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### Course: AP®︎/College Calculus BC > Unit 9

Lesson 4: Defining and differentiating vector-valued functions# Vector-valued functions differentiation

Visualizing the derivative of a position vector valued function. Created by Sal Khan.

## Want to join the conversation?

- So, would the magnitude of the tangent vector essentially be infinite, since as the h approaches zero, the magnitude gets larger? Or am I missing something?(30 votes)
- Remember, as h approaches zero then r(t+h) approaches r(t) such that r(t+h)-r(t) is an infinitesimally small tangent vector. when you divide a very small quantity with another comparable quantity you get a reasonably sized quantity. e.g. 0.0000000000024 / 0.0000000000006 = 4.

So, you won't get an infinitely large tangent vector.(106 votes)

- at12:03, he says "horizontal", doesn't he mean "vertical"?(18 votes)
- Yep, he misspoke. He does name it correctly elsewhere in the video, though.(26 votes)

- What's the difference between taking a gradient and the derivative of a position vector?(10 votes)
- There are several differences. First, the gradient is acting on a scalar field, whereas the derivative is acting on a single vector. Also, with the gradient, you are taking the partial derivative with respect to x, y, and z: the coordinates in the field, while with the position vector, you are taking the derivative with respect to a single parameter, normally t. Finally, the result of a gradient is a vector field while the result of a derivative of a position vector is just another single vector.(23 votes)

- Where is the next video (giving intuition on magnitude) that Sal is talking about?(20 votes)
- Anyone know when and where this normally covered in the academic track?(8 votes)
- Vectors are generally introduced as early as advanced high school mathematics but are not covered in this capacity until Calculus 2 (or equivalent course). They are heavily used in Calculus 3 (or equivalent) as well as Physics.(18 votes)

- Is anyone else concerned about Sal's functions failing the vertical line test?(3 votes)
- Both x and y are functions of a variable t, which isn't plotted. What's plotted is a curve the function makes as t varies in some interval.(27 votes)

- can't we just say that

dr/dt=d/dt(r)=d/dt(x(t)i+y(t)j)=d/dt(x(t)i)+d/dt(y(t)j)=id/dt(x)+jd/dt(y)=dx/dt i +dy/dt j(6 votes)- Yeah, I thought this whole video was pretty self-explanatory as well. But I guess it shows it more rigorously.(4 votes)

- what does dr/dt actually denotes in the graph mentioned in the video ,which in the case of the other usual graphs denotes the slope of the tangent at that point?(3 votes)
- This is a good question, but it's actually not possible to see dr/dt in the graph here, because we only see r(t) graphed in terms of x(t) and y(t) and not directly in terms of t. If we were instead to graph r(t) on the y-axis and t on the x-axis, we would then be able to visualize dr/dt as the slope of the tangent at a given point t. (However, we then couldn't easily visualize the relationship between x(t) and y(t), which this graph allows us to do.)

Conceptually, dr/dt doesn't just mean the slope; it means the instantaneous rate of change of r(t). So for instance, if r(t) is the position of something with respect to time (t), dr/dt would tell us its velocity at any given moment.(5 votes)

- I'm getting some loose connection to the directional derivative, I can't quite picture it clearly enough though (i.e. the connection between the slope of the unseen graph to their vector equivalents).(3 votes)
- Are we ignoring the vertical line test for a function or is there some reason this function is allowed to have two outputs for a particular input?(1 vote)
- The vertical line test fails here, as x and y are both dependent variables. The test only works if you graph a dependent variables against an independent one.

Additionally, if you manage to graph a y vs t or x vs t graph, they would obey the vertical line test(5 votes)

