Lecture Notes
Multivariable Calculus
II Semester 2015
Department of Mathematics
University of the West Indies
Bogotá, Jamaica
Dr. Davide Batic
Contents
1 Scalar and vector fields.
1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Scalar and vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
3
2 Problems
7
3 Line, surface, and volume integrals
3.1 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
8
4 Problems
14
1
Preface
These class notes are the script for the class Multivariable Calculus I held for the first
time to the students in Mathematics, Physics and Actuarial Science at the University
of the West Indies during the second semester 2013. I underline that this manuscript
is by no means a book on multivariable calculus. To be honest, at the very beginning
I decided to write these notes for myself with the aim of presenting the selected topics
in a structured form and with the hope of reducing the time needed to prepare future
lectures. Later on, I realized that it would be good for the students to have such notes
at disposal since they could offer them the possibility to read what we discussed in class
and to check some of the most difficult computations or proofs. The experience shows
that one really understands mathematics only if he is able to reproduce it by its own.
Last but not least, I hope the role played by the students in the improvement process of
the class notes will be important. Together with this manuscript there is a collection of
exercises that can be downloaded at the following link
http : //www.mona.uwi.edu/mathematics/math2403 − multivariable − calculus
2
1
Scalar and vector fields.
Topics: definition of a scalar field, examples of scalar fields and their graphical representation, definition of a vector field, examples of vector fields and their graphical
representation (wind charts), conservation of angular momentum in a central field.
1.1
Introduction.
Multivariable Calculus extends Calculus from the real line to the Euclidean vector space
Rn and instead of working with functions of a real variable we will deal with new mathematical objects such as scalar and vector fields. The Euclidean vector space is simply
the collection of n-tuples
Rn = {(x1 , · · · , xn ) | xi ∈ R ∀i = 1, · · · , n}
equipped with addition and scalar multiplication, where n is a natural number. If
u = (u1 , · · · , un ) is any vector in the Euclidean space we denote its length also called
magnitude or norm as follows
v
u n
q
uX
|u| = t
u2i = u21 + · · · + u2n .
i=1
If not otherwise stated u1 , · · · , un denote the components of the vector u with respect
to the standard basis B = {e1 , · · · , en } of Rn where
e1 = (1, 0, · · · , 0),
e2 = (0, 1, 0 · · · , 0), · · · , en = (0, 0, · · · , 1).
Clearly, u can also be written as a linear combination over the basis vectors, that is
u = u 1 e1 + · · · + u n en =
n
X
u i ei .
i=1
We will derive most of the results for the case n = 2 (Euclidean plane) and n = 3
(Euclidean space).
1.2
Scalar and vector fields
A scalar field is a map assigning a real number to each point in space. Remember that a
point in space can be identified by a vector. This is captured by the following definition
Definition 1.1 A scalar field is a map F : Ω ⊆ Rn −→ I ⊆ R such that
(x1 , · · · , xn ) 7−→ F (x1 , · · · , xn ) ∈ R.
3
For instance, the temperature distribution in a room is an example of a scalar field. Other
examples of scalar fields are the pressure and the density of a material object. Sometimes
it is useful to construct a graphical representation of a scalar field. Let us suppose that
we have a temperature distribution T : R2 −→ [0, +∞) with T (x, y) = x2 + y 2 . If we
look at T (x, y) as to a third variable z, the problem boils down to construct a graphical
representation for the surface z = x2 + y 2 in the Euclidean space R3 . For instance, we
could imagine to cut such a surface with different planes parallel to the xy-plane. Hence,
if z = 0 we get x2 + y 2 = 0 and this equation is satisfied if and only if x = y = 0. This
means that our surface touches the plane z = 0 only at the point (0, 0). If z = 1, we
obtain x2 + y 2 = 1 which is the equation of a circle of radius one and center at (0, 0)
and if z = 4, the intersection of this plane with the given surface will be represented by
the circle x2 + y 2 = 4 with radius 2 and again centered at the origin of the Cartesian
axes. By choosing more planes parallel to the xy-plane we end up with what we call a
contour plot of the scalar field. Another representation of the same scalar field can be
Figure 1: Contour plot of the scalar field T (x, y) = x2 + y 2 .
constructed by keeping in mind the result obtained from the contour plot and adding to
it the information we get by intersecting our surface with the main planes orthogonal to
the xy-plane. For instance, if we consider the intersection with the plane x = 0 which
is orthogonal to the plane xy-plane we obtain the relation z = y 2 which is the equation
4
Figure 2: Plot of the scalar field T (x, y) = x2 + y 2 .
of a parabola. Similarly, if we choose the plane y = 0 we obtain the parabola z = x2 .
