2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS71
2.2
Di¤erentiation and Integration of Vector-Valued
Functions
Simply put, we di¤erentiate and integrate vector functions by di¤erentiating and
integrating their component functions. Since the component functions are realvalued functions of one variable, we can use the techniques studied in calculus
I and II.
2.2.1
Di¤erentiation
De…nition 116 Let !
r (t) be a vector function. The derivative of !
r with re!
d
r
is de…ned to be
spect to t, denoted !
r 0 (t) or
dt
!
r (t + h) !
r (t)
!
r 0 (t) = lim
h!0
h
!
0
Geometrically, r (a) is the vector tangent to the curve at t = a.
De…nition 117 The line tangent to a curve C with position vector !
r (t) at
t = a is the line through !
r (a) in the direction of !
r 0 (a).
De…nition 118 (Unit Tangent Vector) The unit tangent vector, denoted
!
T (t) is de…ned to be
!
!
r 0 (t)
T (t) = !0
(2.4)
k r (t)k
!
Remark 119 Of course, the above de…nition makes sense only if !
r 0 (t) 6= 0 .
The derivative is de…ned in terms of limits. Taking the limit of a vector
function amounts to taking the limits of the component functions. Thus, we
have the following theorem:
Theorem 120 If !
r (t) = hf (t) ; g (t) ; h (t)i then !
r 0 (t) = hf 0 (t) ; g 0 (t) ; h0 (t)i.
There is a similar result for plane curves.
Since the component functions are real-valued functions of one variable, all
the properties of the derivative will hold. We have the following theorem:
Theorem 121 Suppose that !
u and !
v are di¤ erentiable vector functions, c is
a scalar and f is a real-valued function. Then:
1. (!
u (t)
0
!
v (t)) = !
u 0 (t)
!
v 0 (t)
0
2. (c!
u (t)) = c!
u 0 (t)
u (t) + f (t) !
u 0 (t)
3. (f (t) !
u (t)) = f 0 (t) !
0
0
4. (!
u (t) !
v (t)) = !
u 0 (t) !
v (t) + !
u (t) !
v 0 (t)
72
CHAPTER 2. VECTOR FUNCTIONS
0
5. (!
u (t) !
v (t)) = !
u 0 (t) !
v (t) + !
u (t)
0
6. (!
u (f (t))) = f 0 (t) !
u 0 (f (t))
!
v 0 (t)
D
E
2
Example 122 Let !
r (t) = t; et ; sin 2t . Find !
r 0 (t) and the unit tangent
vector at t = 0. Then, …nd the equation of the tangent at t = 0.
1. Computation of !
r 0 (t).
D
E
2
!
r 0 (t) = 1; 2tet ; 2 cos 2t
2. Computation of the unit tangent at t = 0. First, we must …nd the tangent
vector at t = 0 . This vector is !
r 0 (0). From our computation above, we
see that
!
r 0 (0) = h1; 0; 2i
!
The unit tangent vector at t = 0 is T (0).
!
!
r 0 (0)
T (0) =
k!
r 0 (0)k
=
=
h1; 0; 2i
p
5
1
2
p ; 0; p
5
5
3. Computation of the tangent line. The parametric equations of the line
through !
r (0) = h0; 1; 0i with direction vector h1; 0; 2i is
8
< x=t
y=1
:
z = 2t
De…nition 123 (Smooth Curve) A curve C given by a position vector !
r (t)
on an interval I is said to be smooth if the conditions below are satis…ed:
1. !
r 0 (t) is continuous.
!
2. !
r 0 (t) 6= 0 except possibly at the endpoints of I.
Smooth curves will play an important role in the next sections. Geometrically, a curve is not smooth at points where there is a corner also called a
cusp.
Example 124 Consider the curve given by !
r (t) = 1 + t2 ; t3 . Find if it is
1
smooth on R. What about on (0; 1)?
We start by computing !
r 0 (t). !
r 0 (t) = 2t; 3t2 . Wee that !
r 0 (t) is always
!
!
0
continuous. However, r (t) = 0 when t = 0. Hence, this curve is not smooth
on R. It is smooth on (0; 1) since !
r 0 (t) is continuous there and never equal
!
to 0 .
2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS73
We …nish with the proof of a well known result which we state
sition.
Proposition 125 Let C be a curve given by a position vector !
r (t).
c (a constant) then !
r 0 (t) ? !
r (t) for all t.
2
Proof. We know that k!
r (t)k = !
r (t) !
r (t). So, we have !
r (t)
as a propoIf k!
r (t)k =
!
r (t) = c2 .
If we di¤ erentiate each side and use our rules of di¤ erentiation, we get
!
r 0 (t) !
r (t) + !
r (t) !
r 0 (t) = 0
2!
r 0 (t) !
r (t) = 0
!
r 0 (t) !
r (t)
=
0
This says that !
r 0 (t) ? !
r (t).
