MA1521 CALCULUS FOR COMPUTING1
Liu Chunchun
To Wing Keung
August 7, 2023
1 This
notes is exclusively for students taking MA1521 in AY2023/24 Semester 1.
2
Contents
0 Real Numbers and Functions
7
0.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
0.2 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
0.3 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
0.4 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.5 Rational Functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.6 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.7 Exponential and Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . 12
0.8 More on the Domain and Range of a Function . . . . . . . . . . . . . . . . . . . 12
1 Limits and Continuity
15
1.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.3 Evaluation of limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.4 Limits at infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5 More on Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.6 Squeeze (Sandwich) Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.7 Intermediate Value Theorem (IVT) . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.8 Appendix: The Precise Definition of the Limit of a Function . . . . . . . . . . . 25
2 Derivatives
27
2.1 Di↵erentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.2 Standard Derivatives & Di↵erentiation Rules . . . . . . . . . . . . . . . . . . . 32
2.3 Implicit Di↵erentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.4 Derivatives of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.5 Higher-order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3
2.6 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.7 Miscellaneous examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3 Applications of Di↵erentiation
45
3.1 Tangents and Normals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.2 Increasing and Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . 47
3.3 Concave Upward and Concave Downward Functions . . . . . . . . . . . . . . . 48
3.4 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.5 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.6 Applied Maximum and Minimum Problems . . . . . . . . . . . . . . . . . . . . 57
3.7 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.8 Rolle’s Theorem and Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . 63
4 Integrals
67
4.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.2 Standard Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
4.3 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.4 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.5 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
4.6 Riemann Sums and Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . 76
4.7 Fundamental Theorem of Calculus (FTC) . . . . . . . . . . . . . . . . . . . . . . 78
4.8 Miscellaneous Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.9 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
5 Applications of Integration
89
5.1 Area Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
5.2 Volume of Solid of Revolution by Disk Method . . . . . . . . . . . . . . . . . . 94
5.3 Cylindrical Shell Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
5.4 Arc Length of a curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
6 Sequences and Series
6.1 Sequences
105
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
6.2 Finding the Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.3 Limit Laws for Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.4 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
4
6.5 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6.6 The Comparison Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.7 The Ratio Test and Root Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
6.8 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.9 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
6.10 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
6.11 Power Series Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
6.12 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
6.13 Appendix: The Precise Definition of the Limit of a Sequence and some Proofs 131
7 Vectors and Geometry of Space
133
7.1 The 3D-Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
7.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
7.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
7.4 Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
7.5 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
7.6 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
7.7 Planes
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
8 Functions of Several Variables
151
8.1 Vector Functions of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . 151
8.2 Calculus of Vector Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
8.3 Tangent Vector and Tangent Line to a Curve . . . . . . . . . . . . . . . . . . . . 154
8.4 Arc Length of a Space Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
8.5 Functions of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
8.6 Cylinders and Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
8.7 Functions of Three Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
8.8 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
8.9 Higher Order Partial Derivatives
. . . . . . . . . . . . . . . . . . . . . . . . . . 169
8.10 Tangent Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
8.11 Di↵erentiability and Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
8.12 Implicit Di↵erentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
8.13 Increments and Di↵erentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
8.14 Directional Derivatives and the Gradient Vector . . . . . . . . . . . . . . . . . . 181
8.15 Extrema of Functions of Two Variables . . . . . . . . . . . . . . . . . . . . . . . 187
5
9 Double Integrals
195
9.1 Riemann Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
9.2 Volume and Double Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
9.3 Iterated Double Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
9.4 A Special Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
9.5 Double Integral over General Region . . . . . . . . . . . . . . . . . . . . . . . . 205
9.6 Decomposing Domain into Smaller Domains . . . . . . . . . . . . . . . . . . . 212
9.7 Properties of Double Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
9.8 An Application – Finding Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
9.9 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . 215
9.10 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
10 Ordinary Di↵erential Equations
223
10.1 First Order Ordinary Di↵erential Equations . . . . . . . . . . . . . . . . . . . . 223
10.2 Reduction to Separable Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
10.3 Linear First Order ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
10.4 The Bernoulli Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
10.5 Applications of ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
11 More on ODE
235
11.1 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
11.2 2nd Order Linear Equations with Constant Coefficients . . . . . . . . . . . . . 238
11.3 Method of Undetermined Coefficients . . . . . . . . . . . . . . . . . . . . . . . 240
11.4 Appendix: Malthus Model of Population . . . . . . . . . . . . . . . . . . . . . . 243
6
Chapter 0
Real Numbers and Functions
Read Thomas’ Calculus, Chapter 1.
0.1 Numbers
The collection of all real numbers is denoted by R. Thus R includes the integers
. . . , 2, 1, 0, 1, 2, 3 . . . ,
the
p rational numbers, p/q, where p and q are integers (q , 0), and the irrational numbers, like
2, ⇡, e, etc.
a 2 R means a is a member of the set R. In other words, a is a real number. Given two real
numbers a and b with a < b, the closed interval [a, b] consists of all x such that a  x  b,
and the open interval (a, b) consists of all x such that a < x < b. Similarly, we may form the
half-open intervals [a, b) and (a, b].
0.2 Absolute Value
The absolute value of a number a 2 R is written as |a| and is defined as
(
a if a 0
|a| =
a if a < 0.
For example, |2| = 2, | 2| = 2.
7
Some properties of |x| are summarized as follows:
1. | x| = |x|, for all x 2 R.
2. |xy| = |x||y|, for all x, y 2 R.
3.
|x|  x  |x|, for all x 2 R.
4. For a fixed r > 0, |x| < r if and only if x 2 ( r, r).
p
5. x2 = |x|, x 2 R.
6. (Triangle Inequality) |x + y|  |x| + |y| for all x, y 2 R.
Example 0.1. Solve the inequality
2x 1
< 1.
2x + 1
Solution.
2x 1
<1
2x + 1
2x 1
,0<1
2x + 1
2x + 1 2x + 1
,0<
2x + 1
2
,0<
2x + 1
, 0 < 2x + 1
1
,
< x.
2
⌅
Example 0.2. Solve the inequality |x + 1|  |2x 1|.
Solution.
|x + 1|  |2x 1|
, |x + 1|2  |2x 1|2
, x2 + 2x + 1  4x2 4x + 1
, 0  3x2 6x
, 0  3x(x 2)
, x  0 or x 2
, x 2 ( 1, 0] [ [2, 1).
⌅
8
Exercise 0.1. Let r > 0. Prove that |x a| < r if and only if x 2 ( r + a, a + r).
Exercise 0.2. Prove the triangle inequality |x + y|  |x| + |y|.
Exercise 0.3. Prove that for any x, y 2 R, ||x| |y||  |x y|.
0.3 Functions
A function f : A ! B is a rule that assigns to each a 2 A one specific member f (a) of B.
Symbolically we may denote the function by a 7! f (a). We can specify a function f by
giving the rule for f (x).
Example 0.3. f (x) = x2 /(1 x) assigns the number x2 /(1 x) to each x , 1 in R.
The set A is called the domain of f and B is the codomain of f .
The range of f is the subset of B consisting of all the values of f . That is, the range of f =
{f (x) 2 B | x 2 A}.
Given f : A ! R, it means that f assigns a value f (x) in R to each x 2 A.
Such a function is called a real-valued function.
For a real-valued function f : A ! R defined on a subset A of R, the graph of f consists of
all the points (x, f (x)) in the xy-plane.
If f : A ! B and g : B ! C, then the composite function of f and g is the function g f : A !
C given by g f (x) = g(f (x)).
9
Example 0.4. Let f (x) =
1
x
and g(x) = x2 1. Find g f and f
Solution. g f (x) = g(f (x)) = g( 1x ) = ( 1x )2 1 =
f
g(x) = f (g(x)) = f (x2 1) =
1
x2
g.
1.
1
.
x2 1
⌅
Let f : A ! B. If g : B ! A is a function such that f (g(x)) = x for all x 2 B and g(f (x)) = x
for all x 2 A, then g is called the inverse of f . Similarly, f is the inverse of g. The inverse
function of f is usually denoted by f 1 .
Let f : A ! B. f is called an injective function if for any x, y 2 A, f (x) = f (y) ) x = y. f is
called a surjective function if for any z 2 B, there is an x 2 A such that f (x) = z. f is called a
bijective function if f is injective and surjective.
Exercise 0.4. Prove that if f
1
exists, then f is a bijective function.
0.4 Polynomials
A function of the form p(x) = an xn + an 1 xn
called a polynomial of degree n.
1 + ··· + a
1 x + a0 ,
where a0 , . . . , an are constants, is
For example, a quadratic function p(x) = ax2 + bx + c is a polynomial of degree 2.
A polynomial of degree n can be factored as a product of linear and quadratic factors.
For example, x4 1 = (x2 + 1)(x + 1)(x 1).
In general a polynomial of degree n has at most n real roots.
For example, x4 1 has only two real roots 1 and 1.
0.5 Rational Functions
A rational function is a function of the form
domain of
p(x)
, where p(x) and q(x) are polynomials. The
q(x)
p(x)
consists of all real numbers except the roots of q(x).
q(x)
For example, the domain of
x3 + 3
is R \ { 1, 1}.
x4 1
0.6 Trigonometric Functions
The 6 trigonometric functions are sin x, cos x, tan x, csc x, sec x, cot x. They are periodic functions of period 2⇡.
