8/10/2019 Ch13S3

1/5

Instructor: Longfei Li Math 243 Lecture Notes

13.3 Arc Length and Curvature

Arc LengthWe learned the length of a curve in 2D is the limit of lengths of inscribed polygons, and we obtainedthe formula:

L=

ba

[f(t)]2 + [g(t)]2dt

if the curve has the parametric equations x= f(t) and y = g(t) and both f and g are continuous.

Similarly, the length for a curve in space with a vector equation r(t) =< f(t), g(t), h(t) >, a t bis given by the formula

L=

ba

[f(t)]2 + [g(t)]2 + [h(t)]2dt

provided f, g, h are all continuous.

We can write the formula into a more compact form:

L=

ba|r(t)|dt

Example: Find the length of the arc of the circular helix with vector equation r(t) = cos ti +sin tj + tkfrom the point (1, 0, 0) to (1, 0, 2).

Solution:

The curve from the point (1, 0, 0) to (1, 0, 2) is corresponding to the parameter t from 0 to 2, i.e.0 t 2. So

L=

2

0

|r(t)|dt=

2

0

[ sin(t)]2 + [cos t]2 + [1]2dt=

2

0

1 + 1dt= 2

2

1

8/10/2019 Ch13S3

2/5

Figure 1: This is the graph that the equation r(t) = cos ti + sin tj + tk fromthe point (1, 0, 0) to (1, 0, 2) represents

A single curve Ccan be represented by a vector equation

r(t) =< f(t), g(t), h(t) >,

and the vector equation is called a parametrization of the curve C.

Remark: Parametrizations of a curve is not unique.For instance, the twisted cubic

r(t) =< t, t2, t3 >, 1 t 2can also be represented by the parametrization

r(u) =< eu, e2u, e3u >, 0 u ln 2

provided t= eu.Remark: Reparametrize a curve is similar to change of variables.

The arc length arises naturally from the shape the curve and does not depend on parametrization.Arc Length Function: Suppose Cis curve given by a vector function

r(t) =f(t)i +g(t)j +h(t)k, a t b

where r is continuous and Cis traversed once as t increases from a to b.

DefinitionThe arc length function s is defined by

s(t) =

ta|r(u)|du

Remark: s(t) is the arc length of the part Cbetween r(a) and r(t).According to the Fundamental Theorem of Calculus, we have

ds

dt = |r(t)|

Often times, its useful to parametrize a curve with respect to its arc length because arc length isindependent on parametrizations.

2

8/10/2019 Ch13S3

3/5

Example: Reparametrize the liner(t) =with respect to arc length measured from(0, 1, 5) in the direction of increasing t.

Solution:The arc length measured from the point (0, 1, 5) corresponding to t = 0. To reparametrize the curve

w.r.t. the arc length, we need the relation between s and t. Sinceds

dt = |r(t)| = | | =

29 s(t) =

t0

|r(u)|du= t0

29du=

29t

Therefore,

t= s

29,

and the reparametrization is obtained by substituting for t

r(s) =

Curvature

A parametrization r(t) is called smooth on an interval I ifr(t) is continuous and nonzero.

A curve is called smooth if it has a smooth parametrization (No sharp corners or cusps; tangent vectorturns smoothly.)

Note that when Cis nearly straight, the direction of tangent vector changes slowly; when Cbends ortwists more sharply, the direction of tangent vector changes more quickly.

Curvature at a point is defined to measure how quickly the curve changes direction at that point.Definition: The curvature of a curve is

=dT

ds

where Tis the unit tangent vector. Remark: differentiate with respect to arc length.

Its easier to compute a curvature if it is expressed in terms oft instead ofs.

3

8/10/2019 Ch13S3

4/5

From Chain Rule:dT

dt =

dT

ds

ds

dt dT

ds =

dT

dt

dsdt

We knowds

dt = |r(t)|,

So

(t) =|T(t)||r(t)|

The curvature of the curve given by the vector function r:

(t) =|r(t) r(t)|

|r(t)|3

(Read theorem 10 in the textbook for the proof)

For a plane curve, y= f(x), curvature can be given by

(x) = |f(x)|

[1 + (f(x))2]3/2

(This formula is derived from the vector equation formula by using r=< x, f(x), 0>)

Example:Find the curvature of the curve given by r(t) =< a cos t, a sin t, 0>.Solution:

r(t) =< a sin t, a cos t, 0>r(t) =

So|r(t) r(t)| = |(a2 sin2 t+a2 cos2 t)k| = |a2k| =a2

and|r(t)| =a

Therefore, from the formula given by the vector function:

(t) =|r(t) r(t)|

|r(t)|3 = a2

a3 =

1

a

Remark: r(t) in this example represents a circle with radius a. The results shows larger circles have

smaller curvature while smaller circles have lager curvature.

4

8/10/2019 Ch13S3

5/5

The Normal and Binormal Vectors

Sometimes its useful to have a moving reference frame.Remark: Reference frame is a set of orthogonal vectors

We have already introduced the unit tangent vector T(t) = r(t)

|r(t)|. We need a vector orthogonal to

T. Recall from the example of previous lecture that|T| = 1 T T= 0, i.e., T is orthogonal to T.T is not necessarily a unit vector, but ifr is smooth, and r =0, we can find the corresponding unitvector ofT.

Definition: The principle unit normal vector is defined by

N(t) = T(t)

|T(t)|Remark: N(t) indicates the direction where the curve is turning.

We need another unit vector that is orthogonal to both T and N to get a reference frame in 3D space.Definition: The binormal vector is defined by

B(t) =T(t) N(t)

Remark: Bis a unit vector because|B| = |T N| = |T||N| sin 2

= 1

We call the plane containing N and Bat a point Pthe normal plane of the curve at P(It consists ofall lines orthogonal to T)

Formula Summary:

T(t) = r(t)

|r(t)| , N(t) = T(t)

|T(t)| , B(t) =T(t) N(t)

(t) =dT

ds

=|T(t)||r(t)| =|r(t) r(t)|

|r(t)|3

5