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# Line integrals and vector fields

## Video transcript

one of the most fundamental ideas in all of physics is the idea of work and when you first learn work you just say oh that's just forced I'm distancing later on when you learn a little bit about vectors you realize that the force isn't always going in the same direction as your a displacement so you learned that it's work is really the magnitude I'll let me write this the magnitude of the force magnitude of force of the force in the direction or the component of the force in the direction in direction of a displacement displacement is just distance with some direction in direction of displacement displacement times the magnitude of the displacement or you could just kind of say times the distance displaced times the distance under not being too particular about it and the classic example maybe you have an ice cube or some type of block I just say ice so that's not a lot of friction maybe it's standing on a bigger sheet a lake or ice of something and maybe you're pulling on that ice cube at an angle let's say you're pulling at an angle like that that is my force right there let's say my force is equal to well that that's my force vector let's say the magnitude of my force vector so I'll say the magnitude so let me put two brackets around there the magnitude of my force vector we can say is let's say it's ten Newtons and let's say the direction of my force vector right any vector has to have a magnitude and a direction and the direction let's say it's 30 it has a 30 degree angle let's say it's 60 degree angle above horizontal so that's the direction I'm pulling in and let's say I displace it let's say I display this is all a bit of review hopefully if you're displacing it let's say you displace it five Newtons so let's say the displacement that's the displacement vector right there and the magnitude of it is equal to five sorry not five Newton five meters so you've learned in you know from the definition of just say oh I have I'm pulling with ten Newton's of force and I'm moving at five meters you can't just multiply the ten Newton's times the five meters you have to find that can put the magnitude of the component going in the same direction as my displacement so what I I need essentially need to do is the length if you imagine the length of this vector being ten that's that's the total force but you need to figure out the length of the vector that's the component of the force going in the same direction as my displacement and a little simple trigonometry you know that this is ten times the cosine of 60 degrees ten times the cosine of sixty degrees or that's equal to cosine of sixty degrees is one-half so that's just equal to 5 so this magnitude the magnitude of the force going in the same direction of the displacement in this case is five Newton's five Newtons and then you can figure out the work you could say that the work is the work is equal to five Newton's times all right well I'll just try 2.4 times I don't even think it's cross-product times 5 meters which is 25 Newton meters or you could even say 25 25 joules of work have been done this is all you know really a review of somewhat basic physics but just think about what happened here what was the work if I wrote it right in the abstract the work is equal to the 5 Newton's that was the magnitude of my force vector so it's the magnitude of my force vector times the cosine of this angle times the cosine of the so you know let's call that theta let's say it a little generally so times the cosine of the angle this is the amount of my force in the direction of the displacement the cosine of the angle between them times the magnitude of the displacement so times the magnitude of the displacement or if I wanted to rewrite that I could just write that as the magnitude of the displacement times the magnitude of the force times the cosine of theta and I've done multiple videos of in the linear algebra playlist in the physics playlist where I talk about the dot product and the cross product and all of that but this is this is the dot product this is the dot product of D and F of the vectors D and F so in general if you're trying to find the work for a constant displacement and you have a constant force you just take the dot product of those two vectors and if the dot product is a completely foreign concept to you you might want to watch I think I've made multiple four or five videos on the dot product and it's intuition and how it compares but just to give you a little bit of that intuition right here the dot product when I take F dot d or D F what it's giving me is I'm multiplying the magnitude well you could I could just read this out but the idea of the dot product is take how much of this vector is going in the same direction of as this vector and in this case two this much and then multiply the two magnitudes and that's what we did right here so the work is going to be the force vector dot taking the dot product of the force vector with the displacement vector and this of course is a scalar value and we'll work out some examples in the future where you'll see that that's true so this is all a review of fairly elementary physics now let's take a more complex example but I it's really the same idea let's define a vector field let's define a vector field so let's say that I have a vector field F and we're going to think about what this means in a second it's a function of x and y and it's equal to some scalar function of x and y times the I unit vector or the horizontal unit vector plus some other function scalar function of x and y times the vertical unit vector so what would something like this be this is a vector field this is a vector field in two dimensional space or on the XY plane this is a vector field on xy-plane plane or you can even say on or on are to either way I don't want to get too much into the to the math eunice of it but what is this do well if I were to draw my XY plane so that is my again having trouble drawing a straight line alright there we go that's my y-axis and that's my x-axis I'm just drawing the first quadrant but I could you could go negative in either direction if you like what does this thing do well it's essentially saying look you give me any X any Y you give any XY in the XY plane and these are going to end up with some numbers right these are gonna that when you would XY here you're gonna get some value and when you put XY here gonna some value so you're gonna get some combination of the I and J unit vector so you're gonna get some vector so what this does is defines a vector that's associated with every point on the XY plane so you could take you could say hey you know if I take at this point on the XY plane and I were to pop it into this I'll get you know something times I plus something times J and when you add those two maybe I get a vector that looks like something like that maybe and you can do it on every point I'm just taking random samples maybe when I go here the vector looks something like that maybe when I go here the vector looks like this maybe when I go here the vector looks like that maybe when I go up here the vector goes like that I'm just randomly picking points it defines a vector on all of the XY coordinates where where these were these functions these scalar functions are properly defined and that's why it's called a vector field it defines what a potential may be force would be or and