9. For a constant vector c = 0, the function f( t) = tc represents a line parallel to c.
(a) What kind of curve does g( t) = t 3c represent? Explain.
(b) What kind of curve does h( t) = etc represent? Explain.
(c) Compare f ′(0) and g ′(0). Given your answer to part (a), how do you explain the
difference in the two derivatives?
d
df
d 2f
10. Show that
f ×
= f ×
.
dt
dt
dt 2
11. Let a particle of (constant) mass m have position vector r( t), velocity v( t), acceleration
a( t) and momentum p( t) at time t. The angular momentum L( t) of the particle with
respect to the origin at time t is defined as L( t) = r( t) × p( t). If F( t) is the force acting on the particle at time t, then define the torque N( t) acting on the particle with respect to
the origin as N( t) = r( t) × F( t). Show that L′( t) = N( t).
d
df
dg
dh
12. Show that
(f · (g × h)) =
· (g × h) + f ·
× h + f · g ×
.
dt
dt
dt
dt
13. The Mean Value Theorem does not hold for vector-valued functions: Show that for f( t) =
(cos t, sin t, t), there is no t in the interval (0, 2 π) such that
f(2 π) − f(0)
f ′( t) =
.
2 π − 0
C
14. The Bézier curve b3( t) for four noncollinear points b , b , b , b in R3 is defined by the
0
0
1
2
3
following algorithm (going from the left column to the right):
b1( t) = (1 − t)b
b2( t) = (1 − t)b1( t) + tb1( t)
b3( t) = (1 − t)b2( t) + tb2( t)
0
0 + tb1
0
0
1
0
0
1
b1( t) = (1 − t)b
b2( t) = (1 − t)b1( t) + tb1( t)
1
1 + tb2
1
1
2
b1( t) = (1 − t)b
2
2 + tb3
(a) Show that b3( t) = (1 − t)3b
.
0
0 + 3 t(1 − t)2b1 + 3 t 2(1 − t)b2 + t 3b3
(b) Write the explicit formula (as in Example 1.40) for the Bézier curve for the points
b0 = (0,0,0), b1 = (0,1,1), b2 = (2,3,0), b3 = (4,5,2).
15. Let r( t) be the position vector for a particle moving in R3. Show that
d (r×(v × r))= r 2a+(r·v)v−( v 2+r·a)r.
dt
16. Let r( t) be the position vector in R3 for a particle that moves with constant speed c > 0
in a circle of radius a > 0 in the xy-plane. Show that a( t) points in the opposite direction
as r( t) for all t. ( Hint: Use Example 1.37 to show that r( t) ⊥ v( t) and a( t) ⊥ v( t), and hence a( t) ∥ r( t).)
17. Prove Theorem 1.20(g).
1.9 Arc Length
59
1.9 Arc Length
Let r( t) = ( x( t), y( t), z( t)) be the position vector of an object moving in R3. Since v( t) is the speed of the object at time t, it seems natural to define the distance s traveled by the object
from time t = a to t = b as the definite integral
b
b
s =
v( t) dt =
x ′( t)2 + y′( t)2 + z ′( t)2 dt ,
(1.40)
a
a
which is analogous to the case from single-variable calculus for parametric functions in R2.
This is indeed how we will define the distance traveled and, in general, the arc length of a
curve in R3.
Definition 1.13. Let f( t) = ( x( t), y( t), z( t)) be a curve in R3 whose domain includes the interval [ a, b]. Suppose that in the interval ( a, b) the first derivative of each component function
x( t), y( t) and z( t) exists and is continuous, and that no section of the curve is repeated. Then
the arc length L of the curve from t = a to t = b is
b
b
L =
f ′( t) dt =
x ′( t)2 + y′( t)2 + z ′( t)2 dt
(1.41)
a
a
A real-valued function whose first derivative is continuous is called continuously differ-
entiable (or a C 1 function), and a function whose derivatives of all orders are continuous
is called smooth (or a C ∞ function). All the functions we will consider will be smooth. A
smooth curve f( t) is one whose derivative f ′( t) is never the zero vector and whose component
functions are all smooth.
Note that we did not prove that the formula in the above definition actually gives the
length of a section of a curve. A rigorous proof requires dealing with some subtleties, nor-
mally glossed over in calculus texts, which are beyond the scope of this book.16
Example 1.41. Find the length L of the helix f( t) = (cos t,sin t, t) from t = 0 to t = 2 π.
Solution: By formula (1.41), we have
2 π
2 π
2 π
L =
(−sin t)2 + (cos t)2 + 12 dt =
sin2 t + cos2 t + 1 dt =
2 dt
0
0
0
= 2(2 π − 0) = 2 2 π
Similar to the case in R2, if there are values of t in the interval [ a, b] where the derivative
of a component function is not continuous then it is often possible to partition [ a, b] into
subintervals where all the component functions are continuously differentiable (except at
the endpoints, which can be ignored). The sum of the arc lengths over the subintervals will
be the arc length over [ a, b].
16In particular, Duhamel’s principle is needed. See the proof in TAYLOR and MANN, § 14.2 and § 18.2.