10
11
Exercise 0.5. Sketch the graph of csc(x).
0.7 Exponential and Logarithmic Function
A function of the form f (x) = ax , where a > 0 is called an exponential function. It’s inverse
function, denoted by loga x is called the logarithmic function to the base a. (a > 0 and a , 1.)
Let e = 2.718281828459045235360287 · · · be the Euler number. Then the inverse of the
exponential function ex is the natural logarithm ln x.
We have eln x = x for x > 0 and ln ex = x for all x.
The domain of e x is R and the range is the set R+ of all positive real numbers.
0.8 More on the Domain and Range of a Function
If the domain of a function is not specified, then it is understood that we will take the
domain to be as large as possible. This is called the maximal domain of the function.
12
In general it is not so easy to determine the range of a function. In some simple cases, basic
algebraic techniques can be used to find the range of a function.
Example 0.5. Find the maximal domain and the range of f (x) =
1
x 1.
Solution. The maximal domain of f is R \ {1}.
Recall that the range of f = {f (x) 2 R | x , 1}.
To find the range of f , let y = f (x). That is y =
this we see that if y , 0 then we may choose x =
R \ {0}.
1
x 1.
1 + y1
Solving for x, we get x = 1 + y1 . From
to get f (x) = y. Thus the range of f is
⌅
Example 0.6. Find the maximal domain and range of f (x) = x2 x + 1.
Solution. The maximal domain of f is R.
To find the range of f , let y = f (x). That is y = x2 x + 1. Solving for x, we get
p
p
x = 12 (1 ± 1 4(1 y)). That is x = 12 (1 ± 4y 3)).
From this we see that if y
the range of f is [ 34 , 1).
3
4
p
then we may choose x = 12 (1 ± 4y
3)) to get f (x) = y. Thus
⌅
Exercise 0.6. Let f (x) = x + 5 and g(x) = x2 3. Find the maximal domain and range of g(f (x)).
Ans: Maximal domain is R, range is [ 3, 1).
13
14
Chapter 1
Limits and Continuity
Read Thomas’ Calculus, Chapter 2.
1.1 Limits
Let f be a real-valued function defined on some interval I (e.g. (a, b), or (a, b] or (a, 1)).
Let c be a point in I .
• lim f (x) is the value that f (x) approaches when x approaches c from the left.
x!c
• lim+ f (x) is the value that f (x) approaches when x approaches c from the right.
x!c
• Let c be an interior point (i.e. not an end point). If lim f (x) = lim+ f (x) = L 2 R, we
say that lim f (x) exist and has value L.
x!c
15
x!c
x!c
c
0
2
4
-3
6
Left Limit
lim f (x)
x!c
Right Limit
lim+ f (x)
x!c
Limit
lim f (x)
x!c
f (x)
1.2 Continuity
Let f be a real-valued function defined on some interval I (e.g. (a, b), or (a, b] or (a, 1)).
Let c be a point in I .
Continuity at a point
Case 1 c is an interior point
• f is continuous at x = c if
(i) lim f (x) exists,
x!c
(ii) lim f (x) = f (c).
x!c
Case 2 c is the left end-point
• f is continuous at x = c if
(i) lim+ f (x) exists,
x!c
(ii) lim+ f (x) = f (c).
x!c
Case 3 c is the right end-point
• f is continuous at x = c if
(i) lim f (x) exists,
x!c
(ii) lim f (x) = f (c).
x!c
Continuity on an interval
• f is continuous on an interval if f is continuous at x = c for all points c in I .
Remark. Roughly speaking, f is continuous at x = c means that the values of f near x = c
all become very close to f (c) when x is very close to c, so that there is no sudden jump in the
values of f at x = c.
16
Example 1.1. Find the points of discontinuity of the function f whose graph on ( 3, 6] is given
below.
Solution.
Point of discontinuity
Reason
x=2
lim f (x) , f (2)
x!2
x=4
lim f (x) does not exist
x!4
1.3 Evaluation of limits
Results (Law of limits)
The following results are true provided all the limits involved exist. The limit could be
one-sided or two-sided. The number k is a constant.
1. lim(f (x) ± g(x)) = lim f (x) ± lim g(x)
x!c
x!c
x!c
2. lim kf (x) = k lim f (x)
x!c
x!c
3. lim(f (x)g(x)) = (lim f (x))(lim g(x))
x!c
x!c
x!c
f (x)
f (x) lim
x!c
=
x!c g(x)
lim g(x)
4. lim
x!c
5. If g is continuous at the point b and lim f (x) = b, then lim g(f (x)) = g(b) = g(lim f (x))).
x!c
17
x!c
x!c
p
f (x)g(x) + g(x)
Example 1.2. Suppose lim f (x) = 3 and lim g(x) = 4. Find lim
.
x!2
x!2
x!2
g(x) f (x)
Solution.
p
p
f (x)g(x) + g(x) 3 · 4 + 4
lim
=
= 7.
x!2
g(x) f (x)
4 2
⌅
From the Laws of Limits, we have corresponding results on continuity of functions.
Results
• If f and g are continuous at x = c, then for any constant k and any positive constant n,
each of the following functions is continuous at x = c.
(i) f ± g, (ii) f n , (iii) kf , (iv) f g, (v) f /g provided g(c) , 0.
• If g is continuous at x = c and f is continuous at x = g(c), then the composite function
f g is continuous at x = c.
(Note (f
g)(x) = f (g(x))).
[ For example, if f and g are continuous at x = c, then we have lim f (x) = f (c) and lim g(x) =
g(c). Thus
x!c
x!c
lim(f (x) + g(x)) = lim f (x) + lim g(x) = f (c) + g(c),
x!c
x!c
x!c
which implies that f + g is continuous at x = c. ]
Result. The following functions are continuous on any interval contained in their maximal
domain.
1. Polynomials
2. Trigonometric Functions
3. Exponential Functions
4. Logarithmic Functions
5. A combination of any of the above on the domain it is defined.
For example, a rational function
P(x)
is continuous at all points x where Q(x) , 0.
Q(x)
In particular, the rational function
1
is continuous on R \ {1, 2}.
(x 1)(x 2)
18
Example 1.3. Show that the function ln(x + 3) is continuous on the interval ( 3, 1).
Solution. Take any point c 2 ( 3, 1). Then 0 < c + 3 < 1. Then the polynomial g(x) = x + 3
is continuous at x = c, and the logarithmic function f (x) = ln x is continuous at x = c + 3 > 0.
Thus the composition
f g(x) = f (g(x)) = f (x + 3) = ln(x + 3)
is continuous at x = c.
⌅
Remark. Since lim f (x) = f (c) when f is continuous at x = c, finding the limit at x = c of any
x!c
of the above functions is a matter of evaluating f at x = c.
x + ln(x + 3)
.
p
2
x+6
Example 1.4. Evaluate lim
x!
Solution. The function f (x) =
x + ln(x + 3)
is continuous at x = 2, which implies that
p
x+6
lim f (x) = f ( 2).
x! 2
Thus
lim
x! 2
x + ln(x + 3)
2 + ln( 2 + 3)
=
= 1.
p
p
x+6
2+6
⌅
Exercise 1.1. Evaluate lim tan3 (sin x).
x!0
Ans: 0.
1.4 Limits at infinity
Let f be defined on R.
• lim f (x) is the value f (x) approaches as x tends to positive infinity.
x!1
• lim f (x) is the value f (x) approaches as x tends to negative infinity.
x! 1
Graphically, if lim f (x) = c 2 R or lim f (x) = c, then the line y = c is a horizontal asympx!1
tote of the graph of f (x).
x! 1
19
1
1
Example 1.5. From the graph of y = , one sees that approaches 0 as x tends to positive infinity.
x
x
1
Thus we have lim = 0.
x!1 x
1
= 0.
1x
Similarly we have lim
x!
Then for any positive integer n, we also have
✓
◆
1
1 n
=
lim
= 0n = 0,
x!1 x n
x!1 x
lim
and
1
=
1 xn
lim
x!
✓
x!
Example 1.6. Using the result in the previous example, we have
lim (2
x!1
1
) = 2 0 = 2,
x
and thus y = 2 is a horizontal asymptote of the graph of y = 2
Also, lim (2
x! 1
1
) = 2.
x
20
1
1x
lim
1
.
x
◆n
= 0n = 0.
Example 1.7. Evaluate lim
✓
x!1
Solution. lim
x!1
✓
3 p
+ 4 e
2x
x
3 p
+ 4 e
2x
◆2
x
◆2
.
p
= (0 + 4)2 = 4.
⌅
✓
◆
4
Exercise 1.2. Evaluate lim ln 3 2 sin .
x! 1
x
Ans: ln 3.
1.5 More on Limits
Indeterminate forms.
f (x)
where f (x) ! 0 and g(x) ! 0 as x ! c is called an
x!c g(x)
0
indeterminate form of the type .
0
(a) A limit of the form lim
f (x)
where f (x) ! 1 and g(x) ! 1 as x ! c is called an
x!c g(x)
1
indeterminate form of the type .
