Cookies   I display ads to cover the expenses. See the privacy policy for more information. You can keep or reject the ads.

# Constructing a unit normal vector to a curve | Multivariable Calculus | Khan Academy

- [Voiceover] So let's say that we've got the curve R
defined, so this is our curve R.
It's X of T times I
plus Y T times J, it's a curve
in two dimensions on the XY plane.
And let's graph it, just graph it in kind of a
a generalized form.
So that's our Y-axis.
This is our X-axis.
Our curve R might look something like this.
It might look something, let me draw a little
bit more of a.
Maybe it looks something like this.
Maybe that's just part of it.
And as T increases, we're going in that direction
right over there.
What I want to do in this video,
this is really more vector algebra than vector calculus.
Is think about at any given point here
whether we can figure out a normal vector.
In particular, a unit normal vector.
Obviously you can figure out a normal vector
you can just divide it by its magnitude
and you will get the unit normal vector.
So I want to figure out at any given point
a vector that's popping straight out in that direction.
And has a magnitude one.
So that would be our unit normal vector.
And to do that, first we'll think about
what a tangent vector is.
And from a tangent vector we can figure out
the normal vector.
And it really goes back to some of what you might
have done in algebra one, or algebra two
of if you have a slope of a line
the negative reciprocal of that slope is going to be
the slope of that negative line.
And we'll see a very similar thing
when we do it right over here with the vector
with this vector algebra.
So the first thing I want to think about is
how do we construct
how do we construct a tangent line.
Well, you could imagine at sum T
at sum T
this is what our position vector is going to look like.
So call that R one.
R one right over there.
And then if we wait, if we allow T to go up a little bit,
if T is time, we'll wait a little while
a few seconds, or however we were measuring things.
And then R two might look something like this.
This is when T has gotten a little bit larger.
We're further down the path.
And so one way that you can approximate
the slope of the tangent line
or the slope between these two points, for now,
is essentially the difference between these two vectors.
The difference between these two vectors is
you could view that
you could view that as
delta
delta R.
This vector plus that vector is equal to that vector.
Or, R two minus R one is going to give you
this delta R right over here.
And as R two, as that increment
between R one and R two gets smaller
and smaller and smaller.
As we have a smaller and smaller T increment
as we get a smaller and smaller T increment.
So we get a smaller and smaller T increment
the slope of that delta R is going to more
and more approximate the slope
of the tangent line.
All the way to the point that if you have
an infinitely small change in T.
So you have a DT.
So you go from R then you just
you change T a very small amount
that delta R, and we can kind of conceptualize that,
as DR, that does approximate
the A tangent vector.
So if you have a very small change in T
then your very small
del DR I'll call it
because now we're talking about a differential.
Your very small differential.
Right over here.
That is a tangent
that is a tangent vector.
So DR
DR is
a tangent
tangent vector at any
at any given point.
And once again, all of this is a little bit of review.
But DR, we can write as
DR is equal to DX
times I plus the
infinite small change in X
times the I unit vector
plus the infinite small change in Y
times the J unit vector.
And you see that, you see that
if I were to draw
if I were to draw a curve.
Let me just draw another one.
Actually, I don't even have to draw the axis.
If our DR looks like that
if that is our DR
then, we can break that down
into its vertical and horizontal components.
This right over here is DY.
And that right over there
that right over there is
that is DX.
And so we see that DX times I.
Actually, this is DX times I.
And this is DY
this is DY times J.
DY is the magnitude,
J gives us the direction.
DX is the magnitude.
I tells us that we're moving
in the horizontal direction.
Over here, this actually would be a negative.
This must be a negative value right over here
and this must be a positive value
based on the way that I drew it.
So that gives us a tangent vector.
And now we want to from that tangent vector
figure out a normal vector.
A vector that is essentially perpendicular
to this vector right over here.
And there's actually going to be two
vectors like that.
There's going to be the vector
that kind of is perpendicular in the right direction
because we care about direction.
Or the vector that's perpendicular in the left direction.