You may not recognize this result the way it is stated. Think of a circle.
The position vector of a circle is its radius. The theorem stated in the case of
a circle says that the radius of a circle is perpendicular to the tangent to the
circle.
2.2.2
Integration
De…nition 126 If !
r (t) = hf (t) ; g (t) ; h (t)i then
*Z
Z b
Z b
Z
b
!
r (t) dt =
f (t) dt;
g (t) dt;
a
and
a
Z
!
r (t) dt =
Z
a
f (t) dt;
Z
b
h (t) dt
a
g (t) dt;
We have similar de…nitions for plane curves.
Z
+
h (t) dt
R
!
1
. Find R (t) = !
r (t) dt
Example 127 Let !
r (t) = cos 2t; 2 sin t;
2
1+t
!
which satis…es R (0) = h3; 2; 1i.
Z
!
!
R (t) =
r (t) dt
Z
Z
Z
1
=
cos 2tdt;
2 sin tdt;
dt
1 + t2
1
=
sin 2t + C1 ; 2 cos t + C2 ; tan 1 t + C3
2
!
Since we want R (0) = h3; 2; 1i, we must have
h3; 2; 1i = h0 + C1 ; 2 + C2 ; 0 + C3 i
Thus, C1 = 3, C2 =
4 and C3 = 1. It follows that
!
R (t) =
1
sin 2t + 3; 2 cos t
2
4; tan
1
t+1
74
CHAPTER 2. VECTOR FUNCTIONS
2.2.3
Velocity and Acceleration
In this section, we look at direct applications of the derivative and the integral
of a vector function.
De…nition 128 (Velocity and Acceleration) Consider an object moving along
C, a smooth curve, twice di¤ erentiable, with position vector !
r (t).
1. The velocity of the object, denoted !
v (t) is de…ned to be
!
v (t) = !
r 0 (t)
(2.5)
2. The acceleration of the object, denoted !
a (t) is de…ned to be
!
a (t) = !
v 0 (t) = !
r 00 (t)
(2.6)
3. The speed of the object is k!
v (t)k.
Example 129 (Finding the velocity and acceleration of a moving object)
An object is moving along the curve !
r (t) = t; t3 ; 3t for t
0. Find !
v (t),
!
a (t) and sketch the trajectory of the object as well as the velocity and acceleration when t = 1.
Velocity:
Acceleration:
!
v (t) = 1; 3t2 ; 3
!
a (t) = h0; 6t; 0i
Sketch: When t = 1, we have
!
v (1) = h1; 3; 3i
and
!
a (1) = h0; 6; 0i
The trajectory of the object as well as the velocity and acceleration at t = 1
are shown in …gure 2.2.
In many applications, we do not know the position function. Instead, we
know the acceleration and we must …nd the velocity and position function. The
next example illustrates this.
Example 130 (Finding a Position Function by Integration) A moving particle starts at position !
r (0) = h1; 0; 0i with initial velocity !
v (0) = h1; 1; 1i.
!
Its acceleration is a (t) = h4t; 6t; 1i. Find its velocity and position function at
time t.
2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS75
Figure 2.2: Motion of an object along !
r (t) = t; t3 ; 3t
Velocity: Since !
a (t) = !
v 0 (t), it follows that
Z
!
v (t) = !
a (t) dt
Thus
!
v (t) = 2t2 + C1 ; 3t2 + C2 ; t + C3
We …nd the constants by using the initial condition.
!
v (0) = h1; 1; 1i = hC1 ; C2 ; C3 i
Thus, C1 = 1, C2 =
1 and C3 = 1. It follows that
!
v (t) = 2t2 + 1; 3t2
1; t + 1
Position Function: Since !
v (t) = !
r 0 (t), it follows that
Z
!
!
r (t) =
v (t) dt
=
2t3
+ t + C1 ; t3
3
t + C2 ;
t2
+ t + C3
2
We …nd the constants by using the initial condition.
!
r (0) = h1; 0; 0i = hC1 ; C2 ; C3 i
76
CHAPTER 2. VECTOR FUNCTIONS
Thus, C1 = 1 and C2 = C3 = 0. It follows that
!
r (t) =
2t3
+ t + 1; t3
3
t;
t2
+t
2
Remark 131 The above problem can be set up as a di¤ erential equation, that
is an equation which involves the derivatives of an unknown function. Solving the equation amounts to …nding the unknown function. We were given
the acceleration !
a (t) and we had to …nd the position function !
r (t). Since
!
!
00
a (t) = r (t), the above problem could have been stated as: Find !
r (t) given
!
!
!
00
that r (t) = h4t; 6t; 1i, r (0) = h1; 0; 0i and v (0) = h1; 1; 1i. The last two
conditions are called the initial conditions because they are the conditions
when t = 0. Of course, we would solve the problem the same way in this particular example. However, many di¤ erential equations which appear in applied
mathematics are far more challenging to solve.
