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## Multivariable calculus

### Course: Multivariable calculus > Unit 2

Lesson 6: Curvature# Curvature formula, part 4

After the last video made reference to an explicit curvature formula, here you can start to get an intuition for why that seemingly unrelated formula describes curvature. Created by Grant Sanderson.

## Want to join the conversation?

- Isn't cross product only for 3D vectors?(25 votes)
- Yes, you're right, it's a 3D only club. Think of it this way; if you add the third component to the vectors (naturally being zeros) and then take there cross product, you'll see that the first two terms cancel out, and you're left with a z component that's identical to what Grant showed.

If it's confusing why the answer is now a vector in the third dimension, you can watch Sal' video on how to interpret this in terms of physics : https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/v/cross-product-and-torque

or if this doesn't make sense, then watch the video on cross products again and you'll see that by definition, the cross product produces a vector in 3d(12 votes)

- At7:57Grant says the cross product is interpreted as the area of the parallelogram formed by the two vectors. I don't see how this corresponds to the area of the parallelogram without taking the magnitude of the cross product, because crossing the two vectors would result in another vector.(12 votes)
- You are right. The geometric meaning of the cross product is that the magnitude of it is equal to the area of a parallelogram formed by two vectors. I think Grant meant "the magnitude" of the cross product here.(17 votes)

- Why is curvature defined as the vector change between the unit vectors rather than the change in the angle between them? If it was the angle between them, then with a circle going 1/4 around the circumference would be a change of pi/2 over the arc length, (pi/2)*r, therefor satisfying our definition of curvature as 1/r.(5 votes)
- why the cross product of those two vectors represent the curvature of a short length of a curve(3 votes)
- Curvature is always positive, right? Shouldn't we take an absolute value of a numerator?(2 votes)
- I believe it will end up being positive no matter what since we take the magnitude of the top divided by the magnitude of the bottom. My understanding is that magnitude is always a positive number (unless perhaps it was imaginary).(2 votes)

- For anyone who is confused about cross product at around7:57, this articles might be helpful :

cross products https://www.khanacademy.org/math/multivariable-calculus/thinking-about-multivariable-function/x786f2022:vectors-and-matrices/a/determinants-mvc

determinants : https://www.khanacademy.org/math/multivariable-calculus/thinking-about-multivariable-function/x786f2022:vectors-and-matrices/a/determinants-mvc(2 votes) - What I think about the cross product is the sin of the angle between the two vectors (head to tail) times the length of each vector. I think that will help more in visualization.(2 votes)
- I think this is pretty confusing. The cross product is only defined for two vectors in three dimensions. When having this kind of operation extended to other dimensions, it should technically be called an exterior product. But more important: is the cross product of two vectors not supposed to be just another vector?(2 votes)
- Actually the cross product is defined for 2 dimensions as well. When taking the cross product of 2 2-dimensional vectors, you just take the discriminant of the vectors in matrix form, as shown below:

[a, b] x [c, d] = ad - bc

Also, yes, the cross product of two vectors is just another vector, but Grant takes the magnitude of the vector in the video, so it becomes a number.(1 vote)

- The cross product of those two vectors is maximal when the change between two first derivative vectors is 45 degrees because the sin of 90 degrees is equal to 1. But what if that change was more than 45 degrees? Then the first derivative would be larger and the curvature should increase. But in this case the cross product of the first derivative times the second derivative will be smaller because the angle between them is less than 90 degrees, hence the curvature would decrease. Please tell me what I understand wrong, thanks.(2 votes)
- At1:48, Grant says absolute value. Should it be magnitude?(2 votes)

