Arc Length and
Curvature
By Dr. Julia Arnold
Objectives:
1. Find the arc length of a space curve.
2. Use the arc length parameter to describe a plane curve
or space curve.
3. Find the curvature of a curve at a point on the curve.
4. Use a vector-valued function to find frictional force.
Objective 1
1.
Find the arc length of a space
curve.
Given a smooth plane curve C that has parametric equations
x = x(t) and y = y(t) where a  t  b, the arc length s is given by
b
s

2
(See Section 10.3)
2
 x  ( t )    y  ( t )  d t
a
In vector form, where C is given by r(t)=x(t)i + y(t)j, the above
b
equation can be written as
s   r  ( t ) dt
a
We can extend this formula to space quite naturally as
follows:
If C is a smooth curve given by r(t)= x(t)i + y(t)j +z(t)k on an
interval [a,b], then the arc length C on the interval is
b
s

a
b
2
2
2
 x  ( t )    y  ( t )    z  ( t )  dt 

a
r  ( t ) dt
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve
and imagine taking steps from point to point.
Now let the points get closer and closer together and sum
them up.
Let’s look a a few examples:
Example 1: Find the length of the space curve over the given
interval.
r ( t )  2 sin t , 5 t , 2 cos t  0,  

r ( t )  2 sin t , 5 t , 2 cos t

Set up the integral
 0,  
r  ( t )  2 cos t , 5,  2 sin t
Find the derivative of the vector.


4 cos t  25  4 sin t dt
2
2
Substitute into the formula.
0


4(sin t  cos t )  25 dt
2
2
Simplify
0


0

4  25 dt 

0
29 dt 
29 t

0

29 
Integrate and evaluate.
Objective 2
2. Use the arc length parameter
to describe a plane curve or
space curve.
Curves can be represented by vector-valued functions in different
ways depending on the choice of parameter.
For example the following two representations are equivalent.
r (t )  t , t , t
2
3
on 1  t  2 and r ( u )  e , e
u
2u
,e
3u
on 0  u  ln 2
P ro o f
L et u  ln t , then e  t , e
u
2u
t , e
2
3u
t
3
and ln 1  ln t  ln 2  0  u  ln 2
thus
r (u )  e , e
u
2u
,e
3u
on 0  u  ln 2
For motion along a curve the most convenient parameter is time t.
However, for studying the geometric properties of a curve, the
convenient parameter is often arc length s.
If C is a smooth curve given by r(t)= x(t)i + y(t)j +z(t)k on an
interval [a,b], then the arc length of C on the interval [a,b] , with
a<t<b is C = s(t) which is
t
s (t ) 

a
t
r  ( u ) du 

2
2
2
 x  ( u )    y  ( u )    z  ( u )  du
a
The arc length s is called the arc length parameter.
Example 2: Consider the curve represented by the vector-valued
function
4  sin t  t cos t  , 4  cos t  t sin t  ,
r (t ) 
3
t
2
2
A. Write the length of the arc s as a function of t by
evaluating the integral:
t
s

2
2
2
 x  ( u )    y  ( u )    z  ( u )  du
0
Solution:
r  ( t )  4  cos t  t sin t  cos t  , 4   sin t  t cos t  sin t  , 3 t
r  ( t )  4 t sin t , 4 t cos t , 3 t
t
s

16 u sin u  16 u cos u  9 u du
2
2
2
2
2
0
t
s

16 u  9 u du
2
2
0
t
s

0
t
25 u du  s 
2
 5 udu  5
0
u
2
t
2 0

5
2
t
2
Example 2: Consider the curve represented by the vector-valued
function
4  sin t  t cos t  , 4  cos t  t sin t  ,
r (t ) 
3
t
2
2
B. Solve for t in part A and substitute the result into the original set
of parametric equations. This yields a parameterization of the curve
in terms of the arc length parameter s.
Solution:
s
5
t 
2
r ( t ( s )) 
2
2s
t t
2
5

