University of Zimbabwe
HMTH111 CALCULUS 2
Calculus of Several Variables
Department:
Mathematics
Author:
P. Sibanda
March 3, 2019
Chapter 1
Vectors and the Geometry of Space
A vector is a quantity that is characterized by magnitude and direction. We also use the term
length for magnitude. A scalar, on the other hand, is a quantity which has magnitude only. To
differentiate the types of quantities, let’s consider a typical vector, displacement or change of
position. In order to specify displacement, we need to know two things: how far? and in what
direction? In other words, we need to specify distance and the direction. Thus we see that
distance is a scalar whereas displacement is a vector.
Consider the following two situations:
1. A complete stranger to Zimbabwe is in Harare and wants to travel to Masvingo. Is it sufficient
to simply tell them that Masvingo is 350 kilometers from Harare?
2. A well-informed Zimbabwean wants to travel from Harare to Masvingo. Will it be sufficient
to tell them that Masvingo is 350 kilometers from Harare?
Clearly, the information in the first case is not sufficient as the stranger would also want to know
the direction in which to travel. However, in the second case, it is assumed that the person already
has some idea of the location of Masvingo relative to Harare and so specifying only distance would
suffice.
The following are examples of vectors: force, displacement, acceleration, momentum and velocity .
However, the following quantities are scalars and not vectors: area, volume, distance, speed, energy,
work, electrical resistance, temperature, mass and time.
1
1.1
Basic Definitions and Notation
−→
Graphically a vector is represented by an arrow OP defining the direction, the magnitude of the
vector being indicated by the length of the arrow. The tail end O of the arrow is called the origin
or initial point of the vector, and the head P is called the terminal point or terminus. This arrow
−→
representing the vector is called a directed line segment. The length |OP | is the magnitude of the
line segment from O to P .
terminal point
P
3
initial point
O
Figure 1.1: Directed line segment
Vectors can be represented in text by bold-case letters, such as A, B, C and so on or lower-case
boldface letters such as a, b, c and so on. When written by hand, however, vectors are often
−
→
→
denoted by letters with arrows above them, such as −
a , b and so on or a bar above, such as a, b
and so on or a bar below, such as a, b and so on. When the initial point of the vector is fixed, it is
called a fixed or localized vector, otherwise, it is a free vector.
1.2
Unit Vectors
A unit vector is a vector of unit length. A unit vector is sometimes denoted by eb or b
e . Therefore,
|b
e| = 1.
Any vector can be made into a unit vector by dividing it by its length, that is,
b
e=
So
u
.
|u|
u
is a unit vector in the direction of the vector u.
|u|
2
In three-dimensional space R3 , we denote the unit vectors in the positive x -axis, positive y-axis and
positive z -axis by i, j, k respectively. Thus the position vector of a point (x, y, z) is
xi + yj + zk.
In a similar way, the position vector r of a point (x, y) in two-dimensional space R2 is
xi + yj.
In the notation above, the numbers x, y, z are the components of the vectors.
We also denote vectors
in R3 and R2 using column vectors. For example the vector xi + yj + zk
 
( )
x
x
in R3 is denoted by y  and the vector xi + yj in R2 is denoted by
.
y
z
1.3
Magnitude or Length of a Vector
The magnitude or length of a vector r is denoted by |r|. If a vector r = xi + yj + zk, then it can
be easily shown by use of Pythagoras’ theorem that
√
|r| = x2 + y 2 + z 2 .
√
√
√
For example if r = i − 2j + 2k, then the magnitude |r| = 12 + (−2)2 + 22 = 1 + 4 + 4 = 9 = 3.
Example 1.3.1. Given A = 3i − 2j + k, B = 2i − 4j − 3k and C = −i + 2j + 2k, find the magnitudes
of (i) C, (ii) A + B + C and (iii) 2A − 2B − 5C.
√
Solution: (i) |C| = | − i + 2j + 2k| = (−1)2 + 22 + 22 = 3.
(ii) A + B + C = 3i − 2j + k + 2i − 4j − 3k − i + 2j + 2k
√ = (3 + 2 − 1)i +
√(−2 − 4√+ 2)j + (1 − 3 + 2)k =
4i − 4j + 0k. Then |A + B + C| = |4i − 4j + 0k| = 42 + (−4)2 = 32 = 4 2.
(iii) 2A − 2B − 5C = 2(3i − 2j + √
k) − 2(2i − 4j − 3k) − 5(−i
√ + 2j + 2k) = 7i − 6j − 2k. Then
2
2
2
|2A − 2B − 5C| = |7i − 6j − 2k| = 7 + (−6) + (−2) = 89.
Example 1.3.2. Find the component form and magnitude of the vector A having initial point
(−2, 3, 1) and terminal point (0, −4, 4). Then find a unit vector in the direction of A.
Solution: The component form of A is
A = (0 − (−2), −4 − 3, 4 − 1) = (2, −7, 3)
which implies that its magnitude is
|A| =
√
√
22 + (−7)2 + 32 = 62.
The unit vector in the direction of A is
1
A
= √ (2, −7, 3) =
U=
|A|
62
3
(
2
−7
3
√ ,√ ,√
62 62 62
)
.
1.4
Parallel Vectors
Two non-zero vectors A and B are parallel if there is some scalar c such that A = cB.
Example 1.4.1. Vector A has initial point (2, −1, 3) and terminal point (−4, 7, 5). Which of the
following vectors is parallel to A? (i) B = (3, −4, −1) and (ii) C = (12, −16, 4).
Solution: Writing A in component form
A = (−4 − 2, 7 − (−1), 5 − 3) = (−6, 8, 2).
(i) Because B = (3, −4, −1) = − 21 (−6, 8, 2) = − 21 A, then B is parallel to A.
(ii) In this case, you want to find a scalar c such that
(12, −16, 4)
12 = −6c
−16 = 8c
4 = 2c
=
⇒
⇒
⇒
c(−6, 8, 2)
c = −2
c = −2
c = 2.
Because there is no c for which the equation has a solution, the vectors are not parallel.
Definition
Two or more vectors are said to be collinear vectors, when they are along the same lines or parallel
lines.
Theorem 1.4.1. Let a and b be non-zero and non-collinear vectors. Then xa + yb = 0 implies
that x = y = 0.
Proof. The proof is by contradiction. Suppose xa + yb = 0 where x ̸= 0. This means that
a = −( xy )b. Thus the vectors a and b are parallel. In other words they are parallel to the same
line or are collinear. Contradiction. Hence x must be equal to zero and so yb = 0. Therefore y = 0
as b ̸= 0.
Theorem 1.4.2. Let a and b be non-zero and non-collinear vectors. Then x1 a + y1 b = x2 a + y2 b
implies that x1 = x2 and y1 = y2 .
The proof of this theorem is left as an exercise for you.
1.5
Laws of Vector Algebra
If A, B and C are vectors and m and n are scalars, then
4
(i) A + B = B + A
Commutative Law for Addition.
(ii) A + (B + C) = (A + B) + C
(iii) m(nA) = mnA = n(mA)
(iv) (m + n)A = mA + nA
Associative Law for Multiplication.
Distributive Law.
(v) m(A + B) = mA + mB
1.6
Associative Law of Addition.
Distributive Law.
The Dot or Scalar Product
So far we have studied two operations with vectors, vector addition and multiplication by a scalar,
each of which yield another vector. In this section you will study a third vector operation, called
the dot product, this product yields a scalar, rather than a vector.
The dot or scalar product of two vectors A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k, denoted
by A · B (read A dot B) is defined as the product of the magnitudes of A and B and the cosine of
the angle θ between them, that is,
A · B = |A||B| cos θ = A1 B1 + A2 B2 + A3 B3
where 0 ≤ θ ≤ π.
Example 1.6.1. Given A = 2i + 4j + 6k and B = i − 3j + 2k. Compute the scalar product of A
and B.
Solution: From the definition, the scalar product is given by
A · B = (2) · (1) + (4) · (−3) + (6)(2) = 2 − 12 + 12 = 2.
The following laws are valid
(i) A · B = B · A
Commutative Law for Dot Products.
(ii) A · (B + C) = A · B + A · C
Distributive Law.
(iii) m(A · B) = (mA) · B = A · (mB) = (A · B)m, where m is a scalar.
(iv) i · i = j · j = k · k = 1, i · j = j · k = k · i = 0.
(v) If A · B = 0 and A and B are not null vectors, then A and B are perpendicular.
5
1.6.1
Angle Between Two Vectors
If θ is the angle between two non-zero vectors A and B, then
cos θ =
A·B
.
|A||B|
A
9
θ
-
B
Figure 1.2: The angle between two vectors
1.6.2
Definition of Orthogonal Vectors
The vectors A and B are orthogonal if A · B = 0. Two non-zero vectors are orthogonal if and
π
only if the angle between them is θ = .
2
Example 1.6.2. For A = 3i − j + 2k, B = −4i + 2k, C = i − j − 2k and D = 2i − k, find the angles
between the following pairs of vectors. (i) A and B (ii) A and C (iii) B and D.
Solution:
−12 + 4
A·B
−8
−4
= √ √ = √ √ = √ . Because A · B < 0,
(i) cos θ =
14 20
2 14 5
70
( |A||B|
)
−4
= 2.069radians.
θ = cos−1 √
70
A·C
3+1−4
0
(ii) cos θ =
= √ √ = √
= 0. Because A · C = 0, A and C are orthogonal.
|A||C|
14 6
84
π
Furthermore, θ = .
2
B·D
−8 + 0 − 2
−10
(iii) cos θ =
= √ √
=√
= −1. Consequently, θ = π.
|B||D|
20 5
100
6
Exercise
Prove that a parallelogram ABCD is a rhombus if and only if its diagonals are orthogonal.
1.7
The Cross or Vector Product
Many applications in physics, engineering and geometry involve finding a vector in space that is
orthogonal to two given vectors. In this section we will study a product that will yield such a vector.
The cross or vector product of A and B is a vector C = A × B (read A cross B),
A × B = |A||B| sin θn,
where θ is the angle between the vectors, and the unit vector n is perpendicular to both A and B,
with A, B and n forming a right-handed system.
The following laws are valid.
(i) A × B = −B × A
Anticommutative Law.
(ii) A × (B + C) = A × B + A × C
Distributive Law.
(iii) m(A × B) = (mA) × B = A × (mB) = (A × B)m, where m is a scalar.
(iv) i × i = j × j = k × k = 0, i × j = k, j × k = i, k × i = j.
(v) If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k, then
i
j
k
A × B = A1 A2 A3 .
B1 B2 B3
(vi) The magnitude of A × B is the same as the area of a parallelogram with sides A and B.
(vii) If A × B = 0 and A and B are not null vectors, then A and B are parallel.
Example 1.7.1. Given A = i − 2j + k and B = 3i + j − 2k, find the following. (i) A × B
(ii) B × A (iii) B × B.
Solution:
(i)
i j
k
−2 1
1 1
1 −2
A × B = 1 −2 1 =
i−
j+
k = 3i + 5j + 7k.
1 −2
3 −2
3 1
3 1 −2
(ii)
i j
k
1 −2
3 −2
3 1
B × A = 3 1 −2 =
i−
j+
k = −3i − 5j − 7k.
−2 1
1 1
1 −2
1 −2 1
7
B
y
|B| sin θ
θ
-
A
Figure 1.3: The vectors A and B form adjacent sides of a parallelogram
(iii)
i j k
B × B = 3 1 −2 = 0.
3 1 −2
Example 1.7.2. Find the area of the parallelogram determined by A = i+j−3k and B = −6j+5k.
Solution:
i j
k
A × B = 1 1 −3 = −13i − 5j − 6k.
0 −6 5
Therefore
|A × B| =
√
√
(−13)2 + 52 + 62 = 230
which is the desired area.
Example 1.7.3. Find a unit vector that is orthogonal to both A = i − 4j + k and B = 2i + 3j.
Solution: The cross product A × B is orthogonal to both A and B.
i j k
A × B = 1 −4 1 = −3i + 2j + 11k.
2 3 0
√
√
Because |A × B| = (−3)2 + 22 + 112 = 134, a unit vector orthogonal to both A and B is
A×B
3
2
11
= −√
i+ √
j+ √
k.
|A × B|
134
134
134
1.8
The Scalar Triple Product
For vectors A, B and C in space, the dot product of A and B × C
A · (B × C)
is called the scalar triple product. The following laws are valid.
8
(i) A · (B × C) = B · (C × A) = C · (A × B) = volume of a parallelopiped having A, B and C
as edges.
If A = A1 i + A2 j + A3 k, B = B1 i + B2 j + B3 k and C = C1 i + C2 j + C3 k, then
A1 A2 A3
A · (B × C) = B1 B2 B3 .
C1 C2 C3
(ii) As a consequence, the volume of the parallelopiped is 0 if and only if the three vectors are
coplanar. That is, if the vectors A = (A1 , A2 , A3 ), B = (B1 , B2 , B3 ) and C = (C1 , C2 , C3 )
have the same initial point, then they lie in the same plane if and only if
A1 A2 A3
A · (B × C) = B1 B2 B3 = 0.
C1 C2 C3
Example 1.8.1. Find the volume of the parallelopiped having A = 3i − 5j + k, B = 2j − 2k and
C = 3i + j + k as adjacent edges.
Solution:
3 −5 1
0 2
0 −2
2 −2
= 3(4)+5(6)+1(−6) = 36.
+(1)
−(−5)
V = |A · (B × C)| = 0 2 −2 = 3
3 1
3 1
1 1
3 1
1
Example 1.8.2. Determine whether the four points A(−2, 0, 3), B(1, 0, 0), C(1, −3, 3) and D(4, 1, −2)
are coplanar.
−−→
Solution: We construct three vectors from the four points, a = AD = (6, 1, −5),
−→
−→
b = AB = (3, 0, −3), c = AC = (3, −3, 0). The scalar product is
6 1 −5
a · (b × c) = 3 0 −3 = 6(−9) − (1)(9) + (−5)(−9) = −18 ̸= 0.
3 −3 0
Hence not coplanar.
1.9
The Vector Triple Product
Let A, B and C be a triple of vectors. Then the vector
A × (B × C)
is called the triple vector product of A, B and C in that order. The evaluation of a vector triple
product can be made easier using the vector identity
A × (B × C) = (A · C)B − (A · B)C.
9
Example 1.9.1. Given the vectors A = i + 3j − k, B = −2i + j − 5k and C = 3i − 2j + 7k. Verify
the vector identity
A × (B × C) = (A · C)B − (A · B)C.
Solution: Starting with the left hand side
i
j
k
B × C = −2 1 −5 = −3i − j + k.
3 −2 7
A × (B × C) =
i
j
k
1
3 −1 = 2i + 2j + 8k.
−3 −1 1
The right hand side gives
(A · C) = 3(1) + 3(−2) + 7(−1) = −10, (A · C)B = −10(−2i + j − 5k) = 20i − 10j + 50k.
(A · B) = (−2)(1) + (3)(1) + (−5)(−1) = 6, (A · B)C = 6(3i − 2j + 7k) = 18i − 12j + 42k.
Then
(A · C)B − (A · B)C = (20 − 18)i + (−10 + 12)j + (50 − 42)k = 2i + 2j + 8k.
10
Chapter 2
Lines and Planes in Space
In the plane, slope is used to determine an equation of a line. In space, it is more convenient to
use vectors to determine the equation of a line.
2.1
Vector Equation of a Straight Line
Just as in finding the equation of a straight line in Cartesian coordinates, we will find that there
are two cases to consider when finding the equation of a straight line in terms of vectors.
Case I:
One case is when you are given a fixed point on the straight line and a vector parallel to the straight
line. This case is analogous to the situation in Cartesian coordinates when one is given a point
through which a straight line passes and the gradient of the line is known.
Suppose the straight line l passes through a point A with position vector a with respect to our
reference point O. Further, suppose the vector b is parallel to the straight line. Let R be any point
on the straight line and let its position vector relative to O be r. We have
−→ −→ −→
OR = OA + AR.
−→
−→
Since AR is parallel to the vector b, AR = λb where λ is a real number. Hence the vector equation
of a straight line is given by
r = a + λb.
(2.1)
Case II:
Suppose you are given two points through which the straight line passes.
Suppose a straight line l passes through two points A and B with position vectors a and b respectively relative to the reference point O. Let R be any point on the straight line and let its position
11
vector relative to O be r. We have
−→ −→ −→
OR = OA + AR.
−→
−→
−→
But AR = λAB where AB = b − a.
Thus r = a + λ(b − a) = (a − λa) + λb = (1 − λ)a + λb. Therefore the vector equation of a straight
line in this case is given by
r = (1 − λ)a + λb.
