In this set of supplemental notes, we are going to define a frame of mutually orthogonal unit vectors that always travel with a body moving along a curve in space.

The frame has three vectors. (See figure 1) The first one is the unit tangent vector T, the second is N which is the unit vector that gives the directions of dT/ds, and the third is the Binormal vector, B = T ´ N.

The magnitude of dT/ds tells us how much a vehicle's path turns to the left or right as it moves along.

figure 1

The number (dB/ds) · N tells us how much a vehicle's path rotates or twists out of its plane of motion as the vehicle moves along the curve. This is called the torsion of the vehicle's path.

figure 2

Suppose a particle is moving along a smooth curve in the xy-plane. As it moves, the related unit tangent vector, T, turns as the curve bends. Remember that T is a unit vector, therefore its length remains constant. So the only trait that is going to change is the direction of T as the particle moves. We are interested in the rate at which T is turning (per unit length). This rate is called the curvature and is denoted by k . (See figure 2)

FACT:

If T is the unit tangent vector of a smooth curve, the curvature function of the curve is

If the magnitude of dT/ds is large, then T is turning sharply at P, and the curvature of P is large.

If the magnitude of dT/ds is close to zero, then T is turning slowly and the curvature at P is small.

If the particle's path is a straight line, then the direction of T remains the same and the curvature is zero.

By definition, T has constant length. Therefore, the vector dT/ds is orthogonal to T. If we divide this new vector, dT/ds, by the length k , we will get a new unit vector that is orthogonal to T.

FACT:

At a point where k ¹ 0, the principal unit normal vector for a curve in the plane is

Remember that the vector dT/ds always points the direction in which T turns as the curve bends. So when we are looking at the curve in the direction of increasing arc length, dT/ds will point toward the right if T is turning clockwise, and toward the left if T is turning counterclockwise. Therefore, N will always point toward the concave side of the curve. (See figure 3)

How do we determine N given r (t)? Well, let us approach it this way. Given that the

figure 3

arc length parameter for a smooth curve r (t) = f (t)i + g (t)j is defined with ds/dt positive, ds/dt = | ds/dt |, we can use the chain rule to find N.

EXAMPLE 1:

Find T and N for r (t) = (3 - 2t) i + (6 - t ² ) j.

SOLUTION:

First we will find the unit tangent vector T.

v = -2i -2t j

Now to find N.

FACT:

The circle of curvature at a point P on a plane curve where k ¹ 0 is the circle in the plane of the curve that (1) is tangent to the curve at P. (2) It has the same curvature as the curve has at the point P, and (3) lies toward the inner side of the curve. (See figure 4)

figure 4

FACT:

The radius of curvature of a given curve at a point P is the radius of the circle of curvature, which is defined to be 1/ k and is denoted by r.

Notice that we have been talking about what the curvature tells us about a vector function, but we have not told you how to find it. Recall that curvature is defined to by the following.

We can find dT/ds by using the chain rule.

Now from supplemental notes 10, we defined dt/ds = 1/ | v |, so dT/ds now becomes the following.

Therefore, the formula for curvature is the following.

EXAMPLE 2:

Find T, N and k for r (t) = (2t + 3) i + (5 - t ² ) j.

SOLUTION:

First we will find the unit tangent vector T.

v = 2i -2t j

Now to find N.

The formulas for T, N and k for curves in the plane carry over to curves in space.

EXAMPLE 3:

Find T, N and k for r (t) = (e ^t cos t) i + (e ^t sin t) j + 2k.

SOLUTION:

v = (e ^t cos t - e ^t sin t) i + (e ^t sin t + e ^t cos t) j

FACT:

The binormal vector of a curve in space is, B = T ´ N, is a unit vector orthogonal to both T and N.

figure 5

The whole TNB frame (See figure 5) defines a moving right-handed vector frame. This concept is used extensively in calculating the flight paths of space vehicles.

Suppose I want to find the derivative dB/ds. How does dB/ds behave in relation to T, N, and B? First of all, let us define what dB/ds is. Using the concept of the cross product rule from supplemental notes 8, here is the derivative with respect to s.

(Since N is in the direction of dT/ds, then (dT/ds) ´ N = 0.)

dB/ds is orthogonal to T since a cross product is orthogonal to its factors. dB/ds is orthogonal to B implies dB/ds is orthogonal to the plane that contains B and T. Therefore dB/ds is parallel to N, so dB/ds is a scalar multiple of N.

Let

where t is the torsion along the curve. So,

Therefore, t = - (dB/ds) · N.

FACT:

Torsion may be positive, negative, or zero.

figure 6

B is perpendicular to the osculating plane.

N is perpendicular to the rectifying plane.

T is perpendicular to the normal plane. (See figure 6)

EXAMPLE 4:

Find r, T, N, B for r (t) = (cos t) i + (sin t) j + t k at t = 0. Then find the equations for the osculating, normal, and rectifying planes at that time value.

SOLUTION:

r (0) = (cos 0) i + (sin 0) j + 0 k = i

v = (-sin t) i + (cos t) j + k

Now to find the point on the curve. Use the components of r to find this point when t = 0. P (1, 0, 0). Let Q (x, y, z) be another point in the plane, so the vector PQ = (x - 1)i + y j + z k.

Remember that the osculating plane is perpendicular to the vector B, so dot this vector with the vector PQ.

For the normal plane, we will us the vector T. To find the plane, dot T with PQ.

Finally, the rectifying plane. We will dot the vector N to the vector PQ.

When gravity, brakes, etc. accelerate a body, we usually want to know how much of the acceleration acts to move the bodies straight-ahead in the direction of motion (the tangential direction T). Here are the formulas that we can use to rewrite the acceleration in terms of its tangential and normal components.

EXAMPLE 5:

Write a in the form a = a _TT + a _NN at the given value of t without finding T or N.

SOLUTION:

CURVATURE:

NOTATION:

TORSION:

If v ´ a ¹ 0, then

EXAMPLE 6:

Find T, N, B, k , and t for r (t) = (cos³ t) i + (sin³ t) j, 0 < t < p /2.

SOLUTION:

v = (3cos² t)(-sin t)i + (3sin² t)(cos t)j = (-3cos² t sin t)i + (3sin² t cos t)j

a = ((-6cos t)(-sin t)(sin t) - 3cos² t (cos t))i + (6sin t cost cos t - 3sin² t sin t)j = (6cos t sin² t - 3cos³ t)i + (6sin t cos² t - 3sin³ t)j