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# Constructing a unit normal vector to a curve

Figuring out a unit normal vector at any point along a curve defined by a position vector function. Created by Sal Khan.

## Want to join the conversation?

- Calculating the normal unit vector seems pretty straightforward. What is the purpose of finding it? Why kind of applications would you use this in?(15 votes)
- I realize it was two years ago but I'll answer the question anyways if someone is wondering. The purpose of a unit vector is to find the direction in which a vector is traveling in (its magnitude is one.) With this, you can manipulate it and other vectors to have them travel in same direction or different directions easier. A calculus IV concept I recently learned was with a gradient of a function, knowing that a unit vector in the same direction as the gradient would give the maximum possible change of the gradient.(8 votes)

- I am thinking of a proof of the unit normal vector being equaled to that expression using dot product. Actually, I have done it using mspaint lol.

Here is my proof:

http://postimg.org/image/nuj1gp3xb/

http://postimg.org/image/bjnislcmh/

I think it is fairly easy to do it and to understand it once you have learnt vector calculus. (Although It took me an hour of trial and error to figure out this proof, I tried using gradient to prove it but failed. ) May I ask if you want to or not to include this or any similar proofs in your video guides? I guess this will be helpful for the audience.(10 votes) - Does it really matter which component you make negative? There is a khan academy article on constructing unit normal vectors to curves in the section about vector line integrals. This article makes the opposite component (the i component) negative.(4 votes)
- Isn't a tangent vector dr/dt rather than just dr? Why specifially dr/dt cannot be a tangent vector to a curve?(3 votes)
- at ~0:58, I thought that the normal vector always points in the direction the curve is curving. So wouldnt the unit normal vector point in the opposite direction (towards the origin)?(3 votes)
- The normal vector is defined as any vector which is perpendicular to the curve. Hence the vector you're suggesting which points to the origin would also be described as a normal vector. In this case he is simply taking the outward pointing vector without having disambiguated as one would expect if we were to be strict. You can read more about this on MathWorld: http://mathworld.wolfram.com/NormalVector.html(1 vote)

- Why is -dxj going up when it should be going down since it is negative?(3 votes)
- Why only -dx
*j_ ? Why dy_i*doesn't get a minus sign, where normal vector = dy*i_ - dx _j*?(2 votes)- You actually get both, +dy î - dx ĵ and -dy î + dx ĵ are both normal vector to the curve.(2 votes)

- A couple things: Transforming dxi + dyj into dyi - dxj seems very much like taking a determinant. What's the relation? And two, couldn't you find a unit normal vector by finding the unit tangent vector, then making a vector perpendicular to it? i.e., using dot product to find perpendicular vector, or using a different vector and subtracting its projection onto the previous one?(2 votes)
- Yes you can do the transformation (rotation) using the rotational matrix. We can write dx î + dy ĵ as row vector, and cross it with the rotational matrix. 𝜃=-𝜋/2 if the curve is positively oriented (anti-clockwise), 𝜃=𝜋/2 if the curve is negatively oriented (clockwise).

So for positively oriented curve,

| dx dy | X | cos(-𝜋/2) -sin(-𝜋/2) | = | dx dy | X | 0 1 | = | -dy dx |*________*| sin(-𝜋/2) cos(-𝜋/2) |*__________*| -1 0 |

So we obtain, -dy î + dx ĵ

Similarly, for negatively oriented curve , we obtain dy î - dx ĵ

Check http://en.wikipedia.org/wiki/Rotation_matrix for more details

EDIT: I have used dot product to find perpendicular vector.

Here is my proof:

http://postimg.org/image/nuj1gp3xb/

http://postimg.org/image/bjnislcmh/(2 votes)

- Does anyone know what this symbol means?

ψ

I found it in an equation that ran like so: a ψ b = a^2 - 2b. Any help would be appreciated, as I've asked everyone I know and no one seems to have any idea. Thanks.(1 vote)- That sounds like a question that you see on the SAT every now and then. They just define a new operation (along the lines of +, *, ^, etc.) and tell you what it does when you perform the operation with two given numbers. In other words, this symbol used for this purpose is unique to this problem and is not something where you could ask a random math professor at Princeton or something and they would tell you, "Oh yeah, that means a^2 - 2b."

It is the Greek letter psi, though. It makes a sound of "ps," and is the first letter of, for example, the Greek word from which the English word "psychology" is derived.(4 votes)

- So this is what I interpreted at the end ... The vector integral along the curve is equal to the area integral of divergence of the vector for the area formed by the curve … it would seem like the divergence would account for more stuff leaving because it’s the whole area while the line integral accounts for only what is happening along the curve?(2 votes)

## Video transcript

- [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. 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.