60
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Notice that the curve traced out by the function f( t) = (cos t,sin t, t) from Example 1.41 is also traced out by the function g( t) = (cos2 t,sin2 t,2 t). For example, over the interval [0, π],
g( t) traces out the same section of the curve as f( t) does over the interval [0, 2 π]. Intuitively,
this says that g( t) traces the curve twice as fast as f( t). This makes sense since, viewing the
functions as position vectors and their derivatives as velocity vectors, the speeds of f( t) and
g( t) are f ′( t) = 2 and g′( t) = 2 2, respectively. We say that g( t) and f( t) are different parametrizations of the same curve.
Definition 1.14. Let C be a smooth curve in R3 represented by a function f( t) defined on an
interval [ a, b], and let α : [ c, d] → [ a, b] be a smooth one-to-one mapping of an interval [ c, d]
onto [ a, b]. Then the function g : [ c, d] → R3 defined by g( s) = f( α( s)) is a parametrization of C with parameter s. If α is strictly increasing on [ c, d] then we say that g( s) is equivalent to f( t).
s
t
f( t)
α
f
[ c, d]
[ a, b]
R3
g( s) = f( α( s)) = f( t)
Note that the differentiability of g( s) follows from a version of the Chain Rule for vector-
valued functions (the proof is left as an exercise):
Theorem 1.21. Chain Rule: If f( t) is a differentiable vector-valued function of t, and t =
α( s) is a differentiable scalar function of s, then f( s) = f( α( s)) is a differentiable vector-valued function of s, and
df
df dt
=
(1.42)
ds
dt ds
for any s where the composite function f( α( s)) is defined.
Example 1.42. The following are all equivalent parametrizations of the same curve:
f( t) = (cos t,sin t, t) for t in [0,2 π]
g( s) = (cos2 s,sin2 s,2 s) for s in [0, π]
h( s) = (cos2 πs,sin2 πs,2 πs) for s in [0,1]
To see that g( s) is equivalent to f( t), define α : [0, π] → [0,2 π] by α( s) = 2 s. Then α is smooth, one-to-one, maps [0, π] onto [0, 2 π], and is strictly increasing (since α ′( s) = 2 > 0 for all s).
Likewise, defining α : [0, 1] → [0,2 π] by α( s) = 2 πs shows that h( s) is equivalent to f( t).
1.9 Arc Length
61
A curve can have many parametrizations, with different speeds, so which one is the best
to use? In some situations the arc length parametrization can be useful. The idea behind
this is to replace the parameter t, for any given smooth parametrization f( t) defined on [ a, b],
by the parameter s given by
t
s = s( t) =
f ′( u) du.
(1.43)
a
In terms of motion along a curve, s is the distance traveled along the curve after time t
has elapsed. So the new parameter will be distance instead of time. There is a natural
correspondence between s and t: from a starting point on the curve, the distance traveled
along the curve (in one direction) is uniquely determined by the amount of time elapsed, and
vice versa.
Since s is the arc length of the curve over the interval [ a, t] for each t in [ a, b], then it is a
function of t. By the Fundamental Theorem of Calculus, its derivative is
ds
d
t
s ′( t) =
=
f ′( u) du = f′( t) for all t in [ a, b].
dt
dt a
Since f( t) is smooth, then f ′( t) > 0 for all t in [ a, b]. Thus s ′( t) > 0 and hence s( t) is strictly increasing on the interval [ a, b]. Recall that this means that s is a one-to-one mapping of the
interval [ a, b] onto the interval [ s( a), s( b)]. But we see that
a
b
s( a) =
f ′( u) du = 0 and s( b) =
f ′( u) du = L = arc length from t = a to t = b
a
a
So the function s : [ a, b] → [0, L] is a one-to-one, differentiable
α( s)
s
t
mapping onto the interval [0, L]. From single-variable calculus,
[0, L]
[ a, b]
we know that this means that there exists an inverse function
α : [0, L] → [ a, b] that is differentiable and the inverse of s : [ a, b] →
s( t)
[0, L]. That is, for each t in [ a, b] there is a unique s in [0, L] such
that s
Figure 1.9.1 t
= s( t) and t = α( s). And we know that the derivative of α is
= α( s)
1
1
α ′( s) =
=
s ′( α( s))
f ′( α( s))
So define the arc length parametrization f : [0, L] → R3 by
f( s) = f( α( s)) for all s in [0, L].
Then f( s) is smooth, by the Chain Rule. In fact, f( s) has unit speed:
f ′( s) = f′( α( s)) α′( s) by the Chain Rule, so
1
= f′( α( s))
, so
f ′( α( s))
f ′( s) = 1 for all s in [0, L].
So the arc length parametrization traverses the curve at a “normal” rate.
62
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
In practice, parametrizing a curve f( t) by arc length requires you to evaluate the integral
s = t f′( u) du in some closed form (as a function of t) so that you could then solve for t in
a
terms of s. If that can be done, you would then substitute the expression for t in terms of s
(which we called α( s)) into the formula for f( t) to get f( s).
Example 1.43. Parametrize the helix f( t) = (cos t,sin t, t), for t in [0,2 π], by arc length.
Solution: By Example 1.41 and formula (1.43), we have
t
t
s =
f ′( u) du =
2 du = 2 t fo