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### Course: Multivariable calculus > Unit 2

Lesson 4: Differentiating parametric curves# Differential of a vector valued function

Understanding the differential of a vector valued function. Created by Sal Khan.

## Want to join the conversation?

- at01:45, Sal said "derivative" instead of "vector", Its very minor issue, but an annotation would be much appreciated. Thanks(5 votes)
- You help create annotations yourself by clicking on Report a mistake above, and writing out what the mistake is and give a time stamp. Then, if it is approved by the guardians, it becomes an annotation.(11 votes)

- why are differentials "hand wavy"? and if they are so, is there another way to get to the same result without all the waving?(4 votes)
- When he says "hand wavy" I think he's referring to how he treats dr/dt as a fraction, i.e. in the sense that you can just multiply out the dt.

To be honest though, I don't see why that's hand wavy at all. The way I see it rate of change is that ratio so you should be able to treat it like a fraction. Full disclose: I'm no genius.(3 votes)

- Why is the multiplication by a differential "dt" considered to be hand wavy (like Sal does at2:19)? What would be a more rigorous way of showing this? I use/see this trick so often it makes me worry that I don't know the answer to this question.(3 votes)
- As I said to another user:

"To be honest though, I don't see why that's hand wavy at all. The way I see it rate of change is that ratio so you should be able to treat it like a fraction. Full disclose: I'm no genius."

Basically I agree with you, let me know if you worked it out.(1 vote)

- Please start videos for higher level maths...this is superb and systematic(3 votes)
- Why is it not rigorous to multiply by dt. Doesn't "dt" just represent [lim (x -> 0) of t]?(2 votes)
- I knew it, it does have some connection to the directional derivative, this can also be rewritten as a dot product.(2 votes)
- why are differentials "hand wavy"? and if they are so, is there another way to get to the same result without all the waving?(2 votes)
- Sorry for the stupid question, but something is bothering me.

When we calculate the r(t) from one t to another how can we calculate the distance of the x and y sides from one point to the other?

It should be dxi + dyj but it is not.

Example:

r(t) = x(t)i + y(t)^2j

t=1 1 + 1

t=2 2 + 4

so the distance in x is 1 and the distance in y is 3

If we calculate the derivative we will get that r'(t) = 1i + 2y(t)j

t= 1 1 + 2

t=2 1 + 4

Intuitively, these should be the differences of the x and y sides form one point to the other.

They are not.

How to get the exact differences of the vector component sides?(1 vote)- so is dr the magnitude of the derivative of the vector?(1 vote)

- At03:58, does dr mean an ordinary(?) vector, not a position vector?(1 vote)
- why are differentials "hand wavy"? and if they are so, is there another way to get to the same result without all the waving?(1 vote)

## Video transcript

In the last couple of videos
we saw that we can describe a curves by a position
vector-valued function. And in very general terms, it
would be the x position as a function of time times
the unit vector in the horizontal direction. Plus the y position as a
function of time times the unit victor in the
vertical direction. And this will essentially
describe this-- though, if you can imagine a particle and
let's say the parameter t represents time. It'll describe where the
particle is at any given time. And if we wanted a particular
curve we can say, well, this only applies for some curve--
we're dealing, it's r of t. And it's only applicable
between t being greater than a and less than b. And you know, that would
describe some curve in two dimensions. Just me just draw it here. This is all a review of
really, the last two videos. So this curve, it might look
something like that where this is where t is equal to a. That's where t is equal to b. And so r of a will be this
vector right here that ends at that point. And then as t or if you can
imagine the parameter being time, it doesn't have to be
time, but that's a convenient one to visualize. Each corresponding as t gets
larger and larger, we're just going to different-- we're
specifying different points on the path. We saw that two videos ago. And in the last video we
thought about, well, what does it mean to take the derivative
of a vector-valued function? And we came up with this idea
that-- and it wasn't an idea, we actually showed
it to be true. We came up with a
definition really. That the derivative-- I could
call it r prime of t-- and it's going to be a vector. The derivative of a
vector-valued function is once again going to be a derivative. But it was equal to-- the way
we defined it-- x prime of t times i plus y prime
of t times j. Or another way to write that
and I'll just write all the different ways just so
you get familiar with-- dr/dt is equal to dx/dt. This is just a
standard derivative. x of t is a scalar function. So this is a standard
derivative times i plus dy/dt times j. And if we wanted to think about
the differential, one thing that we can think about-- and
whenever I do the math for the differential it's a
little bit hand wavy. I'm not being very rigorous. But if you imagine multiplying
both sides of the equation by a very small dt or this exact dt,
you would get dr is equal to-- I'll just leave it like this. dx/dt times dt. I could make these cancel
out, but I'll just write it like this first. Times the unit vector i
plus dy/dt times dt. Times the unit vector j. Or we could rewrite this. And I'm just rewriting it in
all of the different ways that one can rewrite it. You could also write this as dr
is equal to x prime of t dt times the unit vector i. So this was x prime of t dt. This is x prime of t right
there times the unit vector i. Plus y prime of t. That's just that right there. Times dt. Times the unit vector j. And just to, I guess, complete
the trifecta, the other way that we could write this is
that dr is equal to-- if we just allowed these to cancel
out, then we get is equal to dx times i plus dy
times dy y times j. And that actually makes a
lot of intuitive sense. That if I look at any dr,
so let's say I look at the change between this
vector and this vector. Let's say the super small
change right there, that is our dr, and it's made up of-- it's
our dx, our change in x is that right there. You can imagine it's that right
there times-- but we're vectorizing it by multiplying
it by the unit vector in the horizontal direction. Plus dy times the unit vector
in the vertical direction. So when you multiply this
distance times the unit vector, you're essentially
getting this vector. And when you multiply this
guy-- and actually our change in y here is negative--
you're going to get this vector right here. So when you add those together
you'll get your change in your actual position vector. So that was all a little
bit of background. And this might be somewhat
useful-- a future video from now. Actually, I'm going to leave it
there because really I just wanted to introduce this
notation and get you familiar with it. In the next video, what I'm
going to do is give you a little bit more intuition
for what exactly does this thing mean? And how does it change
depending on different parameterizations. And I'll do it with two
different parameterizations for the same curve.