some other type of feet well we could say force at any point at any point if you happen to have something there maybe that's what the function is and I could keep doing this forever and filling in all the gaps but I think you get the idea it associates a vector with every point on the XY plane now you know this is called a vector field so it probably makes a lot of sense that this could be used to describe any type of field it could be a gravitation field it could be in an electric field it could be a magnetic field and this could be essentially telling you how much force there would be on some particle in that field that's exactly what this would describe now let's say that in in this field I have some particle traveling on the XY plane and let's say it starts here let's say it starts there and by virtue of all of these crazy forces that are acting on it and maybe maybe it's on some maybe it's on some tracks or something so can all it can't it won't always move exactly in the direction that the field is trying to move it at so let's say it moves in a path that moves something like this and let's say that this path or this curve is defined by a a position vector function so let's say that that's defined by R of T which is just X of T times I plus y of T times our unit vector J that's R of T right there well in order for this to be a finite path this is true before T is greater than or equal to a and less than or equal to B this is the path that the particle just happens to take due to all of these wacky forces so when the particle is right here maybe the vector field acting on it maybe it's putting a force like that but since the thing is on some type of tracks it moves in this direction and then when it's here maybe the vector field is like that but it moves in that direction because it's on some type of tracks now everything I've done in this video is to build up to a fundamental question what was the work done on the particle by the field work done on the particle what was the work done on the particle by the field to answer that question we could zoom in a little bit let's say we go let's say I'm gonna zoom in on only you know a little small snippet of our of our of our path and let's say let's just try to figure out what the work is done in a very small part of our path because it's constantly changing the field is changing directions my object is changing direction so let's say when I'm here when I'm here and let's say I move a small amount of my path so let's say I move let's say I move this is an infinite - as elyse infinitesimally small D R right I have a differential it's a differential vector infinitely small displacement and let's say over the course of that I had some the vector field is acting in the local area let's say it looks something like that it's providing a force that looks something like that so that's the vector field in that area or the force directed on that particle right when it's at that point right there's an infinitesimally small amount of time and space you could say okay ever that little small point we have this constant force what would what was the work done over over this small period you could say what's this small interval of work you could say D work or a differential of work well by the same exact logic that we did it with a simple problem it's the magnitude of the force in the direction of our of our of our displacement times the magnitude of our displacement and we know what that is just from the this example up here that's the dot product it's the dot product of the force and our super small displacement so that's equal to the dot product of our force and and our our super small displacement now just by doing this we're just figuring out the work over maybe like a really small super small D R but what we want to do is we want to sum them all up we want to sum up all of the DRS to figure out the total that all of the f dot dr is to figure out the total work done and that's where the integral comes in we will do a line integral over I mean you could think of it two ways you could write just DW there but we could say we'll do a line integral that says along a along this curve C we could call that C or along R whatever you want to say it of DW that'll give us the total work so let's say work is equal to that or we could also write it over the integral or the same over the same curve of F of F dot d R and this might seem like a like you know what go really you know gee this is really abstract Sal how do we actually calculate something like this especially because we have everything parametric permit rised in terms of T how do we get this in terms of T and if you just think about it what is f dot R or what is f dot dr actually to answer that let's think about let's remember what dr looked like if you remember if you remember dr/dt is equal to is equal to X prime of T I'm writing it's like I could have writen DX DT if I wanted to times the I unit vector plus y prime of T times the J unit vector and if we just wanted dr dr we could multiply both sides if we're being a little bit and maybe more hand-wavy with the differentials not too rigorous we get dr is equal to X prime of T DT times the unit vector I plus y prime of T times the differential DT times the unit vector J so this is our dr right here that is our dr and remember what our what our what our force our vector field was it was this thing up here let me copy and paste it and then we'll see that the dot product is actually not not so crazy so i'm gonna copy and then let me paste it down here alright paste it down there so what's this integral going to look like this integral right here that seems well you know gives us the total work done by the field on the particle as it moves along that path which is super fundamental to pretty much any serious physics that you might and eventually find yourself doing so you could say well gee it's going to be the integral well out well let's just say from T is equal to a to T is equal to B right a is where we started off on the path T is equal to a the T is equal to B you could imagine it as being time as a particle moving as time increases and then what is f dot dr what is f dot dr well if you remember from just what the dot product is you can essentially just take you can just put you take the product of the corresponding components of your vector and add them up so this is going to be the integral from t equals a to t equals B of P of P of X really instead of writing X Y is X of T right X is a function of T Y is a function of T so that's that times this thing right here times this component right we're multiplying the AI components so times X prime of T DT and then that plus plus we're gonna do the same thing with the Q function so this is Q plus I'll go to another line hopefully realize I could have just kept writing but I'm running out of space plus Q of X of T y of T times the component of our dr times the y component of the J component Y prime of T DT and we're done and we're done this might still seem a little bit abstract but we're gonna see in the next video that this is actually everything is now in terms of T so this is just a straight-up integration with respect to DT if we want we could take the DTS outside of the equation and it'll look a little bit more normal for you but this is essentially all that we have to do and we're gonna see some concrete examples of taking a line integral through a vector field or on or using a vector functions in the next video