1
(b) A limit of the form lim
Replacement rule. Let I be an open interval containing the point x = c. Suppose f (x) =
g(x) for all x 2 I, except possibly at x = c. Then lim f (x) = lim g(x).
x!c
x!c
x2 7x + 6
.
x!6 36 x 2
Example 1.8. Evaluate lim
(x 1)(x 6)
x2 7x + 6
x 1
5
= lim
== lim
=
.
x!6 36 x 2
x!6 (6 + x)(x 6)
x!6 (6 + x)
12
p
p
x + 12
6 x
Exercise 1.3. Evaluate lim
.
2
x! 3
18 2x
Solution. lim
Ans:
1
36 .
Result. Limits of the form lim
P(x)
x!±1 Q(x)
, where P(x) and Q(x) are polynomials in x.
21
⌅
leading term
z}|{
Ax↵
P(x)
= lim
x!±1 Q(x)
x!±1
lim
Bx
|{z}
leading term
8
>
>
>
>
>
+··· >
<
=>
>
+··· >
>
>
>
:
0
A
B
1 or 1
| {z }
if ↵ <
if ↵ =
if ↵ >
depends on the question
p
(18x2 + 5x 1)(2 x 1)3
.
x!1
(3x 1)4
Example 1.9. Evaluate lim
p
7
(18x2 + 5x 1)(2 x 1)3
144x 2 + · · ·
Solution. lim
= lim
= 0.
x!1
x!1 81x 4 + · · ·
(3x 1)4
Exercise 1.4. Evaluate lim p
x! 1
(1 + 2x)3
16x6 + 9x 1
.
Ans: 2.
Useful results.
If lim g(x) = 0, then
x!c
sin(g(x))
g(x)
= lim
= 1,
x!c
x!c sin(g(x))
g(x)
• lim
tan(g(x))
g(x)
= lim
= 1.
x!c
x!c tan(g(x))
g(x)
• lim
In particular, when c = 0 and g(x) = x,
sin x
x
= lim
= 1,
x!0 x
x!0 sin x
• lim
tan x
x
= lim
= 1.
x!0 x
x!0 tan x
• lim
sin(e x )
sin 3x
ln x
= 1, lim
= 1, lim
= 1.
x!1
x!0 3x
x!1 tan(ln x)
e x
For example, lim
22
⌅
Example 1.10. Evaluate lim
x!0
sin2 x
.
sin(3x2 ) + x tan(2x)
Solution.
sin2 x
sin x
x2
lim
= lim
2
2
x!0 sin(3x ) + x tan(2x)
x!0
sin(3x )
tan(2x)
3x2
+ 2x2
2
2x
3x
⇣ sin x ⌘2
12
1
x
= lim
=
=
2
x!0 sin(3x )
tan(2x) 3 + 2 5
3
+2
2
2x
3x
x2
2
⌅
⇣
p ⌘
Exercise 1.5. Evaluate lim+ x2 cot(2x) csc2 (3 x) .
x!0
Ans:
1
18 .
p
tan( x 2)
Exercise 1.6. Evaluate lim
.
x!4 sin(16 x 2 )
Ans:
1
32 .
1.6 Squeeze (Sandwich) Theorem
Squeeze Theorem. Suppose g(x)  f (x)  h(x) for all x in some open interval containing
a point c, except possibly at x = c. If lim g(x) = lim h(x) = L, then lim f (x) = L.
x!c
x!c
23
x!c
Example 1.11. It is given that 3 x2  f (x)  1 + 2ex for all x. Find lim f (x).
x!0
Solution. As lim 3 x2 = 3 and lim 1 + 2e x = 3, we have by Squeeze Theorem that lim f (x) = 3.
x!0
x!0
x!0
⌅
Exercise 1.7. Use Squeeze Theorem to show that lim |f (x)| = 0 ) lim f (x) = 0.
x!c
x!c
Remark The converse of the above result, namely lim f (x) = 0 ) lim |f (x)| = 0 is true.
x!c
x!c
Hence, we have
Result. lim f (x) = 0 , lim |f (x)| = 0.
x!c
x!c
✓
◆
2
3
Example 1.12. Evaluate lim x cos(
) .
x!0
sin x
Solution. Notice that for all x, we have
1  cos(
2
2
)  1 =) cos(
) 1
sin x
sin x
2
=) x3 cos(
)  |x3 | = |x|3
sin x
2
=) |x|3  x3 cos(
)  |x|3 .
sin x
Notice that lim |x|3 = 03 = 0 and thus also lim |x|3 = 0.
x!0
x!0
✓
Hence by Squeeze Theorem, we have lim x3 cos(
x!0
◆
2
) = 0.
sin x
!
sin(2 ln x)
1
Exercise 1.8. Evaluate lim
) + 2x sin( ) .
x!1
ln x
x
Ans: 2.
24
⌅
1.7 Intermediate Value Theorem (IVT)
If a real-valued function f is continuous on [a, b] and k is a number between f (a) and f (b),
then f (c) = k for some c 2 [a, b].
Example 1.13. Show that the equation x3 e x = 10 has a solution between 1 and 1.5.
Solution. Let f (x) = x3 ex . f is continuous on R. We have f (1) = e = 2.718 and f (1.5) =
1.53 e 1.5 = 15.126. Thus f (1) < 10 < f (1.5). By Intermediate Value Theorem, there is a solution to f (x) = 10 between 1 and 1.5.
Exercise 1.9. Show that the equation 10 = x + 2 tan(2x) has a solution between 3 and 4.
1.8 Appendix: The Precise Definition of the Limit of a Function
Remark: Section 1.8 will be excluded from the assessments (quizzes and the Final Exam).
Remark: The precise definition of the limit of a function introduced in this section will not
be needed in MA1521, and it will be studied in detail in MA2108 Mathematical Analysis I,
where many results in calculus will be proved rigorously.
Let f (x) be defined on an open interval containing the point c, except possibly at c itself.
We say that the limit of f (x) as x approaches c is the number L, and write
lim f (x) = L,
x!c
if, for every number ✏ > 0, there exists a corresponding number
0 < |x c| < ) |f (x) L| < ✏.
25
> 0 such that for all x,
Example 1.14. Prove from definition that lim 5x 3 = 2.
x!1
Solution. Note that |(5x 3) 2| = 5|x 1|. Given ✏ > 0, we choose
0 < |x 1| < ) 5|x 1| < 5 ) |(5x 3) 2| < ✏.
x
+ 3 = 1.
63
Exercise 1.10. Prove from definition that lim
x!
26
= ✏/5. Then for all x,
⌅
Chapter 2
Derivatives
Read Thomas’ Calculus, Chapter 3.
2.1 Di↵erentiability
Definition 2.1. The derivative of a function f at the point x0 , denoted by f 0 (x0 ), is given by
the following limit
f (x0 + h) f (x0 )
.
h
h!0
f 0 (x0 ) = lim
Consider the tangent to the curve y = f (x) at the point (x0 , f (x0 )). From the diagram, as
h ! 0, Q approaches P and hence, the gradient of the chord PQ, namely
f (x0 + h) f (x0 )
,
h
27
approaches the limiting value
f (x0 + h) f (x0 )
= f 0 (x0 ).
h
h!0
lim
When f 0 (x0 ) exists, we say that f is di↵erentiable at x = x0 . Geometrically in this case, f 0 (xo )
is equal to the slope of the tangent line to the curve y = f (x) at the point (x0 , f (x0 )).
Remark. An alternative formula for f 0 (x0 ) is
f 0 (x0 ) = lim
x!x0
f (x) f (x0 )
.
x x0
(This can be seen by letting x = x0 + h, and noting that ‘h ! 0’ is the same as ‘x ! x0 ’.)
Example 2.1. Let f (x) = x2 . Find f 0 (3) (that is, x0 = 3).
Solution.
f (3 + h) f (3)
h
h!0
2
(3 + h) 32
= lim
h
h!0
9 + 6h + h2 9
= lim
h
h!0
= lim (6 + h)
f 0 (3) = lim
h!0
= 6.
Remark. Thus the slope of the tangent line to the curve y = x2 at the point (3, 9) is equal to
6.
⌅
Definition 2.2. Suppose the derivative f 0 (x) exists for all x in an open interval I. We can
then treat f 0 (x) as a function defined on I. The process of finding the derivative of a function
dy
df
d
is called di↵erentiation. If y = f (x) , we can also write dx
f (x), dx or dx to denote f 0 (x).
In summary, we have
dy df
f (x + h) f (x)
d
f (x) =
=
= f 0 (x) = lim
.
dx
dx dx
h
h!0
28
Example 2.2. (a) Use the above definition of derivative to di↵erentiate the function
(b) Find the equation of the tangent to the curve y =
Ans: (a)
p 1p ,
2 x(2+ x)2
(b) (y
1
3) =
1
18 (x
1p
2+ x
1
p .
2+ x
at the point (1, 13 ).
1).
Solution. (a)
1
1
p
p
d
1
2+ x+h 2+ x
p = lim
dx 2 + x h!0
h
p
p
1 (2 + x) (2 + x + h)
= lim
p
p
h!0 h (2 + x + h)(2 + x)
p p
1
x
x+h
= lim
p
p
h!0 h (2 + x + h)(2 + x)
p p
p p
1
x
x+h
x+ x+h
= lim
p
p ·p p
h!0 h (2 + x + h)(2 + x)
x+ x+h
1
h
= lim
p
p p p
h!0 h (2 + x + h)(2 + x)( x + x + h)
1
= lim
p
p p p
h!0 (2 + x + h)(2 + x)( x + x + h)
1
= p
p .