## Video transcript

In the last video, we hopefully
got ourselves a respectable understanding of how a
vector-valued function works, or even better, a position
vector-valued function, that is, in some ways, a
replacement for traditional parameterization to
describe a curve. And what I want to do in this
video is just get a little bit of gut sense of what it means
to take of a derivative of a vector-valued function. In this case, it'll be with
respect to our parameter t. So let me draw some
new stuff right here. So let's say I have the
vector-valued function r of t, and this is no different than
what I did in last video. x of t times unit vector i plus y of
t times the unit vector j. If we were doing it in 3
dimensions, we'd add a z of t times k, but let's keep things
relatively simple, and let's say that this describes a
curve, and let's say the curve we're dealing with, t is
between a and b, and this curve will look something like, let
me do my best effort to draw the curve, I'll just draw some
random curve here, so let's say the curve looks
something like that. This is when t is equal
to a, so it's going to go in this direction. This is when t is equal to b
right here, this is t is equal to a, so this right here would
be x of a, this right here is y of a, and similarly, this up
here, this is x of b, and this over here is y of b. Now, we saw in the last video
that the endpoints of these position vectors are what's
describing this curve. So r of a we saw in the
last video, it describes that point right there. I don't want to review
that too much. But what I want to do is
think about, what is the difference between 2 points? So let's say that we take
some random point here. Let's say, some random t here. Let's call that r of t. Well actually, I'm going to do
a different point, just because I want to make it a
little bit clearer. So let's say-- I'm going to
switch colors-- let's say that that right there
is r of some t. Some particular t, right there. That is r of t. It's going to be, you
know, a plus something. So that's some a particular t. And let's say that we want to
figure out, and let's say we increase t by a little bit. By h. So let's say that r of t plus
h, well, if we view the parameter t as time, we've
moved in forward in time by some amount, so our little
particle has moved a little bit. And let's say that
we're over here. So that is that right there,
in yellow, is r of t plus h. Just a slightly
larger value for h. Now, one question we might ask
ourselves, is how quickly is r changing with respect to t? So the first thing we might
want to say, well, what's the difference between these two? If I were to take, and I
want to visualize it. If I were to take r, the
position vector, that we get by evaluating r at t plus h, and
from that, I would subtract r of t. What do we get? Well, you might want to review
some of your vector algebra but we're essentially just
going to get this vector. Let me do it in a
nice, vibrant color. We're going to get this
vector right there, that I'm doing in magenta. So that magenta vector right
there is, let me do it, that magenta one right there,
is the vector r of t plus h minus r of t. And it should make sense,
because when you add vectors, you go heads to tails. You could alternatively write
this as r of t plus this character right here, plus r
of t plus h minus r of t. When you add two vectors,
you're adding, let me make it very clear, I'm adding
this vector to this vector right here. You put the tail of the
second vector at the head of the first. So this is the first vector,
and I put the tail of the second there, and then the sum
of those two, as we predicted, should be equal to
this last one. It should be equal
to r of t plus h. And we see that is the case,
and algebraically, you would see that obviously this
guy and that guy are going to cancel out. So hopefully that
satisfies you. And I want to be clear. This, all of a sudden, this
isn't a position vector. We're not saying that hey,
let's nail this guy's tail at the origin and use this guy to
describe a unique position. Now all of a sudden he's, it's
just kind of a pure vector. It's describing just
a change between two other position vectors. So this guy is right out here. But this vector literally
describes the change. But say we care, and how would
this look algebraically if we were to expand it like that? So this is going to be equal
to, what's r of t plus h? That's the same thing as x
of, let me do it over here. This is the same thing as x of
t plus h times the unit vector i plus y of t plus h times the
unit vector j, that's just that piece, that piece right there
is that piece, minus this piece, so minus, I'll do it in
the second line, I could have done it out here, but I'm
running out of space. Minus x of t, right r of t is
just x of t times i, plus, but I'll just distribute the minus
sign, so it's minus y of t times j. Actually let me write it,
this would be minus, let me write this way, plus this. So you realize that this
is really just this guy right here. I'm just evaluating at t. So you have x of t and y
of t, and then later we can distribute, right? If you distribute this minus
sign, you get a minus x of t and a minus y of t. And in vector addition, you
might need a little review on this if you haven't seen it in
a while, you know that you can just add the corresponding
components. You can add the x-components
and you can add the y-components. So this is going to be equal
to, let me rewrite it over here, because I think I'm going
to need some space later on. So let me rewrite it over here. So I have r of t plus h minus r
of t is equal to, and I'm just going to group the x- and the
y-components, this is equal to the x-components added
together, but this is a negative, so we're going to
subtract this guy from that guy. So x of t plus h minus x of t,
and then all of that times our unit vector in the x-direction,
and then we'll have plus y of t plus h minus y of t times a
unit vector the j-direction, I'm just rearranging things
right now, and this will tell us what is our change between
any 2 r's for given change in distance. And our change in distance
here is h between any 2 position vectors. Now, what I set out at the
beginning of this video, I said, well, I wanted to figure
out the change, and we're going to think about the
instantaneous change with respect to t. So I want to see, well,
how much did this change over a period of h? Instead of writing h we could
