144
2.3
CHAPTER 2. VECTOR FUNCTIONS
Arc Length
In this section, we discuss the notion of curve in greater detail and introduce
the very important notions of arc length and curvature.
2.3.1
Parametrizations of a Curve
We have described the curve C with position vector !
r (t) = hf (t) ; g (t) ; h (t)i
for t 2 [a; b] as the set of points (x; y; z) where
8
< x = f (t)
y = g (t)
:
z = h (t)
as t varies in the interval [a; b]. In fact, a curve is a little more than just a set
of points. It is a succession of points traversed in a certain order. As t varies
from a to b, the points obtained will trace the curve in a certain order. We say
that a parametrized curve is an oriented curve.
Many curves are parametrized with time as their parameter. One reason for
this is that the equations of these curves are often derived from principles in the
sciences. These principles often involve time. However, time may not always be
the best parameter. Consider the following example. Suppose that you are an
explorer, think of the path you follow as a curve. One day you are on a curve
that is so fascinating that you want to record what you see so other people
can follow your steps. You want to describe the path you follow (the curve)
from a known point. Suppose you use statements like: "For 2 hours, we went
straight east, then after 2 hours, we turned left 45 degrees, we continued on
this path for three hours, ...". Would that be a precise description? Of course
not. It would depend how fast you were moving. So, in this case, time would
not be a good parameter to use. Or if you use it, you would also need to add
velocity. Or, instead of time and velocity, you could use distance travelled as
your parameter. If instead, you had used statements like: "we travelled straight
east for 10 miles, then turned left 45 degrees. We stayed on that path for 15
miles....". This statement no longer depends on how fast you are moving along
the path. For a curve, distance travelled is called arc length. It is often used
as the parameter of a curve. We will learn how to change the parameter of a
curve (this is called reparametrize the curve) from one parameter to another.
Let us start with some precise de…nitions.
De…nition 142 (Reparametrization of a Curve) Let I and J be two intervals. Suppose that C is a smooth curve with position vector
!
r (t) for t 2 I
Let ' : J ! I. Consider the curve C 0 given by the position vector
!
R (u) = !
r (' (u)) for all u 2 J
2.3. ARC LENGTH
145
In other words C 0 is obtained from C by performing the change of variable
t 7! ' (u) (replace t by ' (u)) in the position vector of C.
!
1. If '0 (u) > 0 for all u 2 J then R and !
r take on the same values in the
same order. In this case, C and C 0 are the same oriented curve. The only
di¤ erence between the two curves is that C and C 0 may not be traversed
at the same speed
!
2. If '0 (u) < 0 for all u 2 J then R and !
r take on the same values but in
opposite order. In this case, C and C 0 have the same path, but they are
traversed in opposite order. In addition, they may not be traversed at the
same speed.
3. The change of parameter when '0 (u) > 0 is said to be sense-preserving.
The change of parameter when '0 (u) < 0 is said to be sense-reversing.
In both cases, we say that C 0 is a reparametrization of C. If in addition
j'0 (u)j = 1, the curves will be traversed at the same speed.
Example 143 The parametrization
!
r (t) = hcos t; sin t; ti , t 2 [0; 2 ]
gives one spiral of the helix. Let t = ' (u) = u, u 2 [0; 2]. Then, ' maps [0; 2]
onto [0; 2 ] and '0 (u) = > 0. Then the curve
!
R (u)
= !
r (' (u))
= hcos u; sin u; ui , u 2 [0; 2]
is precisely the same curve with the same orientation. Finally, if we let t =
(v) = 2 v, v 2 [0; 1]. Then maps [0; 1] onto [0; 2 ], In addition 0 (v) =
2 > 0. Then the curve
! (v)
= !
r ( (v))
=
hcos 2 v; sin 2 v; 2 vi , v 2 [0; 1]
is precisely the same curve with the same orientation. The only di¤ erence is the
speed at which the curve is being traversed. Figures 2.4, 2.5and 2.6 show the
!
curves corresponding to !
r , R , and ! respectively. We also plotted the position
vector when the parameter was 1. It is easy to see that the curves are the same.
It is also easy to see that for the same value of the parameter, the position vector
has moved the furthest in the last curve. Indeed, from the derivatives of ' and
we see that the curve which will be traversed the fastest is the one corresponding
to ! and the slowest is the one corresponding to !
r.