Remark 132 In general, we can recover velocity by integration when acceleration is known. Similarly, we can recover position by integration when velocity
is known. More precisely, if !
a (t0 ) and !
v (t0 ) are known, then
!
v (t) = !
v (t0 ) +
Z
t
t0
!
a (u) du
This can be derived from the de…nition of acceleration. Since a (t) = v 0 (t),
Rt
Rt
integrating from t0 to t gives t0 a (u) du = t0 v 0 (u) du. Using the fundamental
Rt
theorem of calculus, we get t0 a (u) du = v (t) v (t0 ) hence the result. Similarly,
if !
v (t) and !
r (t0 ) are known, then
!
r (t) = !
r (t0 ) +
Z
t
t0
!
v (u) du
This can be proven the same way as the formula for velocity.
Remark 133 When studying the motion of an object, the time at which we
start tracking the object is usually set to 0. The acceleration at time t = 0 is
called initial acceleration and denoted !
a 0 . Similarly, the velocity at time
t = 0 is called initial velocity and usually denoted !
v 0 . The position at time
t = 0 is called initial position and usually denoted !
r 0 . The term initial
means at time t = 0.
2.2.4
Projectile Motion
If you wondered how we can know the acceleration and not the other quantities,
this section will hopefully answer your questions. We begin with a little bit of
!
physics. Newton’s second law of motion states that if at time t a force F (t)
2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS77
acts on an object of mass m, this action will produce an acceleration !
a (t) of
the object satisfying
!
F (t) = m!
a (t)
Thus, if we know the force acting on the object, we can …nd the acceleration.
We will then be in the situation of the previous example. We illustrate this
with a classical problem in physics, the problem of …nding the trajectory of an
object being thrown in the air and subject to the laws of physics.
Example 134 A projectile is …red with an angle of elevation and initial velocity !
v 0 as shown in …gure 2.3. If we neglect the air resistance, the only force
acting on the object is gravity.
1. Find the position of the object !
r (t).
2. Express the range d in terms of
3. Find the value of
.
which maximizes the range d.
Answer to 1. We begin by setting the axes so that y is the vertical direction
and x the horizontal direction. In addition, the motion of the projectile
begins at the origin. The only force the object is subject to is gravity. Since
the force of gravity acts downward, we have
!
a (t) = h0; gi
where g = 9:81m=s2 . We can now compute !
v (t).
Z
!
!
v (t) =
a (t) dt
= hC1 ; gt + C2 i
!
Let v 0 = hv0x ; v0y i. Then, we have
!
v (0) = hv ; v i = hC ; C i
0x
Thus
0y
1
2
!
v (t) = hv0x ; gt + v0y i
Elementary trigonometry tells us that
v0x = k!
v 0 k cos
!
v
= k v k sin
0y
0
Thus
!
v (t) = hk!
v 0 k cos ; gt + k!
v 0 k sin i
!
Now, we can compute r (t).
Z
!
!
r (t) =
v (t) dt
=
(k!
v 0 k cos ) t + C1 ;
1 2
gt + (k!
v 0 k sin ) t + C2
2
78
CHAPTER 2. VECTOR FUNCTIONS
Since !
r (0) = h0; 0i, both C1 and C2 are 0 hence
!
r (t) =
(k!
v 0 k cos ) t;
1 2
gt + (k!
v 0 k sin ) t
2
So, we can see that the motion in the x direction and in the y direction
follow di¤ erent laws. This can be made more obvious by writing the parametric equations of the trajectory of the object.
(
x = (k!
v 0 k cos ) t
1 2
gt + (k!
v 0 k sin ) t
y=
2
Answer to 2. d, the range, is the horizontal distance corresponding to y = 0.
We can …nd d by …nding the value of t for which y = 0 and plugging this
value in the equation for x.
y
=
()
1 2
gt + (k!
v 0 k sin ) t = 0
2
1
gt + k!
v 0 k sin
=0
2
0 ()
t
2 k!
v 0 k sin
. t = 0 corresponds
g
2 k!
v 0 k sin
to the initial position. The value of t we want is t =
. This
g
value gives us
This happens when either t = 0 or t =
d =
d =
2 k!
v 0 k sin
(k!
v 0 k cos )
g
2
!
k v k sin 2
0
g
Answer to 3. We see that d is maximum when sin 2 = 1 that is when
=
4
.
Make sure you have read, studied and understood what was done above
before attempting the problems.
2.2.5
Problems
Do odd # 1 - 25 at the end of 10.1 in your book.
Do odd # 1 - 7, 13, 15, 17 at the end of 10.2 in your book.
2.3. ARC LENGTH AND CURVATURE
79
Figure 2.3: Trajectory of a Projectile
2.3
Arc Length and Curvature
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