# Differential of a vector valued function | Multivariable Calculus | Khan Academy

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.