2.2
(2.2)
The Cartesian Equation of a Straight Line
We know that the general equation of a straight line in 2 dimensions has the form ax + by + c = 0
or the equivalent form y = mx + c.
You might expect the general equation of a straight line in 3 dimensional space to have the form
ax + by + cz + d = 0. In fact this is the general equation of a plane and not a straight line. In
general, there are two Cartesian coordinate forms of the equation of a straight line in 3 dimensional
space namely:
1. parametric equations form
2. symmetric equations form.
They are all derived from the vector equation (2.1).
Parametric Equations
     
x
x1
x2
Suppose vectors r, a, b have column vectors y ,  y1 ,  y2  respectively and suppose
z
z1
z2
r=a+
λb
is
the
vector
equation
of
straight
line
l.
 
   
x
x1
x2
Then y  =  y1  + λ  y2 .
z2
z
z1
Thus, we have
x = x1 + λx2 ,
y = y1 + λy2 ,
z = z1 + λz2
(2.3)
These are the parametric equations of the straight line l with parameter λ in 3 dimensional space.
Symmetric Equations
From the equations in (2.3), we obtain
x − x1
y − y1
z − z1
=
=
= λ.
x2
y2
z2
12
Hence, we have
x − x1
y − y1
z − z1
=
=
.
x2
y2
z2
(2.4)
These are the symmetric form of the Cartesian equations of a straight line in 3 dimensional space.
Example 2.2.1. Find parametric and symmetric equations of the line L that passes through the
point (1, −2, 4) and is parallel to A = (2, 4, −4).
Solution: To find a set of parametric equations of the line, use the coordinates x1 = 1, y1 = −2
and z1 = 4 and the direction numbers x2 = 2, y2 = 4 and z2 = −4.
x = 1 + 2λ,
y = −2 + 4λ,
z = 4 − 4λ.
Because x2 , y2 and z2 are all non-zero, a set of symmetric equations is
x−1
y+2
z−4
=
=
.
2
4
−4
Neither the parametric equations nor the symmetric equations of a given line are unique.
Example 2.2.2. Find a set of parametric equations of the line that passes through the points
(−2, 1, 0) and (1, 3, 5).
Solution: Begin by letting P = (−2, 1, 0) and Q = (1, 3, 5). Then a direction vector for the line
passing through P and Q is given by
A = (1 − (−2), 3 − 1, 5 − 0) = (3, 2, 5) = (x2 , y2 , z2 ).
Using the direction numbers x2 = 3, y2 = 2 and z2 = 5, with the point P = (−2, 1, 0) you can obtain
the parametric equations
x = −2 + 3λ, y = 1 + 2λ, z = 5λ.
2.3
Planes in Space
We have seen how an equation of a line in space can be obtained from a point on the line and a
vector parallel to it. Now we will see that an equation of a plane in space can be obtained from a
point in the plane and a vector normal (perpendicular) to it.
Consider the plane containing the point P = (x1 , y1 , z1 ) and having a non-zero normal vector
−→
n = (a, b, c). This plane consists of all points Q = (x, y, z) for which vector P Q is orthogonal to n.
Using the dot product, we have the following
−→
n·P Q = 0
(a, b, c) · (x − x1 , y − y1 , z − z1 ) = 0
a(x − x1 ) + b(y − y1 ) + c(z − z1 ) = 0.
The third equation of the plane is said to be in standard form.
13
2.3.1
Standard Equation of a Plane in Space
Theorem 2.3.1. The plane containing the point (x1 , y1 , z1 ) and having a normal vector n = (a, b, c)
can be represented, in standard form by the equation
a(x − x1 ) + b(y − y1 ) + c(z − z1 ) = 0.
By regrouping terms, we obtain the general form of the equation of a plane in space,
ax + by + cz + d = 0.
Given the general form of the equation of a plane, it is easy to find a normal vector to the plane.
simply use the coefficients of x, y and z and write n = (a, b, c).
Example 2.3.1. Find the general equation of the plane containing the points (2, 1, 1), (0, 4, 1) and
(−2, 1, 4).
Solution: We need a point in the plane and a vector that is normal to the plane. There are three
choices for the point, but no normal vector is given. To obtain a normal vector, use the cross
product of vectors B and C extending from the point (2, 1, 1) to the points (0, 4, 1) and (−2, 1, 4).
The component forms of B and C are
B = (0 − 2, 4 − 1, 1 − 1) = (−2, 3, 0)
C = (−2 − 2, 1 − 1, 4 − 1) = (−4, 0, 3)
and it follows that
i j k
n = B × C = −2 3 0 = 9i + 6j + 12k = (a, b, c)
−4 0 3
is normal to the given plane. Using the direction numbers for n and the point (x1 , y1 , z1 ) = (2, 1, 1),
you can determine an equation of the plane to be
a(x − x1 ) + b(y − y1 ) + c(z − z1 )
9(x − 2) + 6(y − 1) + 12(z − 1)
9x + 6y + 12z − 36
3x + 2y + 4z − 12
=
=
=
=
0
0 (Standard f orm)
0
0. (General f orm)
Remark: Check to see that each of the three points satisfies the equation 3x + 2y + 4z − 12 = 0.
2.4
Angle Between Two Planes
Two distinct planes in three-dimensional space either are parallel or intersect in a line. If they
intersect, you can determine the angle between them from the angle between their normal vectors.
14
Specifically, if vectors n1 and n2 are normal to two intersecting planes, the the angle θ between the
normal vectors is equal to the angle between the two planes and is given by
cos θ =
n1 · n2
.
|n1 ||n2 |
Consequently, two planes with normal vectors n1 and n2 are
1. perpendicular, if n1 · n2 = 0.
2. parallel, if n1 is a scalar multiple of n2 .
Example 2.4.1. Find the angle between the two planes
x − 2y + z = 0
2x + 3y − 2z = 0
Equation f or P lane 1
Equation f or P lane 2
and find parametric equations of their line of intersection.
Solution: The normal vectors for the planes are n1 = (1, −2, 1) and n2 = (2, 3, −2). Consequently,
the angle between the two planes is as follows
n1 · n2
|n1 ||n2 |
−6
= √ √
6 17
−6
= √
102
≈ −0.59409 (θ ≈ cos−1 −0.59409).
cos θ =
You can find the line of intersection of the two planes by simultaneously solving the two linear
equations representing the planes. Multiply the first equation by −2 and add the result to the
second equation.
−2x + 4y − 2z = 0
2x + 3y − 2z = 0
0x + 7y − 4z = 0.
4z
z
So y =
. Substituting this into one of the original equations, we have x = . Finally, letting
7
7
z
t = , we obtain the parametric equations
7
x = t,
y = 4t,
z = 7t (Line of Intersection).
15
2.5
Distances Between Points, Planes and Lines
In this section we want to find two basic types of distance problems in space,
1. Finding the distance between a point and a plane.
2. Finding the distance between a point and a line.
The solutions of these problems illustrate the versatility and usefulness of vectors in coordinate
geometry. The first problem uses the dot product of two vectors, and the second problem uses the
cross product. The distance D between a point Q and a plane is the length of the shortest line
segment connecting Q to the plane.
2.5.1
Distance Between a Point and a Plane
Theorem 2.5.1. The distance between a plane and a point Q (not in the plane) is
−→
|P Q · n|
.
D=
|n|
To find a point in the plane given by ax + by + cz + d = 0 (a ̸= 0), let y = 0 and z = 0. Then,
from the equation ax + d = 0, we conclude that the point (− ad , 0, 0) lies in the plane.
Example 2.5.1. Find the distance between Q = (1, 5, −4) and the plane given by 3x − y + 2z = 6.
Solution: We know that n = (3, −1, 2) is normal to the given plane. To find a point in the plane,
let y = 0 and z = 0, and obtain the point P = (2, 0, 0). The vector from P to Q is given by
−→
P Q = (1 − 2, 5 − 0, −4 − 0) = (−1, 5, −4).
Using the distance formula,
−→
|P Q · n|
|(−1, 5, −4) · (3, −1, 2)|
√
D=
=
|n|
9+1+4
| − 3 − 5 − 8|
√
=
14
16
= √ .
14
Remark: The choice of the point P is arbitrary in the Example. Try choosing a different point to
verify that we obtain the same distance.
16
The distance between the point Q = (x0 , y0 , z0 ) and the plane given by ax + by + cz + d = 0 is
|a(x0 − x1 ) + b(y0 − y1 ) + c(z0 − z1 )|
√
a2 + b 2 + c 2
|ax0 + by0 + cz0 + d|
√
D =
a2 + b 2 + c 2
D =
where P = (x1 , y1 , z1 ) is a point on the plane and d = −(ax1 + by1 + cz1 ).
2.6
Finding the Distance Between Two Parallel Planes
Example 2.6.1. Find the distance between the two parallel planes given by 3x − y + 2z − 6 = 0
and 6x − 2y + 4z + 4 = 0.
Solution: To find the distance between the planes, choose a point in the first plane, say (x0 , y0 , z0 ) =
(2, 0, 0). Then, from the second plane, we determine that a = 6, b = −2, c = 4 and d = 4, and
conclude that the distance is
|ax0 + by0 + cz0 + d|
√
a2 + b 2 + c 2
|6(2) + (−2)(0) + (4)(0) + 4|
√
=
62 + (−2)2 + 42
16
8
= √ = √ ≈ 2.14.
56
14
D =
2.7
Distance Between a Point and a Line in Space
The formula for the distance between a point and a line in space resembles that for the distance
between a point and a plane, except that we replace the dot product by the cross product and
replace the normal vector n by a direction vector for the given line.
Theorem 2.7.1. The distance between a point Q and a line in space is given by
−→
|P Q × u|
D=
|u|
where u is the direction vector for the line and P is a point on the line.
Example 2.7.1. Find the distance between the point Q = (3, −1, 4) and the line given by
x = −2 + 3t,
y = −2t,
17
z = 1 + 4t.
Solution: Using the direction numbers, 3, −2 and 4, we know that the direction vector for the line
is u = (3, −2, 4). To find a point on the line, let t = 0, and obtain P = (−2, 0, 1). thus,
−→
P Q = (3 − (−2), −1 − 0, 4 − 1) = (5, −1, 3)
and we form the cross product
i j k
−→
P Q × u = 5 −1 3 = 2i − 11j − 7k = (2, −11, −7).
3 −2 4
Finally
√
−→
|P Q × u|
174 √
D=
= 6 ≈ 2.45.
= √
|u|
29
18
Chapter 3
Vector-Valued Functions
This chapter introduces the concept of vector-valued functions. Vector-valued functions can be used
to study curves in the plane and in space. These functions can also be used to study the motion of
an object along a curve.
3.1
Plane Curves and Parametric Equations
Until now, we have been representing a graph by a single equation involving two variables. Consider
the path followed by an object that is propelled into the air at an angle of 45◦ . If the initial velocity
of the object is 48 meters per second, the object travels the parabolic path given by
y=−
x2
+ x,
72
Rectangular form.
However, this equation does not tell the whole story. Although it does tell you where the object
has been, it doesn’t tell you when the object was at a given point (x, y). To determine this time,
you introduce a third variable t, called a parameter. By writing both x and y as functions of t,
you obtain the parametric equations
√
√
x = 24 2t and y = −16t2 + 24 2t Parametric equations.
From this set of equations, you can determine that at√time t√= 0, the object is at the point (0, 0).
Similarly, at time t = 1, the object is at the point (24 2, 24 2 − 16), and so on.
Theorem 3.1.1. If f and g are continuous functions of t on an interval I, then the equations
x = f (t)
and y = g(t)
are called parametric equations and t is called a parameter.
19
y
6
v
r(t2 )
vs
r(t0 )
x
1
C
r(t1 )
7
*
I
-
x
Figure 3.1: Curve in plane
A space curve is the set of all ordered triples (f (t), g(t), h(t)) together with their defining parametric equations
x = f (t), y = g(t) and z = h(t)
where f, g and h are continuous functions of t on the interval I.
Theorem 3.1.2. A function of the form
r(t) = f (t)i + g(t)j Plane
or
r(t) = f (t)i + g(t)j + h(t)k Space
is a vector-valued function, where the component functions f, g and h are real-valued functions of the parameter t. Vector-valued functions are sometimes denoted as r(t) = (f (t), g(t)) or
r(t) = (f (t, g(t), h(t)).
Vector-valued functions serve dual roles in the representation of curves. By letting the parameter
t represent time, we can use a vector-valued function to represent motion along a curve. Or, in
the more general case, we can use a vector-valued function to trace the graph of a curve. Unless
stated otherwise, the domain of a vector-valued function r is considered to be the intersection of
the domains of
√ the component functions f, g and h. For example, the domain of
r(t) = ln ti + 1 − tj + tk is the interval (0, 1].
20
3.2
Limits and Continuity
Many techniques and definitions used in calculus of real-valued functions can be applied to vectorvalued functions. For example, we can add and subtract vector-valued functions, multiply a vectorvalued function with a scalar, take the limit of a vector-valued function, differentiate a vector-valued
function, and so on.
r1 (t) + r2 (t) = [f1 (t)i + g1 (t)j] + [f2 (t)i + g2 (t)j]
= [f1 (t) + f2 (t)]i + [g1 (t) + g2 (t)]j.
3.3
Definition of the Limit of a Vector-Valued Function
If r is a vector-valued function such that r(t) = f (t)i + g(t)j, then
[
]
[
]
lim r(t) = lim f (t) i + lim g(t) j
Plane
t→a
t→a
t→a
provided f and g have limits as t → a.
If r is a vector-valued function such that r(t) = f (t)i + g(t)j + h(t)k, then
[
]
[
]
[
]
lim r(t) = lim f (t) i + lim g(t) j + lim h(t) k
Space
t→a
t→a
t→a
t→a
provided f, g and h have limits as t → a.
If r(t) approaches the vector L as t → a, the length of the vector r(t) − L approaches 0, that is
|r(t) − L| → 0 as t → a.
The next theorem extends the notion of continuity of vector-valued functions.
Theorem 3.3.1. A vector-valued function r is continuous at a point given by t = a if the limit
of r(t) exists as t → a and
lim r(t) = r(a).
t→a
A vector-valued function r is continuous on an interval I if it is continuous at every point in
the interval. It follows that a vector-valued function is continuous at t = a if and only if each of its
component functions is continuous at t = a.
Example 3.3.1. Discuss the continuity of the vector-valued function given by
r(t) = ti + aj + (a2 − t2 )k
21
at t = 0.
Solution: As t approaches 0, the limit is
[
]
[
]
[
]
lim r(t) = lim t i + lim a j + lim (a2 − t2 ) k
t→0
t→0
t→0
t→0
= 0i + aj + a2 k
= aj + a2 k.
Because
r(0) = (0)i + (a)j + (a2 )k
= aj + a2 k.
We can conclude that r is continuous at t = 0.
3.4
3.4.1
Differentiation and Integration of Vector-Valued Functions
Differentiation of Vector-Valued Functions
The definition of the derivative of a vector-valued function parallels the definition given for realvalued functions.
Theorem 3.4.1. The derivative of a vector-valued function r is defined by
r(t + △t) − r(t)
△t→0
△t
r′ (t) = lim
for all t for which the limit exists. If r′ (t) exists, then r is differentiable at t. In addition to
r′ (t), other notations for the derivative of a vector-valued function are
Dt [r(t)],
d
[r(t)]
dt
and
dr
.
dt
Differentiation of vector-valued functions can be done on a component-by-component basis. To see
why this is true, consider the function given by
r(t) = f (t)i + g(t)j.
22
Applying the definition of the derivative produces the following
r′ (t) =
=
=
=
=
r(t + △t) − r(t)
△t→0
△t
f (t + △t)i + g(t + △t)j − f (t)i − g(t)j
lim
△t→0
△t
{[
]
[
] }
f (t + △t) − f (t)
g(t + △t) − g(t)
lim
i+
j
△t→0
△t
△t
] } {
[
] }
{
[
g(t + △t) − g(t)
f (t + △t) − f (t)
lim
i + lim
j
△t→0
△t→0
△t
△t
f ′ (t)i + g ′ (t)j.
lim
Note that the derivative of the vector-valued function r is itself a vector-valued function. r′ (t) is a
vector tangent to the curve given by r(t) and pointing in the direction of increasing t-values.
Theorem 3.4.2. If r(t) = f (t)i + g(t)j, where f and g are differentiable functions of t, then
r′ (t) = f ′ (t)i + g ′ (t)j.
If r(t) = f (t)i + g(t)j + h(t)k, where f, g and h are differentiable functions of t, then
r′ (t) = f ′ (t)i + g ′ (t)j + h′ (t)k.