Example 144 The curve
!
r (t) = hcos t; sin t; 0i for t 2 [0; 2 ]
146
CHAPTER 2. VECTOR FUNCTIONS
Figure 2.4: Curve given by hcos t; sin t; ti, t 2 [0; 2 ]
Figure 2.5: Curve given by hcos t; sin t; ti, t 2 [0; 2]
Figure 2.6: Curve given by hcos 2 t; sin 2 t; 2 ti, t 2 [0; 1]
2.3. ARC LENGTH
147
traces a unit circle in the xy-plane counterclockwise. The curve
!
R (u) = hcos (2
u) ; sin (2
u) ; 0i for u 2 [0; 2 ]
traces the same circle clockwise.
This last example is important to remember, we state a more general result
as a proposition.
Proposition 145 Given a curve C by its position vector !
r (t) for t 2 [a; b],
!
the curve with position vector R (u) = !
r (a + b u) for u 2 [a; b] corresponds
to the same set of points, traversed in opposite order.
That the order is reversing is easy to see. The change of parameter is ' which
maps [a; b] onto [a; b] de…ned by ' (u) = a + b u. Clearly, '0 (u) = 1 < 0, so
the order is reversed.
2.3.2
Arc Length
The following de…nition gives the length of the portion of a curve !
r (t) between
t = a and t = b. The formula given is valid for 2-D curves as well as 3-D curves.
Though it is given here as a de…nition, the formula can actually be proven.
De…nition 146 (Arc Length) Let C be a smooth curve with position vector
!
r (t) for t 2 [a; b]. The length of the portion of the curve between t = a and
t = b is
Z b
L=
k!
r 0 (t)k dt
(2.7)
a
It can also be proven that L as computed above does not depend on the
parametrization used.
Example 147 Compute the circumference of a circle of radius a, a > 0.
Since all the circles having the same radius have the same circumference, we can
assume our circle is centered at the origin. In this case, one parametrization of
such a circle is
!
r (t) = ha cos t; a sin ti , t 2 [0; 2 ]
From formula 2.7, we see that
L=
Z
0
2
k!
r 0 (t)k dt
So, we need to compute k!
r 0 (t)k.
!
r 0 (t) = h a sin t; a cos ti
148
CHAPTER 2. VECTOR FUNCTIONS
So
k!
r 0 (t)k
p
=
p
=
a2 sin2 t + a2 cos2 t
a2
= a since a > 0
Thus
Z
L =
2
adt
0
=
2
atj0
=
2 a
Which is the formula we all know.
Example 148 Compute the length of the arc of the circular helix !
r (t) =
hcos t; sin t; ti from the point (1; 0; 0) to the point (1; 0; 2 ).
Let us …rst notice that this corresponds to the arc between t = 0 and t = 2 (t
is the z-coordinate of a point on the helix). From formula 2.7, we see that
L=
Z
0
2
k!
r 0 (t)k dt
So, we need to compute k!
r 0 (t)k.
!
r 0 (t) = h sin t; cos t; 1i
So
k!
r 0 (t)k
=
=
p
sin2 t + cos2 t + 1
p
2
(2.8)
Thus
L =
Z
2
p
2dt
0
=
=
p
2
2t
p0
2 2
We saw in the previous section that curves could be parametrized di¤erent
ways. For motion along a curve, time is often the parameter of choice. However,
to describe the geometric properties of a curve time is not the most appropriate
parameter. The parameter of choice is arc length. We explain how this is done
and why.
2.3. ARC LENGTH
149
De…nition 149 (Arc length Function) Let C be a smooth curve with position vector !
r (t) for t 2 [a; b] and suppose C is traversed only once as t increases
from a to b. We de…ne the arc length function from t = a, denoted s or s (t)
to be
Z t
k!
r 0 (u)k du
(2.9)
s (t) =
a
Geometrically, s (t) is the length of the arc of C between !
r (a) and !
r (t).
So, it is always a positive quantity which increases as t increases. From the
de…nition of t and using the fundamental theorem of calculus, we have the
following theorem.
Theorem 150
0
ds
= !
r (t)
dt
(2.10)
We now show how a curve can be parametrized with respect to arc length.
Goal: Given a smooth curve C with position vector !
r (t), t 2 I for some
interval I, we want to reparametrize C with respect to arc length that is we
need to …nd a function ' : J ! I such that t = ' (s) where s is arc length. This
amounts to …nding a relationship between t and s. To do this, we compute s (t)
using formula 2.9. We illustrate this procedure with a few examples.
Example 151 Reparametrize the helix !
r (t) = hcos t; sin t; ti with respect to
arc length measured from the point (1; 0; 0).
