Chapter 13 – Vector Functions
13.3 Arc Length and Curvature
Objectives:
 Find vector, parametric,
and general forms of
equations of lines and
planes.
 Find distances and angles
between lines and planes
13.3 Arc Length and Curvature
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Arc Length

Assuming that the space curve is traversed exactly
once on [a,b], and the component functions are
differentiable on [a,b], then arc length is given by
the integrals below.
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Note

Plane curves are described in 2
while space curves are defined in
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3.
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Example 1 – pg. 860 #4

Find the length of the curve.
r(t )  cos ti  sin tj  ln cos tk
0t  / 4
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Parameterization in Terms of
Arc Length
The relation below allows us to use
distance along the curve as the parameter.
 This parameterization does not depend on
coordinate system.
 Replace r(t) with r(t(s)) .

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Example 2 – pg.860 #14

Reparametrize the curve with respect
to arc length measured from the
point where t = 0 in the direction of
increasing t.
r(t )  e cos 2ti  2j  e sin 2tk
2t
2t
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Recall

If C is a smooth curve defined by the
vector r, recall that the unit tangent
is given by
T t  
r ' t 
r ' t 
and indicates the direction of the
curve.
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Visualization

The Unit Tangent Vector
T(t) changes direction very slowly
when C is fairly straight, but it
changes direction more quickly when
C bends or twists more sharply.
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Definition - Curvature
13.3 Arc Length and Curvature
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Curvature and the Chain Rule

If we use the Chain rule on
curvature, we will have:
dT dT ds

dt
ds dt

and
dT dT / dt


ds
ds / dt
where ds / dt  r '(t )
So we have:
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Note:
Small circles have large curvature.
 Large circles have small curvature.
 Curvature of a straight line is always
0 because the tangent vector is
constant.

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Theorem - Curvature

We can always use equation 9 to
compute curvature, but the below
theorem is easier to apply.
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Example 3

Use Theorem 10 to find the
curvature.
r (t )  ti  tj  1  t  k
2
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Definition – Unit Normal
T '(t )
N(t ) 
T '(t )
As you can see, N is
perpendicular to T(t).
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Definition – Binormal Vector

The Binormal Vector is perpendicular
to both T and N. It is also a unit
vector and is defined as:
B(t )  T(t )  N(t )
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Visualization

The TNB Frame
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Example 4 – pg.861 # 48

Find the vectors T, N, and B at the
given point.
r(t )  cos t,sin t,ln cos t
(1,0,0)
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Other Definitions

The normal plane is determined by the
vectors N and B at a point P on the curve C.
It consists of all lines that are orthogonal to
the tangent vector.

The osculating plane of C and P is
determined by the vectors T and N.

An osculating circle is a circle that lies in
the oculating place of C at P, has the same
tangent as C at P, lies on the concave side of
C (towards N), and has radius =1/.
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Visualization

Osculating Circle
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Summary of Formulas
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In groups, work on the following
problems

Problem 1 – pg. 860 #6
Find the arc length of the curve.
r(t )  12ti  8t j  3t k
3/2
2
0  t 1
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In groups, work on the following
problems

Problem 2 – page 860 #16
Reparametrize the curve below with respect to
arc length measured from the point (1,0) in the
direction of increasing t. Express the
reparametrization in its simplest form. What can
you conclude about the curve?
2t
 2

r (t )   2
 1 i  2
j
 t 1  t 1
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In groups, work on the following
problems

Problem 3
a) Find the unit tangent and unit
normal vectors.
b) Use formula 9 to find curvature.
r(t )  2sin t,5t,2cos t
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In groups, work on the following
problems

Problem 4 – pg. 860 #31
At what point does the curve have a
maximum curvature? What happens
to the curvature as x.
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In groups, work on the following
problems

Problem 5 – pg. 861 #50
Find equations of the normal plane
and osculating plane of the curve at
the given point.
x  t,
y t , z t ;
2
3
(1,1,1)
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More Examples
The video examples below are from
section 13.3 in your textbook. Please
watch them on your own time for
extra instruction. Each video is
about 2 minutes in length.
◦ Example 1
◦ Example 3
◦ Example 7
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Demonstrations
Feel free to explore these
demonstrations below.
TBN Frame
 Circle of Curvature

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