Example 3.4.1. For the vector-valued function given by r(t) = ti + (t2 + 2)j, find r′ (t).
Solution: r′ (t) = i + 2tj.
Higher-order derivatives of vector-valued functions are obtained by successive differentiation of each
component function.
Example 3.4.2. For the vector-valued function given by r(t) = cos ti + sin tj + 2tk, find each of the
following (i) r′ (t) (ii) r′′ (t) (iii) r′ (t) · r′′ (t) (iv) r′ (t) × r′′ (t).
Solution: (i) r′ (t) = − sin ti + cos tj + 2k.
(ii) r′′ (t) = − cos ti − sin tj + 0k = − cos ti − sin tj.
(iii) r′ (t) · r′′ (t) = sin t cos t − sin t cos t = 0.
i
j
k
− sin t cos t
cos t 2
− sin t 2
′
′′
k
j+
i−
(iv) r (t) × r (t) = − sin t cos t 2 =
− cos t − sin t
− cos t 0
− sin t 0
− cos t − sin t 0
= 2 sin ti − 2 cos tj + k.
The parametrization of the curve represented by the vector-valued function
r(t) = f (t)i + g(t)j + h(t)k
is smooth on an open interval I if f ′ , g ′ and h′ are continuous on I and r′ (t) ̸= 0 for any value
of t in the interval I.
23
Example 3.4.3. Find the intervals on which the epicycloid C given by
r(t) = (5 cos t − cos 5t)i + (5 sin t − sin 5t)j,
0 ≤ t ≤ 2π
is smooth.
Solution: The derivative of r is
r′ (t) = (−5 sin t + 5 sin 5t)i + (5 cos t − 5 cos 5t)j.
In the interval [0, 2π], the only values of t for which
r′ (t) = 0i + 0j
π
3π
are t = 0, , π, , 2π. Therefore, we conclude that C is smooth in the intervals
2
2
(
)
( π ) ( π ) ( 3π )
3π
,
, π , π,
and
, 2π .
0,
2
2
2
2
3.5
Properties of the Derivative
Let r and u be differentiable vector-valued functions of t, let w be a differentiable real-valued
function of t, and let c be a scalar.
1. Dt [cr(t)] = cr′ (t).
2. Dt [r(t) ± u(t)] = r′ (t) ± u′ (t).
3. Dt [w(t)r(t)] = w(t)r′ (t) + w′ (t)r(t).
4. Dt [r(t) · u(t)] = r(t) · u′ (t) + r′ (t) · u(t).
5. Dt [r(t) × u(t)] = r(t) × u′ (t) + r′ (t) × u(t).
6. Dt [r(w(t))] = r′ (w(t))w′ (t).
7. If r(t) · r(t) = c, then r(t) · r′ (t) = 0.
To prove property 4, let r(t) = f1 (t)i + g1 (t)j and u(t) = f2 (t)i + g2 (t)j, where f1 , f2 , g1 and g2 are
differentiable functions of t. Then
r(t) · u(t) = f1 (t)f2 (t) + g1 (t)g2 (t)
and it follows that
Dt [r(t) · u(t)] = f1 (t)f2′ (t) + f1′ (t)f2 (t) + g1 (t)g2′ (t) + g1′ (t)g2 (t)
= [f1 (t)f2′ (t) + g1 (t)g2′ (t)] + [f1′ (t)f2 (t) + g1′ (t)g2 (t)]
= r(t) · u′ (t) + r′ (t) · u(t).
24
Example 3.5.1. For the vector-valued function given by
1
r(t) = i − j + ln tk and
t
find (i) Dt [r(t) · u(t)]
u(t) = t2 i − 2tj + k
(ii) Dt [u(t) × u′ (t)].
1
1
Solution: (i) Because r′ (t) = − 2 i + k and u′ (t) = 2ti − 2j, we have
t
t
Dt [r(t) · u(t)] = r(t) · u′ (t) + r′ (t) · u(t)
(
)
(
)
1
1
1
i − j + ln tk · (2ti − 2j) + − 2 i + k · (t2 i − 2tj + k)
=
t
t
t
1
= 2 + 2 + (−1) +
t
1
= 3+ .
t
(ii) Because u′ (t) = 2ti − 2j and u′′ (t) = 2i, we have
Dt [u(t) × u′ (t)] = [u(t) × u′′ (t)] + [u′ (t) × u′ (t)]
i
j k
= t2 −2t 1 + 0
2
0 0
= 2j + 4tk.
3.6
Integration of Vector-Valued Functions
If r(t) = f (t)i + g(t)j, where f and g are continuous on [a, b], then the indefinite integral (antiderivative) of r is
]
]
[∫
[∫
∫
g(t) dt j
f (t) dt i +
r(t) dt =
and its definite integral over the interval a ≤ t ≤ b is
]
]
[∫ b
[∫ b
∫ b
g(t) dt j.
r(t) dt =
f (t) dt i +
a
a
a
If r(t) = f (t)i+g(t)j+h(t)k, where f, g and h are continuous on [a, b], then the indefinite integral
(anti-derivative) of r is
[∫
]
[∫
]
[∫
]
∫
r(t) dt =
f (t) dt i +
g(t) dt j +
h(t) dt k
and its definite integral over the interval a ≤ t ≤ b is
[∫ b
]
[∫ b
]
[∫ b
]
∫ b
r(t) dt =
f (t) dt i +
g(t) dt j +
h(t) dt k.
a
a
a
25
a
The anti-derivative of a vector-valued function is a family of vector-valued functions all differing by
a constant vector C.
Example 3.6.1. Find the indefinite integral
∫
(ti + 3j) dt.
Solution: Integrating on a component-by-component basis produces
∫
t2
(ti + 3j) dt = i + 3tj + C.
2
Example 3.6.2. Evaluate the integral
∫ 1
∫
r(t) dt =
0
Solution:
∫
0
3.7
1
1
(
0
(∫
)
1
−t
ti +
j + e k dt.
t+1
)
(∫ 1
)
1
−t
t dt i +
dt j +
e dt k
r(t) dt =
0 t+1
0
0
[
]1
[
]1
[
]1
3 4
−t
=
t 3 i + ln |t + 1| j + −e
k
4
0
0
0
(
)
1
3
i + ln 2j + 1 −
k.
=
4
e
1
√
3
)
√
3
(∫
1
Velocity and Acceleration
We begin by looking at the motion of an object in the plane. As an object moves along a curve in
the plane, the coordinates x and y of its center of mass are each functions of time t. Rather than
using the letters f and g to represent these two functions, it is convenient to write x = x(t) and
y = y(t). So, the position vector r(t) takes the form
r(t) = x(t)i + y(t)j.
Theorem 3.7.1. If x and y are twice-differentiable functions of t, and r is a vector-valued function
given by r(t) = x(t)i + y(t)j, then the velocity vector, acceleration vector and speed at time t are as
follows
1. Velocity = v(t) = r′ (t) = x′ (t)i + y ′ (t)j.
2. Acceleration = a(t) = r′′ (t) = x′′ (t)i + y ′′ (t)j.
√
3. Speed = |v(t)| = |r′ (t)| = [x′ (t)]2 + [y ′ (t)]2 .
26
Example 3.7.1. Find the velocity vector, speed and acceleration vector of a particle that moves
t
t
along the plane curve C given by r(t) = 2 sin i + 2 cos j.
2
2
Solution: The velocity vector is
t
t
v(t) = r′ (t) = cos i − sin j.
2
2
The speed (at any time) is
√
′
|r (t)| =
The acceleration vector is
cos2
t
t
+ sin2 = 1.
2
2
1
t
1
t
a(t) = r′′ (t) = − sin i − cos j.
2
2
2
2
So far, we have concentrated on finding the velocity and acceleration by differentiating the position
vector. Many practical applications involve the reverse problem, finding the position function for a
given velocity or acceleration.
Example 3.7.2. An object starts from rest at the point P = (1, 2, 0) and moves with an acceleration
of a(t) = j + 2k where |a(t)| is measured in meters per second per second. Find the location of the
object after t = 2 seconds.
Solution: From the description of the object’s motion, we can deduce the following initial conditions. Because the object starts from rest, we have v(0) = 0. Moreover, because the object starts at
the point (x, y, z) = (1, 2, 0), we have
r(0) = x(0)i + y(0)j + z(0)k = 1i + 2j + 0k = i + 2j.
To find the position function, we should integrate twice, each time using one of the initial conditions
to solve for the constant of integration. The velocity vector is
∫
∫
v(t) = a(t) dt = (j + 2k) dt = tj + 2tk + C
where C = C1 i + C2 j + C3 k. Letting t = 0 and applying the initial condition v(0) = 0, we obtain
v(0) = C1 i + C2 j + C3 k = 0 =⇒ C1 = C2 = C3 = 0.
Thus, the velocity at any time t is
v(t) = tj + 2tj.
Integrating once more produces
r(t) =
∫
∫
v(t) dt =
(tj + 2tj) dt =
t2
j + t2 k + C
2
where C = C4 i + C5 j + C6 k. Letting t = 0 and applying the initial condition r(0) = i + 2j, we have
r(0) = C4 i + C5 j + C6 k = i + 2j =⇒ C4 = 1, C5 = 2, C6 = 0.
Thus, the position vector is
)
t2
+ 2 j + t2 k.
r(t) = i +
2
The location of the object after 2 seconds is given by r(2) = i + 4j + 4k.
(
27
3.8
Tangent Vectors and Normal Vectors
In the previous section, we learned that the velocity vector points in the direction of motion.
3.8.1
Definition of Unit Tangent Vector
Let C be a smooth curve represented by r on an open interval I. The unit tangent vector T(t)
at t is defined to be
r′ (t)
T(t) = ′
, r′ (t) ̸= 0.
|r (t)|
Example 3.8.1. Find the unit tangent vector to the curve given by r(t) = ti + t2 j when t = 1.
Solution: The derivative of r(t) is r′ (t) = i + 2tj. Thus, the unit tangent vector is
T(t) =
r′ (t)
1
(i + 2tj).
=√
′
|r (t)|
1 + 4t2
When t = 1, the unit tangent vector is
1
T(1) = √ (i + 2j).
5
The tangent line to a curve at a point is the line passing through the point and parallel to the
unit tangent vector.
Example 3.8.2. Find T(t) and then a set of parametric equations for the tangent line to the helix
π
given by r(t) = 2 cos ti + 2 sin tj + tk at the point corresponding to t = .
4
Solution:
is r′ (t) = −2 sin ti + 2 cos tj + k, which implies that
√ The2 derivative of r(t) √
′
|r (t)| = 4 sin t + 4 cos2 t + 1 = 5. Therefore, the unit tangent vector is
T(t) =
When t =
r′ (t)
1
= √ (−2 sin ti + 2 cos tj + k).
′
|r (t)|
5
π
, the unit tangent vector is
4
( √
)
√
(π )
√
√
1
2
2
1
T
=√
−2
i+2
j = k = √ (− 2i + 2j + k).
4
2
2
5
5
There are infinitely many vectors that are orthogonal to the tangent vector T(t).
28
Definition of Principal Unit Normal Vector
Let C be a smooth curve represented by r on an interval I. If T′ (t) ̸= 0, then the principal unit
normal vector at t is defined to be
T′ (t)
.
N(t) = ′
|T (t)|
Example 3.8.3. Find N(t) and N(1) for the curve represented by
r(t) = 3ti + 2t2 j.
Solution: By differentiating, we obtain
r′ (t) = 3i + 4tj and
|r′ (t)| =
√
9 + 16t2
which implies that the unit tangent vector is
T(t) =
r′ (t)
1
√
=
(3i + 4tj).
|r′ (t)|
9 + 16t2
Then differentiating T(t) with respect to t to obtain
12
1
16t
T′ (t) = √
(4j) −
3 (3i + 4tj) =
3 (−4ti + 3j).
2
2
9 + 16t
(9 + 16t ) 2
(9 + 16t2 ) 2
√
9 + 16t2
12
|T′ (t)| = 12
=
.
2
3
(9 + 16t )
9 + 16t2
Therefore, the principal unit normal vector is N(t) =
1
the principal unit normal is N(1) = (−4i + 3j).
5
29
1
T′ (t)
√
=
(−4ti + 3j). When t = 1,
|T′ (t)|
9 + 16t2
Chapter 4
Functions of Several Variables
So far we have dealt only with functions of single (independent) variables. Many familiar quantities,
however, are functions of two or more variables. For instance, the work done by a force (W = F D)
and the volume of a right circular cylinder (V = πr2 h) are both functions of two variables. The
volume of a rectangular solid (V = lwh) is a function of three variables. The notation for a function
of two or three variables is as follows
z = f (x, y) = x2 + xy
| {z }
2 variables
and
w = f (x, y, z) = x + 2y − 3z.
| {z }
3 variables
4.1
Definition of a Function of Two Variables
Let D be a set of ordered pairs of real numbers. If to each ordered pair (x, y) in D there corresponds
a unique real number f (x, y), then f is called a function of x and y. The set D is the domain
of f and the corresponding set of values for f (x, y) is the range of f . For the function given by
z = f (x, y), we call x and y the independent variables and z the dependent variable.
As with functions of one variable, the most common way to describe a function of several variables
is with an equation, and unless otherwise restricted, we can assume that the domain is the set of
all points for which the equation is defined. for example, the domain of the function given by
f (x, y) = x2 + y 2
is assumed to be the entire xy−plane.
Example 4.1.1. Find the domains of the following functions.
√
x2 + y 2 − 9
x
(ii) g(x, y, z) = √
(i) f (x, y) =
.
2
x
9 − x − y2 − z2
30
Solution: (i) The function f is defined for all points (x, y) such that x ̸= 0 and x2 + y 2 ≥ 9. Thus,
the domain is the set of all points lying on or outside the circle x2 + y 2 = 9.
(ii) The function g is defined for all points (x, y, z) such that x2 + y 2 + z 2 < 9. Consequently, the
domain is the set of all points (x, y, z) lying inside a sphere of radius 3 that is centred at the origin.
Functions of several variables can be combined in the same ways as functions of single variables. For
instance, we can form the sum, difference, product and quotients of two functions of two variables
as follows
1. (f ± g)(x, y) = f (x, y) ± g(x, y)
Sum or Difference.
2. (f g)(x, y) = f (x, y)g(x, y)
Product.
3.
f
f (x, y)
(x, y) =
,
g
g(x, y)
4.2
g(x, y) ̸= 0
Quotient.
The Graph of a Function of Two Variables
As with functions of single variables, we can learn a lot about the behaviour of a function of two
variables by sketching its graph. The graph of a function f of two variables is the set of all points
(x, y, z) for which z = f (x, y) and (x, y) is in the domain of f . The graph can be interpreted as a
surface in space.
4.2.1
Level Curves
A second way to visualize a function of two variables is to use a scalar field in which the scalar
z = f (x, y) is assigned to the point (x, y). A scalar field can be characterized by level curves (or
contour lines) along which the value of f (x, y) is constant. For example, the weather map shows
level curves of equal pressure called isobars. In weather maps for which the level curves represent
points of equal temperature, the level curves are called isotherms. Another common use of level
curves is in representing electrical potential fields, in this type of map, the level curves are called
equipotential lines.
4.2.2
Level Surfaces
The concept of a level curve can be extended by one dimension to define a level surface. If f is
a function of three variables and c is a constant, then the graph of the equation f (x, y, z) = c is a
level surface of the function f .
31
4.3
4.3.1
Limits and Continuity
Neighborhoods in the Plane
In this section, we will study limits and continuity involving functions of two or three variables. We
begin our discussion of the limit of a function of two variables by defining a two-dimensional analog
to an interval on the real line. Using the formula for the distance between two points (x, y) and
(x0 , y0 ) in the plane, we can define the δ-neighborhood about (x0 , y0 ) to be the disc centered at
(x0 , y0 ) with radius δ > 0
√
{(x, y) : (x − x0 )2 + (y − y0 )2 < δ}.
closed. When this formula contains the less than inequality, <, the disc is called open, and when
6
sq
δ
(x0 , y0 )
-
Figure 4.1: A closed disc
it contains the less than or equal to inequality, ≤, the disc is called closed.
A point (x0 , y0 ) in a plane region R is an interior point of R if there exists a δ-neighborhood
about (x0 , y0 ) that lies entirely in R. If every point in R is an interior point, then R is an open
region. A point (x0 , y0 ) is a boundary point of R if every open disc centered at (x0 , y0 ) contains
points inside R and points outside R. If a region contains all its boundary points, then the region
is closed.