2 x(2 + x)2
(b) From part (a),
d
1
1
p = p
p .
dx 2 + x 2 x(2 + x)2
At x = 1, derivative is
1
=
.
p
18
2 1(2 + 1)2
p
1
1
y
1
3 = 1.
Thus the equation of the tangent at (1, ) is
3
x 1
18
Di↵erentiability implies continuity.
Theorem 2.1. If f is di↵erentiable at x = x0 , then f is continuous at x = x0 .
29
⌅
Proof. It is given that f 0 (x0 ) exists. For all x near x0 , we have
f (x) =
f (x) f (x0 )
· (x x0 ) + f (x0 ).
x x0
Letting x ! x0 , we have
f (x) f (x0 )
· (x x0 ) + f (x0 ))
x0
x0 ) + f (x0 )
= f (x0 ).
lim f (x) = lim (
x!x0
x!x0
x
0
= f (x0 ) · (x0
Hence f is continuous at x = x0 .
⌅
Remark. The converse of the above result is not true in general. For example the absolute
value function f (x) = |x| is continuous at x = 0 but not di↵erentiable at x = 0.
Di↵erentiability on Intervals.
Definition 2.3. A function f is said to be di↵erentiable on an interval I if it is di↵erentiable
at every point in I .
Remark. If the interval has endpoints, then the limit in defining the derivative at an endpoint should be replaced by the appropriate one-sided limit.
Exercise 2.1. Show that f (x) = |x2 2x| is not di↵erentiable at x = 2.
Example 2.3. Let n be a positive integer. Show that
Solution.
(x + h)n xn
d n
x = lim
dx
h
h!0
30
d n
x = nxn 1 .
dx
= lim
(xn + nxn 1 h +
n(n 1) n 2 2
h + · · · + hn )
2 x
xn
h
h!0
(using binomial theorem)
= lim
nxn 1 h +
h
h!0
= lim nxn
h!0
n(n 1) n 2 2
h + · · · + hn
2 x
1 + n(n 1) x n 2 h + · · · + hn 1
2
= nxn 1 .
⌅
Example 2.4. Show that
d
sin x = cos x.
dx
Solution.
2 cos
sin(x + h) sin x
d
sin x = lim
= lim
dx
h
h!0
h!0
2x + h
h
sin
2
2
h
A B
(using sin A sin B = 2 cos A+B
2 sin 2 )
h
sin
h
2
= lim cos(x + ) · lim
2 h!0 h
h!0
2
= (cos x) · 1 = cos x.
⌅
31
2.2 Standard Derivatives & Di↵erentiation Rules
Table 1
Function
Derivative
xn
nxn
cos(x)
1
sin(x)
sin(x)
cos(x)
tan(x)
sec2 (x)
sec(x)
sec(x) tan(x)
csc(x)
csc(x) cot(x)
cot(x)
csc2 (x)
ex
ex
ln(x)
1
x
sin 1 (x)
p 1
1 x2
cos 1 (x)
p 1
1 x2
tan 1 (x)
1
1+x2
1
1+x2
cot 1 (x)
p1
,
|x| x2 1
sec 1 (x)
p1
,
|x| x2 1
csc 1 (x)
32
|x| > 1
|x| > 1
Table 2
Function
Derivative
ng 0 (x)(g(x))n
(g(x))n
1
cos(g(x))
g 0 (x) sin(g(x))
sin(g(x))
g 0 (x) cos(g(x))
tan(g(x))
g 0 (x) sec2 (g(x))
sec(g(x))
g 0 (x) sec(g(x)) tan(g(x))
csc(g(x))
g 0 (x) csc(g(x)) cot(g(x))
cot(g(x))
g 0 (x) csc2 (g(x))
eg(x)
g 0 (x)eg(x)
ln(g(x))
g 0 (x)
g(x)
0
p g (x)
sin 1 (g(x))
1 g(x)2
0
cos 1 (g(x))
p g (x)
tan 1 (g(x))
g 0 (x)
1+g(x)2
1 g(x)2
g 0 (x)
1+g(x)2
cot 1 (g(x))
0
g (x)
p
sec 1 (g(x))
|g(x)| g(x)2 1
, |g(x)| > 1
0
g (x)
p
csc 1 (g(x))
|g(x)| g(x)2 1
, |g(x)| > 1
Note that when g(x) = x, the formulae in Table 2 reduce to those in Table 1.
The formulae in Table 2 follow readily from those in Table 1 and the Chain Rule (which we
will see very soon).
33
Rules of Di↵erentiation
Let u and v be di↵erentiable functions of x, and let c be a constant.
Constant Rule
d
(c) = 0
dx
Constant Multiple Rule
d
du
(cu) = c
dx
dx
Sum Rule
d
du dv
(u + v) =
+
dx
dx dx
Product Rule
d
du
dv
(uv) =
v +u
dx
dx
dx
d u
( )=
dx v
Quotient Rule
du
dx v
v2
u dv
dx
Example 2.5. Di↵erentiate x3 + x2 tan x with respect to x.
Solution. By the sum rule and product rule,
d 3
d 3 d 2
(x + x2 tan x) =
x + (x tan x)
dx
dx
dx
d 2
d
2
= 3x + (x ) · tan x + x2 ·
tan x
dx
dx
= 3x2 + 2x tan x + x2 sec2 x.
⌅
Example 2.6. Di↵erentiate
x2
with respect to x.
ln x
Solution. By the quotient rule,
d
d
( x2 ) · ln x x2 ·
ln x
d x2
dx
= dx
dx ln x
(ln x)2
1
2x ln x x2 ·
x
=
(ln x)2
2x ln x x
=
.
(ln x)2
⌅
34
Let f (u) be di↵erentiable at u = g(x), and let g be a di↵erentiable function of x.
d
(f (g(x))) = f 0 (g(x)) · g 0 (x)
dx
Chain Rule
Write y = (f (g(x)) and u = g(x) as functions of x. Then the Chain Rule can be abbreviated as
dy dy du
=
·
,
dx du dx
dy
with the understanding that
is evaluated at u = g(x), so that all three expressions are
du
regarded as functions of x.
Example 2.7. Di↵erentiate sin(x3 + x + 2) with respect to x.
Solution. Let u = x3 + x + 2, so that
y = sin(x3 + x + 2) = sin u,
and u = x3 + x + 2.
Then by the Chain Rule,
dy dy du
=
·
dx du dx
= cos u · (3x2 + 1)
= (cos(x3 + x + 2)) · (3x2 + 1)
=)
d
sin(x3 + x + 2) = (3x2 + 1) cos(x3 + x + 2).
dx
⌅
Example 2.8. Use the Chain Rule to derive the formula
from a corresponding formula in Table 1.
d
(g(x))n = ng 0 (x)(g(x))n
dx
1
(in Table 2)
Solution. Let u = g(x), so that
y = (g(x))n = u n ,
and u = g(x).
Then by the Chain Rule and the first formula in Table 1,
dy dy du
=
·
dx du dx
= nu n 1 · g 0 (x)
= n(g(x))n
=)
1
· g 0 (x)
d
(g(x))n = ng 0 (x)(g(x))n 1 .
dx
⌅
35
!
x2
Example 2.9. Di↵erentiate ln
with respect to x.
(6x 7)2
Ans:
2
x
12
6x 7 .
Solution. First we have
!
!
x2
x
ln
= 2 ln
= 2 ln(x) 2 ln(6x 7).
(6x 7)
(6x 7)2
!
d
x2
2
1
2
12
Thus
ln
=
2·
·6 =
.
2
dx
x
6x 7
x 6x 7
(6x 7)
⌅
Exercise 2.2. Di↵erentiate with respect to x.
p
(a) (x + 1)2 tan 1 ( x)
sin 1 (2x)
(b) p
1 4x2
p
4x sin 1 (2x)
x+1
2
Ans: (a) 2(x + 1) tan 1 ( x) + p , (b)
+
.
3
1 4x2
2 x
(1 4x2 ) 2
2.3 Implicit Di↵erentiation
The diagram shows the curve given implicitly by the equation
x3 + y 3 9xy = 0
known as the folium of Descartes.
36
Suppose we wish to find the gradient of the curve at the point (2, 4). For this example, getting
an explicit expression for y in terms of x is challenging. It turns out that it is possible to find
dy
dx by a method known as implicit di↵erentiation. This consists of di↵erentiating both sides
of the given equation with respect to x and solving the resulting equation for
di↵erentiating a function in y with respect to x, we need the following result.
dy
dx .
When
dy
d
g(y) = g 0 (y) .
dx
dx
Example 2.10. Consider the curve x3 + y 3 9xy = 0. Find
Ans:
dy
dx
=
dy
dx .
3x2 9y
.
3y 2 9x
Solution. Di↵erentiating both sides of the equation, we have
3x2 + 3y 2
That is
Thus
dy
dx
9y
9x
dy
= 0.
dx
dy
(3y 2 9x) = 3x2 + 9y.
dx
dy
3x2 9y
=
.
dx
3y 2 9x
⌅
Example 2.11. Find
dy
dx
for points on the curve x3 e y + cos(xy) = 2023.
