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## Surface integral preliminaries (videos)

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# Partial derivatives of vector-valued functions

## Video transcript

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.