32
4.3.2
Limit of a Function of Two Variables
Let f be a function of two variables defined on an open disc centered at (x0 , y0 ), except possibly at
(x0 , y0 ) and let L be a real number. Then
lim
f (x, y) = L
(x,y)→(x0 ,y0 )
if for each ε > 0 there corresponds a δ > 0 such that
|f (x, y) − L| < ε whenever 0 <
√
(x − x0 )2 + (y − y0 )2 < δ.
Definition
A function f (x, y) has a limit L as (x, y) approaches (x0 , y0 ) if given any ϵ > 0 there exists
δ > 0 (depending on ϵ and (x0 , y0 )) such that whenever (x − x0 )2 + (y − y0 )2 < δ 2 , then
|f (x, y) − L| < ϵ.
The definition of the limit of a function of two variables is similar to the definition of the limit
of a function of a single variable, yet there is a critical difference. For a function of two variables,
the statement (x, y) → (x0 , y0 ) means that the point (x, y) is allowed to approach (x0 , y0 ) from any
direction. If the value of
lim
f (x, y)
(x,y)→(x0 ,y0 )
is not the same for all possible approaches or paths, to (x0 , y0 ), then the limit does not exist. We
usually choose convenient paths. Some of these are
1. along the x or y axis,
2. along straight lines,
3. along well-defined curves such as parabolae, for e.g., y = x3 .
Example 4.3.1. Show that
lim
x = a.
(x,y)→(a,b)
Solution: Let f (x, y) = x and L = a. We need to show that for each ε > 0, there exists a
δ−neighborhood about (a, b) such that
|f (x, y) − L| = |x − a| < ε
whenever (x, y) ̸= (a, b) lies in the neighborhood. We can observe that from
√
0 < (x − a)2 + (y − b)2 < δ
it follows that
|f (x, y) − a| = |x − a| =
√
(x − a)2 ≤
thus, we can choose δ = ε, and the limit is verified.
33
√
(x − a)2 + (y − b)2 < δ.
Example 4.3.2. Prove that
x2 + 2y = 5.
lim
(x,y)→(1,2)
Solution: Using the definition of limits, we must show that, given ε > 0, we can find a δ > 0
such that |x2 + 2y − 5| < ε when 0 < |x − 1| < δ and 0 < |y − 2| < δ. If 0 < |x − 1| < δ and
0 < |y − 2| < δ, then
1−δ <x<1+δ
and
2−δ <y <2+δ
excluding x = 1 and y = 2. Thus,
1 − 2δ + δ 2 < x2 < 1 + 2δ + δ 2
and
4 − 2δ < 2y < 4 + 2δ.
Adding
5 − 4δ + δ 2 < x2 + 2y < 5 + 4δ + δ 2
or
−4δ + δ 2 < x2 + 2y − 5 < 4δ + δ 2 .
If δ ≤ 1, it follows that
−5δ < x2 + 2y − 5 < 5δ
i.e., |x2 + 2y − 5| < 5δ whenever 0 < |x − 1| < δ and 0 < |y − 2| < δ. Choosing 5δ = ε i.e., δ = 5ε or
δ = 1 which ever is smaller, it follows that |x2 + 2y − 5| < ε when 0 < |x − 1| < δ and 0 < |y − 2| < δ
i.e.,
lim x2 + 2y = 5.
(x,y)→(1,2)
Example 4.3.3. Show that the following limit does not exist.
)2
( 2
x − y2
.
lim
(x,y)→(0,0)
x2 + y 2
(
)2
x2 − y 2
consists of all points in the
Solution: The domain of the function given by f (x, y) =
x2 + y 2
xy-plane except for the point (0, 0). To show that the limit as (x, y) approaches (0, 0) does not exist,
consider approaching (0, 0) along two different paths. along the x-axis, every point is of the form
(x, 0) and the limit along this approach is
(
lim
(x,0)→(0,0)
x2 − y 2
x2 + y 2
)2
=
(1)2 = 1.
lim
(x,0)→(0,0)
However, if (x, y) approaches (0, 0) along the line y = x, we obtain
(
lim
(x,x)→(0,0)
x2 − y 2
x2 + y 2
)2
(
=
lim
(x,x)→(0,0)
Hence, f does not have a limit as (x, y) → (0, 0).
34
0
2x2
)2
= 0.
Example 4.3.4. Show that
lim
(x,y)→(0,0) x2
xy
does not exist.
+ y2
Solution: The fact that the limit taken along the x and y−axis exists and equal zero may lead us
to suspect that the
lim f (x, y) exists. We have not examined every path to (0, 0). We now try
(x,y)→(0,0)
any line through the origin given by y = mx,
lim
f (x, y) =
(x,y)→(0,0)
mx2
m
=
.
2
2
2
(x,y)→(0,0) x + m x
1 + m2
lim
This limit changes as the gradient m changes. For example (i) on y = 2x,
lim
f (x, y) =
(x,y)→(0,0)
2
and
5
5
(ii) on y = 5x, lim f (x, y) = . There is no single number L that we can call the limit of f
(x,y)→(0,0)
26
as (x, y) → (0, 0). So the limit does not exist.
4.3.3
Limit Combinations
In general, if
lim
(x,y)→(x0 ,y0 )
f (x, y) = L1 and
lim
(x,y)→(x0 ,y0 )
g(x, y) = L2 , then
1. lim[f (x, y) ± g(x, y)] = lim f (x, y) ± lim g(x, y) = L1 ± L2 .
2. lim[f (x, y) · g(x, y)] = (lim f (x, y))(lim g(x, y)) = L1 L2 .
3. lim cf (x, y) = c lim f (x, y) = cL1 , where c is any number.
4. lim
lim f (x, y)
L1
f (x, y)
=
=
, if L2 ̸= 0.
g(x, y)
lim g(x, y)
L2
Example 4.3.5. Evaluate
lim
(2x + 5xy − 3y 2 ).
(x,y)→(2,1)
Solution:
lim
(2x + 5xy − 3y 2 ) =
(x,y)→(2,1)
lim
2x +
(x,y)→(2,1)
lim
(x,y)→(2,1)
5xy +
(−3y 2 )
lim
(x,y)→(2,1)
= lim 2x + (lim 5x)(lim y) − 3 lim y 2
x→2
x→2
y→1
y→1
= 2 lim x + 5(lim x)(lim y) − 3(lim y)(lim y)
x→2
x→2
y→1
y→1
y→1
= 2 · 2 + 5 · 2 · 1 − 3 · 1 · 1 = 4 + 10 − 3 = 11.
4.4
Continuity of a Function of Two Variables
Notice that the limit of
f (x, y) =
35
5x2 y
x2 + y 2
as (x, y) → (1, 2) can be evaluated by direct substitution. That is, the limit is f (1, 2) = 2. In such
cases the function f is said to be continuous at the point (1, 2).
Definition of Continuity of a Function of Two Variables
A function f of two variables is continuous at a point (x0 , y0 ) in an open region R if
1. f is defined at (x0 , y0 ),
2.
lim
f (x, y) exists, and
(x,y)→(x0 ,y0 )
3.
lim
(x,y)→(x0 ,y0 )
f (x, y) = f (x0 , y0 ).
The function f is continuous in the open region R if it is continuous at every point in R.
Properties of Continuous Functions of Two Variables
If k is a real number and f and g are continuous at (x0 , y0 ), then the following functions are
continuous at (x0 , y0 ),
1. Scalar Multiple kf .
2. Sum and difference f ± g.
3. Product f g.
4. Quotient
f
, if g(x0 , y0 ) ̸= 0.
g
Polynomials and rational functions in two variables are continuous at any point at which they are
defined.
4.5
4.5.1
Partial Derivatives
Partial Derivatives of a Function of Two Variables
In the application of functions of several variables, the question often arises, “How will a function
be affected by a change in one of its independent variables?”. You can answer by considering the
36
independent variables one at a time. The process is called partial differentiation, and the result
is referred to as the partial derivative of f with respect to the chosen independent variable 1 .
Definition of Partial Derivatives of a Function of Two Variables
If z = f (x, y), then the first partial derivatives of f with respect to x and y are the functions fx
and fy defined by
f (x + ∆x, y) − f (x, y)
∆x→0
∆x
f (x, y + ∆y) − f (x, y)
fy (x, y) = lim
∆y→0
∆y
fx (x, y) =
lim
provided the limits exist. This definition indicates that if z = f (x, y), then to find fx we consider
y constant and differentiate with respect to x. Similarly, to find fy , we consider x constant and
differentiate with respect to y.
Example 4.5.1. Find fx and fy for f (x, y) = 3x − x2 y 2 + 2x3 y.
Solution: Considering y to be constant and differentiating with respect to x produces
fx (x, y) = 3 − 2xy 2 + 6x2 y.
Considering x to be constant and differentiating with respect to y produces
fy (x, y) = −2x2 y + 2x3 .
Notation for Partial Derivatives
For z = f (x, y), the partial derivatives fx and fy are denoted by
∂
∂z
f (x, y) = fx (x, y) = zx =
∂x
∂x
and
∂
∂z
f (x, y) = fy (x, y) = zy =
.
∂y
∂y
The first partials evaluated at the point (a, b) are denoted by
∂z
∂x
= fx (a, b)
(a,b)
1
The introduction of partial derivatives followed Newton’s and Leibniz’s work in calculus by several years. Between
1760, Leonhard Euler and Jean Le Rond d’Alembert (1717-1783) separately published several papers on dynamics,
in which they established much of the theory of partial derivatives
37
and
∂z
∂y
x2 y
Example 4.5.2. For f (x, y) = xe
= fy (a, b).
(a,b)
, find fx and fy and evaluate each at the point (1, ln 2).
Solution: Because
2
fx (x, y) = xex y (2xy) + ex
2y
the partial derivative of f with respect to x at (1, ln 2) is
fx (1, ln 2) = eln 2 (2 ln 2) + eln 2 = 4 ln 2 + 2.
Because
2
fy (x, y) = xex y (x2 ) = x3 ex
2y
the partial derivative of f with respect to y at (1, ln 2) is
fy (1, ln 2) = eln 2 = 2.
4.5.2
Partial Derivatives of a function of Three or More Variables
The concept of a partial derivative can be extended naturally to functions of three or more variables.
For instance, if w = f (x, y, z), then there are three partial derivatives, each of which is formed by
holding two of three variables constant.
f (x + ∆x, y, z) − f (x, y, z)
∂w
= fx (x, y, z) = lim
∆x→0
∂x
∆x
∂w
f (x, y + ∆y, z) − f (x, y, z)
= fy (x, y, z) = lim
∆y→0
∂y
∆y
f (x, y, z + ∆z) − f (x, y, z)
∂w
= fz (x, y, z) = lim
.
∆z→0
∂z
∆z
Example 4.5.3. (i) To find the partial derivative of f (x, y, z) = xy + yz 2 + xz with respect to z,
consider x and y to be constant and obtain
]
∂ [
xy + yz 2 + xz = 2yz + x.
∂z
(ii) to find the partial derivative of f (x, y, z) = z sin(xy 2 + 2z) with respect to z, consider x and y
to be constant. Then, using the Product rule, we obtain
]
∂ [
∂
∂
z sin(xy 2 + 2z) = (z) [sin(xy 2 + 2z)] + sin(xy 2 + 2z) [z]
∂z
∂z
∂z
2
2
= (z)[cos(xy + 2z)](2) + sin(xy + 2z)
= 2z cos(xy 2 + 2z) + sin(xy 2 + 2z).
x+y+z
(iii) To find the partial derivative of f (x, y, z, w) =
with respect to w, consider x, y and
w
z to be constant and obtain
[
]
x+y+z
∂ x+y+z
=−
.
∂w
w
w2
38
4.5.3
Higher-Order Partial Derivatives
It is possible to take second, third and higher partial derivatives of a function of several variables,
provided such derivatives exist. Higher-order derivatives are denoted by the order in which the
differentiation occurs. For instance, the function z = f (x, y) has the following second partial
derivatives.
1. Differentiate twice with respect to x:
∂
∂x
(
∂f
∂x
)
=
∂ 2f
= fxx .
∂x2
=
∂ 2f
= fyy .
∂y 2
2. Differentiate twice with respect to y:
∂
∂y
(
∂f
∂y
)
3. Differentiate first with respect to x and then with respect to y:
( )
∂ ∂f
∂ 2f
=
= fxy .
∂y ∂x
∂y∂x
4. Differentiate first with respect to y and then with respect to x:
( )
∂ ∂f
∂ 2f
=
= fyx .
∂x ∂y
∂x∂y
The third and fourth cases are called mixed partial derivatives.
Example 4.5.4. Find the second partial derivatives of f (x, y) = 3xy 2 − 2y + 5x2 y 2 and determine
the value of fxy (−1, 2).
Solution: Begin by finding the first partial derivatives with respect to x and y.
fx (x, y) = 3y 2 + 10xy 2
and
fy (x, y) = 6xy − 2 + 10x2 y.
Then, differentiate each of these with respect to x and y.
fxx (x, y) = 10y 2 ,
fyy (x, y) = 6x + 10x2 ,
fxy (x, y) = 6y + 20xy
and
fyx (x, y) = 6y + 20xy.
At (−1, 2), the value of fxy is fxy (−1, 2) = 12 − 40 = −28.
Notice that the two mixed partials are equal. Sufficient conditions for this occurrence are given in
the next theorem.
39
Theorem 4.5.1 (Equality of Mixed Partial Derivatives). If f is a function of x and y such that fx
and fy are continuous on an open disc R, then for every (x, y) in R,
fxy (x, y) = fyx (x, y).
The order of differentiation of the mixed partial derivatives is irrelevant.
Example 4.5.5. Show that fxz = fzx and fxzz = fzxz = fzzx for the function given by
f (x, y, z) = yex + x ln z.
Solution: First partials:
fx (x, y, z) = yex + ln z,
fz (x, y, z) =
x
.
z
Second partials: (Note the first two are equal)
1
1
fxz (x, y, z) = , fzx (x, y, z) = ,
z
z
Third partials: (Note that all three are equal)
fxzz (x, y, z) = −
4.6
1
,
z2
fzxz (x, y, z) = −
fzz (x, y, z) = −
1
,
z2
x
.
z2
fzzx (x, y, z) = −
1
.
z2
Increments and Differentials
We generalize the concepts of increments and differentials to functions of two or more variables. So
∆x and ∆y are the increments of x and y, and the increment of z is given by
∆z = f (x + ∆x, y + ∆y) − f (x, y).
4.6.1
Definition of Total Differential
If z = f (x, y) and ∆x and ∆y are increments of x and y, then the differentials of the independent
variables x and y are
dx = ∆x and dy = ∆y
and the total differential of the dependent variable z is
dz =
∂z
∂z
dx +
dy = fx (x, y)dx + fy (x, y)dy.
∂x
∂y
This definition can be extended to a function of three or more variables. for example,
if w = f (x, y, z, u), then dx = ∆x, dy = ∆y, dz = ∆z, du = ∆u, and the total differential of w is
dw =
∂w
∂w
∂w
∂w
dx +
dy +
dz +
du.
∂x
∂y
∂z
∂u
40
Example 4.6.1.
(i) The total differential dz for z = 2x sin y − 3x2 y 2 is
dz =
∂z
∂z
dx +
dy = (2 sin y − 6xy 2 )dx + (2x cos y − 6x2 y)dy.
∂x
∂y
(ii) The total differential dw for w = x2 + y 2 + z 2 is
dw =
∂w
∂w
∂w
dx +
dy +
dz = 2xdx + 2ydy + 2zdz.
∂x
∂y
∂z
Approximation by Differentials
For small ∆x and ∆y we can use the approximation ∆z ≈ dz. The approximation of ∆z by dz is
called a linear approximation.
Example 4.6.2. Use the differential dz to approximate the change in
√
z = 4 − x2 − y 2
as (x, y) moves from the point (1, 1) to (1.01, 0.97).
Solution: Letting (x, y) = (1, 1) and (x + ∆x, y + ∆y) = (1.01, 0.97) produces
dx = ∆x = 0.01 and
dy = ∆y = −0.03.
Thus, the change in z can be approximated by
∆z ≈ dz
∂z
∂z
=
dx +
dy
∂x
∂y
−x
−y
= √
∆x + √
∆y.
4 − x2 − y 2
4 − x2 − y 2
When x = 1 and y = 1, you have
1
1
0.02 √
∆z ≈ − √ (0.01) − √ (−0.03) = √ = 2(0.01) ≈ 0.0141.
2
2
2
Theorem 4.6.1. If a function of x and y is differentiable at (x0 , y0 ), then it is continuous at
(x0 , y0 ).
Note that the existence of fx and fy is not sufficient to guarantee differentiability.
