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Let's have the vector valued function r of s and t is equal to-- well, x is going to be a function of s and t. So we'll just write it as x of s and t times the x unit vector, or i, plus y of s and t times the y unit factor, or j, plus x of s and t times the z unit vector, k. So given that we have this vector valued function, let's define or let's think about what it means to take the partial derivative of this vector valued function with respect to one of the parameters, s or t. I think it's going to be pretty natural, nothing completely bizarre here. We've taken partial derivatives of non-vector valued functions before, where we only vary one of the variables. We only take it with respect to one variable. You hold the other one constant. We're going to do the exact same thing here. And we've taken regular derivatives of vector valued functions. The path in those just ended up being the regular derivative of each of the terms. And we're going to see, it's going to be the same thing here with the partial derivative. So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you can just swap it with t, and you're going to get the same exact result. I'm going to define it as being equal to the limit as delta s approaches 0 of r of s plus delta s. Only finding the limit with respect to a change in s comma t. We're holding t, as you can imagine, constant for given t, minus r of s and t. All of that over delta s. Now, if you do a little bit of algebra here, you literally, you know-- r of s plus delta s comma t, that's the same thing as x of s plus delta s t i, plus y of s plus delta s t j, plus z. All that minus this thing. If you do a little bit of algebra with that, and if you don't believe me, try it out. This is going to be equal to the limit of delta s approaching 0-- and I'm going to write it small because it'd take up a lot of space-- of x of s plus delta s comma t minus x of s and t, I think you know where I'm going. This is all a little bit monotonous to write it all out, but never hurts. Times s or divided by delta s times i-- and then I'll do it in different colors, so it's less monotonous-- plus y. Where every-- those limited delta s [? approaches ?] 0 applies to every term I'm writing out here. y of s plus delta s comma t minus y of s comma t, all of that over delta s times j. And then finally, plus z of s plus delta s comma t minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition. If you literally just put s plus delta s in place for s-- you evaluate all this, do a little algebra-- you're going to get the exact same thing. And this, hopefully, pops out at you as, gee, we're just taking the partial derivative of each of these functions with respect to s. And these functions right here, this x of s and t, this is a non-vector valued function. This y, this is also a non-vector valued function. z is also a non-vector valued function. When you put them all together, it becomes a vector valued function, because we're multiplying the first one times a vector. The second one times another vector. The third one times another vector. But independently, these functions are non-vector valued. So this is just the definition of the regular partial derivatives. Where we're taking the limit as delta s approaches 0 in each of these cases. So this is the exact same thing. This is equal to-- this is the exact same thing as the partial derivative of x with respect to s times i plus the partial derivative y with respect to s times j plus the partial derivative of z with respect to s times k. I'm going to do one more thing here and this is pseudo mathy, but it's going to come out-- the whole reason I'm even doing this video, is it's going to give us some good tools in our tool kit for the videos that I'm about to do on surface integrals. So I'm going to do one thing here that's a little pseudo mathy, and that's really because differentials are these things that are very hard to define rigorously, but I think it'll give you the intuition of what's going on. So this thing right here, I'm going to say this is also equal to-- and you're not going to see this in any math textbook, and hard core mathematicians are going to kind of cringe when they see me do this. But I like to do it because I think it'll give you the intuition on what's going on when we take our surface integrals. So I'm going to say that this whole thing right here, that that is equal to r of s plus the differential of s-- a super small change in s-- t minus r of s and t, all of that over that same super small change in s. So hopefully you understand at least why I view things this way. When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small. And in my head, that's how I imagine differentials. When someone writes the derivative of y with respect to x-- and let's say that they say that that is 2-- and we've done a little bit of math with differentials before. You can imagine multiplying both sides by dx, and you could get dy is equal to 2dx. We've done this throughout calculus. The way I imagine it is super small change in y-- infinitely small change in y-- is equal to 2 times-- though, you can imagine an equally small change in x. So it's a-- well, if you have a super small change in x, your change in y is going to be still super small, but it's going to be 2 times that. I guess that's the best way to view it. But in general, I view differentials as super small changes in a variable. So with that out of the way, and me explaining to you that many mathematicians would cringe at what I just wrote, hopefully this gives you a little-- this isn't like some crazy thing I did. I'm just saying, oh, delta s as delta approaches 0, I kind of imagine that as ds. And the whole reason I did that, is if you take this side and that side, and multiply both sides times this differential ds, then what happens? The left hand side, you get the partial of r with respect to s is equal to this times ds. I'll do ds in maybe pink. Times ds-- this is just a regular differential, super small change in s. This is a kind of a partial, with respect to s. That's going to be equal to-- well, if you multiply this side of the equation times ds, this guy's going to disappear. So it's going to be r of s, plus our super small change in s, t minus r of s and t. Now let me put a little square around this. This is going to be valuable for us in the next video. We're going to actually think about what this means and how to visualize this on a surface. As you can imagine, this is a vector right here. You have 2 vector valued functions and you're taking the difference. And we're going to visualize it in the next video. It's going to really help us with surface integrals. By the same exact logic, we can do everything we did here with s, we can do it with t, as well. So we can define the partial-- I'll draw a little-- I can define the partial of r with respect-- let me do it in a different color, completely different color. It's orange. The partial of r with respect to t-- the definition is just right here. The limit as delta t approaches 0 of r of s t plus delta t minus r of s and t. In this situation we're holding the s, you can imagine, in constant. We're finding its change in t, all of that over delta t. And the same thing falls out. This is equal to the partial of x with respect to ti plus y with respect to tj, plus z with respect to tk. Same exact thing, you just kind of swap the s's and the t's. And by that same logic, you'd have the same result but in terms of t. If you do this pseudo mathy thing that I did up here, then you would get the partial of r with respect to t times a super small change in t. dt, our t differential, you could imagine, is equal to r of st plus dt minus r of s and t. So let's box these two guys away. And in the next video, we're going to actually visualize what these mean. And sometimes, when you kind of do a bunch of like, silly math like this, you're always like, all right, what is this all about? Remember, all I did is I said, what does it mean to take the derivative of this with respect to s or t? Played around with it a little bit, I got this result. These 2 are going to be very valuable for us, I think, in getting the intuition for why surface integrals look the way they do.