have written delta t, it would've been the same thing. So I want to divide this by h. So I want to say, look. My vectors changed this much,
but I want to say it's over a period of h. And this is analogous
to when we do slope. We say rise over run, over
delta y, or change in y, over change in x. This is kind of the change in
our function per change in x. Let's just divide everything,
or I shouldn't say change x, per change in t. So here, our change
in t is h, right? The difference between t plus h
and t is just going to be h. And so we're going to
divide everything by h. When you multiply a vector by
some scalar, or divide it by some scalar, or you're just the
taking each of its components and multiplying or dividing
by that scalar, and we get that right there. So this, for any finite
difference right here, h, this'll tell us how much
our vector changes per h. But if we want to find the
instantaneous change, right, just like what we did when we
first learned differential calculus, we said, ok. This is kind of
analogous to a slope. This would be good, this would
work out well for us, if the path under question looked
something like this. If it was a linear path. If our path looked
something like this. We could just calculate this,
and we'll essentially have the average change in our position
vectors, so you could imagine, 2 position vectors,
that's one of them. Well, actually, they'd
all be parallel. Well, the position vectors,
they don't have to be parallel. They could be like that. And then, this would just
describe the change between these 2 per h, or how quickly
are the position vectors changing per our change
in our parameter, right? This is, the h, you could also
consider, is kind of a delta t. Sometimes people find the
h simpler, or sometimes they find the delta t. But anyway, I'm concerned
with the instantaneous. We're dealing with curves,
we're dealing with calculus. This would have been OK
if we were just in an algebraic, linear world. So what do we do? Well maybe, we can just take
the limit as h approaches 0. Let me scroll this over. So let's just take the limit,
let me do this in a nice vibrant color, let's take, I'm
running out of colors, the limit as h approaches 0
of both sides of this. So here, too, I'm going to take
the limit as h approaches 0, and here, too, I'm going to
take the limit as h approaches 0. So I just want to say, well,
what happens, how much do I change per a change in my
parameter t, but what's kind of the instantaneous change, as
the difference gets smaller and smaller and smaller? This is exactly what we first
learned when we learned about instantaneous slope, or
instantaneous velocity, or slope of a tangent line. Well, this thing looks a little
bit undefined to me, right now. We haven't defined limits for
vector-valued functions, we haven't defined derivatives
for vector-valued functions. But lucky for us, all of
this stuff here looks pretty familiar. This is actually the
definition of our derivative. And these are scalar-valued
functions right here. They're multiplied by vectors,
in order for us to get vector-valued functions. But this right here, by
definition, this is the derivative, this
is x prime of t. Or this is dx dt. This right here is y
prime of t, or we could write that as dy dt. So all of a sudden we can
define, we can say, and I'm being a little hand-wavy here,
but I want to give you the intuition, more than anything. We can say that the derivative,
we can call this expression right here, as the derivative
of my vector-valued function r with respect to t, or we could
call it dr dt, notice I keep the vector signs there. This is its derivative, and all
it's going to be equal to, r prime of t, is going to be
equal to, well, this is just the derivative of x with
respect to t, is equal to x prime of t times the x-unit
vector, the horizontal unit vector, plus y prime of t,
times the y-unit vector, times j, the unit vector in the
horizontal direction. That's a pretty nice
and simple outcome. But the hard thing may be
to a kind of visualize what it represents. So if we think about what
happens, let me draw a big graph, just to get the
visualization in a healthy way. So let's say my curve looks
something like this. That's my curve. And let's say that this is,
we want to figure out the instantaneous change at
this point right here. So that is r of t. And then if we take r of t plus
h, we saw this already, you know, t plus h might be
something like right there. So this is r of t plus h. Right now, the difference
between these two, and this is just the numerator when you
take the difference, or how fast we're changing from this
vector to that vector in terms of t, and it's hard
to visualize here. And I'm going to do a whole
video so we can think about the magnitudes here. That might be some vector. Well, the difference
between these two is just going to be that. But then when you divide it by
h, it's going to be a larger vector, right, if we assume
that h is a small number. Let's say h is less than one. We're going to get a
larger vector, right? But this is kind of the average
change over this time. But as h gets smaller and
smaller and smaller, this r prime of t is going to, its
direction is going to be tangential to the curve. And I think you can
visualize that, right? As these two guys get closer
and closer and closer, the dr's get smaller, so the change, the
dr, the difference between the two, the delta r's, get smaller
and smaller, you can imagine if h was even smaller, if
it was right here. Then all of a sudden, the
difference between those two vectors is getting smaller. And it's getting more and more
tangential to the curve. But then we're also dividing by
a smaller h, so the actual derivative, as the limit of h
approaches 0, it might be you know, maybe it's even a
bigger number there. And actually, the magnitude
of this vector, it's a little hard to visualize. It's going to be dependent our
parameterization for the curve. it's not dependent on
the shape of the curve. The direction of this vector is
dependent on the shape of this curve, and the direction, so
the direction, this will be tangent to the curve. Or you could imagine that this
vector is on the tangent line to the curve. The magnitude of it is a little
bit hard to understand. I'll try to give you a
little bit of intuition on that in the next video. But this is what I want you to
understand right now, because we're going to be able to use
this in the future, when we do the line integral over
vector-valued functions.