Example 4.6.3. Show that fx (0, 0) and fy (0, 0) both exist, but that f is not differentiable at (0, 0)
where f is defined as

 −3xy
,
(x, y) =
̸ (0, 0)
f (x, y) =
x2 + y 2
 0,
(x, y) = (0, 0),
41
Solution: You can show that f is not differentiable at (0, 0) by showing that it is not continuous at
this point. to see that f is not continuous at (0, 0), look at the values of f (x, y) along two different
approaches to f (x, y). Along the line y = x, the limit is
lim
f (x, y) =
(x,x)→(0,0)
−3x2
3
=−
2
(x,x)→(0,0) 2x
2
lim
whereas along y = −x we have
lim
f (x, y) =
(x,−x)→(0,0)
3x2
3
=
.
(x,−x)→(0,0) 2x2
2
lim
Thus, the limit of f (x, y) as (x, y) → (0, 0) does not exist, and we can conclude that f is not
continuous at (0, 0). Hence f is not differentiable at (0, 0). On the other hand, by the definition of
the partial derivatives fx and fy , we have
f (∆x, 0) − f (0, 0)
0−0
= lim
=0
∆x→0
∆x→0 ∆x
∆x
fx (0, 0) = lim
and
f (0, ∆y) − f (0, 0)
0−0
= lim
= 0.
∆y→0
∆y→0 ∆y
∆y
fy (0, 0) = lim
Thus, the partial derivatives at (0, 0) exist.
4.7
Chain Rules for Functions of Several Variables
Theorem 4.7.1. Let w = f (x, y), where f is a differentiable function of x and y. If x = g(t) and
y = h(t), where g and h are differentiable functions of t, then w is a differentiable function of t,
and
∂w dx ∂w dy
dw
=
+
.
dt
∂x dt
∂y dt
Example 4.7.1. Let w = x2 y − y 2 , where x = sin t and y = et . Find
Solution: By the Chain Rule for one independent variable, we have
∂w dx ∂w dy
dw
=
+
dt
∂x dt
∂y dt
= 2xy(cos t) + (x2 − 2y)et .
When t = 0, x = 0, and y = 1, it follows that
dw
= 0 − 2 = −2.
dt
42
dw
at t = 0.
dt
∂w
∂x
w
∂w
∂y
y
x
dy
dt
dx
dt
t
t
Figure 4.2: Chain rule: one independent variable
The Chain Rule can be extended to any number of variables. for example, if each xi is a differentiable
function of a single variable t, then for w = f (x1 , x2 , . . . , xn ), we have
dw
∂w dx1
∂w dx2
∂w dxn
=
+
+ ··· +
.
dt
∂x1 dt
∂x2 dt
∂xn dt
Another type of composite function is one in which the intermediate variables are themselves functions of more than one variable. For example, if w = f (x, y), where x = g(s, t) and y = h(s, t), then
it follows that w is a function of s and t, and we consider the partial derivatives of w with respect
to s and t. One way to find these partial derivatives is to write w as a function of s and t explicitly
by substituting the equations x = g(s, t) and y = h(s, t) into the equation w = f (x, y), then find
the partial derivatives in the usual way.
Example 4.7.2. Find
∂w
s
∂w
and
for w = 2xy, where x = s2 + t2 and y = .
∂s
∂t
t
s
into the equation w = 2xy to obtain
t
( 3
)
(s)
s
2
2
w = 2xy = 2(s + t )
=2
+ st .
t
t
Solution: Begin by substituting x = s2 + t2 and y =
Then, to find
∂w
, hold t constant and differentiate with respect to s.
∂s
( 2
)
∂w
3s
6s2 + 2t2
=2
+t =
.
∂s
t
t
Similarly, to find
∂w
, hold s constant and differentiate with respect to t to obtain
∂t
( 3
)
( 3
)
∂w
s
−s + st2
2st2 − 2s3
=2 − 2 +s =2
=
.
∂t
t
t2
t2
43
The following theorem gives an alternative method for finding the partial derivatives without explicitly writing w as a function of s and t.
Theorem 4.7.2 (Chain Rule: Two Independent Variables). Let w = f (x, y), where f is a differ∂x ∂x ∂y
entiable function of x and y. If x = g(s, t) and y = h(s, t) such that the first partials
, ,
∂s ∂t ∂s
∂y
∂w
∂w
all exist, then
and
exist and are given by
and
∂t
∂s
∂t
∂w
∂w ∂x ∂w ∂y
=
+
∂s
∂x ∂s
∂y ∂s
∂w
∂w ∂x ∂w ∂y
=
+
.
∂t
∂x ∂t
∂y ∂t
and
w
∂w
∂y
#
"!
∂w
∂x
y
x
∂x
∂t
t
'$
&%
∂x
∂s
s
∂y
∂t
∂y
∂s
t
s
Figure 4.3: Chain Rule: Two independent variables
∂w
∂w
s
Example 4.7.3. Use the Chain rule to find
and
for w = 2xy where x = s2 + t2 and y = .
∂s
∂t
t
Solution: We can hold t constant and differentiate with respect to s to obtain,
∂w
∂w ∂x ∂w ∂y
=
+
∂s
∂x ∂s
∂y ∂s
( )
1
= 2y(2s) + 2x
t
( 2)
2s2 + 2t2
s
+
= 4
t
t
2
2
6s + 2t
=
.
t
44
Similarly, holding s fixed gives
∂w
∂w ∂x ∂w ∂y
=
+
∂t
∂x ∂t
∂y ∂t
( )
−s
= 2y(2t) + 2x
t2
( )
(s)
−s
2
2
= 2
(2t) + 2(s + t )
t
t2
2s3 + 2st2
= 4s −
t2
2
3
4st − 2s − 2st2
=
t2
2
2st − 2s3
.
=
t2
∂w
∂w
Example 4.7.4. Find
and
when s = 1 and t = 2π for the function given by w = xy+yz+xz
∂s
∂t
where x = s cos t, y = s sin t and z = t.
Solution: By extending the theorem, we have
∂w ∂x ∂w ∂y ∂w ∂z
∂w
=
+
+
∂s
∂x ∂s
∂y ∂s
∂z ∂s
= (y + z)(cos t)+)(x + z)(sin t) + (y + x)(0)
= (y + z)(cos t) + (x + z)(sin t)
When s = 1 and t = 2π, we have x = 1, y = 0 and z = 2π. Therefore
∂w
= 2π(1) + (1 + 2π)(0) + 0 = 2π.
∂s
Furthermore
∂w ∂x ∂w ∂y ∂w ∂z
∂w
=
+
+
∂t
∂x ∂t
∂y ∂t
∂z ∂t
= (y + z)(−s sin t) + (x + z)(s cos t) + (y + x)(1)
and for s = 1 and t = 2π it follows that
∂w
= (0 + 2π)(0) + (1 + 2π)(1) + (0 + 1)(1) = 2 + 2π.
∂t
Example 4.7.5. If u = x + y and v = xy, find
∂x ∂x ∂y ∂y
,
,
,
.
∂u ∂v ∂u ∂v
Solution: Clearly du = dx + dy and dv = ydx + xdy, and hence −ydx = xdy − dv and
xdx = xdu − xdy. Adding these two equations, yield (x − y)dx = xdu − dv. Also, xdy = dv − ydx
and −ydy = ydx − ydu, hence (x − y)dy = dv − ydu. Thus
dx =
x
1
du −
dv
x−y
x−y
45
and
dy =
1
y
dv −
du.
x−y
x−y
Consequently, we have
∂x
x
=
,
∂u
x−y
∂x
−1
=
∂v
x−y
∂y
−y
=
∂u
x−y
∂y
1
=
.
∂v
x−y
Example 4.7.6. Parabolic co-ordinates (u, v) are defined implicitly in terms of the Cartesian coordinates (x, y) by the pair of equations
x=
Obtain expressions for
u2 − v 2
,
2
y = uv.
∂u ∂v
∂v
,
and
in terms of u and v and verify that
∂y ∂x
∂y
∂u ∂v ∂u ∂v
+
= 0.
∂x ∂x ∂y ∂y
∂f
∂f
∂ϕ ∂ϕ
and
in terms of
,
, u and v, and
∂x
∂y
∂u ∂v
[( )
( )2 ( )2
( )2 ]
2
∂f
∂f
1
∂ϕ
∂ϕ
+
= 2
+
.
2
∂x
∂y
u +v
∂u
∂v
Given that f (x, y) = ϕ(u, v), obtain expressions for
deduce that
u2 − v 2
, y = uv, then dx = udu − vdv and dy = vdu + udv. Multiplying the
Solution: Since x =
2
first by u and the second by v, we have
udx = u2 du − uvdv,
vdy = v 2 du + uvdv.
On adding them, we obtain
(u2 + v 2 )du = udx + vdy.
(4.1)
Also multiplying the first by v and the second by u, we get
vdx = uvdu − v 2 dv,
udy = uvdu + u2 dv.
Subtracting them, we obtain
(u2 + v 2 )dv = udy − vdx.
(4.2)
From (4.1) and (4.2), we have
du =
u2
u
v
dx + 2
dy,
2
+v
u + v2
and
u
v
dy − 2
dx.
2
+v
u + v2
∂u
∂u
∂v
∂v
Comparing these equations with du =
dx +
dy and dv =
dx +
dy, we get
∂x
∂y
∂x
∂y
dv =
u
∂u
= 2
,
∂x
u + v2
u2
v
∂u
= 2
,
∂y
u + v2
u
∂v
= 2
,
∂y
u + v2
46
v
∂v
=− 2
.
∂x
u + v2
Now
∂u ∂v ∂u ∂v
−uv
uv
+ 2
= 0.
+
= 2
2
∂x ∂x ∂y ∂y
u +v
u + v2
Given that f (x, y) = ϕ(u, v), it follows that u and v are implicit functions of x and y and so
∂ϕ ∂u ∂ϕ ∂v
u ∂ϕ
v
∂ϕ
∂f
=
+
= 2
− 2
2
2
∂x
∂u ∂x ∂v ∂x
u + v ∂u u + v ∂v
and
Now
Hence
∂f
∂ϕ ∂u ∂ϕ ∂v
v
∂ϕ
u ∂ϕ
=
+
= 2
+ 2
.
2
∂y
∂u ∂y ∂v ∂y
u + v ∂u u + v 2 ∂v
)2
( )2
( )2
∂ϕ
∂ϕ ∂ϕ
∂ϕ
∂f
u2
2uv
v2
= 2
− 2
+ 2
2
2
2
2
2
2
∂x
(u + v )
∂u
(u + v ) ∂u ∂v (u + v )
∂v
(
(
( )2
)
)2
2
∂ϕ
∂ϕ ∂ϕ
∂ϕ
∂f
v2
2uv
u2
= 2
+ 2
+
.
∂y
(u + v 2 )2 ∂u
(u + v 2 )2 ∂u ∂v (u2 + v 2 )2 ∂v
(
(
∂f
∂x
)2
(
+
∂f
∂y
)2
1
= 2
u + v2
[(
∂ϕ
∂u
)2
(
+
∂ϕ
∂v
)2 ]
.
Example 4.7.7. Let w = f (x, y), where x and y are given in polar coordinates by the equations
∂w ∂w
∂ 2w
x = r cos θ and y = r sin θ. Calculate
,
and
in terms of r and θ and the partial
∂r ∂θ
∂r2
derivatives of w with respect to x and y.
Solution: Here x and y are intermediate values, while the independent variables are r and θ. First
note that
∂y
∂x
∂y
∂x
= cos θ,
= sin θ,
= −r sin θ and
= r cos θ.
∂r
∂r
∂θ
∂θ
Then
∂w
∂w ∂x ∂w ∂y
∂w
∂w
=
+
=
cos θ +
sin θ
∂r
∂x ∂r
∂y ∂r
∂x
∂y
and
∂w ∂x ∂w ∂y
∂w
∂w
∂w
=
+
= −r
sin θ + r
cos θ.
∂θ
∂x ∂θ
∂y ∂θ
∂x
∂y
Next,
( )
(
)
∂ 2w
∂ ∂w
∂ ∂w
∂w
∂wx
∂wy
=
=
cos θ +
sin θ =
cos θ +
sin θ,
2
∂r
∂r ∂r
∂r ∂x
∂y
∂r
∂r
∂w
∂w
where wx =
and wy =
. Therefore
∂x
∂y
)
(
)
(
∂ 2w
∂wy ∂x ∂wy ∂y
∂wx ∂x ∂wx ∂y
+
cos θ +
+
sin θ
=
∂r2
∂x ∂r
∂y ∂r
∂x ∂r
∂y ∂r
( 2
)
( 2
)
∂ w
∂ w
∂ 2w
∂ 2w
=
sin θ cos θ +
cos θ +
cos θ +
sin θ sin θ.
∂x2
∂y∂x
∂x∂y
∂y 2
Finally, because wyx = wxy , we get
∂ 2w
∂ 2w 2
∂ 2w
∂ 2w
2
cos
θ
sin
θ
+
=
cos
θ
+
2
sin θ.
∂r2
∂x2
∂x∂y
∂y 2
47
4.8
The Gradient of a Function of Two Variables
The gradient of a function of two variables is a vector-valued function of two variables.
Let z = f (x, y) be a function of x and y such that fx and fy exist. The gradient of f , denoted by
∇f (x, y), is the vector
∇f (x, y) = fx (x, y)i + fy (x, y)j.
We read ‘∇f ’ as “del f ”. Another notation for the gradient is grad f (x, y).
Example 4.8.1. Find the gradient of f (x, y) = y ln x + xy 2 at the point (1, 2).
Solution: Using fx (x, y) =
y
+ y 2 and fy (x, y) = ln x + 2xy, we have
x
)
(y
2
∇f (x, y) =
+ y i + (ln x + 2xy)j.
x
At the point (1, 2), the gradient is
)
2
2
+ 2 i + (ln 1 + 2(1)(2))i
∇f (1, 2) =
1
= 6i + 4j.
(
Example 4.8.2. Find ∇f (x, y, z) for the function given by
f (x, y, z) = x2 + y 2 − 4z.
Solution: The gradient vector is given by
∇f (x, y, z) = fx (x, y, z)i + fy (x, y, z)j + fz (x, y, z)k
= 2xi + 2yj − 4k.
4.9
4.9.1
Tangent Planes and Normal Lines
Tangent Plane and Normal Line to a Surface
So far we have represented surfaces in space primarily by the equations of the form
z = f (x, y).
In general, the representation is
F (x, y, z) = 0.
48
For a surface S given by z = f (x, y), we can convert to the general form by defining F as
F (x, y, z) = f (x, y) − z.
Then, because f (x, y) − z = 0, we can consider S to be the level surface of F given by
F (x, y, z) = 0.
In the process of finding a normal line to a surface, we are also able to solve the problem of finding
a tangent plane to the surface. Let S be a surface given by F (x, y, z) = 0, and p = (x0 , y0 , z0 )
be the point on S. Let C be a curve on S through P that is defined by the vector-valued function
r(t) = x(t)i + y(t)j + z(t)k. Then, for all t
F (x(t), y(t), z(t)) = 0.
If F is differentiable and x′ (t), y ′ (t) and z ′ (t) all exist, it follows from the Chain Rule that
0 = F ′ (t) = Fx (x, y, z)x′ (t) + Fy (x, y, z)y ′ (t) + Fz (x, y, z)z ′ (t).
At (x0 , y0 , z0 ), the equivalent vector form is
0 = ∇F (x0 , y0 , z0 ) ·
|
{z
}
r′ (t)
|{z}
.
T angentV ector
Gradient
This result means that the gradient at P is orthogonal to the tangent vector of every curve on S
through P .
Theorem 4.9.1 (Definition of Tangent Plane and Normal Line). Let F be differentiable at the
point P = (x0 , y0 , z0 ) on the surface S given by F (x, y, z) = 0 such that ∇F (x0 , y0 , z0 ) ̸= 0.
1. The plane through P that is normal to ∇F (x0 , y0 , z0 ) is called the tangent plane to S at
P.
2. The line through P having the same direction of ∇F (x0 , y0 , z0 ) is called the normal line to
S at P .
Theorem 4.9.2 (Equation of Tangent Plane). If F is differentiable at (x0 , y0 , z0 ) then an equation
of the tangent plane to the surface given by F (x, y, z) = 0 at (x0 , y0 , z0 ) is
Fx (x0 , y0 , z0 )(x − x0 ) + Fy (x0 , y0 , z0 )(y − y0 ) + Fz (x0 , y0 , z0 )(z − z0 ) = 0.
Example 4.9.1. Find an equation of the tangent plane to the hyperboloid given by
z 2 − 2x2 − 2y 2 = 12
at the point (1, −1, 4).
Solution: Begin by writing the equation of the surface as
z 2 − 2x2 − 2y 2 − 12 = 0.