In order to do this, we must compute s (t). The initial point corresponds to
t = 0. Thus, by formula 2.9, we have
s (t) =
Z
t
0
k!
r 0 (u)k du
p
From example 148, formula 2.8, we know that k!
r 0 (u)k = 2. Thus,
s (t)
=
Z tp
2du
0
=
p
=
p
2u
t
0
2t
s
Thus, t = p . It follows that the reparametrization of the helix with respect to
2
arc length is
!
R (s)
s
p
2
s
s
s
cos p ; sin p ; p
2
2
2
= !
r
=
150
CHAPTER 2. VECTOR FUNCTIONS
Remark 152 If we had been given an interval for t such as t 2 [0; 2 ], we could
have found
p the corresponding interval for s by using theprelation between s and
t: s = 2t. When t = 0, s = 0 and when t = 2 , s = 2 2 .
Example 153 Reparametrize a circle of radius a centered at the origin given
by !
r (t) = ha cos t; a sin ti, t 2 [0; 2 ] with respect to arc length.
By formula 2.9, we have
Z t
s (t) =
k!
r 0 (u)k du
0
The reader will check that k!
r 0 (u)k = a. Thus
Z t
adu
s (t) =
0
=
at
s
Hence, t = . Thus, a reparametrization of the circle of radius a centered at
a
the origin is
!
s
R (s) = !
r
D a s
sE
=
a cos ; a sin
a
a
Remark 154 We invite the reader to check that in both examples above, we
!
have R 0 (s) = 1. In fact, it can be proven that this is always true. We state
this result as a theorem we will not prove.
Theorem 155 Let C be a smooth curve with position vector !
r (s) where s is
the arc length parameter. Then
k!
r 0 (s)k = 1
Furthermore, if t is any parameter such that k!
r 0 (t)k = 1, then t must be the
arc length parameter.
ds
Proof. From equation 2.10, we see that for any parameter t we have
=
dt
ds
k!
r 0 (t)k. If t is arc length, then equation 2.10 becomes
= k!
r 0 (s)k ()
ds
1 = k!
r 0 (s)k. Conversely, if k!
r 0 (t)k = 1 then from the same equation, it
ds
follows that
= 1 or t = s+constant. In other words, t is the arc length
dt
parameter from some point.
Corollary 156 Let C be a smooth curve with position vector !
r (s) where s is
the arc length parameter. Then
!
T (s) = !
r 0 (s)
!
!
r 0 (s)
Proof. By de…nition, T (s) = !0
. From the theorem, k!
r 0 (s)k = 1,
k r (s)k
hence the result.
2.3. ARC LENGTH
2.3.3
151
Problems
1. Find the unit tangent vector and the lengthpof the curve if the position
vector of the curve is !
r (t) = 2 cos t; 2 sin t; 5t , for 0 t
.
2. Find the unit tangent vector and
of the curve if the position
E
D the 3length
!
2 2
vector of the curve is r (t) = t; 0; 3 t , for 0 t 8.
3. Find the unit tangent vector and the length of the curve if the position
vector of the curve is !
r (t) = 0; cos3 t; sin3 t , for 0 t
2.
4. Find the unit tangent vector and
E curve if the position
D the length pof the
3
.
vector of the curve is !
r (t) = t cos t; t sin t; 2 3 2 t 2 , for 0 t
5. Find the point on the curve !
r (t) = h5 sin t; 5 cos t; 12ti at a distance 26
units from the point (0; 5; 0) in the direction of increasing arc length.
6. Find the arc length parameter and then the length of the indicated portion
of the curve for !
r (t) = h4 cos t; 4 sin t; 3ti, 0 t
2.
7. Find the arc length parameter and then the length of the indicated portion
of the curve for !
r (t) = het cos t; et sin t; et i, ln 4 t 0.
8. Find the arc length parameter
p and
p then the length of the indicated
p pportion
of the curve for !
r (t) =
2t; 2t; 1 t2 , from (0; 0; 1) to
2; 2; 0 .
9. To illustrate that arc length does not depend on the parametrization of a
curve, we give three di¤erent parametrization of the same curve. In each
case, …nd the length of the given portion of the curve. You should …nd
the same result three times.
(a)
(b)
(c)
!
r (t) = hcos 4t; sin 4t; 4ti 0 t
2
!
r (t) = cos 2t ; sin 2t ; 2t 0 t 4
!
r (t) = hcos t; sin t; ti 2
t
0
10. The involute of a circle. If a string wound around a …xed unit circle
is unwound while held taught in the plane of the circle, its end P traces
an involute of the circle. Derive the equation of the involute !
r (t) =
hcos t + t sin t; sin t t cos ti for t > 0.
