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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
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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.
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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.
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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.
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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
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