The surface represented in Fig. 2 is called a paraboloid of revolution because it can
be obtained by rotating a parabola with vertex at the the origin around the z-axis. The
circles appearing in Fig. 2 are the same circles you see also in Fig. 1 where the third
dimension along z has been suppressed. Finally, notice that such circles represents points
in space characterized by the same value of the temperature. A vector field is a map
sending vectors into vectors. More precisely we have the following definition
Definition 1.2 A vector field is a map V : Ω ⊆ Rn −→ Γ ⊆ Rm such that
(x1 , · · · , xn ) 7−→ V(x1 , · · · , xn ) = (V1 (x1 , · · · , xn ), · · · , Vm (x1 , · · · , xn )),
where the natural numbers n and m do not necessarily need to coincide.
Note that the above vector field associates to each vector in Rn a vector (V1 , · · · , Vm )
belonging to Rm . Furthermore, each component of the vector field can be interpreted as
a scalar field since for any i = 1, · · · , n the function Vi maps the vector (x1 , · · · , xn ) into
a real number! An example of a vector field is the velocity
v=
5
dr
dt
of an object at the time t whose vector position r = r(t) is some vector-valued function
of t. Clearly, v and r are vector fields with n = 1 and m = 3 since they associate to a
certain time t a vector in the Euclidean space. Other examples of vector fields are the
acceleration of a material object a = dv/dt, the force field F, the electric and magnetic
fields E and B appearing in Maxwell’s theory of electromagnetism and so on. In the
case of simple vector fields, we can construct graphical representations of vector fields
which are called wind charts. For instance, let us suppose that a certain wind velocity
is described by the vector field V : R2 −→ R2 such that V(x, y) = (y, x). We can choose
a sequence of points in the domain of definition of V and compute the corresponding
vectors. Then, we draw these vectors as arrows emanating from the corresponding points.
For instance, at the point (1, 0) we have the vector (0, 1), at the point (1, 1) the vector
(1, 1), and choosing more points we end up with the wind chart represented by Fig. 3.
The next example concerning the conservation of the angular momentum of a particle
Figure 3: Wind chart of V(x, y) = (y, x).
subject to a central force is an important application of vector fields.
Example Suppose we have a particle of mass m at the time t located at the point
(x, y, z) ∈ R3 . We further suppose that the particle follows a certain trajectory as time
evolves. This means that the coordinates of the particle are some functions of time, say
x = x(t), y = y(t), and z = z(t). We will further suppose that these real functions are
6
differentiable. At this point the position of the particle can be identified by the position
vector r(t) = (x(t), y(t), z(t)) and its velocity and acceleration are given by
v=
dr
,
dt
a=
dv
.
dt
Let us imagine that our particle experiences the influence of a central force, that is a
force represented by the vector
p field F = −f (r)r, where f is some positive continuous
function of the distance r = x2 + y 2 + z 2 of the particle from the origin of a Cartesian
system of coordinates. Note that the minus sign in the expression of the force signalizes
that the force has opposite direction with respect to the position vector r. This means
that the force attracts the particle to the origin. In physics the angular momentum of a
particle of mass m with velocity v and position vector r is defined as the cross product
L = mr × v.
We want to show that under the above hypotheses the angular momentum of the particle
is conserved in time, that is it remains constant as the time varies. This will be the case
if we can show that dL/dt = 0. To do that we first observe that
dL
dr
dv
= m × v + mr ×
= mv × v + mr × a = r × (ma) = r × F,
dt
dt
dt
where we used Newton’s law F = ma and the fact that v × v = 0 since the vector v is
parallel to itself. Taking into account that we have a central force, we finally obtain
dL
= −r × (f (r)r) = −f (r)r × r = 0.
dt
2
Problems
Exercise 2.1 (First incourse test AY 2012-13) Construct the contour plot of the scalar
field T : R2 −→ R such that T (x, y) = x2 − y. Sketch also the wind chart for the vector
field V : R2 −→ R2 such that V(x, y) = (x + y, −x).
Exercise 2.2 (First incourse test AY 2013-14) Sketch the following surfaces in R3
1. z = −y 2 ,
2. z = x2 + y 2 − 4x − 6y + 13.
7
3
Line, surface, and volume integrals
Topics: smooth curve in R2 , examples of smooth curves, arclength of a smooth curve,
example of a computation of the length of a smooth curve, line integral of a scalar
field along a smooth curve, examples, line integral of a scalar field with respect to a
coordinate along a smooth curve, example, a vector-valued integral of a scalar field along
a smooth curve, example, line integral of a vector field, example, a vector-valued line
integral of a vector field along a smooth curve, parametric surfaces, examples, parametric
representation of a sphere, normal vector to a surface, construction of a tangent plane to
a surface at a given point, definition of surface area, examples, surface integrals of scalar
fields, examples, surface integrals of vector fields, examples, volume integrals, physical
motivation, examples.