Solution. Di↵erentiating both sides of the equation, we have
Apply product rule to x3 ey
z
{
!
dy
dy
2 y
3 y
3x e + x e
sin(xy) x
+ y = 0.
dx
dx
Solving for
}|
dy
, we obtain
dx
dy
= 3x2 ey + y sin(xy)
dx
dy
3x2 e y + y sin(xy)
=)
= 3 y
.
dx
x e x sin(xy)
(x3 ey
x sin(xy))
⌅
In general, for a given equation of the form f (x, y) = 0, let
37
f x denote the expression obtained by di↵erentiating f (x, y) with respect to x, treating y as a
constant,
f y denote the expression obtained by di↵erentiating f (x, y) with respect to y, treating x as a
constant.
Then,
dy
f
= x.
dx
fy
2.4 Derivatives of Inverse Functions
Recall that
(i) a bijective function (or one-one function) f has an inverse f
1
defined on the range of f .
(ii) increasing or decreasing functions are bijective.
Theorem 2.2. Let f be bijective and di↵erentiable on an open interval I . Then
(f
1 0
) (a) =
1
f
0 (f 1 (a))
.
Let b = f 1 (a), so that a = f (b). At the point (a, f 1 (a)) = (a, b) on the curve y = f
gradient of the tangent is
1
1
(f 1 )0 (a) = 0 1
= 0 .
f (f (a)) f (b)
Proof. Let b = f 1 (a), so that a = f (b). As f
x for all x. By chain rule,
1
is the inverse function of f , we have f
38
1 (x),
1 (f
the
(x)) =
(f
1 )0 (f
(x)) · f 0 (x) = 1
) (f
1 )0 (f
(b)) · f 0 (b) = 1
) (f
1 )0 (a) · f 0 (b) = 1
) (f
1 )0 (a) =
1
f
0 (b)
=
1
f
0 (f 1 (a))
.
⌅
Remark. Using a = f (b) and letting b vary over I , one obtains the formula
1 0
(f
relating the derivative functions of f
) (f (x)) =
1
1
f
0 (x)
and f . This formula is often abbreviated as follows:
Let y = y(x) be a function. Suppose y has an inverse function x = x(y). Then
dx
1
=
.
dy dy
dx
with the understanding that
as functions of x.
dx
is evaluated at y = y(x), and both expressions are regarded
dy
This can also be obtained directly as follows: Note that
x = x(y(x))
for all x. Di↵erentiating both sides (and applying the abbreviated Chain Rule on the right
hand side), we get
dy
dx
d
dx dy
1
=
(x(y(x)) =) 1 =
·
=)
=
dx dx
dy dx
dx dy
dx
with the understanding that
Example 2.12. Show that
(Recall that sin 1 (x) 2 [
dx
is evaluated at y = y(x).
dy
d
1
sin 1 (x) = p
.
dx
1 x2
⇡ ⇡
, ] for all x 2 [1, 1].)
2 2
39
⌅
Solution. Let y = sin 1 (x), so that x = sin y. Then
dy
1
d
1
1
=
=)
sin 1 (x) =
=
.
d
dx dx
dx
cos y
(sin y)
dy
dy
Note that
q
p
0 =) cos y = 1 sin2 y = 1 x2 .
⇡ ⇡
y = sin (x) 2 [ , ] =) cos y
2 2
1
Thus
d
1
.
sin 1 (x) = p
dx
1 x2
Exercise 2.3. Let f (x) = x5 2e
⌅
2x
+ 3.
(a) Show that f is bijective by showing that f is increasing on R.
(b) Find the gradient of the tangent to the curve y = f 1 (x) at the point (1, 0).
1
(c) Let g(x) =
. Find the value of g 0 (1).
1
2x + 3f 1 (x)
Ans: (b) 14 , (c)
5
16 .
2.5 Higher-order Derivatives
Given a di↵erentiable function f , we can find the derivative of its derivative function f 0
with respect to x to get a new function, denoted by f 00 or f (2) , called the second derivative of
f provided that f 0 is di↵erentiable. That is,
f 00 (x) =
d 0
f (x).
dx
Notations. Let y = f (x).
f (2) (x) = f 00 (x) =
d 2y
= y 00 = D 2 f (x).
dx2
In general, we can define the nth order derivative of f for any positive integer n provided
the derivative exists. For example, the third derivative of y = f (x) is defined by
f
(3)
(x) = f
000
!!
d (2)
d d d
(f (x)) .
(x) =
f (x) =
dx
dx dx dx
We shall denote the nth order derivative of f by
f (n) (x) =
d ny
= D n y = D n f (x).
dxn
40
Example 2.13. Let n be a positive integer. For an non-negative integer k, find
dk n
x .
dxk
Solution.
k = 1:
k = 2:
k = 3:
d n
x = nxn 1 .
dx
d2 n
x = n(n 1)xn 2 .
dx2
d3 n
x = n(n 1)(n 2)xn 3 .
dx3
..
.
k  n:
dk n
x = n(n 1)(n 2) · · · (n k + 1)xn k .
dxk
In particular, when k = n, we have
k = n + 1:
dn n
x = n(n 1)(n 2) · · · (1) = n!.
dxn
d n+1 n
d dn n
d
x
=
( nx ) =
(n!) = 0.
n+1
dx dx
dx
dx
Similarly, we have
dk n
x = 0 for all k
dxk
n + 1.
2.6 Parametric Equations
A curve defined by the parametric equations
x = f (t) and y = g(t), (t is the parameter)
is di↵erentiable at a point where t = t0 if both f and g are di↵erentiable at t = t0 . Usually we
also assume f 0 (t0 ) , 0 or g 0 (t0 ) , 0.
By chain rule,
dy dy dx
dy dy dx g 0 (t)
=
·
=)
=
÷
=
,
dt dx dt
dx dt dt f 0 (t)
and
41
!
d 2y
d dy
dx
=
÷
=
2
dt dx
dt
dx
⇣
0
d g (t)
dt f 0 (t)
f 0 (t)
⌘
=
g 00 (t)f 0 (t) g 0 (t)f 00 (t)
.
f 0 (t)3
Examples of parametric curves
• Ellipses
x = a cos t + x0 and y = b sin t + y0 ,
where a > 0, b > 0, x0 and y0 are fixed constants and 0  t < 2⇡.
• Circles
x = r cos t + x0 and y = r sin t + y0 ,
where r > 0, x0 and y0 are fixed constants and 0  t < 2⇡.
• Hyperbolas
x = a sec t + x0 and y = b tan t + y0 ,
or
x = a tan t + x0 and y = b sec t + y0 ,
where a > 0, b > 0, x0 and y0 are fixed constants and ⇡  t  ⇡, t ,
⇡ ⇡
2, 2.
Example 2.14. For the parametric curve given by
x = 2t t 2 , y = t t 3 ,
find the point(s) on the curve at which the tangent is parallel to the line 2y = x + 2023.
8
Ans: (0, 0) and ( 59 , 27
).
Solution. We are given x = 2t t 2 , y = t t 3 .
First
dy
dy
1 3t 2
= dt =
.
dx dx
2 2t
dt
1
The gradient of the line 2y = x + 2023 is .
2
So we look for the value(s) of t such that
dy 1
= . Thus
dx 2
1 3t 2 1
1
= () 1 3t 2 = 1 t () t(3t 1) = 0 () t = 0, .
2 2t
2
3
42
When t = 0, (x, y) = (2 · 0 02 , 0 03 ) = (0, 0).
1
1
When t = , (x, y) = (2 ·
3
3
1 1
( )2 ,
3 3
1
5 8
( )3 ) = ( , )
3
9 27
5 8
The corresponding points on the curve are (0, 0) and ( , ).
9 27
2.7 Miscellaneous examples
Functions of the form f (x)g(x) .
The derivative of y = f (x)g(x) can be found by first finding the derivative of ln y and then
dy
solving for
.
dx
Example 2.15. Di↵erentiate with respect to x.
(a) (x2 e3x )4 tan x ,
(b) 5x ln x .
Ans: (a)
(x2
e3x )4 tan x
4 sec2 x ln(x2
e3x ) +
!
2x 3e3x
· 4 tan x ,
x2 e3x
(b) 5x ln x ln 5(1 + ln x).
Solution. (a) Let y = (x2 e3x )4 tan x , so that
ln y = ln(x2 e3x )4 tan x = 4 tan x · ln(x2 e3x ).
Di↵erentiating both sides with respect to x, we have
(x2 e3x )0
1 dy
= (4 tan x)0 ln(x2 e 3x ) + 2 3x (4 tan x)
y dx
(x e )
2x 3e3x
= 4 sec2 x ln(x2 e 3x ) + 2 3x · 4 tan x
x e
!
dy
2x 3e3x
2
2
3x
=)
= y · 4 sec x ln(x e ) + 2 3x · 4 tan x
dx
x e
= (x
2
3x 4 tan x
e )
2
4 sec x ln(x
43
2
!
2x 3e 3x
e ) + 2 3x · 4 tan x .
x e
3x
⌅
(b) Let y = 5x ln x . Then
ln y = ln(5x ln x ) = x ln x · ln 5.