4  sin


2s
5
2s
5

2s
5
cos
2s  
 , 4  cos
5  
2s
5

2s
5
sin
2s  3 
, 
5  2 
2s 

5 
2
Example 2: Consider the curve represented by the vector-valued
function
4  sin t  t cos t  , 4  cos t  t sin t  ,
r (t ) 
3
t
2
2
C. Find the coordinates of the point on the curve for arc lengths
s  5 so t 
2 5
5
Solution:
r ( t ( 5 )) 

4  sin


2 5

5
r ( t ( 5 ))  1.030, 5.408,1.342
2 5
5
cos
2 5  
 , 4  cos
5  
 
2 5
5

2 5
5
sin
2 5  3
,
5  2





2 5 

5 

2
Example 2: Consider the curve represented by the vector-valued
function
r (t ) 
4  sin t  t cos t  , 4  cos t  t sin t  ,
3
t
2
2
C. Find the coordinates of the point on the curve for arc lengths
2(4)
s  4 thus t 

5
8
5
Solution:
r ( t (4)) 

4  sin



r ( t (4))  4  sin


8

8
5
5
8
8
5

5
r ( t (4))  2.291, 6.029, 2.4
cos
cos
8  
 , 4  cos
5  
8  
 , 4  cos
5  
8

8
5
5
8
8
5

5
sin
8  3
, 
5  2 
sin
8 
 , 2.4
5 
8 

5 
2
Example 2: Consider the curve represented by the vector-valued
function
r (t ) 
4  sin t  t cos t  , 4  cos t  t sin t  ,
D. Verify that
3
t
2
2
r ( s )  1
Solution:

4  sin


r ( t ( s )) 
r ( s ) 
r ( s ) 

4  cos


4
sin
5
r ( s ) 
2s
cos
5
2s  1

5  2
2 

5 s 
2s  
 , 4  cos
5  
2s
5

2 s  16
2
cos 


5  25

2s  9


5  25
2s
sin
2s

5
2s  1

5  2
2s 4
, cos
5 5

2
sin 

25

16

5
2s
sin
5
2  1

5 s   2
2s  3
,
5  2
2 
 cos
5 s 
2s 3
,
5 5
16
25

9
25




1
2s 

5 
2
2s  
 , 4   sin
5  
 
2s  1

5  2
2 

5 s 
2s
5
cos
2s  1

5  2
2  1

5 s   2
2 
 sin
5 s 
2s  3
,
5  5

This brings us to a Theorem about the arc length
parameter, namely
If C is a smooth curve given by
r (s)  x(s)i  y(s) j
or
r ( s )  x ( s ) i  y ( s ) j + z(s)k
Where s is the arc length parameter, then
r ( s )  1
Moreover, if t is any parameter for the
vector-valued function r such that r ( t )  1
Then t must be the arc length
parameter.
This theorem is stated without proof.
Objective 3
3. Find the curvature of a curve
at a point on the curve.
Curvature
An important use of the arc length parameter is to find curvature.
Curvature is the measure of how sharply the curve
bends.
For example, in this helix we get more
bend
here
Than here.
We can calculate curvature by calculating the magnitude of the rate of
change of the unit tangent vector T with respect to the arc length s.
T2
T3
T1
Definition of Curvature
Let C be a smooth curve ( in the plane or in space) given
by r(s), where s is the arc length parameter. The
curvature K at s is given by
K 
dT
ds
 T (s)
Example 3: Find the curvature using s is the
arc length parameter, for
3 2
r ( t )  4(sin t  t cos t ) i  4(cos t  t sin t ) j + t k
2
Solution: This was the problem we did earlier and found the arc
length parameter to be: t  2 s
and the function to be
r ( t ( s )) 
5

