Using a line integral to find work

let's apply what we learned in the last video and do a concrete example of the work done by a vector field on something going through some type of path through the field so let's say that I have a vector field it's defined over r2 over the XY plane so it's a function of x and y it Associates a vector with every point on the plane and let's say my vector field is Y times the unit vector I let's say - - x times the unit vector J and so you could imagine if we were to draw let's draw our x and y axes I'll do it over here if we were to draw our x and y axes this associates this Associates a vector a force vector let's say this is actually a force vector with every point with every point in our XY plane so this is X this is y so if we're at the point for example 1 0 what will the vector look like that's associated with that point well at 1 0 y is 0 so this will be 0 I minus 1 J minus 1 J looks like this so minus 1 J minus 1 J will look like that at X is equal to 2 I'm just picking points at random once it'll be easy Y is still 0 and now the force vector here would be minus 2 J so it would look something like this minus 2 J something like that likewise if we were to go here where Y is equal to 1 and X is equal to 0 when y is equal to 1 we have 1 I 1 I plus 0 minus 0 J so then our vector is going to look like that at this point our vector will look like that and if we were to go to 2 you could get the picture you can keep plotting these points you just want to get a sense of what it looks like if you go here the vector is going to look like that if you go maybe this point right here the vector is going to look like that I think you get the general idea I could keep filling in the space for this entire field all over you know just to make it symmetric if I was he year if I was right over here the vector is going to look like that you get the idea I could just fill in all of the points if I had to now in that field I have some particle moving and let's say it's path is described by the curve C and the parametrizations of it is X of T is equal to cosine of T and Y of T y of T is equal to sine of T and the path will occur from T let's say 0 is less than or equal to T it's less than or equal to 2 pi you might already recognize what this would be this parametrizations is essentially a counterclockwise circle so the path that this guy is going to go is going to start here where you can imagine T in this case you could almost you know it's it's you can imagine is it's the angle of the circle but you could also imagine T is x at time equals zero we're going to be over here then at time of PI over 2 we're going to have traveled we're going to have traveled quarter of the circle to there so we're moving in that direction and then at time at after pi seconds we would have gotten right there and then all the way after 2 pi seconds we would have gotten all the way around the circle so we're doing our path our curve is one counterclockwise rotation around the circle so to speak so what is the work done by this field on this curve so the work done so the work we learned in the previous video is equal to the line integral over this contour over this contour of our field of our vector field dotted with dotted with the differential of our movement so dotted with the differential of our movement dr well i haven't even defined R yet I mean I kind of have just the parametrizations a vector function we need to have some R that defines this path this is just a standard parameterization but if I wanted to write it as a vector function of T we would write that R of T is equal to X of T which is cosine of T times I plus y of T times J which is just sine of T sine of T times J and likewise this is for 0 is less than or equal to T which is less than or equal to 2 pi this and this are equivalent the reason why I took the pain of doing this is so now I can take its its vector function derivative and I can figure out its differential and then I can actually take the dot product with this thing over here so let's do all of that and actually calculate this line integral and figure out the work done by this field and one thing might already pop at your mind we're going in a counterclockwise direction but at every point where we're passing through it looks like the field is going exactly opposite the direction of our motion for example here we're moving upwards the field is pulling us backwards here we're moving to the top left the field is moving us to the bottom right here we're moving exactly the left field is pulling us to the right so it looks like the field is always doing the exact opposite of what we're trying to do it's it's hindering our ability to move so I'll give you a little intuition this will probably deal with negative work for example if I lift something off the ground I have to apply force to fight gravity I'm doing positive work but gravity is doing negative work on that we're just going to do the math here just to make you comfortable with this idea but it's interesting to think about what's exactly going or even here the field is let me the field I'm doing that pink color so let me stick to that the field is pushing in that direction so it's always going opposite the motion but let's just do the math to make everything in the last video a little bit more concrete so a good place to start is the derivative of our of our position vector function with respect to T so switch so we have d R DT which we could also write as R prime of T this is equal to the derivative of X of T with respect to T which is minus sine of T times I plus the derivative of y of T with respect to T that's derivative of sine of T is just cosine of T cosine of T times J and if we want the differential we just multiply everything times DT so we get dr dr is equal to we could write it this way we could actually even just put the the d well let me just do it so it's minus sine of T DT I'm just multiplying each of these terms by DT distributive property times the unit vector I plus cosine of T DT times the unit vector J so we have this piece now and then we want to take the dot product with this over here but let me rewrite our vector our vector field in terms of in terms of in terms of T so to speak so what's our field going to be doing at any point T we don't have to worry about every point we don't have to worry for example that over here the vector field is going to be doing something like that because that's not on our path that that force never had an impact on the particle we only care about what happens along our path so we can find a function that set what we can essentially substitute Y and X for their relative functions with respect to T and then we'll have the force from the field at any point or any time T so let's do that so this guy right here if I were to write it as a function of T this is going to be equal to Y of T right y is a function of T so it's sine of T sine of T right that's that sine of T times I times I plus or actually minus minus X or X of T X as a function of T so minus cosine of t minus cosine of T times J and now all of it seems a little bit more straightforward if we want to find this line integral this line integral is going to be the same thing as the integral let me pick a nice soothing color maybe this is a nice one the integral from T is equal to 0 to t is equal to 2 pi of f dot dr of f dot dr now when you take the dot product you just take you just multiply the corresponding components and add it up so we take the product of the minus sign and the sine of T or the sine of T with the minus sine of T DT I get you're going to get minus sign squared T DT and then you're going to add that too so you're gonna have that plus let me write that DT a little bit no so wacky looking DT DT and then you're going to have that plus these two guys multiplied by each other so that's well there's a minus sign here so plus let me just change this to - minus cosine squared cosine squared DT and if we factor out a minus sign and a DT what is this going to be equal to this is going to be equal to the integral from 0 to 2 pi of we could say sine squared plus I want to put the T there sine square root of T plus cosine squared of T actually let me take the minus sign out to the front so if we just factored the minus times we put a minus there to make this a plus make this a plus so the minus sign out there and we factor DT out I did a couple of steps in there but I think you got it this is just algebra at this point factoring out a minus sign so this becomes positive and then you have a DT and a DT factor that out and you get this you could multiply this out and you'll get what we originally have if that confuses you at all and the reason why I did that we know what sine squared of anything plus cosine squared of that same anything is that's them that falls right out of the unit circle definition of our trig function so this is just 1 so our whole integral has been reduced to the minus integral from 0 to 2 pi of DT of DT and this is we've seen this before we could sell you say this is or 1 if you want to put something there too then the antiderivative of 1 is just so this is just going to be equal to minus and that minus sign is just the same minus sign that we're carrying forward the antiderivative of 1 is just T and we're going to evaluate it from 2 pi to 0 or from 0 to 2 pi so this is equal to minus that minus sign right there 2 pi 2 pi minus T that's 0 so minus 0 so this is just equal to minus 2 pi and there you have it we figured out the work that this field that this field did on the particle or whatever as the whatever thing was moving around in this counterclockwise fashion in our intuition held up we actually got a negative number for the work done and that's because at all times the field was actually going exactly opposite or was actually opposing the movement of if we think of it as a particle in its counterclockwise direction anyway hopefully you found that helpful