3.1
Line integrals
Given a curve C in R2 or R3 we want to know how to compute integrals of scalar and
vector fields along the given curve. First we need to give a rigorous definition of curve.
Examples of curves in R2 are presented in Fig. 4 Each point P on the curve C1 can be
y
B
P = (x, y)
C1
C2
r
A
x
0
Figure 4: Examples of smooth curves.
identified by a position vector r that in turn can be used to describe the whole curve if
the coordinates of the points belonging to C1 are suitably parameterized. The curve C2
is an example of a closed curve, i.e. a curve where the initial and final points coincide.
Definition 3.1 A smooth curve in the Euclidean plane is a vector-valued function
r : [a, b] ⊂ R −→ R2 such that
8
1. t ∈ [a, b] 7−→ r(t) = (x(t), y(t)),
2. the following derivative exists and is unique for all t ∈ [a, b]
dr
dx dy
=
,
.
dt
dt dt
We call parametric representation of the curve C the set of equation x = x(t) and
y = y(t). We will say that the curve C is closed if r(a) = r(b).
As we will see in the next examples, the second condition in the above definition ensures
that the curve is smooth, that is it does not exhibit corners or cusps.
Example Consider
√ the curve C in the Euclidean plane described by the relation
√ y =
2
2−x with x ∈ [0, 2]. The curve starts at A = (0, 2) and ends at the point B = ( 2, 0).
See Fig. 5. We want to find a parametric representation of C. The simplest choice is to
y
A
C
0
x
B
Figure 5: Graphical representation of the curve y = 2 − x2 .
identify x with a parameter t ∈ [0,
representation is given by
x = t,
√
2]. Then, y = 2 − x2 = 2 − t2 and the parametric
y = 2 − t2 ,
t ∈ [0,
√
2].
This implies that the vector-valued function r identifying points on C will be
√
r(t) = (t, 2 − t2 ), t ∈ [0, 2].
It is interesting to observe that the parametric representation of a curve does not need
to be unique. If we decide instead to parameterize the coordinate y according to the
relation y = t, since y varies on the interval [0, 2] (see Fig. 5), we must require that
t ∈ [0, 2]. Then, using the Cartesian equation of C we find that t = 2 − x2 . Solving for x
9
√
√
yields
x
=
±
2
−
t.
Since
x
∈
[0,
2], we have to choose the positive root and therefore
√
x = 2 − t. Hence, in this case we obtained a different but equivalent parameterization
given by
√
x = 2 − t, y = t, t ∈ [0, 2].
In the next example we look at a curve that fails to be smooth.
Example Let us consider a path C obtained by joining the line segments C1 and C2 as
in Fig. 6. It is easy to see that C1 is parameterized as
y
C1
A
D
C2
0
B
x
Figure 6: Example of a non smooth curve with initial point A and final point B.
x = t,
y = 1,
t ∈ [0, 1]
and C2 can be represented as
x = 1,
y = 1 − t,
t ∈ [0, 1].
Therefore, C1 and C2 are described for t ∈ [0, 1] by the vector-valued functions r1 (t) =
(t, 1) and r2 (t) = (1, t − 1), respectively. If we compute the derivatives of r1 and r2 at
the point D = (1, 1), we find that such derivatives exist but they do not coincide since
dr1
= (1, 0),
dt
dr2
= (0, −1).
dt
Our curve fails to be smooth due to the presence of a corner at D.
Example We want to construct the parametric representation of the upper half-circle
x2 + y 2 = 1 depicted in Fig. 8. Let x = cos θ with θ ∈ [0, π]. Then, from the equation
x2 + y 2 = 1 we get
y 2 = 1 − x2 = 1 − cos2 θ = sin2 θ.
10
y
P (x, y)
C
r
θ
B
A
0
x
Figure 7: Representation of the upper half-circle x2 + y 2 = 1 with start and end points
at A and B, respectively.
Taking the square root we obtain y = ±| sin θ|. Since θ ∈ [0, π], it follows that sin θ ≥ 0
there and hence | sin θ| = sin θ. Furthermore, we are considering the upper half-circle
where y ≥ 0. This implies that we have to take the root with the plus sign. At this
point we can conclude that y = sin θ. Thus, the parametric representation of the curve
C is
x = cos θ, y = sin θ, θ ∈ [0, π].
The corresponding vector-valued function is r(θ) = (cos θ, sin θ) with θ ∈ [0, π].
Example Let us derive the parametric representation of the line segment in Fig. 8. The
initial and final points are identified by the vectors rA = (xA , yA ) and rB = (xB , yB ).