4  sin


2s
5

2s
cos
5
2s  
 , 4  cos
5  
2s
5

2s
5
sin
2s  3
,
5  2




2s 

5 
2
Using the formula for curvature K in terms of arc length s, namely
r ( s )
and knowing that T ( s ) 
and r  ( s )  1 we get:
r ( s )
T ( s )  r ( s ) 
4
5
sin
2s 4
, cos
5 5
2s 3
,
5 5
Since curvature K is K 
dT
ds
 T (s)
Example 3: Find the curvature using s is the
arc length parameter, for
3 2
r ( t )  4(sin t  t cos t ) i  4(cos t  t sin t ) j + t k
2
Solution:
dT
 T (s)
K 
ds
4
T ( s )  r ( s ) 
2s 4
, cos
5 5
sin
5
T  ( s )  r  ( s ) 
T  ( s )  r  ( s ) 
T ( s ) 
4

1 4

2 5
2
2
5
5s
2
cos
cos
2
25 5 s
T ( s ) 
T ( s ) 

2
 cos
125 s 
125 s

2s
2 s 2
,
5 5

5
8
8
2 s 1 4
,

5 2 5
cos
5s
2
2 2
5 5s
2s
2s 3
,
5 5
4

sin
 sin
2
5s
5s

2s 

5 
2 10 s
25 s
sin
5s
,0
5
2
2s
5
2s
5
2s
sin
5s
25 5 s
5

2
2
2
0
,0
Using winplot, this
is the curve in
question.
Since s =
5
t
2
2
In terms of t the
curvature would be
y
x
K 
2 10 s
25 s
2 10 
5
t
2

25 
5
2
t
2

2
2 5  5t
125
2
t
2
2

4 25 t
125 t
2
2

20 t
125 t
2

4
25 t
We can see that the curvature of a circle is the same
everywhere and reason it to be a constant which turns
out to be 1/r where r is the radius of the circle.
See example 4 in your text.
Other formulas for curvature.
Since the previous definition depends on the arc length
parameter, it might be good to have some alternative
definitions which depend on an arbitrary parameter t.
Two formulas for curvature
Theorem 12.8
If C is a smooth curve given by r(t), then
the curvature K of C at t is given by
K 
T ( t )
r ( t )

r  ( t )  r  ( t )
r ( t )
3
Example 4: using the alternative curvature formula
on the same vector-valued function
3 2
r ( t )  4(sin t  t cos t ) i  4(cos t  t sin t ) j + t k
2
We can compare our answers. From Example 2A we already know
that r ( t )   4 t sin t , 4 t cos t , 3t
r ( t )  5 t
Next we need to find T(t).
T (t ) 
r ( t )
r ( t )

 4 t sin t , 4 t cos t , 3 t
5t
Now we need T’(t)

 4 sin t 4 cos t 3
,
,
5
5
5
T (t ) 
 4 sin t 4 cos t 3
,
,
5
5
5
T ( t ) 
 4 cos t  4 sin t
,
,0
5
5
Using the formula
Now we find
K 
r ( t )
r  ( t )  r  ( t )

r ( t )
3
T ( t )
 4 cos t  4 sin t
,
,0
5
5
T ( t ) 
T ( t ) 
16
cos t 
2
25
K 
T ( t )
T ( t )
r ( t )
16
25
sin t 
2
4
5
4
4
 5 
5t
2 5t
Which is what we got back on slide 37
Click on the purple crayon to get back
to this slide.
Using the other formula
We have
K 
r  ( t )  r  ( t )
K 
3
r ( t )
T ( t )
r ( t )
4
4
 5 
5t
2 5t
r  ( t )   4 t sin t , 4 t cos t , 3 t
r ( t )  5 t
w e need r  ( t )   4 t cos t  4 sin t ,  4 t sin t  4 cos t , 3
i
j
 4 t sin t
4 t cos t
 4 t cos t  4 sin t
 4 t sin t  4 cos t
r  ( t )  r  ( t ) 
k
3t 
3
12 t cos t i  3 t (  4 t cos t  4 sin t ) j  (  4 t sin t )(  4 t sin t  4 cos t ) k
  3 t (  4 t sin t  4 co s t ) i  (  12 t sin t ) j  4 t cos t (  4 t cos t  4 sin t ) k  
(12 t cos t  12 t sin t  12 t cos t ) i  (  12 t cos t  12 t sin t  12 t sin t ) j
2
2
 (16 t sin t  16 t sin t cos t  16 t cos t  16 t cos t sin t ) k 
2
2
2
2
(12 t sin t ) i  (  12 t cos t ) j  (16 t sin t  16 t cos t ) k  (12 t sin t ) i  (  12 t cos t ) j  (16 t ) k
2
2
2
2
2
2
2
2
w e need r  ( t )  r  ( t )
144 t sin t  144 t cos t  256 t
4
2
4
2
K 
4

