about pass integral or line integral of scalar function. Now I'm talking about uh pass integral of vector function. Why? Because it's very important for physics. Okay. So, how do we define it? Well, basically I have already done the whole thing here. Obviously, if you have some kind of a line and how can line in multiple dimensions be defined? Well, as I was saying before, x of param some parameter um s y parameter s z parameter s for three dimensional case or we can say this is vector r of parameter s right because three points define a vector or n points in n dimensional case. All right fine. So how do we calculate this particular pass integral? So this is the key. So you basically have to multiply um as a scalar product uh vector f and and infinite decimal piece of the curve obviously uh okay in infinite decimal piece of curve we can we can use dr let's say now what is dr well it's this is a vector right vector has three components three component components are dx, dy and dz in threedimensional case. I'm not talking about n dimensional. Three dimensional is enough actually. Right? So this is an infinite decimal increment along the curve where r is the vector r which depends on certain parameter and obviously these things also depends on the same parameter. Okay. Now f is a vector. Now f also depends on on on three in this particular three arguments x y and z. Now but this is a vector which means it has three components f_sub_x which depends on xyz f_sub_y and fz. Each one of them depends on all these x y and z. and x y and z in in their turn. This is the coordinate along the curve along this path. They all depends again on parameter s. So what should I do to calculate the work? Well, this is the key. So the whole integral of vector f time dr along some kind of a pass from a to b. What is this? Well, this is a scalar product. This is a scalar product. Do you remember how uh scalar product actually is supposed to be calculated in coordinate fashion? It's x coordinate * x coordinates plus y * y + z * z. So the whole thing actually can be divided in from uh uh from zero to s. The s is the maximum parameter which corresponds to the end point. zero is the beginning corresponds to the beginning of the point. And now I just have to really do very very simple thing. F of X of S Y of S Z of S time DX. So the first coordinate times the first coordinate plus and X is also depends on S, right? plus second component f_sub_y * d y + f_sub_z * dz and all of them are functions of x y and z which in turn are functions of s all smooth functions and the whole thing is now this is just a function of s and obviously dx can be uh changed into dx by d s * ds and now we have ds which can actually be taken out completely. So it would be x component of vector function times um basically velocity along the x component and then uh velocity along y etc. So altogether you can actually say that this is a scalar product of uh the force times the velocity and that's where we are merging with physics in this particular case. So this is how you calculate the work of let's say the wind which goes in one direction along the path of the sailboat which goes to slightly different direction not necessarily straight uh along the where wherever the wind is blowing. No we are going somewhere else but we're using the sail to position it properly. So our movement would be wherever we want and so we are moving along certain paths and that's how we can calculate the work of the um uh of the wind. So in any case the um pass integral of either scalar functions or or vector functions in all these cases we finally reduce them to a regular integral by regular real parameter from from from zero to whatever the value is. So this is all about definition of the pass integral and the pass integral in this case is defined using all these concepts which basically lead you to regular uh definite integral of uh real functions uh on real arguments. Well that's it. Now I suggest you to go to um notes for this lecture because they basically uh describe the same thing but maybe in some other words maybe better words than I'm doing right now. So you go to unisur.com, you go to math4's uh course. Now this is um u continuation I think it's called uh integrals part two and that's where you will find the pass integrals. That's it. Thank you very much and good luck.