11. Distance along a line. Show that if !
u is a unit vector, then the arc
length parameter along the line !
r (t) = P + t!
u from the point P at
0
0
t = 0 is t itself.
12. Find the length of the curve given by !
r (t) = h2 sin t; 5t; 2 cos ti,
t 10.
13. Find the length of the curve given by !
r (t) = 1; t2 :t3 , 0 t 1
10
14. Reparametrize the curve given by !
r (t) = h2t; 1 3t; 5 + 4ti with respect
to arc length measured from the point corresponding to t = 0.
152
CHAPTER 2. VECTOR FUNCTIONS
2.3.4
Answers
1. Find the unit tangent vector and the lengthpof the curve if the position
vector of the curve is !
r (t) = 2 cos t; 2 sin t; 5t , for 0 t
!
T (t) =
*
p +
5
2
2
sin t; cos t;
3
3
3
L=3
2. Find the unit tangent vector and
of the curve if the position
D the 3length
E
!
2 2
vector of the curve is r (t) = t; 0; 3 t , for 0 t 8.
!
T (t) =
p
1
t
p
; 0; p
1+t
1+t
52
3
L=
3. Find the unit tangent vector and the length of the curve if the position
vector of the curve is !
r (t) = 0; cos3 t; sin3 t , for 0 t
2.
!
T (t) = h0;
cos t; sin ti
L=
3
2
4. Find the unit tangent vector and
D the length pof the
E curve if the position
!
2 2 32
vector of the curve is r (t) = t cos t; t sin t; 3 t , for 0 t
.
!
T (t) =
*
p +
cos t t sin t sin t + t cos t
2t
;
;
t+1
t+1
t+1
2
L=
+
2
5. Find the point on the curve !
r (t) = h5 sin t; 5 cos t; 12ti at a distance 26
units from the point (0; 5; 0) in the direction of increasing arc length.
Rt
We need to …nd t0 such that 0 0 k!
r 0 (t)k dt = 26 . Thus, we need to solve
26 = 13t0
Thus t0 = 2 . It follows that the point is
!
r (2 ) = h0; 5; 24 i
2.3. ARC LENGTH
153
6. Find the arc length parameter and then the length of the indicated portion
of the curve for !
r (t) = h4 cos t; 4 sin t; 3ti, 0 t
2.
s (t) = 5t
Hence
s
2
=
5
2
7. Find the arc length parameter and then the length of the indicated portion
of the curve for !
r (t) = het cos t; et sin t; et i, ln 4 t 0.
p
p
3
s (t) = 3et
4
Hence
p
3 3
s (0) =
4
8. Find the arc length parameter
p and
p pportion
p then the length of the indicated
of the curve for !
r (t) =
2t; 2t; 1 t2 , from (0; 0; 1) to
2; 2; 0 .
s (t) = 2
Hence
p
1
tp
1 + t2
ln t + 1 + t2 +
2
2
s (1) = ln 1 +
p
2 +
p
2
9. To illustrate that arc length does not depend on the parametrization of a
curve, we give three di¤erent parametrization of the same curve. In each
case, …nd the length of the given portion of the curve. You should …nd
the same result three times.
(a) !
r (t) = hcos 4t; sin 4t; 4ti 0
t
2
L=2
(b) !
r (t) = cos 2t ; sin 2t ; 2t 0
t
sin t; ti
2
2
p
2
4
L=2
(c) !
r (t) = hcos t;
p
t
0
p
L=2 2
10. The involute of a circle. If a string wound around a …xed unit circle is
unwound while held taught in the plane of the circle, its end P traces an
involute of the circle. Derive the equation of the involute
!
r (t) = hcos t + t sin t; sin t t cos ti for t > 0:
154
CHAPTER 2. VECTOR FUNCTIONS
11. Distance along a line. Show that if !
u is a unit vector, then the arc
length parameter along the line !
r (t) = P0 + t!
u from the point P0 at
t = 0 is t itself.
Just compute s using the formula in the notes.
12. Find the length of the curve given by !
r (t) = h2 sin t; 5t; 2 cos ti,
t 10.
p
L = 20 29
13. Find the length of the curve given by !
r (t) = 1; t2 :t3 , 0
L
t
10
1
1: 439 7
14. Reparametrize the curve given by !
r (t) = h2t; 1 3t; 5 + 4ti with respect
to arc length measured from the point corresponding to t = 0.
!
R (s) =
2s
p ;1
29
3s
4s
p ;5 + p
29
29