Di↵erentiating both sides with respect to x, we have
1 dy
= ln 5 · (x ln x)0
y dx
1
= ln 5 · (x · + 1 · ln x)
x
= ln 5 · (1 + ln x)
dy
=)
= y · ln 5 · (1 + ln x)
dx
d x ln x
(5
) = 5x ln x ln 5 (1 + ln x) .
dx
=)
⌅
Change of base formulae
loga x =
ln x
, a > 0 and a , 1.
ln a
p
Example 2.16. Di↵erentiate log(2+x2 ) 1 + 2x with respect to x.
Ans:
1
(ln(2+x2 ))2
✓
ln(2+x2 )
1+2x
x ln(1+2x)
2+x2
◆
.
Solution.
p
d
log(2+x2 ) 1 + 2x
dx
p
d ln 1 + 2x
d ln(1 + 2x)
=
=
dx ln(2 + x2 ) dx 2 ln(2 + x2 )
=
1 (ln(1 + 2x))0 ln(2 + x2 ) ln(1 + 2x)(ln(2 + x2 ))0
2
(ln(2 + x2 ))2
2
2x
2
1 1 + 2x ln(2 + x ) ln(1 + 2x) 2 + x2
=
2
(ln(2 + x2 ))2
ln(2 + x2 )
1
=
1 + 2x
(ln(2 + x2 ))2
!
x ln(1 + 2x)
.
2 + x2
⌅
44
Chapter 3
Applications of Di↵erentiation
Read Thomas’ Calculus, Chapter 4.
3.1 Tangents and Normals
• The tangent at the point (x0 , f (x0 )) on the graph of a di↵erentiable function f has
equation
y f (x0 ) = m(x x0 ).
• The normal at the point (x0 , f (x0 )) on the graph of a di↵erentiable function f has equation
1
y f (x0 ) =
(x x0 ),
m
where m = f 0 (x0 ).
Example 3.1. The curve C has equation x2 + y 2 + 3xy = 5.
(a) Find the equations of the tangent and normal at the point (1, 1).
45
(b) Find (if any) the equations of the tangents that are parallel to the axes.
Ans: (a) tangent: y = x + 2, normal: y = x.
Solution. (a) Di↵erentiating the equation x2 + y 2 + 3xy = 5 with respect to x, we have
2x + 2yy 0 + 3y + 3xy 0 = 0.
Thus y 0 =
2x + 3y
.
3x + 2y
5
At (1, 1), we have y 0 =
= 1. Thus the gradient of the tangent at (1, 1) is 1 and the
5
gradient of the normal is 1.
Therefore, the equation of tangent is y
The equation of normal is y
(b) Recall that y 0 =
1 = (x 1). That is y = x + 2.
1 = x 1. That is y = x.
2x + 3y
.
3x + 2y
If a tangent is parallel to the x-axis, then its gradient is 0. Thus we shall find the points on
2x + 3y
2x
the curve such that y 0 = 0. That is
= 0 () 2x + 3y = 0 () y =
.
3x + 2y
3
Substituting this into the equation of the curve x2 + y 2 + 3xy = 5, we obtain
x2 + (
That is
2x 2
2x
) + 3x(
) = 5.
3
3
5x2
= 5, which has no solution.
9
⌅
If a curve is defined parametrically by
x = x(t) and y = y(t),
then
• the equation of the tangent at the point where t = t0 is
y
y(t0 ) = m(x x(t0 )),
• the equation of the normal at the point where t = t0 is
y
where m is the value of
y(t0 ) =
1
(x x(t0 )),
m
dy dy dx
=
÷
at t = t0 .
dx dt dt
46
Example 3.2. A curve is defined by
x = t 2 t and y = (t + 1)2 .
The tangent at the point A on the curve passes through ( 1, 0) and (1, 8). Find the equation of the
normal at the point A.
Ans: y =
x
4
+ 4.
Solution. The gradient of the line through ( 1, 0) and (1, 8) is
Also
8 0
= 4.
1 ( 1)
dy y 0 2t + 2
= =
.
dx x0 2t 1
Solving
dy
2t + 2
= 4, that is
= 4, gives t = 1.
dx
2t 1
Thus the point A on the curve has coordinates (12 1, (1 + 1)2 ) = (0, 4).
The equation of the normal through A is
y 4
1
=
x 0
4
or equivalently,
y=
x
+ 4.
4
⌅
3.2 Increasing and Decreasing Functions
Definition 3.1. (a) The function f is increasing on an interval I if f (x2 ) > f (x1 ) for x1 , x2 2 I
with x2 > x1 .
(b) The function f is decreasing on an interval I if f (x2 ) < f (x1 ) for x1 , x2 2 I with x2 > x1 .
Theorem 3.1. Let f be di↵erentiable on (a, b) and continuous on [a, b].
(a) f is increasing on [a, b] if f 0 (x) > 0 for all x in (a, b).
(b) f is decreasing on [a, b] if f 0 (x) < 0 for all x in (a, b).
Thus if f 0 (x) > 0(< 0) on (a, b) except possibly at a finite number of points at which f 0 (x) = 0,
then f is increasing (decreasing) on [a, b].
Example 3.3. Show that the function f (x) = 3x3
(0, 1) to its range R.
47
3e
x
4
x
is bijective (one-one) on the interval
Solution. f 0 (x) = 9x2 + 3e
x
+
4
> 0 for all x 2 (0, 1).
x2
Thus f is increasing on (0, 1). It follows that f is injective on (0, 1).
Note that lim+ 3x3 3e
x!0
x
4
= 1 and lim 3x3 3e
x!+1
x
x
4
= +1.
x
As f is continuous on (0, 1), the range of f is R.
Thus f (x) is bijective from (0, 1) onto its range R.
3.3 Concave Upward and Concave Downward Functions
Definition 3.2. (a) The graph of f is concave upward (downward) at (c, f (c)) if f 0 (c) exists
and there is an open interval I containing c such that for all x , c in I , the point (x, f (x)) on
the graph of f is above (below) the tangent line to the graph of f at x = c.
(b) The graph of f is concave upward (downward) on (a, b) if it is concave upward (downward)
at every point in (a, b).
The following theorem gives a test for concavity.
Theorem 3.2. Let f be di↵erentiable on (a, b). Let c 2 (a, b).
(a) If f 00 (c) > 0, then the graph of f is concave upward at (c, f (c)).
(b) If f 00 (c) < 0, then the graph of f is concave downward at (c, f (c)).
48
⌅
Note that the converse of the theorem is not true. For example, if f (x) = x4 , then the graph
of f is concave upward at (0, 0), but f 00 (0) = 0.
Definition 3.3. A point (c, f (c)) where the graph of a function f has a tangent line and where
the concavity changes is called a point of inflection.
Point of inflection: change of concavity
Theorem 3.3. Let f be di↵erentiable on (a, b). Let c 2 (a, b). If (c, f (c)) is a point of inflection
of the graph of f and f 00 (c) exists, then f 00 (c) = 0.
Example 3.4. Determine the intervals on which the function
f (x) = 2x3 + 15x2 24x + 7
is
(i) increasing, (ii) decreasing, and its graph is (iii) concave upward (iv) concave downward. (v)
Find the point(s) of inflection of the graph of f .
Ans: (i) [1, 4], (ii) ( 1, 1] [ [4, 1), (iii) ( 1, 52 ), (iv) ( 52 , 1), (v) ( 52 , 19
2 ).
Solution. f 0 (x) = 6x2 + 30x 24 = 6(x 1)(x 4).
49
x
f 0 (x)
x<1
x=1
0
1<x<4
+
x=4
0
4<x
Thus f is increasing on [1, 4], and decreasing on ( 1, 1] [ [4, 1),
We have f 00 (x) = 12x + 30 = 12(x
x
f
5
2 ).
00 (x)
x < 52
+
x=
0
5
2
5
2
<x
Thus the graph of f is concave upward on ( 1, 52 ) and concave downward on ( 52 , 1).
5 19
There is a change of concavity of the graph of f at ( 52 , 19
2 ). Thus ( 2 , 2 ) is a point of inflection
of the graph of f .
Summarizing, (i) increasing on [1, 4], (ii) decreasing on ( 1, 1], [4, 1), (iii) concave upward
on ( 1, 52 ), (iv) concave downward ( 52 , 1), (v) inflection point at x = 52 .
3.4 Related Rates
Let y = f (x) and let x and y be functions of a third variable t that represents, for example,
time. By the Chain Rule,
dy dy dx
=
.
dt dx dt
Example 3.5. Water is flowing at a rate of 100 cm3 per second into an inverted conical flask of
height 16cm and base radius 4cm. At the instant when the height of water level is 12cm, the water
level is rising at the rate of 3cm per sec. Calculate the rate at which water is leaking from the flask.
50
Ans: 15.18 cm3 per second.
Solution. By similar triangles,
r
4
h
=
) r = . Then
h 16
4
1
1 h
V = ⇡r 2 h = ⇡( )2 h.
3
3 4
Thus V =
⇡ 3
h . Di↵erentiating with respect to t,
48
dV dV dh
⇡
dh
=
·
=
· 3h2 · .
dt
dh dt 48
dt
At h = 12,
dh
= 3 cm per second. Thus
dt
dV
⇡
=
⇥ 3 ⇥ (12)2 ⇥ 3 = 27⇡ = 84.82 cm3 per second.
dt
48
Therefore, leaking rate is 100 27⇡ = 15.18 cm3 per second.
