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let's say I 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 vector or J plus Z 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 pass and that was just ended up being the regular derivative of each of the terms 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 the partial derivative of R with respect to s and everything I do with respect to s you could 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 the limit as Delta s approaches 0 of R of S Plus Delta s we're only finding the limit with respect to a change in s comma T we're holding T you can imagine constant for given T minus R of s and T all of that over 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 TJ plus C all of 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 to be equal to this is all going to be equal to let me write it here it's going to be equal to the limit of Delta s approaching zero and I'm going to write it small because and take up a lot of space of X of S Plus Delta s comma t minus X of 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 the limit as Delta X approaches zero 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 tool it allows we're going to get this exact same thing and this hopefully makes pops out at us G 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 an on vector valued function this Y this is also an on 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 we're taking the limit as Delta X 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 of y with respect to s times J plus the partial derivative of Z with respect to s times K now 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 doing even doing this video is it's going to give give us some good tools or in our toolkit 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 Matthew 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 hardcore 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 this whole thing right here that that is equal to 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 smooth 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 zero these Delta s are going to get super duper duper small and in my head that's how I imagine differentials when someone says when someone writes D the the derivative of Y with respect to X and let's say that they you know and let's say that they say that is 2 and we've done a little bit of math with differentials before multi you can imagine multiplying both sides by DX and you get dy is equal to 2 DX we've done this throughout calculus but the way I imagine it's 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 that 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 two 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 you know this doesn't this isn't like some crazy thing I did I'm just saying Oh Delta s is Delta s approaches zero I kind of imagined that as D s and the whole reason I did that is if you take this side and that side and multiply both sides times this differential D s then what happens the left hand the left hand side you get the partial of R with respect to s is equal to this oh let me times D s I'll do D s and maybe pink times D s this is just a regular differential super small change in s this was the kind of a partial with respect to s that's going to be equal to that's going to be equal to well if you multiply this side of the equation times D s this guy is 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 and 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 right you have two vector valued functions 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 we can define the partial I'll draw a little well 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 zero 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 it constant it's 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 T I plus y with respect to t J plus Z with respect to TK same exact thing you just kind of swap the SS 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 Matthew thing that I did up here and 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 s T 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 always like write whatever it is all about it remember all I did is I say what does it mean to take the derivative of this with respect to s or T play it around with it a little bit I got this result this is going to be soup these two are going to be very valuable for us I think I'm getting the intuition for why surface integrals look the way they do