We look for a vector-valued function r = r(t) such that r(0) = rA and r(1) = rB . This
means that the general expression for r must depend on rA , rB , and the parameter t. A
comparison with the equation of a line in the Euclidean plane suggests that it would be
reasonable to assume that r is linear in t and therefore we consider the following guess,
namely
r(t) = (at + b)rA + (ct + d)rB ,
where a, b, c, d are real constants to be determined by means of the conditions r(0) = rA
and r(1) = rB . Employing such conditions we end up with the following system of
equations
brA + drB = rA , (a + b)rA + (c + d)rB = rB .
We see that it must be b = 1, d = 0, a + b = 0, and c + d = 1. Solving this system yields
a = −1,
b = 1,
11
c = 1,
d = 0.
y
B
C
A
rB
rA
x
0
Figure 8: Representation of a line segment with start and end points at A and B,
respectively.
Finally, the vector-valued equation describing the line segment is
r(t) = (1 − t)rA + trB .
(1)
The corresponding parametric equations are
x(t) = (1 − t)xA + txB ,
y(t) = (1 − t)yA + tyB .
Suppose that C is an arbitrary smooth curve with initial point A and end point B,
respectively. Let r = r(t) with t ∈ [tA , tB ] be a vector-valued function describing C. If
we consider an infinitesimal line element of the curve, then its length ds can be computed
by Pythagoras’s theorem and we find that
p
ds = (dx)2 + (dy)2 .
Let x = x(t) and y = y(t) be some parametric representation of C. Then, the Chain
Rule applied to ds yields
s
2 2 s 2 2
dy
dy
dx
dx
ds =
dt +
dt =
+
dt.
dt
dt
dt
dt
Hence, the length of C is
L=
Z
ds =
C
Z
tB
dt
tA
s
dx
dt
2
+
dy
dt
2
.
Example We want to compute the length of the curve described by the equation
r
3 9 2
y(x) = 1 +
x , y ∈ [1, 4].
4
12
(2)
y
A
B
C
E
x
0
Figure 9: Representation of y(x) = 1 +
√
and (2 3, 4), respectively.
q
3
9 2
4x
√
with start and end points at (−2 3, 4)
Fig. 9 represents the above function for y ∈ [1, 4]. Even though our curve is not smooth
due to the presence of a cusp at the point E = (0, 1), we observe that the function is
even, that is y(−x) = y(x) and hence the total lenght L will be simply twice the length
of the smooth curve with start point E and end point B. To construct a parametric
representation of that portion of the curve, let y = t with t ∈ [1, 4]. Then, x as a function
of t can be found by solving the equation
r
3 9 2
x ,
t=1+
4
whose solutions are
2
x = ± (t − 1)2/3 .
3
Since x ≥ 0 for the portion of curve between the points E and B, we conclude that
2
x = (t − 1)2/3 .
3
In order to apply (2) observe that
dx √
= t − 1,
dt
Hence, we have that
L=2
Z
C
ds = 2
E
Z
13
dy
= 1.
dt
4
1
√
28
dt t = .
3
We give the definition of line integral of a scalar field.
Definition 3.2 Let F = F (x, y) be a scalar field and C a smooth curve described by the
vector-valued function r(t) = (x(t), y(t)) with t ∈ [a, b]. The line integral of F along C is
defined as
s 2
Z b
Z
dy
dx 2
dt F (x(t), y(t))
ds F (x, y) =
+
.
(3)
dt
dt
a
C
Example Let us evaluate the line integral of the scalar field F (x, y) = xy 4 along the
smooth curve C represented by the positive quarter of the circle x2 + y 2 = 1 followed in
the counterclockwise direction and then along the line segment with start point (−2, 1)
and end point (1, 2). For the quarter of circle we use the parameterization x = cos θ
and y = cos θ with θ ∈ (0, π/2). Then, the infinitesimal length ds along the curve C is
obtained by applying (2) which gives ds = dt. Hence,
Z
ds F (x, y) =
C
Z
π/2
0
1
dθ cos θ sin4 θ = .
5
Concerning the integration along the line segment we apply (1) which
leads to the
√
representation r(t) = (3t − 2, t + 1) with t ∈ [0, 1]. In this case ds = 10dt and
√
Z
√ Z 1
10
4
ds F (x, y) = 10
.
dt (3t − 2)(t + 1) =
2
C
0
4
Problems
Exercise 4.1 Determine
the√length of the curve x = y 2 /2 for x ∈ [0, 1/2]. Assume y
√
positive. Answer: [ 2 − ln ( 2 − 1)]/2.
14
References
[1] P. C. Matthews, Vector Calculus, Springer Verlag (2005)
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