⌅
Exercise 3.1. A particle is moving horizontally in the x–y plane along the line y = 5 in such a
way that its distance from the origin (0, 0) is increasing at a rate of 1 unit per sec. Calculate the
rate at which the particle is moving horizontally at the instant when it is 13 units from (0, 0).
Ans:
13
12
unit per sec.
3.5 Maximum and Minimum Values
51
Definition 3.4. (Absolute Extrema) A function f has an
(a) absolute/global maximum at x = c if f (x)  f (c) for all x in the domain of f .
(b) absolute/global minimum at x = c if f (x)
f (c) for all x in the domain of f .
Definition 3.5. (Local Extrema) A function f defined on some interval I has a
(a) relative/local maximum at x = c if f (x)  f (c) for x in some open interval containing
x = c.
(b) relative/local minimum at x = c if f (x)
x = c.
f (c) for x in some open interval containing
Theorem 3.4. (Extreme Value Theorem) If f is continuous on a closed interval [a, b], then f
attains an absolute maximum value and an absolute minimum value at some points in [a, b].
Question. What if f is not continuous or if the domain is not a closed interval of the form
[a, b]?
52
Theorem 3.5. If f is di↵erentiable on an open interval containing x = c and f has a local
extremum at x = c, then f 0 (c) = 0.
Definition 3.6. (Critical Point) A number c in the domain of a function f is a critical point
of f if the following 2 conditions hold:
(i) it is not an end-point,
(ii) either f 0 (c) = 0 or f 0 (c) does not exist.
Theorem 3.6. If f has a local minimum/maximum at x = c, then c is a critical point of f .
However, the converse of this result is not true in general.
From the above results, we conclude that an absolute extremum occurs either at the end point or
at a critical point.
Hence, to find absolute extrema of a continuous function f defined on [a, b] , we
(1) find the values of f at all critical points of f on (a, b),
(2) find the values of f (a) and f (b).
The largest and smallest values from steps (1) and (2) are the absolute maximum value and
absolute minimum value of f respectively.
Remark. If the function f is continuous on [a, b] and is considered as defined on (a, b] and
the largest (smallest) value of f obtained from steps 1 and 2 occurs at x = a, then f has no
absolute maximum (minimum) on (a, b].
53
Example 3.6. Find the absolute maximum and minimum values of
g(x) =
x
x2 + 1
on (a) [ 2, 1], (b) ( 2, 1), (c) ( 1, 1].
Ans: (a) absolute maximum value = 12 , absolute minimum value = 12 , (b) no absolute maximum value, absolute minimum value = 12 , (c) absolute maximum value = 12 , no absolute
minimum value.
Solution. g 0 (x) =
1 · (x2 + 1) x(2x) 1 x2 (1 x)(1 + x)
= 2
=
.
x2 + 1
x +1
x2 + 1
g 0 (x) = 0 , x = 1, 1. Thus g has a critical point at x = 1, 1.
x
g(x)
2
2
5
1
1
2
1
1
2
1
1
(a) on [ 2, 1]: absolute maximum value = , absolute minimum value =
,
2
2
1
(b) on ( 2, 1): no absolute maximum value, absolute minimum value =
,
2
1
(c) on ( 1, 1]: absolute maximum value = , no absolute minimum value.
2
Exercise 3.2. Find the absolute maximum and minimum values of
h(x) = x5/3 x2/3
on (a) [ 1, 8], (b) ( 1, 1).
Ans: (a) absolute maximum value = 28, absolute minimum value = 2.
54
⌅
(
Exercise 3.3. Let f (x) =
|x|
x2 6x + 10
5x<2
. Find
2x4
(a) the critical points of f ,
(b) the absolute maximum and minimum values of f .
Ans: (a) critical points at x = 0, 2, 3, (b) absolute maximum value = 5, absolute minimum
= 0.
Theorem 3.7. (First Derivative Test for Absolute Extrema) Let f be di↵erentiable on an open
interval containing a critical point c except possibly at c and f is continuous at c.
(1) If f 0 (x) > 0 for all x < c and f 0 (x) < 0 for all x > c, then f has an absolute maximum at
c.
(2) If f 0 (x) < 0 for all x < c and f 0 (x) > 0 for all x > c, then f has an absolute minimum at
c.
Example 3.7. Find the absolute maximum value of the function
h(x) = 6x1/2 x3/2
on the interval (1, 1).
p
Ans: absolute maximum value = 4 2.
1
Solution. h(x) = 6x 2
3
x 2 . Then h0 (x) = 3x
3 1 3(2 x)
x2 =
.
1
2
2x 2
1
2
Thus h has 1 critical point in the interval (1, 1) given by x = 2.
When 1 < x < 2, h0 (x) > 0. When x > 2, h0 (x) < 0.
Thus by the first derivative test for absolute extrema, h has an absolute maximum on the
p
p
1
3
2
2 =6 2 2 2=
interval
(1,
1)
given
by
x
=
2,
and
the
maximum
value
of
h
is
h(2)
=
6
·
2
2
p
4 2.
⌅
Theorem 3.8. (First Derivative Test for Local Extrema) Let f be di↵erentiable on an open
interval containing a critical point c except possibly at c and f is continuous at c.
(1) If f 0 changes from positive to negative at x = c, then f has a local maximum at c.
(2) If f 0 changes from negative to positive at x = c, then f has a local minimum at c.
55
(3) If f 0 does not change sign at x = c, then f has no local extremum at c.
Example 3.8. Let f (x) = 3x4 8x3 .
(a) Find the critical points of f .
(b) Determine whether a local maximum or local minimum or neither occurs at each of these
points.
Ans: (a) critical point at x = 0, 2, (b) f has a local minimum at x = 2.
Solution. It is given that f (x) = 3x4 8x3 . Then f 0 (x) = 12x3 24x2 = 12x2 (x 2).
Thus f 0 (x) = 0 () 12x2 (x 2) = 0 () x = 0, 2.
Hence f has critical points at x = 0, 2.
x
f
0 (x)
x<0
0
0
0<x<2
2
0
2<x
+
By the first derivative test (for local extrema), f has no local extremum at x = 0, and f has a
local minimum at x = 2.
⌅
Theorem 3.9. (Second Derivative Test) Let f be a twice di↵erentiable function defined in an
open interval containing c.
(1) If f 0 (c) = 0 and f 00 (c) < 0, then f has a local maximum at c.
(2) If f 0 (c) = 0 and f 00 (c) > 0, then f has a local minimum at c.
(3) No conclusion can be drawn if f 00 (c) = 0.
Remark. The functions x4 , x4 , x3 , has a local min, max, neither a max nor a min at x = 0,
respectively, but have 0 second derivative at x = 0.
56
Example 3.9. Let f (x) = (x 4)ex .
(a) Find the critical points of f .
(b) Determine whether a local maximum or local minimum or neither occurs at each of these
points.
Ans: (a) critical point at x = 3, (b) f has a local minimum at x = 3.
Solution. It is given that f (x) = (x 4)ex . Then f 0 (x) = 1 · e x + (x 4)e x = (x 3)ex .
Thus f 0 (x) = 0 () (x 3)ex = 0 () x = 3. Hence f has a critical point at x = 3.
Then f 00 (x) = 1 · ex + (x 3)ex = (x 2)e x . Thus f 00 (3) = (3 2)e3 = e 3 > 0.
By the second derivative test, f has a local minimum at x = 3.
(
Exercise 3.4. Let f (x) =
⌅
x2 ,
2x
0x<2
.
2
(x 2) , x 2
(a) Find the critical points of f .
(b) Determine whether a local maximum or local minimum or neither occurs at each of these
points.
Ans: (a) critical points at x = 1 and x = 2, (b) f has a local maximum at x = 1 and a local
minimum at x = 2.
3.6 Applied Maximum and Minimum Problems
Example 3.10. A light bulb is placed at the top of a vertical pole to illuminate a circular field of
fixed radius R metres. The intensity of illumination, I , is proportional to
x
3
(x2 + R2 ) 2
,
where x is the height of the pole. Find the value of x that maximises I. Justify that your answer
indeed corresponds to the absolute maximum value of I .
57
Ans: x =
pR .
2
Solution. I =
Kx
3
(x2 + R2 ) 2
, where K is a positive constant.
The domain of I is x > 0.
1
3
x (x2 + R2 ) 2 2x]
2
(x2 + R2 )3
3
I0 =
K[(x2 + R2 ) 2
1
K(x2 + R2 ) 2 [(x2 + R2 ) 3x2 ]
(x2 + R2 )3
p
p
K(R2 2x2 ) K(R + 2x)(R
2x)
=
=
.
5
5
(x2 + R2 ) 2
(x2 + R2 ) 2
=
Thus I 0 = 0 () x = ± pR .
2
Since x > 0, the only critical point is at x =
When x <
pR , I 0
2
> 0. When x >
pR , I 0
2
pR .
2
< 0.
Thus I has an absolute maximum at x =
pR .
2
⌅
Exercise 3.5. Determine the coordinates of the points on the curve y = 16
x that are closest to the
origin. With the aid of a graph, justify that your answer indeed gives the shortest distance.
Ans: (4, 4), ( 4, 4).
Solution Let d be the distance from (x,
By Pythagoras’ theorem, d 2 = x2 + (
16
) to the origin.
x
16 2
) .
x
58
Note that d attains a minimum value L at (x0 , y0 ) if and only if d 2 attains a minimum value
L2 at (x0 , y0 ).