49
Then, considering
F (x, y, z) = z 2 − 2x2 − 2y 2 − 12
we have
Fx (x, y, z) = −4x,
Fy (x, y, z) = −4y,
Fz (x, y, z) = 2z.
At the point (1, −1, 4), the partial derivatives are
Fx (1, −1, 4) = −4,
Fy (1, −1, 4) = 4,
Fz (1, −1, 4) = 8.
Therefore, an equation of the tangent plane at (1, −1, 4) is
−4(x − 1) + 4(y + 1) + 8(z − 4)
−4x + 4 + 4y + 4 + 8z − 32
−4x + 4y + 8z − 24
x − y − 2z + 6
=
=
=
=
0
0
0
0.
To find the equation of the tangent plane at a point on a surface given by z = f (x, y), we can define
the function F by
F (x, y, z) = f (x, y) − z.
Then S is given by the level surface F (x, y, z) = 0 and an equation of the tangent plane to S at the
point (x0 , y0 , z0 ) is
fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) − (z − z0 ) = 0.
Example 4.9.2. Find the equation of the tangent plane to the paraboloid
z =1−
1 2
(x + 4y 2 )
10
at the point (1, 1, 12 ).
Solution: From z = 1 −
fx (x, y) = −
1 2
(x + 4y 2 ), we obtain
10
1
x
=⇒ fx (1, 1) = −
5
5
and
fy (x, y) = −
4y
4
=⇒ fy (1, 1) = − .
5
5
Therefore, an equation of the tangent plane at (1, 1, 12 ) is
(
)
1
fx (1, 1)(x − 1) + fy (1, 1)(y − 1) − z −
= 0
2
(
)
4
1
1
= 0
− (x − 1) − (y − 1) − z −
5
5
2
1
4
3
− x− y−z+
= 0.
5
5
2
The gradient ∇F (x, y, z) gives a convenient way to find equations of normal lines.
50
Example 4.9.3. Find a set of symmetric equations for the normal line to the surface given by
xyz = 6 at the point (2, 3, 1).
Solution: Begin by writing F (x, y, z) = xyz − 6. Then, the gradient is given by
∇F (x, y, z) = Fx (x, y, z)i + Fy (x, y, z)j + Fz (x, y, z)k
= yzi + xzj + xyk
and at point (2, 3, 1) we have
∇F (2, 3, 1) = (3)(1)i + (2)(1)j + (2)(3)k
= 3i + 2j + 6k.
The normal line at (2, 3, 1) has direction numbers 3, 2 and 6, and the corresponding set of symmetric
equations is
x−2
y−3
z−1
=
=
.
3
2
6
4.10
Extrema of Functions of Two Variables
4.10.1
Absolute Extrema and Relative Extrema
Theorem 4.10.1 (Extreme Value Theorem). Let f be a continuous function of two variables x and
y defined on a closed bounded region R in the xy-plane.
1. There is at least one point in R where f takes on a minimum value.
2. There is at least one point in R where f takes on a maximum value.
A minimum value is also called an absolute minimum and a maximum is also called an absolute
maximum.
Definition of Relative Extrema
Let f be a function defined on a region R containing (x0 , y0 ).
1. The function f has a relative minimum at (x0 , y0 ) if
f (x, y) ≥ f (x0 , y0 )
for all (x, y) in an open disc containing (x0 , y0 ).
51
2. The function f has a relative maximum at (x0 , y0 ) if
f (x, y) ≤ f (x0 , y0 )
for all (x, y) in an open disc containing (x0 , y0 ).
To locate relative extreme of f , we can investigate the points at which the gradient of f is 0. Such
points are called critical points of f .
Definition of Critical Point
Let f be defined on an open region R containing (x0 , y0 ). The point (x0 , y0 ) is a critical point of
f if one of the following is true.
1. fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0.
2. fx (x0 , y0 ) and fy (x0 , y0 ) does not exist.
Theorem 4.10.2. If f has a relative extremum at (x0 , y0 ) on an open region R, then (x0 , y0 ) is a
critical point of f .
Example 4.10.1. Determine the relative extrema of
f (x, y) = 2x2 + y 2 + 8x − 6y + 20.
Solution: Begin by finding the critical points of f . Because fx (x, y) = 4x + 8 and fy (x, y) = 2y − 6
are defined for all x and y, the only critical points are those for which both first partial derivatives
are 0. To locate these points, let fx (x, y) and fy (x, y) be 0, and solve the system of equations
4x + 8 = 0
and 2y − 6 = 0
to obtain the critical point (−2, 3). By completing the square, we can conclude that for all
(x, y) ̸= (−2, 3),
f (x, y) = 2(x + 2)2 + (y − 3)2 + 3 > 3.
Therefore, a relative minimum of f occurs at (−2, 3). The value of the relative minimum is
f (−2, 3) = 3.
The above example shows a relative minimum occurring at one type of critical point, the type for
which both fx (x, y) and fy (x, y) are 0. The next example concerns a relative maximum that occurs
at the other type of critical point, the type for which either fx (x, y) or fy (x, y) is undefined.
52
1
Example 4.10.2. Determine the relative extrema of f (x, y) = 1 − (x2 + y 2 ) 3 .
Solution: Because
fx (x, y) = −
2x
3(x2
+
2
y2) 3
and
fy (x, y) = −
2y
3(x2
2
+ y2) 3
it follows that both partial derivatives are defined for all points in the xy-plane except for (0, 0).
Moreover, because the partial derivatives cannot both be 0 unless both x and y are 0, we can conclude
that (0, 0) is the only critical point. Note that f (0, 0) = 1, for all other (x, y) it is clear that
1
f (x, y) = 1 − (x2 + y 2 ) 3 < 1.
Therefore, f has a relative maximum at (0, 0).
The Second Partials Test
Some critical points yield saddle points, which are neither relative maxima nor relative minima.
Theorem 4.10.3. Let f have continuous second partial derivatives on an open region containing
a point (a, b) for which
fx (a, b) = 0 and fy (a, b) = 0.
To test for relative extrema of f , we define the quantity
d = fxx (a, b)fyy (a, b) − [fxy (a, b)]2 .
1. If d > 0 and fxx (a, b) > 0, then f has a relative minimum at (a, b).
2. If d > 0 and fxx (a, b) < 0, then f has a relative maximum at (a, b).
3. If d < 0, then (a, b, f (a, b)) is a saddle point.
4. The test is inconclusive if d = 0.
A convenient device for remembering the formula for d in the Second Partials Test is given by
the 2 × 2 determinant
f (a, b) fxy (a, b)
d = xx
fyx (a, b) fyy (a, b)
where fxy (a, b) = fyx (a, b).
Example 4.10.3. Find the relative extrema of f (x, y) = −x3 + 4xy − 2y 2 + 1.
Solution: Begin by finding the critical points of f . Because
fx (x, y) = −3x2 + 4y
and
53
fy (x, y) = 4x − 4y
are defined for all x and y, the only critical points are those for which both first partial derivatives
are 0. Solving the equations −3x2 + 4y = 0 and 4x − 4y = 0, we see that from the second equation
that x = y, and by substitution into the first equation, we obtain two solutions, y = x = 0 and
y = x = 34 . Because
fxx (x, y) = −6x, fyy (x, y) = −4, fxy (x, y) = 4
it follows that, for the critical point (0, 0),
d = fxx (0, 0)fyy (0, 0) − [fxy (0, 0)]2 = 0 − 16 < 0
and, by the Second Partials Test, we can conclude that (0, 0) is a saddle point of f . Furthermore,
for the critical point ( 34 , 43 ),
(
d = fxx
4 4
,
3 3
)
(
fyy
4 4
,
3 3
)
[
(
− fxy
4 4
,
3 3
)]2
= −8(−4) − 16 > 0
and because fxx ( 43 , 34 ) = −8 < 0 we can conclude that f has a relative maximum at ( 43 , 43 ).
54
Chapter 5
Multiple Integration
5.1
Iterated Integrals
In the previous chapter, we saw that it is meaningful to differentiate functions of several variables
with respect to one variable while holding the other variable constant, we can integrate functions
of several variables by a similar procedure. For example, if we have the partial derivative
fx (x, y) = 2xy, then by considering y constant, we can integrate with respect to x to obtain
∫
f (x, y) =
fx (x, y) dx
Integrate with respect to x
∫
=
2xy dx
y is held constant
∫
= y 2x dx
Factor out constant y
= y(x2 ) + C(y)
= x2 + C(y).
Anti-derivative of 2x is x2
C(y) is function of y
Note that the constant of integration, C(y) is a function of y. In other words, by integrating with
respect to x, we are able to recover f (x, y) only partially. For example, by considering y constant,
we can apply the Fundamental Theorem of calculus to evaluate
∫
2y
2y
= (2y)2 y − (1)2 y = 4y 3 − y.
2xy dx = x2 y
1
1
Note that the variable of∫ integration cannot appear in either limit of integration. For example, it
x
y dx.
makes no sense to write
0
∫
Example 5.1.1. Evaluate
x
(2x2 y −2 + 2y) dy.
1
55
Solution: Considering x to be constant and integrating with respect to y produces
[
]x
∫ x
−2x2
2 −2
2
+y
(2x y + 2y) dy =
y
1
(
)
)1 (
−2x2
−2x2
2
=
+x −
+1
x
1
= 3x2 − 2x − 1.
Notice that in the above example the integral defines a function of x and can itself be integrated.
]
∫ 2 [∫ x
2 −2
Example 5.1.2. Evaluate
(2x y + 2y) dy dx.
1
1
Solution:
y
6
1≤x≤2
y=x
1≤y≤x
2
1
1
2
-
Figure 5.1: The region of integration
Using the result from the previous example, we have
]
∫ 2 [∫ x
∫ 2
2 −2
(2x y + 2y) dy dx =
(3x2 − 2x − 1) dx
1
1
1
[ 3
]2
= x − x2 − x 1
= 2 − (−1)
= 3.
56
x
The integral in the above example is an iterated integral. Iterated integrals are usually written
simply as
∫ b ∫ g2 (x)
∫ d ∫ h2 (y)
f (x, y) dydx and
f (x, y) dxdy.
a
g1 (x)
c
h1 (y)
The inside limits of integration can be variable with respect to the outer variable of integration.
However, the outside limits of integration must be constant with respect to both variables of
integration. After performing the inside integration, we obtain a definite integral, and the second
integration produces a real number.
One order of integration will often produce a simpler integration problem than the other order. The
order of integration affects the ease of integration, but not the value of the integral.
5.1.1
Comparing Different Orders of Integration
Example 5.1.3. Sketch the region whose area is represented by the integral
∫ 2∫ 4
dxdy.
0
y2
Then find another iterated integral using the order dydx to represent the area, and show that both
integrals yield the same value.
Solution: From the given limits of integration, we know that
y
6
y2
∆y
4
-
x
Figure 5.2: Fig 1
y2 ≤ x ≤ 4
Inner limits of integration
57
which means that the region R is bounded on the left by the parabola x = y 2 and on the right by the
line x = 4. Furthermore, because
0≤y≤2
Outer limits of integration
we know that R bounded below by the x-axis. The value of this integral is
∫ 2∫ 4
∫ 2 ]4
dxdy =
x dy
y2
0
0
∫
y2
2
(4 − y 2 ) dy
0
[
]2
y3
16
= 4y −
= .
3 0
3
=
To change the order of integration to dydx, place a vertical rectangle in the region. From this we
can see that the constant bounds 0 ≤ x ≤ 4 serve as the outer limits of integration.
√ By solving for
2
y in the equation x = y , we can conclude that the inner bounds are 0 ≤ y ≤ x. Therefore, the
area of the region can be represented by
∫ 4 ∫ √x
dydx.
0
0
By evaluating this integral, we can see that it has the same value as the original integral.
y
6
y=
∆x
Figure 5.3: Fig 2
∫
4
∫
√
∫
x
y
dydx =
0
0
0
] √x
4
∫
dx
0
√
x dx
=
0
]4
2 3
16
=
x2 = .
3
3
4
0
58
-
x
√
x
∫
4
∫
2
2
ey dydx as an iterated integral with order of integration reversed
Example 5.1.4. Express
x
2
0
and evaluate.
Solution: From the given limits of integration we see that, for a fixed x, y varies from y = x2 to
y = 2 and x varies from x = 0 to x = 4. We can also describe the region as, for y fixed, x varies
from
∫ ∫ x = 0 to x = 2y and y varies from y = 0 to y = 2. The corresponding iterated integral is
2
2y
2
ey dxdy. Solving, we have
0
0
∫
2
∫
∫
2y
e dxdy =
0
0
]x=2y
2
y2
y2
xe
dy
0
∫
x=0
2
=
0
= ey
2
2yey dy
]2
2
0
= e − 1.
4
∫
2
∫
1
3
yex dxdy.
Example 5.1.5. Evaluate
y
2
0
3
Solution: We cannot integrate first with respect to x, as indicated, because it happens that ex
has no elementary anti-derivative. So we try to evaluate the integral by first reversing the order of
integration.
∫ 2∫ 1
∫ 1 ∫ 2x
3
x3
ye dxdy =
yex dydx
0
y
2
0
∫
0
1
=
0
∫
=
0
1
]2
1 2
3
y xex dx
2
0
3
2x2 ex dx
]1
2 x3
e
3
0
2
(e − 1).
=
3
=
59
5.2
5.2.1
Double Integrals and Volume
Double Integrals and Volume of a Solid Region
Definition of Double Integral
If f is defined on a closed, bounded region R in the xy-plane, then the double integral of f over
R is given by
∫∫
n
∑
f (x, y) dA = lim
f (xi , yi )∆xi ∆yi
|∆|→0
R
i=1
provided the limit exists. If the limit exists, then f is integrable over R.
A double integral can be used to find the volume of a solid region that lies between the xy-plane
and the surface given by z = f (x, y).
Volume of a Solid Region
If f is integrable over a plane region R and f (x, y) ≥ 0 for all (x, y) in R, then the volume of the
solid region that lies above R and below the graph of f is defined as
∫∫
V =
f (x, y) dA.
R
Example 5.2.1. Find the volume of the solid region R bounded by the surface
f (x, y) = e−x
2
and the planes y = 0, y = x and x = 1.
Solution: The base of R in the xy-plane is bounded by the lines y = 0, x = 1 and y = x. The two
possible orders of integration are
∫ 1∫ 1
∫ 1∫ x
2
−x2
e
dydx and
e−x dxdy.
0
0
0
y
By setting ∫
up the corresponding iterated integrals, we can see that the order dxdy requires the anti2
derivative
e−x dx, which is not an elementary function. On the other hand, the order dydx
60
produces the integral
∫
∫
0
x
−x2
e
1
∫
dydx =
0
]x
1
e
−x2
y
0
∫
dx
0
1
=
xe
−x2
dx
0
]1
1 −x2
= − e
2
( 0 )
1 1
−1
= −
2 e
e−1
=
= 0.316.
2e
5.3
Triple Integrals
Definition of Triple Integral
If f is continuous over a bounded solid region Q, then the triple integral of f over Q is defined
as
∫∫∫
n
∑
f (x, y, z) dV = lim
f (xi , yi , zi ) ∆Vi
|∆|→0
Q
i=1
provided the limit exists. The volume of the solid region Q is given by
∫∫∫
Volume of Q =
dV.
Q
Evaluation by Iterated Integrals
Let f be continuous on a solid region Q defined by
a ≤ x ≤ b,
h1 (x) ≤ y ≤ h2 (x),
g1 (x, y) ≤ z ≤ g2 (x, y)
where h1 , h2 , g1 and g2 are continuous functions. Then
∫ b ∫ h2 (x) ∫
∫∫∫
f (x, y, z) dV =
a
Q
h1 (x)
g2 (x,y)
f (x, y, z) dV.
g1 (x,y)
Example 5.3.1. Evaluate the triple iterated integral
∫ 2 ∫ x ∫ x+y
ex (y + 2z) dzdydx.
0
0
0
61
Solution: For the first integration, hold x and y constant and integrate with respect to z.
]x+y
∫ 2 ∫ x ∫ x+y
∫ 2∫ x
ex (y + 2z) dzdydx =
ex (yz + z 2 )
dydx
0
0
0
0
∫
0
2∫
0
x
ex (x2 + 3xy + 2y 2 ) dydx.
=
0
0
For the second integration, hold x constant and integrate with respect to y.
)]x
∫ 2∫ x
∫ 2[ (
3xy 2 2y 3
x 2
2
x
2
e (x + 3xy + 2y ) dydx =
e x y+
+
dx
2
3
0
0
0
0
∫
19 2 3 x
=
x e dx
6 0
[
]2
19 x 3
e (x − 3x2 + 6x − 6)
=
6
0
(
)
19 e2
=
+1 .
