Partial derivatives of vector-valued functions Partial Derivatives of Vector-Valued Functions
Partial derivatives of vector-valued functions
- 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.
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