144 t  256 t
4
r  ( t )  r  ( t )
r ( t )
3

4

20 t
400 t
2
125 t
3

4
4
25 t
 20 t
2
2
Example 5
Solution to question on previous slide.
At (4,0) the curvature would be:
y


S in ce y   1 
x
, th en a t x  4, y   1  2   1
2
y  

1

2

1
K 

1
2
3
[1  (  1) ] 2
2


2
3
[2 ] 2
Thus 25/2 would be the
radius of the circle
which would be
approximately 5.66

1
5
 .1 7 7
22


            

Using winplot I found the
normal to the tangent line
at x = 4 (blue line) and then
found the center to be
approximately at (0,-4).












Figure 12.37





   
Objective 4
Use a vector-valued function to
find frictional force.
Example 6
Solution:
r (t )  2 ti  t j 
2
1
3
t k
3
r (t) = 2 i  2 t j  t k
2
r (t) 
T (t ) 
T ( t ) 
4  4t  t
2
r (t)
2
2
2
2 i  2tj  t k


t
t 2
2
 4 ti + 4  2 t
2
2
 j  4 tk
2


t

 2
2
16 t + 4  2 t
2
2
2 t  4t  4
4
T ( t ) 
( t  2)  t  2
2
r (t)
T ( t ) 

4

2
2

2
 16 t
2
4


2 (t  2)

t 2
2
16 t + 16 - 16t + 4t  16 t
2
2
2
t 2
2

2
2

2

t 2
2
4

2
2
2
t 2
2
Continued
Example 6
Solution continued
r (t)  t  2 w hich
2
ds
is also
dt
T ( t ) 
2
t 2
T (t)
2
2
2
2
t
K 
 22 
2
r ( t )
t 2
t 2


2
ds
t 2
2
dt
2
aT 
d s
dt
2
 2t
2
aN
2
 ds 
 K


2
 dt 
t 2


2

 t 2
2

2
2
Figure 12.38
Hint for HW
Confusion of Mass and Weight
A few further comments should be added about the single force which is a
source of much confusion to many students of physics - the force of gravity.
The force of gravity acting upon an object is sometimes referred to as the
weight of the object. Many students of physics confuse weight with mass. The mass of an
object refers to the amount of matter that is contained by the object; the weight of an object
is the force of gravity acting upon that object. Mass is related to how much stuff is there and
weight is related to the pull of the Earth (or any other planet) upon that stuff. The mass of an
object (measured in kg) will be the same no matter where in the universe that object is located.
Mass is never altered by location, the pull of gravity, speed or even the existence of other
forces. For example, a 2-kg object will have a mass of 2 kg whether it is located on Earth, the
moon, or Jupiter; its mass will be 2 kg whether it is moving or not (at least for purposes of our
study); and its mass will be 2 kg whether it is being pushed upon or not.
On the other hand, the weight of an object (measured in Newtons) will vary according to where
in the universe the object is. Weight depends upon which planet is exerting the force and the
distance the object is from the planet. Weight, being equivalent to the force of gravity, is
dependent upon the value of g.
On earth's surface g is 9.8 m/s2 (often approximated as 10 m/s2) or
32 feet/s2. On the moon's surface, g is 1.7 m/s2. Go to another planet, and
there will be another g value. Furthermore, the g value is inversely
proportional to the distance from the center of the planet. So if we were to
measure g at a distance of 400 km above the earth's surface, then we would
find the g value to be less than 9.8 m/s2.
For comments on this presentation you may
email the author: Dr. Julia Arnold at
jarnold@tcc.edu