6
3
Example 5.3.2. If f (x, y, z) = xy + yz and T consists of those points (x, y, z) in space satisfying
the inequalities −1 ≤ x ≤ 1, 2 ≤ y ≤ 3 and 0 ≤ z ≤ 1, then
∫∫
∫ 1 ∫ 3∫ 1
f (x, y, z) dV =
(xy + yz) dzdydx
−1
T
=
=
=
=
5.4
5.4.1
0
]1
1 2
dydx
xyz + yz
2
−1 2
z=0
)
∫ 1 ∫ 3(
1
xy + y dydx
2
−1 2
]3
∫ 1[
1 2 1 2
xy + y
dx
4
−1 2
y=2
)
∫ 1(
5
5
=
x+
dx
2
4
−1
]1
[
5
5 2 5
x + x
= .
4
4 −1 2
∫
=
2
1
∫
3
[
Change of Variables : Jacobians
Jacobians
The Jacobian is named after the German mathematician Carl Gustav Jacobi (1804-1851). For the
single integral
∫ b
f (x) dx
a
62
you can change variables by letting x = g(u), so that dx = g ′ (u)du, and obtain
∫
∫
b
d
f (x) dx =
a
f (g(u))g ′ (u) du
c
where a = g(c) and b = g(d). Note that the change of variable introduces an additional factor g ′ (u)
into the integrand. This also occurs in the case of double integrals.
∫∫
∫∫
∂x ∂y ∂y ∂x
f (x, y) dA =
f (g(u, v), h(u, v))
−
dudv
∂u ∂v ∂u ∂v
|
{z
}
R
S
Jacobian
where the change of variables x = g(u, v) and y = h(u, v) introduces a factor called the Jacobian
of x and y with respect to u and v.
Definition of the Jacobian
If x = g(u, v) and y = h(u, v), then the Jacobian of x and y with respect to u and v, denoted by
∂(x, y)
is
∂(u, v)
∂x ∂x
∂u ∂v
∂(x, y)
∂x ∂y ∂y ∂x
=
=
−
.
∂(u, v)
∂u ∂v ∂u ∂v
∂y ∂y
∂u ∂v
∂(u, v)
In cases it is more convenient to express u and v in terms of x and y, we can first compute
∂(x, y)
∂(x, y)
explicitly and then find the needed Jacobian
from the formula
∂(u, v)
∂(x, y) ∂(u, v)
·
= 1.
∂(u, v) ∂(x, y)
Example 5.4.1. Find the Jacobian for the change of variables defined by
x = r cos θ
and
y = r sin θ.
Solution: From the definition of a Jacobian, we obtain
∂x
∂r
∂(x, y)
=
∂(r, θ)
∂y
∂r
∂x
∂θ
∂y
∂θ
=
cos θ −r sin θ
= r cos2 θ + r sin2 θ = r.
sin θ r cos θ
63
The above example points out that the change of variables from rectangular to polar coordinates
for a double integral can be written as
∫∫
∫∫
f (x, y) dA =
f (r cos θ, r sin θ) rdrdθ, r > 0
R
S
∫∫
=
f (r cos θ, r sin θ)
∂(x, y)
drdθ
∂(r, θ)
S
where S is the region in the rθ-plane that corresponds to the region R in the xy-plane. In general,
a change of variables is given by a one-to-one transformation T from a region S in the uv-plane
to a region R in the xy-plane, to be given by
T (u, v) = (x, y) = (g(u, x), h(u, v))
where g and h have continuous first partial derivatives in the region S. Note that the point (u, v)
lies in S and the point (x, y) lies in R. In most cases, we are hunting for a transformation for which
the region S is simpler than the region R.
Change of Variables for Double Integrals
Theorem 5.4.1. Let R and S be regions in the xy- and uv-planes that are related by the equations
x = g(u, v) and y = h(u, v) such that each point in R is the image of a unique point in S. If f is
∂(x, y)
continuous on R, g and h have continuous partial derivatives on S, and
is non-zero on S,
∂(u, v)
then
∫∫
∫∫
∂(x, y)
dudv.
f (x, y) dA =
f (g(u, v), h(u, v))
∂(u, v)
R
S
Example 5.4.2. Let R be the region bounded by the lines
x − 2y = 0,
Evaluate the double integral
x − 2y = −4,
x + y = 4,
and x + y = 1.
∫∫
3xy dA.
R
Solution: to begin, let u = x + y and v = x − 2y. Solving this system of equations for x and y
1
1
produces x = (2u + v) and y = (u − v). The partial derivatives of x and y are
3
3
∂x
2
= ,
∂u
3
∂x
1
= ,
∂v
3
∂y
1
=
∂u
3
64
and
∂y
1
=−
∂v
3
which implies that the Jacobian is
∂(x, y)
=
∂(u, v)
=
∂x
∂u
∂x
∂v
∂y ∂y
∂u ∂v
2 1
3 3
1
1
−
3
3
2 1
1
= − − =− .
9 9
3
Therefore, we obtain
∫∫
]
1
1
∂(x, y)
3 (2u + v) (u − v)
dvdu
3
3
∂(u, v)
S
∫ 4∫ 0
1
(2u2 − uv − v 2 ) dvdu
9
1
−4
]0
∫ 4[
1
uv 2 v 3
2
2u v −
−
du
9 1
2
3 −4
)
∫ (
1 4
64
2
8u + 8u −
du
9 1
3
]4
[
64
1 8u3
2
+ 4u − u
9 3
3 1
164
.
9
∫∫
3xy dA =
R
=
=
=
=
=
[
Example 5.4.3. Suppose R is the
∫ ∫ plane bounded by the hyperbolas xy = 1, xy = 3 and
x2 − y 2 = 1, x2 = y 2 = 4. Find
(x2 + y 2 ) dxdy.
R
Solution: The hyperbolas bounding R are u−curves and v−curves if u = xy and v = x2 − y 2 . We
can most easily write x2 + y 2 in terms of u and v by first noting that
4u2 + v 2 = 4x2 y 2 + (x2 − y 2 )2 = (x2 + y 2 )2 ,
so that x2 + y 2 =
√
4u2 + v 2 . Now
∂(u, v)
y
x
=
= −2(x2 + y 2 ).
2x
−2y
∂(x, y)
Hence we have
1
1
∂(x, y)
=−
=− √
.
2
2
∂(u, v)
2(x + y )
2 4u2 + v 2
65
Therefore
∫∫
∫
2
4
∫
2
(x + y ) dxdy =
R
1
3
√
4u2
+
v2
1
1
√
dudv =
2 4u2 + v 2
∫
4
∫
1
1
3
1
dudv = 3.
2
Example 5.4.4. Find the area of the region R bounded by the curves xy = 1, xy = 3 and
xy 1.4 = 1, xy 1.4 = 2.
Solution: Define change of variables transformation by u = xy and v = xy 1.4 . Then
∂(u, v)
y
x
= 1.4
= (0.4)xy 1.4 = (0.4)v.
y
(1.4)xy 0.4
∂(x, y)
So
∂(x, y)
1
2.5
=
=
. Consequently,
∂(u, v)
∂(u, v)
v
∂(x, y)
∫∫
∫
dxdy =
R
1
2
∫
3
1
66
2.5
dudv = 5 ln 2.
v
Chapter 6
Infinite Series
One important application of infinite sequences is in representing infinite summations. If {an } is
an infinite sequence, then
∞
∑
an = a1 + a2 + a3 + · · ·
n=1
is called an infinite series (or simply a series). The numbers a1 , a2 , a3 , . . . are called the terms of
the series. To find the sum of an infinite series, consider the following sequence of partial sums.
S1
S2
S3
..
.
= a1
= a1 + a2
= a1 + a2 + a3
.
..
..
= ..
.
.
Sn = a1 + a2 + a3 + · · · + an .
If this sequence of partial sums converges, then the series is said to converge and has the sum
indicated in the following definition.
6.1
Definition of Convergent and Divergent Series
For the infinite series
∑
an , the nth partial sum is given by
Sn = a1 + a2 + a3 + · · · + an .
∑
If the sequence of partial sums {Sn } converges to S, then the series
an converges. The limit S
is called the sum of the series. If {Sn } diverges, then the series diverges.
67
Example 6.1.1. The series
∞
∑
1
1 1 1
1
=
+
+
+
+ · · · has the following partial sums.
n
2
2
4
8
16
n=1
S1 =
S2 =
s3 =
..
. =
sn =
1
2
1 1
3
+ =
2 4
4
1 1 1
7
+ + =
2 4 8
8
..
..
.. ..
.
.
. .
1 1 1
2n − 1
1
.
+ + + ··· + n =
2 4 8
2
2n
2n − 1
= 1, it follows that the series converges and its sum is 1.
n→∞
2n
Example 6.1.2. The nth partial sum of the series
) (
) (
) (
)
∞ (
∑
1
1
1
1 1
1 1
−
= 1−
+
−
+
−
+ ···
n n+1
2
2 3
3 4
n=1
Because lim
is given by Sn = 1 −
1
. Because the limit of Sn is 1, the series converges and its sum is 1.
n+1
Example 6.1.3. The series
∞
∑
1 = 1 + 1 + 1 + · · · diverges, because Sn = n and the sequence of
n=1
partial sums diverges.
The series in Example (6.1.2) is a telescoping series. That is, it is of the form
(b1 − b2 ) + (b2 − b3 ) + (b3 − b4 ) + (b4 − b5 ) + · · ·
note that b2 is canceled by the second term, b3 is canceled by the third term and so on. Because the
nth partial sum of the series is Sn = b1 − bn+1 , it follows that a telescoping series will only converge
if and only if bn approaches a finite number as n → ∞. Moreover, if the series converges, then its
sum is
S = b1 − lim bn+1 .
n→∞
Example 6.1.4. Find the sum of the series
∞
∑
n=1
2
.
−1
4n2
Solution: Using partial fractions, we can write
an =
2
2
1
1
=
=
−
.
−1
(2n − 1)(2n + 1)
2n − 1 2n + 1
4n2
From the telescoping form, we can see that the nth partial sum is
) (
)
(
)
(
1 1
1
1
1
1 1
−
+
−
+ ··· +
−
=1−
.
Sn =
1 3
3 5
2n − 1 2n + 1
2n + 1
68
Thus, the series converges and its sum is 1. That is,
∞
∑
n=1
6.2
(
)
2
1
= lim Sn = lim 1 −
= 1.
n→∞
4n2 − 1 n→∞
2n + 1
Geometric Series
A geometric series with ratio r is given by
∞
∑
arn = a + ar + ar2 + · · · + arn + · · · , a ̸= 0.
n=0
Theorem 6.2.1. A geometric series with ratio r diverges if |r| ≥ 1. If 0 < |r| < 1, then the series
∞
∑
a
converges to the sum
arn =
, 0 < |r| < 1.
1
−
r
n=0
Example 6.2.1. The geometric series
( )
( )n
( )2
∞
∞
∑
∑
1
1
1
3
=
3(1)
+
3
+ ···
=
3
+
3
n
2
2
2
2
n=0
n=0
has a ratio of r = 21 with a = 3. Because 0 < |r| < 1, the series converges and its sum is
a
3
S=
=
= 6.
1−r
1 − 12
6.2.1
If
∑
Properties of Infinite Series
an = A and
∑
bn = B and c is a real number, then the following series converge to the
∑
∑
∑
∑
indicated sums. (i)
can = cA (ii)
(an ± bn ) =
an ±
bn = A ± B.
n-th Term Test for Divergence
Limit of n−th Term of a Convergent Series
If the series
∑
an converges, then the sequence {an } converges to 0.
If the sequence {an } does not converge to 0, then the series
69
∑
an diverges.
6.3
Test for Convergence or Divergence of Series
In this and the following section, we will study several convergence tests that apply to series with
positive terms.
6.3.1
The Integral Test
If f is positive, continuous, and decreasing for x ≥ 1 and an = f (n), then
∞
∑
n=1
∫
an and
∞
f (x) dx
1
either both converge or both diverge.
Example 6.3.1. Apply the integral test to the series
∞
∑
n=1
Because f (x) =
obtain
x2
n2
n
.
+1
x
satisfies the conditions for the integral test (check this), we can integrate to
+1
∫
1
∞
1
x
dx =
2
x +1
2
∫
∞
1
2x
dx = lim
2
x +1
2 b→∞
1
[
]b
1
=
lim ln(x2 + 1)
2 b→∞
1
1
2
=
lim [ln(b + 1) − ln 2]
2 b→∞
= ∞.
∫
1
b
2x
dx
+1
x2
Thus, the series diverges.
Example 6.3.2. Apply the integral test to the series
∞
∑
n=1
Solution: Because f (x) =
obtain
∫
1
n2
1
.
+1
1
satisfies the conditions for the integral test, we can integrate to
x2 + 1
∞
[
]b
dx
−1
lim
= lim tan x
b→∞ 1 x2 + 1
b→∞
1
−1
−1
lim (tan b − tan 1)
b→∞
π π
π
=
− = .
2
4
4
dx
=
2
x +1
=
∫
b
Thus, the series converges.
70
6.3.2
p− Series and Harmonic Series
A series of the form
∞
∑
1
1
1
1
= p + p + p + ···
p
n
1
2
3
n=1
is a p−series, where p is a positive constant. For p = 1, the series
∞
∑
1 1
1
= 1 + + + ···
n
2 3
n=1
is the harmonic series.
Theorem 6.3.1. The p−series
∞
∑
1
1
1
1
= p + p + p + ···
p
n
1
2
3
n=1
(i) converges if p > 1 and (ii) diverges if 0 < p ≤ 1.
Example 6.3.3. From the Theorem it follows that the harmonic series
∞
∑
1
1 1
= 1 + + + ···
n
2 3
n=1
diverges.
6.4
Comparisons of Series
6.4.1
Direct Comparison Test
This is a test for positive-term series. It allows you to compare a series having complicated terms
with a simpler series whose convergence or divergence is known.
Direct Comparison Test
Theorem 6.4.1. Let 0 ≤ an ≤ bn for all n.
1. If
∞
∑
n=1
bn converges, then
∞
∑
an converges.
n=1
71
2. If
∞
∑
an diverges, then
n=1
∞
∑
bn diverges.
n=1
Example 6.4.1. Determine the convergence or divergence of
∞
∑
n=1
1
.
2 + 3n
∞
∑
1
Solution: This series resembles
(Convergent geometric series). Term-by-term comparison
n
3
n=1
yields
1
1
an =
< n = bn , n ≥ 1.
n
2+3
3
Thus, by the Direct Comparison Test, the series converges.
∞
∑
1
√ .
Example 6.4.2. Determine the convergence or divergence of
2
+
n
n=1
Solution: The series resembles
∞
∑
1
1
n=1
n2
(Divergent p−series). Term-by-term comparison yields
1
1
√ ≤√ ,
2+ n
n
n≥1
which does not meet the requirements for divergence. Still expecting the series to diverge, we can
∞
∑
1
compare the given series with
(Divergent Harmonic series). In this case, term-by-term comn
n=1
parison yields
1
1
√ = bn , n ≥ 4
an = ≤
n
2+ n
and, by the Direct Comparison Test, the given series diverges.
6.4.2
Limit Comparison Test or Quotient Test
Often a given series closely resembles a p−series or a geometric series, yet we cannot establish the
term-by-term comparison necessary to apply the Direct Comparison Test. We can apply a second
comparison test, called the Limit Comparison Test.
Limit Comparison Test
(
)
an
Suppose that an > 0 and bn > 0 and lim
= L where L is finite and positive. Then the two
n→∞
bn
∑
∑
∑
series
an and
bn , either both converge or both diverge. If L = 0 and
bn converges, then
∑
∑
∑
an converges. If L = ∞ and
bn diverges, then
an diverges.
72
Example 6.4.3. Show that the following harmonic series diverges.
∞
∑
n=1
1
,
an + b
a > 0, b > 0.
∞
1
∑
1
n
1
an+b
Solution: By comparison with
we have lim 1 = lim
= . Because this limit is
n→∞
n→∞ an + b
n
a
n
n=1
grater than 0, we can conclude from the Limit comparison Test that the given series diverges.
The limit Comparison Test works well for comparing a messy algebraic series with a p−series. In
choosing an appropriate p−series, we must choose one with an nth term of the same magnitude as
the nth term of the given series.
Given series
∞
∑
1
2
3n − 4n + 5
n=1
∞
∑
1
√
3n − 2
n=1
∞
2
∑ n − 10
4n5 + n3
n=1
6.5
Comparison series
∞
∑
1
n2
n=1
∞
∑
1
√
n
n=1
∞
∞
2
∑n
∑
1
=
n5
n3
n=1
n=1
Conclusion
Both series converge.
