If is a curve, the * osculating plane* is the plane determined by the
velocity and acceleration vectors at a point.

Suppose the point on the curve is . Then a point lies in the osculating plane exactly when the following vectors determine a parallelepiped of volume 0:

That is,

* Example.* Find the osculating plane to

The point is . Compute the velocity and acceleration:

Plug in :

The osculating plane is

The velocity vector is tangent to
the curve . If I divide
the velocity vector by its length, I get a unit vector tangent to the
curve. Thus, the * unit tangent vector* is

I want to find a way of measuring how much a curve is curved. A reasonable way to do this is to measure the rate at which the unit tangent vector changes.

When a curve is gently curved as in the top figure, the unit tangent only changes a little as you move along the curve. When a curve is sharply curved as in the bottom figure, the unit tangent changes a lot as you move along the curve.

One other thing: The "curviness" of a road should not
depend on the speed of the cars which traverse it. Curviness is
*intrinsic* to the road. The curve parameter t has to do with
the rate that a hypothetical particle traces out the curve. The arc
length s depends only on the path of the particle, i.e. the curve
itself. Therefore, I should consider the rate of change of the unit
tangent *with respect to arc length*.

Let s denote arc length measured along the curve from some fixed point on the curve. That is,

This gives s as a function of the parameter t. For any reasonable
curve, arc length measured from increases as you move along the curve. Since an
increasing function is injective (one-to-one), it is invertible.
Therefore, I can solve for t in terms of s. Then I can plug the
result into the curve equations to express in terms of s. The result
is called the * arc length parametrization of the
curve*.

Here are some properties of the arc length parametrization:

* Proposition.* (a) .

(b) .

* Proof.* (a) Differentiate both sides of this
equation with respect to t:

By the Fundamental Theorem of Calculus,

This property says that the rate of change of arc length is equal to the speed of the curve.

(b)

* Example.* Find the arc length parametrization
of .

I'll measure arc length along the curve from . I want to use t as the parameter, so I'll switch to u as the dummy integration variable. Thus, .

Now , so

Solve for t in terms of s:

Therefore, the arc length parametrization of the curve is

* Definition.* The * curvature
vector* is . It measures how
much a curve is curved by finding the rate of change of the unit
tangent with respect to arc length.

The * curvature* is the length of the curvature
vector:

* Remark.* Some people define curvature in a way
that allows it to be positive or negative. Since I've defined
curvature as the length of a vector, my definition requires that it
be positive.

* Proposition.* (a) is perpendicular to .

(b) lies in the plane determined by the velocity and acceleration vectors --- i.e. the osculating plane.

* Proof.* (a) Differentiate with respect to s:

Therefore, is perpendicular to .

(b) By the Chain Rule and the Product Rule,

The last expression is a linear combination of the velocity and acceleration vectors. Therefore, lies in the plane determined by the velocity and acceleration vectors.

* Definition.* The * unit
normal* is given by

Thus, the unit vector is a unit vector perpendicular to the unit
tangent . Moreover, the curvature vector
has *length* equal to the curvature and *direction*
given by the unit normal:

Next, I want to obtain some formulas for the curvature. I'll need a couple of lemmas.

* Lemma.*

* Proof.* Since ,

Differentiate with respect to t:

* Lemma.*

* Proof.*

In the next result, primes will denote differentiation with respect to t (not s). Thus, .

* Theorem.*

* Proof.*

Now is the length of , so by multiplying out the dot products,

For the next to the last equality, I used the vector identity from the preceding lemma. Finally, take the square root of both sides:

* Example.* Find the curvature of the helix

Hence,

Then

The curvature is

Next, I'll obtain formulas for the curvature of a curve in the x-y-plane.

* Theorem.* (a) If is a curve in the
x-y-plane, then

(b) If is a curve in the x-y-plane, then

* Proof.* (a) Regard as a curve in 3 dimensions by writing . Then

So

Since , I have

(b) I may parametrize by setting , . Then

Plugging these into the formula from the first part of the theorem, I get

* Example.* Find the curvature of

I have

Hence,

Also,

The curvature is

* Example.* Find the curvature of at the point .

Thus, and .

The curvature is

In computing the unit normal, a reasonable strategy is to compute the unit tangent . The unit normal has the same direction as , but

The stuff on the bottom is a number; therefore, the unit normal has the same direction as . So I can compute and then divide it by its length; since it has the right direction and right length, it must be the unit normal.

* Example.* Find the unit tangent and unit normal
to the helix .

Hence,

Next,

Therefore,

For curves in the plane, it's often easy to see what the unit normal is by inspection. The unit normal points in the direction in which the curve is curving:

Once you know a tangent vector , there are two obvious vectors which are perpendicular to :

Just pick the one that points in the direction in which the curve is curving, divide by its length, and you have the unit normal.

* Example.* Find the unit tangent and unit normal
to the curve .

Hence,

Remember that the unit normal must be perpendicular to the unit tangent! Ignore the factor of which is a number and doesn't affect the tangent's direction --- the direction of is given by . There are two obvious vectors in the plane perpendicular to it, namely

Here's a picture of the curve:

You can see that the normal vector should always point to the left, so its x-coordinate must be negative. Therefore, points in the direction of the normal. Now

The * binormal* is the cross product of the unit tangent
and unit normal:

Since the cross product of unit vectors is a unit vector, the binormal is a unit vector which is perpendicular to the unit tangent and unit normal.

* Example.* Find the binormal vector for the
helix .

I computed the unit tangent and unit normal in an earlier example:

Therefore,

The * radius of curvature* is defined to be . The * osculating
circle* at a point on a curve is the circle of radius R, tangent
to the curve at the point, whose center lies R units along the unit
normal from the point. The circle lies in
the osculating plane. The parametric equations for the osculating
circle at are

* Example.* Find the osculating circle to the
helix

First, .

The unit tangent, unit normal, and curvature are

The osculating circle is

Simplifying and solving for x, y, and z, I get

* Example.* Find the osculating circle to at .

Parametrizing the curve by , , I have

Thus, , so

Next,

The center is units from in the direction given by :

Therefore, the circle is

Copyright 2018 by Bruce Ikenaga