Vector field line integrals dependent on path direction

let's say I have a position vector function that looks like this R of T is equal to X of T times the unit vector I plus y of T times the unit vector J let me actually graph this so let's say R of T so I'll draw it a little bit straighter than that so that's my y-axis that is my x-axis and let's say R of T and this is for T is less than let me write this so this is for T is between a and B so when T is equal to a where this vector right here so if you actually substitute T is equal to a here you would get a vector a position vector that would point to that point over there and then as T increases it traces out a curve or the end points of our position vectors trace out a curve that looks something like that so when T is equal to B we get a position vector that looks that points to that point right there so this defines a path and the path is going in this upward direction just like that now let's say that we have another another position vector function let me call it o R let me just call it I'll call it R I don't know our reverse our sub R R or sub R of T actually I'm just call it R of T to make it more confusing if I try to differentiate but it's a different one it's the green our R of T instead of being X of T times I it's going to be X of a plus B minus T times I and instead of Y of T it's going to be Y of a plus B minus T times I and we've seen this in the last two videos this the path defined by these position vector by this position vector function is going to look more like this let me draw my axes it's my y-axis that is my x-axis maybe I should label them Y and X Y and X this path is going to look just like this instead of starting here and going there when T is equal to 1 let me make it clear this is also true for a is less than or equal to T which is less than B so T is going to go from A to B but here when T is equal to a you substitute it over here you're going to get this vector you're going to get you're going to start over here you're going to start over there and as you increment T as you make it larger and larger and larger you're going to trace out that same path but in the opposite direction so it's the same path in the opposite direction and so when T is equal to B you put that in here you're actually going to get X of a and Y of a they're right the B's cancel out and so you're going to point right like that so these are the same you can imagine the shape of these paths are the same but we're going in the exact in the exact opposite direction so we're going to do in this video is to see what happens how I guess you could say if I have some vector field let's say I have some vector field f of XY equals P of XY I plus Q of X Y J right this is just a vector field over over the Dover the XY plane how the line integral of this vector field of this vector field over this path over this path compares to the line integral the same vector field over that path how that compares to this to we call this the minus curve so if this is the positive curve this is the positive curve we're going to call this the minus curve so how does it going over the positive curve compared to going over the minus curve f of F dot d R so before I break into the math let's just think about it a little bit let me draw this vector field F so maybe it looks I'm just going to draw a random stuff so you know on every point in the XY plane it has a vector it defines or maps a vector to every point on the XY plane but we really care about the points that are on the curve so maybe the on the curve you know this is the vector field at the points on the curve and let me draw it over here too so all the points on the curve where we care about this is our vector field that is our vector field and let's just get an intuition of what's going to be going we're summing over the dot we're taking each point along the line right we're going to take each point along the line and we're summing let me start over here we're taking each point along the line we did a different color and we're summing the dot product of the value of the vector field at that point the dot product of that with dr or the differential of our position vector function and dr you can kind of imagine as a small an infinitesimally small vector going in the direction of our movement and when we take this dot product here it's essentially it's going to be a scalar value but the dot product if you remember it's the magnitude of F in the direction of D R times the magnitude of D R so it's this you can imagine it's the shadow of F onto D our let me zoom into that because I think it's useful so this little thing that I'm drawing right here let's say that this is my path this is F at that point F at that point looks something like that and then D R at this point looks something like that let me do a different color dr dr looks something like that so that is f and so the dot product of these two says okay let's how much of f is going in the same direction as dr and you can kind of imagine it as a shadow so you take the f that's going the same direction as dr the magnitude of that times the magnitude of dr that is the dot product in this case we're going to get a positive number because this length is positive this length is positive that's going to be a positive number now what if our dr was going in the opposite direction as it is in this case so let me draw maybe that same part of the curve the same part of the curve we have our F our F will look something like that that is our F I'm drawing this exact same part of the curve but now our dr our dr isn't going in that direction our dr that at this point is going to be going in the other direction we're tracing the curve in the opposite direction our dr is now going to be going in that direction so if you do so this is our dr so if you do f dot dr you're taking the shadow or how much of f is going in the direction of dr you take the shadow down here it's going in the opposite direction of dr so you can imagine that when you multiply the magnitudes you should get a negative number our direction is now opposite they're not going into the shadow of f on to the same direction as dr is going in the opposite directions dr in this case it's going in the same direction as dr so the intuition is is that maybe these two things are the negative of each other and now we can