Both series diverge.
Both series converge.
Alternating Series
So far, most series we have dealt with have had positive terms. In this section, we will study series
that contain both positive and negative terms. The simplest such series is an alternating series,
whose terms alternate in sign. For example, the geometric series
)n ∑
∞ (
∞
∑
1 1 1
1
1
1
(−1)n n = 1 − + − +
− ···
−
=
2
2
2
4
8
16
n=0
n=0
is an alternating geometric series with r = − 12 . Alternating series occur in two ways, either the odd
terms are negative or the even terms are negative.
Alternating Series Test
Let an > 0. The alternating series
∞
∑
(−1)n an and
n=1
∞
∑
n=1
conditions are met.
73
(−1)n+1 an converge, if the following two
1. an+1 ≤ an for all n.
2. lim an = 0.
n→∞
Example 6.5.1. Determine the convergence or divergence of
∞
∑
1
(−1)n+1 .
n
n=1
1
1
1
≤
for all n and the limit (as n → ∞) of
is 0, we can apply the
n+1
n
n
Alternating Series Test to conclude that the series converges. (This series is called the alternating
harmonic series)
Solution: Because
Example 6.5.2. Determine the convergence or divergence of
∞
∑
n=1
n
.
(−2)n−1
Solution: To apply the Alternating Series Test, note that, for n ≥ 1,
1
n
≤
2
n+1
n−1
n
2
≤
n
2
n+1
n−1
(n + 1)2
≤ n2n
n
n+1
≤ n−1 .
n
2
2
Hence, an+1 =
n
n+1
≤ n−1 = an for all n. Furthermore, by L’Hopital’s rule,
n
2
2
x
1
n
lim x−1 = lim x−1
= 0 =⇒ lim n−1 = 0.
x→∞ 2
x→∞ 2
n→inf 2
(ln 2)
Therefore, by the Alternating Series Test, the given series converges.
Cases for which the Alternating Series Test Fails
Example 6.5.3. The alternating series
∞
∑
(−1)n+1 (n + 1)
n=1
n
=
2 3 4 5 6
− + − + − ···
1 2 3 4 5
passes the first condition in the alternating series test because an+1 ≤ an for all n. We cannot apply
the Alternating Series Test, because the series does not pass the second condition.
The alternating series
2 1 2 1 2 1 1 1
− + − + − + − + ···
1 1 2 2 3 3 2 4
passes the second condition because an approaches 0 as n → ∞. We cannot apply the Alternating
Series Test, however, because the series does not pass the first condition.
74
6.6
Absolute and Conditional Convergence
Occasionally, a series may have both positive and negative terms and not be an alternating series,
for example, the series
∞
∑
sin 1 sin 2 sin 3
sin n
=
+
+
+ ···
2
n
1
4
9
n=1
has both positive and negative terms, yet it is not an alternating series. One way to obtain some
information about the convergence of this series is to investigate the convergence of the series
∞
∑
sin n
. By direct comparison, we have | sin n| ≤ 1, for all n, so
n2
n=1
1
sin n
≤ 2,
2
n
n
Thus, by the Direct Comparison Test, the series
n ≥ 1.
∞
∑
sin n
n=1
n2
converges. But the question still is “Does
the original series converge?”
Theorem 6.6.1 (Absolute Convergence). If the series
also converges.
∑
|an | converges, then the series
∑
an
The converse of the Theorem is not true. For example, the alternating harmonic series
∞
∑
(−1)n+1
n=1
n
1 1 1
+ − + ···
2 3 4
=1−
converges by the Alternating Series Test. Yet the harmonic series diverges. This type of convergence
is called conditional.
Definition of Absolute and Conditional Convergence
1.
2.
∑
∑
an is absolutely convergent if
∑
an is conditionally convergent if
|an | converges.
∑
an converges but
∑
|an | diverges.
3. An absolutely convergent series converges.
Example 6.6.1. Determine whether the following series are convergent or divergent. Classify any
convergent series as absolutely or conditionally convergent.
n(n+1)
∞
∑
1
1
1 1
(−1) 2
+
− ···.
=− − +
(a)
n
3
3 9 27 81
n=1
75
Solution: This in not an alternating series. However, because
n(n+1)
∞
∞
∑
∑
(−1) 2
1
=
n
3
3n
n=1
n=1
is a convergent geometric series, so the given series is absolutely convergent, hence convergent.
∞
∑
(−1)n
1
1
1
1
(b)
=−
+
−
+
− ···.
ln(n + 1)
ln 2 ln 3 ln 4 ln 5
n=1
Solution: In this case, the alternating series test indicates that the given series converges. However,
the series
∞
∑
(−1)n
1
1
1
=
+
+
+ ···
ln(n
+
1)
ln
2
ln
3
ln
4
n=1
diverges by direct comparison with terms of the harmonic series. Therefore, the given series is
conditionally convergent.
∞
∑
(−1)n
1
1
1
1
√
(c)
= −√ + √ − √ + √ − · · · .
n
1
2
3
4
n=1
Solution: The given series converges by the Alternating Series Test. Moreover, because the
p−series
∞
∑
1
(−1)n
1
1
1
√
= √ + √ + √ + √ + ···
n
1
2
3
4
n=1
diverges, the given series is conditionally convergent.
6.7
The Ratio and Root Tests
6.7.1
The Ratio Test
This is a test for absolute convergence.
Ratio Test
1.
2.
∑
an converges absolutely if lim
n→∞
∑
an diverges if lim
n→∞
an+1
< 1.
an
an+1
an+1
> 1 or lim
= ∞.
n→∞
an
an
76
3. The Ratio Test is inconclusive if lim
n→∞
an+1
= 1.
an
Although the Ratio Test is not a cure for all ills related to tests for convergence, it is particularly
useful for series that converge rapidly. Series involving factorials or exponentials are frequently of
this type.
Example 6.7.1. Determine the convergence or divergence of
∞
∑
2n
n=0
Solution: Because an =
2n
, we can write the following
n!
[ n+1
an+1
2
lim
= lim
n→∞
n→∞ (n + 1)!
an
[ n+1
2
= lim
n→∞ (n + 1)!
2
= lim
n→∞ n + 1
= 0.
n!
.
]
2n
÷
n!
]
n!
· n
2
Therefore, the series converges.
Example 6.7.2. Determine whether the following series converge or diverge.
∞
∞
∑
∑
n2 2n+1
nn
(a)
(b)
.
3n
n!
n=0
n=1
Solution:
an+1
is less than 1.
an
[
( n+2 ) (
)]
2
3n
2
= lim (n + 1)
n→∞
3n+1
n2 2n+1
2(n + 1)2
= lim
n→∞
3n2
2
< 1.
=
3
(a) This series converges because the limit of
lim
n→∞
an+1
an
an+1
is grater than 1.
an
( )]
[
(n + 1)n+1 n!
lim
n→∞
(n + 1)!
nn
(
)]
[
(n + 1)n+1 1
lim
n→∞
(n + 1)
nn
)n
(
(n + 1)n
1
lim
= lim 1 +
n→∞
n→∞
nn
n
e > 1.
(b) This series diverges because the limit of
lim
n→∞
an+1
an
=
=
=
=
77
6.7.2
The Root Test
This test of convergence or divergence of series works especially well for series involving nth powers.
Root Test
Let
∑
1.
2.
an be a series with non-zero terms.
∑
an converges absolutely if lim
√
n
n→∞
∑
an diverges if lim
n→∞
|an | < 1.
√
√
n
|an | > 1 or lim n |an | = ∞.
n→∞
3. The Root Test is inconclusive if lim
n→∞
√
n
|an | = 1.
Example 6.7.3. Determine the convergence or divergence of
∞
∑
e2n
n=1
nn
.
Solution: We can apply the Root Test as follows
√
lim n |an | =
n→∞
√
lim
n→∞
n
e2n
nn
2n
en
= lim n
n→∞ n n
e2
= lim
n→∞ n
= 0 < 1.
Because this limit is less than 1, we can conclude that the series converges absolutely.
6.8
6.8.1
Power Series
Definition of Power Series
If x is a variable, then the infinite series of the form
∞
∑
an x x = a0 + a1 x + a2 x 2 + a3 x 3 + · · · + an x n + · · ·
n=0
78
is called a power series. More generally, a series of the form
∞
∑
an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + · · · + an (x − c)n + · · ·
n=0
is called a power series centered at c, where c is a constant.
Example 6.8.1. (a) The following power series is centered at 0.
∞
∑
xn
n=0
n!
=1+x+
x2 x3
+
+ ···
2
3!
(b) The following power series is centered at −1.
∞
∑
(−1)n (x + 1)n = 1 − (x + 1) + (x + 1)2 − (x + 1)3 + · · ·
n=0
(c) The following power series is centered at 1.
∞
∑
1
1
1
(x − 1)n = (x − 1) + (x − 1)2 + (x − 1)3 + · · ·
n
2
3
n=1
6.8.2
Radius and Interval of Convergence
A power series in x can be viewed as a function of x
f (x) =
∞
∑
an (x − c)n
n=0
where the domain of f is the set of all x for which the power series converges.
Convergence of a Power Series
For a power series centered at c, precisely one of the following is true.
1. The series converges only at c.
2. There exists a real number R > 0 such that the series converges absolutely for |x − c| < R,
and diverges for |x − c| > R.
3. The series converges absolutely for all x.
79
The number R is the radius of convergence of the power series. In the series converges only at
c, then the radius of convergence is R = 0, and if the series converges for all x, then the radius
of convergence is R = ∞. The set of all values of x for which the power series converges is the
interval of convergence of the power series.
Example 6.8.2. Find the radius of convergence of
∞
∑
n!xn .
n=0
Solution: For x = 0, we obtain
f (0) =
∞
∑
n!0n = 1 + 0 + 0 + · · · = 1.
n=0
For any fixed value of x such that |x| > 0, let un = n!xn . Then
lim
n→∞
un+1
un
(n + 1)!xn+1
n→∞
n!xn
= |x| lim (n + 1)
n→∞
= ∞.
=
lim
Therefore, by the Ratio Test, the series diverges for |x| > 0, and converges only at its center, 0.
Hence, the radius of convergence is R = 0.
Example 6.8.3. Find the radius of convergence of
∞
∑
3(x − 2)n .
n=0
Solution: For x ̸= 2, let un = 3(x − 2)n . Then
lim
n→∞
un+1
un
3(x − 2)n+1
n→∞
3(x − 2)n
= lim |x − 2|
=
lim
n→∞
= |x − 2|.
By the Ratio Test, the series converges if |x − 2| < 1 and diverges if |x − 2| > 1. Therefore, the
radius of convergence of the series is R = 1.
Finding the Interval of Convergence
Example 6.8.4. Find the interval of convergence of
∞
∑
xn
n=1
80
n
.
Solution: Letting un =
xn
produces
n
un+1
lim
n→∞
un
=
lim
xn+1
n+1
xn
n
lim
nx
n+1
n→∞
=
n→∞
= |x|.
Therefore, by the Ratio Test, the radius of convergence is R = 1. Moreover, because the series is
centered at 0, it converges in the interval (−1, 1). This interval, however, is not necessarily the
interval of convergence. To determine this, we must test for convergence at each endpoint. When
x = 1, we obtain the divergent harmonic series
∞
∑
1
1 1
= 1 + + + ···
n
2 3
n=1
When x = −1, we obtain the convergent alternating harmonic series
∞
∑
(−1)n
n=1
n
= −1 +
1 1 1
− + − ···
2 3 4
Therefore, the interval of convergence for the series is [−1, 1).
Example 6.8.5. Find the interval of convergence of
∞
∑
(−1)n (x + 1)n
n=0
Solution: Letting un =
2n
.
(−1)n (x + 1)n
produces
2n
un+1
lim
n→∞
un
=
lim
n→∞
(−1)n+1 (x+1)n+1
2n+1
(−1)n (x+1)n
2n
n
2 (x + 1)
n→∞
2n+1
x+1
.
=
2
=
lim
x+1
< 1 or |x+1| < 2. Hence, the radius of convergence
2
is R = 2. Because the series is centered at x = −1, it will converge in the interval (−3, 1).
Furthermore, at the endpoints we have
By the Ratio test, the series converges if
∞
∑
(−1)n (−2)n
2n
n=0
and
=
n=0
∞
∑
(−1)n (2)n
n=0
∞
∑
2n
2n
2n
=
∞
∑
1
(Diverges when x=-3)
n=0
∞
∑
=
(−1)n
(Diverges when x=1)
n=0
both of which diverge. Thus, the interval of convergence is (−3, 1).
81
Example 6.8.6. Find the interval of convergence of
∞
∑
xn
n=1
Solution: Letting un =
n2
.
xn
produces
n2
un+1
lim
= lim
n→∞
n→∞
un
xn+1
(n+1)2
xn
n2
n2 x
= lim
= |x|.
n→∞ (n + 1)2
Thus, the radius of convergence is R = 1. Because the series is centered at x = 0, it converges in
the interval (−1, 1). When x = 1, we obtain the convergent p−series
∞
∑
1
1
1
1
= 1 + 2 + 2 + 2 + ···
2
n
2
3
4
n=1
(Converges when x=1)
When x = −1, we obtain the convergent alternating series
∞
∑
(−1)n
n=1
n2
= −1 +
1
1
1
−
+
− ···
22 32 42
(Converges when x=-1)
Therefore, the interval of convergence for the given series is [−1, 1].
Example 6.8.7. Find the interval of convergence of
∞
∑
nxn .
n=1
Solution: The series is a power series with an = n and c = 0. Let un = nxn , so un+1 = (n+1)xn+1 .
Then
un+1
(n + 1)|x|n+1
n+1
=
=
|x| =⇒ |x| as n → ∞.
n
un
n|x|
n
The limit is less than one whenever |x| < 1. The Ratio Test then shows
∞
∑
un is convergent for
n=1
|x| < 1, and the series diverges for |x| > 1. This means the radius of convergence is R = 1. We
know that the series is convergent for −1 < x < 1. We need to check convergence at the endpoints
∞
∑
n. This series does not approach zero as n → ∞, we
of this interval. When x = 1, we have
n=1
know this series must diverge. Similarly, the series is divergent when x = −1. The interval of
convergence is (−1, 1).
Example 6.8.8. Find the interval of convergence of
∞
∑
(−1)n (x − 2)n
n=1
Solution: Let un =
n4n
.
|x − 2|n
|x − 2|n+1
,
so
u
=
. Then
n+1
n4n
(n + 1)4n+1
|x − 2|n+1 n4n
|x − 2|
un+1
n|x − 2|
=
=⇒
=
n+1
n
un
(n + 1)4
|x − 2|
(n + 1)4
4
82
as
n → ∞.
|x − 2|
|x − 2|
The Ratio Test gives convergence for
< 1 and divergence for
> 1. Solving the first
4
4
inequality, we have
|x − 2| < 4 =⇒ −4 < x − 2 < 4 =⇒ −2 < x < 6.
When x = −2, the series is
∞
∑
(−1)n (−2 − 2)n
n=1
n4n
∞
∑
1
=
,
n
n=1
which is a divergent p−series. When x = 6, we have
∞
∞
∑
(−1)n (6 − 2)n ∑ (−1)n
=
.
n
n4
n
n=1
n=1
The Alternating Series Test shows that the series is convergent. The interval of convergence is
−2 < x ≤ 6.
∞
∑
xn
.
Example 6.8.9. Find the interval of convergence of the series
n
n3
n=1
xn
Solution: With un = n , we find that
n3
lim
n→∞
un+1
un
xn+1
|x|
n|x|
(n + 1)3n+1
=
=
.
= lim
xn
n→∞ 3(n + 1)
3
n3n
|x|
< 1 provided |x| < 3, so the Ratio Test implies that the given series converges absolutely if
3
∑1
|x| < 3 and diverges if |x| > 3. When x = 3, we have the divergent harmonic series
and when
n
∑ (−1)n
x = −3, we have the convergent alternating series
. Thus the interval of convergence of
n
the given power series is [−3, 3).
∞
∑
2n xn
Example 6.8.10. Find the interval of convergence of
.
n!
n=0
Now
Solution: With un =
2n xn
, we find that
n!
2n+1 xn+1
2|x|
(n + 1)!
=0
lim
= lim
n n
2 x
n→∞
n→∞ n + 1
n!
for all x. Hence the Ratio Test implies that the power series converges for all x, and its interval of
convergence is (−∞, ∞).
1
1
“Everybody is a genius. But if you judge a fish by its ability to climb a tree, it will live its whole life believing
that it is stupid.”— Albert Einstein
83