- [Voiceover] So, let's sum up where we are so far.

We're looking at this formula

and trying to understand why it corresponds to curvature.

Why it tells you how much a curve actually curves.

And the first thing we did is we noticed that this numerator

corresponds to a certain cross-product.

The cross-product between the first derivative

and the second derivative of the function

parameterizing the curve.

And the way we started to understand that

is we say, well, the function parameterizing the curve

s of t produces vectors whose tips trace out

that curve itself.

And now if you think about how one tip moves to the next

the direction that it needs to go

for that tip to move to the next one,

that's what the first derivative tells you.

And when you treat that in an infinitesimal way

this is why you get tangent vectors along the curve

and there's an entire series of videos on that

for the derivative of a position vector valued function,

and they explain why the first derivative

of a parametric function,

gives you tangent vectors to that curve,

but if we draw all of those tangent vectors

just in their own space,

their own little s prime of t space,

you get all of these vectors,

and we're rooting them at the same point

to be able to relate them more easily.

The way that you move from the tip of one of those

to the next one, is given by the second derivative,

it kind of plays the same roll for the first derivative

as the first derivative plays

for the original parametric function.

And specifically if you have a circumstance

where the tangent vectors are just turning,

the only thing they're doing is purely turning around,

which is when your curve is actually curving,

this corresponds to a case when the

second derivative function is pretty much perpendicular

as vector.

The vector it produces is perpendicular

to the first derivative vector.

So, this is loosely why the cross-product

is kind of a good measurement of curving

cause it tells you how perpendicular these guys are.

But, there is a bit of a catch.

The original formula for curvature,

the whole reason we're doing it with respect to arc length

and not with respect to the parameter t,

is that curvature doesn't really care about

how you parameterize the function.

If you imagine zipping along it really quickly,

so your first derivative vectors are all super long,

it shouldn't matter,

compared to crawling along it like a turtle.

The curvature should just always be the same.

But, this is a problem if you think back to

the cross-product that we're now looking at,

where you're taking the cross-product between

the first derivative and the second derivative,

because if you were traveling along this curve

twice as quickly, what that would mean is

your first derivative vector,

so I'll kind of draw it again over here,

would be twice as long,

to indicate that your going twice as fast.

And similarly your second derivative vector,

to kind of keep up with that changing rate,

would also be twice as long.

And as a result, the parallelogram that they trace out,

I should actually, kinda going off screen here.

The parallelogram that they trace out

would be actually four times as big.

Right?

Because both of the vectors get scaled up.

So, the way that we really want to be thinking

about this is not the tangent vector due to the derivative,

but normalizing this.

And this should kind of make sense,

because we are thinking in terms of unit tangent vectors

for the curvature as a whole.

So, if you imagine instead kind of cutting of the vector

to make sure that it's got a unit length,

a length of one.

And what that means is,

you're taking the derivative vector

and dividing it by it's own magnitude,

by the magnitude of that derivative vector,

and then similarly we'll want to scale everything else down,

so you're taking this and kinda scaling it down

by what you need.

The resulting parallelogram they trace out

is a more pure measurement of how perpendicular they are,

without caring how long they are.

And with this guy would be, by the way then,

this is the second derivative vector

not normalized with respect to itself,

but we're still dividing,

you know the thing we're diving by

as we scale everything down,

is still just the size of that first derivative vector.

So, this cross-product, if we take the cross-product

between s prime normalized, s double prime,

oh no, no, no, sorry,

s prime itself,

and I should be saying vectors for all of these,

these are all vectors.

If we take the cross-product between that

and s double prime,

scaled down by that same value that's still s prime,

so it's not normalized, this is just scaled down by s prime.

This here is a more pure measurement of how perpendicular

the second derivative vector is to first.

And the reason that we don't really care about

the second derivative being normalized is,

if it was the case,

that you know,

the second derivative was really, really strong

and wasn't necessarily a unit vector,

that's fine that's just telling us that

the tangent vector turns much more quickly

and the curvature should be higher.

And in fact, it turns out that this whole expression is

the derivative of the unit tangent vector t,

that unit tangent vector that I've talked about a lot,

with respect to the parameter t.

So, whatever the parameter of your original function is.

And now, if you think back to the,

I'm not sure if it was the last video

or the one before that,

but I talked about how,

when you take,

when we're looking for this derivative of

the tangent vector, with respect to arc length,

the way that you compute this is to

first take its derivative with respect to the parameter,

which is something we can actually do,

cause everything is expressed in terms of that parameter,

and then dividing it by the,

basically, the change in arc length

with respect to that parameter,

which is the size of that first derivative function.

So, if this whole thing is the derivative

of the tangent vector with respect to t,

what that means is when we take this,

and we divide that whole thing by,

the derivative, by s prime,

that should give us curvature,

and in fact that's just worth writing in its own here.

That's curvature.

Curvature is equal to,

and what I'm going to do is, I'm going to take,

since we three different times,

we see the magnitude of s prime here,

magnitude of s prime here,

magnitude of s prime here.

Since we see that three times,

I'm going to go over here,

and I'm going to put that on the denominator

but cubing it,

so s prime, that derivative vector,

the magnitude of that derivative vector cubed.

And then on the top, we still have s prime,

that vector s prime cross-product with s double prime,

that vector.

And this, I mean, you could think of this as yet another

formula for curvature,

I think I've given you like four at this point.

Or, you could think of it as just kind of the same thing,

and if we look back up at our original one,

that I was trying to justify.

This is just the spelled out version of it,

because what is this bottom component here?

If we take x prime squared plus y prime squared,

and if we think of the square root of that,

so kinda taking it to the one half power,

that would be the magnitude of the derivative,

and I kind of showed that in some of the previous videos.

And what we're doing is we're cubing that,

so that whole formula that's the very explicit,

you know, in terms of x and y's,

what's going on is really just expressing this idea,

you take the cross-product between the first derivative

and the second derivative,

and then because you're normalizing it,

you know normalizing with respect to the first derivative,

you want to scale down the second derivative

by that same amount,

just so the parallelogram we're thinking,

kinda shrinks and everything stays in proportion.

And then once again we're dividing about that s prime,

basically because curvature is supposed to be

with respect to s, and not with respect to t.

So, that's a way of, kinda getting a correction factor

for how wrong you're going to be if you just think

in terms of the parameter t instead of steps

in terms of the arc length.

So, hopefully this makes the original formula

a little bit less random, seeming,

you know, in this two dimensional case.

And it also gives another strong conceptual tool

for understanding yet another way that you can think about

how much a curve itself actually curves.

And I think I've probably done enough videos here going

through all the different formulas,

for what curvatures should be.

And in the next one or two I'll go through some

specific examples, just to see what it looks like

to compute that.