## Video transcript

- [Voiceover] So we've been
talking about curvature. And this means you've got
some sort of parametric curve, you might think as parameterized by a vector valued function S of t. And curvature is supposed to measure just how much this curve actually curves. So here it's curving quite a
bit, curvature should be high. But here it's, you know. relatively straight so curvature should be low. And I brought up in the last video this pretty complicated formula. And to remind you of the circumstances, I was saying that curvature,
denoted with this little kappa, is typically calculated as the derivative of the unit
tangent vector function. So we'll think of some kind of function that gives unit tangent
vectors at every single point. So you know, just a vector with a length of one lying
tangent to the curve. And that's going to be some kind of function of the same parameter. So big T for unit tangent
vector, little t for the parameter, hopefully
that's not too confusing. Then we're taking the derivative
of this, not with respect to t, the parameter, but with
respect to arc length, s. And by arc length, I mean
if you take a tiny step along the curve, and consider
this to be of size ds, so s typically denotes
length along a curve, this will be a tiny change in length. You're wondering how much that
unit tangent vector changes. And specifically what I mean by that, if you kind of imagine
tangent vectors off sitting in their own space somewhere,
where each one of them has unit lengths so they're all just
kind of of the same length and lets say they're all
stemming from the same points. So rather than drawing them
stemming from the curve, to show that they're tangent, I just want to show them in their own space. The derivative, the change in
that tangent vector, would be some, some other vector that
tells you how you move from one to the other, kind of tells
you how much it's turning. So the curvature itself is
given not by that vector, because this would be a vector quantity, but by the absolute value of it. And I said in the last
video that it turns out that this quantity, when you
work it out for a vector-value function with components, with componenets x of t, x of t and y of t, that it happens to equal
this complicated formula. And what I'm gonna do here,
I going to break it down and show why this formula
isn't actually all that random. It's actually kind of
sensible as a description for how much the curve
really curves at a point. So lets start by looking
at the numerator here. This x-prime of t, first
derivative of the first component, times the second derivative of
the second component, minus, and then kind of symmetrically, y-prime times x double-prime. You might be able to recognize this as a certain cross product. So the cross product, if
we take x-prime, y-prime, and these are still
functions of t, as a vector crossed with the vector
containing the double primes. And if you aren't familiar
with the cross products or you're feeling a little shaky and you kind of want to review, now would be a good time pause the video, take a look at the cross product videos, and remind yourself both of how to compute it and what the
underlying intuition is. Because the way that we compute
a cross product like this is you take the components in the
diagonal here, you know that rightward diagonal, and
multiply them together. That's where you get your
x-prime, y double-prime. And then you subtract off
the components in the other diagonal, you know it kind
of feels like a determinant. X double-prime, times y-prime. But the way that you
interpret this vector, well I'll get to the
interpretation in just a moment. First, first let me kind of write out what this is in terms of our function S. This first vector is just
the first derivative of S, so that's S-prime of t,
the first derivative of S. And we're crossing that with this one, which is the second derivative,
vector value derivative, but the second derivative of S. So before we, before we
do anything else lets just start to think about, what do
both of these vectors mean? How do you interpret the
S-prime and the S double-prime? I'm going to go ahead and give ourselves a little bit of room here. So I'll draw the curve again. We're on our x, y plane, and
you have some kind of curve. The function S itself is giving vectors whose tips trace out this curve, right? As T changes, the tips of these
vectors trace out the curve. And now the first derivative,
that first derivative vector, S-prime of t, is telling you how that tip should move
to go along the curve. As you go from one S
vector to another S vector, what direction should that tip move. And what this means is
that at every given point, when you kind of do this
in a limiting fashion and you only look at infinitesimal
changes in the original vector, you always get some
kind of tangent vector, right? So all of these are tangent
vectors, not necessarily unit, you might have like a very
long tangent vector to indicate that you're traveling very
quickly across that space. And now, how would you think about the second derivative vector
here, S double-prime of t? Well the way to do
that, I like to think of all of the tangents
vectors then just kind of living in their own space, right? If this here is representing
S of t, I just want to look in isolation at what
S-prime of t looks like. So each one of those
vectors, you know maybe this first one is just is
very long, gargantuan, indicating you're going very quickly. And then after that you've
got something else here, maybe it's pointing a little bit down. And you're kind of thinking about how all of these derivative vectors change, but I want them all
rooted at the same point. Just see what happens when
they're all rooted at the origin, because that, that gives
us a way to think about what changing the vector should look like. So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction, right? So this second derivative
value is going to tell us how the tip of the
first derivative should move. And then similarly over here it tells us how the tip of that should move. And just as an example and
kind of a hint as to why this has something to do with curvature, if you have a curve
that turns very sharply, turns very sharply, you would
have you know a tangent vector that starts by pointing up
and to the right and then very quickly is pointing down,
down and to the right. So if they were off, you know
we'd kind of draw them on their own, rooted in their own location, you can maybe see how the
second derivative vector is telling it to turn
somehow in that direction. And if you, you know
capture not just those two but infinitesimally what's
going on as you move from one to the next you're gonna get this kind of turning motion
for all of these guys. So the second derivative
vector would be pulling perpendicular, perpendicular to that first derivative vector as
a way of telling it to turn. So just to draw that on
its own, if you have a first derivative vector, and
then the second derivative vector is perpendicular to it, it's telling it how it
should turn in some way. But if it was, you know
lets say it was not purely perpendicular but it was
also pointing against, it would be telling that, that vector, that first derivative vector,
to shrink in some way. So that would be an indication that not only is it turning
but it's getting smaller, meaning the trajectory based
on S is probably slowing down. And if it was kind of turning
it but also pointing away, that would mean it's telling the first derivative vector to grow. So not only should you
be turning on your curve but you should also be speeding up. But the case we care about
most is we're just trying to measure how perpendicular it is, right? And this is where the
cross product comes in, because if you think
about how you interpret that cross product, how you
interpret the cross product, it's basically the area
if you kind of take these two vectors, tip-to-tail as
they are, and think about the parallelogram that they trace out. Lets see, so this blue dotted line should be parallel to the blue vector. The area of this, of this
traced out parallelogram for each of them, that's what tells you that's how you interpret
the cross product. The cross product between
S-prime and S double-prime. So this cross product, by giving
you that area, is kind of a measure of just how perpendicular
these vectors are, right? Because if one of them,
if they point very much in the same direction, right,
and they're only slightly perpendicular, that means that
the parallelogram they trace, they trace out is gonna
have a very small area. It's gonna be a smaller
area in comparison. So with that, I don't want
this video to run too long, so I'm gonna call it in
here, but then I'll just continue on in the next
one through the same line of reasoning to build
to our original formula.