do some math and try to see if that is definitely definitely the case so let us first figure out let's write an expression for the differential dr so in this case dr dr dt is going to be equal to X prime of T times I plus plus y prime of T times J in this other example in the reverse case our dr dr dt is going to be what's it going to be equal to it's the derivative of x with respect to t the derivative of this term with respect to t that's the derivative of the inside which is minus 1 or minus times the derivative the outside with respect to the inside so that's going to be it's going to be derivative the inside is minus 1 times the derivative of the outside with respect to the inside x prime of a plus B minus T times I and then same thing for the second term derivative of Y of this term is riveted inside which is minus 1 times the derivative of the outside with respect to the inside which is Y prime of a plus B minus T so this is going to be the derivative of the inside times y prime of a plus B minus T J so this is dr DT in this case this is dr d t in that case and then if we wanted to write the differential dr and the forward curve example it's going to be equal to X prime of T times I plus y Prime T times J times the scaler DT times the scaler DT I could multiply it down into each of these terms but it keeps it simple just leaving it on the outside same logic over here d R is equal to minus X I change my shade of green but it's at least it still green a plus B minus T I minus y prime a plus B minus T J and I'm multiplying both sides by DT now we're ready to express this as a function of T so this curve right here I'll do it in pink the pink one is going to be equal to the integral from T is equal to a to T is equal to B of f of f of X of f of X of T y of T dot dot this thing over here which is I'll just write it out here I could simplify it later X prime of T I plus y prime of T J and then all of that times the scalar DT this will be a scalar value and then we'll have another scalar value DT over there now what is this going to be equal to if I take this reverse integral the reverse integral is going to be the integral from I'm gonna need a little more space from A to B of F of not X of T but X of a plus B minus T y of a plus B minus T I'm writing it small so I have some space dot this is a vector so dot this guy right here dot dr dot minus x prime of a plus B minus T I minus minus y prime of I'm using up too much space let me scroll go back a little bit actually let me take it make it even simpler let me take this minus sign out of it let me put a plus a plus and I'll put the minus sign out front so the minus sign is just a scalar value so we could put that minus sign out you know when you take a dot product and if you multiply a scalar times a dot product that's you could just take the scalar out that's all I'm saying so we can take that minus sign out to this part right here and then you have X prime of a plus B minus T I plus y prime of a plus B minus T you scroll over a little bit T J D T so this is the forward this is when we're following it along the forward curve this is when we're following it along the reverse curve now let's like we did when the scalar example let's make a substitution I want to make it very clear what I did all I did here is I just took the dot product but this negative sign I just took it out I just said this is the same thing as negative 1 times this thing or negative 1 times this thing is the same thing as that so let's make a substitution on this side because I really just want to show you that this is the negative of that right there because that's what our intuition was going for so let me just focus on that side so let me make a substitution U is equal to a plus B minus T then we get D U is equal to minus DT right just take the derivative of both sides or you get DT is equal to minus D U and then you get when T is equal to when T is equal to a U is equal to a plus B minus a so then U is equal to B and then when T is equal to B U is equal to a right when T is equal to B a plus B minus B is a U is equal to a so this thing using that simple that's that substitution simplifies to this is the whole point that simplifies to minus the integral from u is when T is a USB from B when T is B use a the integral from U is equal to B to use equal to a of f of X of U X of U Y of U - right that is you that is you dot dot X prime of you let me write dot just write this way X prime X prime of u times I that's you right there plus y prime of u times J and then instead of a DT I need to put a D u DT is equal to minus D u so I could write minus D u here or just to not make it confusing I'll put the D u here and take the minus out front I already have a minus out there so they cancel out they will cancel out just like that and so you might say hey Sal with these two things look pretty similar to each other they don't look like the negative each other and I would say well you're almost right except this guy's limits of integration are reversed from this guy so this thing right here if we reverse the limits of integration we have to then make it negative so this is equal to minus the integral from A to B of F dot or F the the vector f of X of U Y of U dot X prime of U I plus y prime of U J D u and now this is identical this integral this definite integral is identical to that definite integral we just have a different variable we're doing DT here we're having D u here but we're going to get the same exact number for any A or B and given this vector F and the path the the position vector path R of T so just to summarize everything up when you're dealing with when you're dealing with line integrals over vector fields the direction matters if you go in the reverse direction you're going to get the negative version of that and that's because at any point we take the dot product you're going to get you're not going into necessarily you're going in the different in the opposite direction so it'll be the negative of each other but when you're dealing with a scalar field we saw in the last video we when you're dealing with the scalar field we saw that it doesn't matter which direction that you traverse the path it that the positive path has the same value as a negative path and that's just because we're just trying to find the area of that curtain hopefully you found that mildly amusing