And we can pick either one.
But for this video, I'm gonna focus on the one
that goes in the right direction.
We're gonna see that that's gonna be useful
in the next video when we start doing a little bit of
vector calculus.
And so let's think about what that might be.
And what I'll do to make it a little bit clearer.
Let me draw a DR again.
I'll draw a DR like this.
This is our DR.
This is DR.
And then this, right over here.
This right over there, we already said this is DY
times I.
And then this, sorry,
this is DY times J.
We're going in the vertical direction.
DY times J.
And then in a different color
this right now if I already used that color.
I haven't used, oh,
I haven't used or had blue yet.
So this right over here is DX
DX times I.
So we know from our algebra courses
you take the negative reciprocal
so there's gonna be something about swapping
these two things around, and then taking the
negative one.
But to figure out, we want the one that goes to the right.
So which one should we use?
So let's think about it a little bit.
If we, if we take DY
times I.
So we take this length
in the I direction,
we're gonna get
we're gonna get this.
We're gonna get that, so this is
DY
times I.
And then if we were to
if we were to take D, if we were to just take
DX
times
J, that would take us down.
'Cause DX it must be negative here
since it's pointed to the left.
So we have to swap the sign of DX to go upwards.
So we swap the sign of DX
to go upwards.
'Cause I was here it was a negative sign.
It went leftwards, we want it to go upwards.
So this is gonna be negative
negative DX times J.
We're now moving in the vertical direction.
And that, at least visually, this isn't kind of a
rigorous proof that I'm giving you.
But this is hopefully good a good visual representation
that that does
that that does get you.
I should have drawn it a little bit.
That does get you pretty
that gets you pretty close, just visually inspecting it
to what looks like the perpendicular line.
It's consistent with what you learned in algebra class,
as well.
That we're taking the negative reciprocal,
we're swapping the X's and the Y's.
Or the change in X and the change in Y.
And we're taking the negative of one of them.
And so we have our normal line
just like that.
Our normal vector.
So a normal vector is going to be
DYI
minus DXJ.
But then if we want a normalize it
we want to divide by
by that magnitude.
So a normal, let me write it this way.
A normal vector.
So let me call this
I'll just call it A.
A normal vector
is going to be DY
times I.
Is going to be
DY
times I.
Minus DX times J.
I'll do that same blue color.
Minus DX
times J.
Now, if we want this to be a unit normal vector
we have to divide it by the magnitude of A.
But what is the magnitude of A?
The magnitude of A
is going to be equal to
it's going to be equal to the square root
of, and I'll just start with the DX squared.
So it's the negative DX squared.
Which is just going to be DX squared.
The same thing as positive DX squared.
It's going to be DX squared
plus DY squared.
Plus DY
squared.
I could have put the negative right in here
but then when you square it, that negative would disappear.
But this thing right over here
and we saw this when we first started exploring arc length.
This thing right over here is the exact same thing
as DS,
and I know there's no DS
that we've shown right over here.
But we've seen it multiple times.
When you're thinking of about if you
if you think about the length of DR as DS
that's exactly what this thing over here is.
So this can also be written as
DS.
So the infinite hasn't really changed
in the arc length
but it's a scaler quantity.
You're not concerned
you're just concerned with the absolute distance.
You're not concerned so much with the direction.
Another way to be do it is
it's the magnitude
it's the magnitude right over here of DR.
So now we have everything we need to construct
our unit normal vector.
Our unit normal vector at any point.
And I'll now write N
and I'll put a hat on top of it.
Say that this is a unit normal vector.
We'll have magnitude one is going to be
equal to A divided by this.
Or we could even write it this way.
So we could write it as
there are multiple ways we can write it.
We can write it as
I'll write it in this color.
As DY
times I
minus DX
times J.
And then all of that times
or maybe not times, divided by,
DS.
Divided by the magnitude of this.
So divided by
divided by
DS.
And obviously I can distribute it on each of these
by on each of these terms.
But this right here, we've been able to construct
a unit normal vector at any point
on this curve.