Thus it suffices to minimize d 2 . Let D = d 2 . we have
D = x2 + (
16 2
) .
x
Note that x , 0.
Hence,
dD
2 · 162 2(x4 44 )
= 2x
=
.
dx
x3
x3
2(x4 44 )
dD
= 0 ()
= 0 () x4 = 44 () x = ±4.
dx
x3
dD
dD
When 0 < x < 4,
< 0. When 4 < x,
> 0.
dx
dx
Thus D attains the absolute minimum at x = 4 on (0, 1).
Similarly, when x < 4,
dD
dD
< 0. When 4 < x < 0,
> 0.
dx
dx
Thus D attains the absolute minimum at x = 4 on ( 1, 0).
The 2 points are (4, 4) and ( 4, 4).
⌅
Exercise 3.6. A triangular plot ABC in which \A = 60 and AB is 80 metres long is to be divided
into two plots of equal areas by a fence built along the line PQ. The fence costs $10 per metre. Let
the lengths of AC and AP be 10b and 10x respectively.
(a) Show that the length of PQ is 10z, where z 2 = x2 + 16b
x2
(b) Show that 4  x  8.
2
4b.
(c) Determine, to the nearest dollar, the minimum cost of fencing, and the corresponding value of
x, when (i) b = 9, (ii) b = 25.
59
Ans: (a) $600, (b) $1096.59.
Exercise 3.7. A bus carries 60 passengers each day from a train station to a shopping mall. It
costs $1.50 per passenger to ride the bus. Research reveals that 4 more (fewer) people would ride
the bus for each 5 cents decrease (increase) in bus fare. Determine the bus fare (to the nearest cent)
per passenger that will maximise revenue of the bus operator.
Ans: $1.12.
y2
2
Exercise 3.8. Find the point P on the ellipse xa2 + b2 = 1 in the first quadrant such that the triangle
bounded by the axes of the ellipse and the tangent to the ellipse at the point P has the least area.
Ans: ( pa , pb ).
2
2
2
y2
Exercise 3.9. Find the point P on the ellipse xa2 + b2 = 1 in the first quadrant such that the line
segment tangent to the ellipse at the point P with endpoints on the axes of the ellipse has the least
length.
60
Ans: (
3
a2
3
1 ,
b2
1
(a+b) 2 (a+b) 2
).
3.7 L’Hôpital’s Rule
Theorem 3.10. Let f and g be di↵erentiable at all points in some open interval containing
x = c (except possibly at c). If lim f (x) = 0 = lim g(x) or lim f (x) = 1 = lim g(x), then
x!c
x!c
x!c
x!c
f (x)
f 0 (x)
= lim 0 ,
x!c g(x)
x!c g (x)
lim
provided the limit on the right exists or equals 1 or 1.
The result also holds
(1) for limits at infinity, i.e. c = ±1.
(2) for one-sided limits.
Sketch of Proof. Here we give a proof in the special case when lim f (x) = 0 = lim g(x), and
x!c
the functions f , g, f 0 and g 0 are continuous at x = c, and g 0 (c) , 0. Then
x!c
f (c) = lim f (x) = 0 and g(c) = lim g(x) = 0.
x!c
x!c
Thus
lim
x!c
f (x)
f (x) 0
= lim
g(x) x!c g(x) 0
f (x) f (c)
= lim
x!c g(x) g(c)
f (x) f (c)
x c
= lim
x!c g(x) g(c)
x c
f 0 (c)
= 0
(by definition of f 0 (c) and g 0 (c) and noting that g 0 (c) , 0)
g (c)
lim f 0 (x)
x!c
=
(by continuity of f 0 , g 0 at x = c)
0
lim g (x)
x!c
f 0 (x)
.
x!c g 0 (x)
= lim
⌅
61
Example 3.11. Evaluate
3x2 + x ln x
x!1 x 2 + 2 ln x
✓
◆
1
(b) lim 2 csc 2x
x!0
x
(c) lim+ x ln x
(a) lim
x!0
Ans: (a) 3, (b) 0, (c) 0.
Solution. (a) We apply L’Hôpital’s Rule twice.
lim
x!1
= lim
x!1
= lim
x!1
=
3x2 + x ln x
x2 + 2 ln x
6x + ln x + 1
2
2x +
x
1
+0
x
2
2
x2
6+
(by L’Hôpital’s Rule)
(by L’Hôpital’s Rule)
6+0
= 3.
2 0
(b) We apply L’Hôpital’s Rule twice.
✓
lim 2 csc 2x
x!0
1
x
◆
= lim
2x sin(2x)
x sin(2x)
= lim
2 2 cos(2x)
sin(2x) + 2x cos(2x)
= lim
0 + 4 sin(2x)
2 cos(2x) + (2 cos(2x) 4x sin(2x))
x!0
x!0
x!0
=
(by L’Hôpital’s Rule)
0
= 0.
2+2 0
62
(by L’Hôpital’s Rule)
(c) Exercise.
⌅
There are 3 indeterminate forms of the following types:
(1) 00 ,
(2) 10 ,
(3) 11 .
Example 3.12.
00 :
x
lim+ x = lim+ e
x!0
ln(xx )
x!0
= lim+ e
x ln x
x!0
=e
lim x ln x
x!0+
=e
ln x
lim
x!0+ x 1
=e
x 1
lim
x!0+ x 2
lim
= ex!0+
x
= e 0 = 1.
2
2x 20
10 : lim+ ( )x = lim+ x =
= 2.
x!0 x
x!0 x
1
11 :
1
x
lim (cos x) = lim e
x!0
x!0
ln cos x
x
=e
lim
x!0
ln cos x
x
=e
1 ( sin x)
lim cos x
1
x!0
lim
= ex!0
tan x
= e 0 = 1.
3.8 Rolle’s Theorem and Mean Value Theorem
Theorem 3.11. (Rolle’s Theorem) Let f be continuous on [a, b] and di↵erentiable on (a, b). If
f (a) = f (b), then there is at least one number c in (a, b) such that f 0 (c) = 0.
Example 3.13. Use Rolle’s Theorem to prove that the equation
ln x + 2x = 3
has at most one positive solution.
63
Solution. Let f (x) = ln x + 2x. Note that f is a di↵erentiable function defined for x > 0. Also
1
f 0 (x) = + 2. Suppose f (x) = 3 has two positive solutions a and b with 0 < a < b. That is
x
f (a) = f (b) = 3. By Rolle’s Theorem applied to f on the interval [a, b], there exists c 2 (a, b)
1
such that f 0 (c) = + 2 = 0, which is not true as c > a > 0. This contradiction shows that
c
f (x) = 3 cannot have two positive solutions. In other words, ln x + 2x = 3 has at most one
positive solution.
Remark. Note that
f (1) = ln(1) + 2 = 2 and
f (e) = ln e + 2e = 1 + 2e > 3.
Thus we have
f (1) < 3 < f (e).
Also f is di↵erentiable and thus continuous on [1, e]. Thus by Intermediate Value Theorem
(see Theorem 1.7 of Lecture Notes), there exists a number c 2 [1, 3] such that f (c) = 3.
Since we have also proved that the equation f (x) = 3 has at most one positive solution, we
knows that the equation f (x) = 3 has exactly one positive solution.
⌅
Exercise 3.10. Use Rolle’s Theorem to prove that the equation
2e x + x2 + 3x = 0
has at most two real solutions.
Rolle’s Theorem can be used to prove the following result.
Theorem 3.12. (Mean Value Theorem) Let f be continuous on [a, b] and di↵erentiable on
(a, b). Then, there is at least one number c in (a, b) such that
f 0 (c) =
f (b) f (a)
.
b a
64
Theorem 3.13. Let f be continuous on [a, b] and di↵erentiable on (a, b). If f 0 (x) > 0(< 0) for
all x 2 (a, b), then f is increasing (decreasing) on [a, b].
Proof. Let’s suppose f 0 (x) > 0 for all x 2 (a, b). Let x1 , x2 2 [a, b] and x1 < x2 . Thus f is continuous on [x1 , x2 ] and di↵erentiable on (x1 , x2 ) as f is continuous on [a, b] and di↵erentiable
on (a, b). By mean value theorem applied to f on [x1 , x2 ], there is a number c in (x1 , x2 ) such
that
f (x2 ) f (x1 )
= f 0 (c).
x2 x1
Since f 0 (c) > 0 and x2 > x1 , we have f (x2 ) > f (x1 ). That is f is increasing on [a, b].
⌅
Exercise 3.11. Use the Mean Value Theorem to prove that for all real numbers x and y,
| cos x cos y|  |x y|.
Exercise 3.12. Use the Mean Value Theorem to show that if f 0 (x) = 0 for all x in (a, b), then f is
a constant function on (a, b).
Exercise 3.13. Use the Mean Value Theorem to prove that there exists a real number ✓ 2 ( ⇡5 , ⇡4 )
such that
sin
Deduce that sin ⇡5 <
p
2
40 (20
⇡
⇡
= sin
5
4
⇡).
65
⇡
cos ✓.
20
66
Chapter 4
Integrals
Read Thomas’ Calculus, Chapter 5.
4.1 Antiderivatives
Definition 4.1. F is an antiderivative of f on an interval I if F 0 (x) = f (x) for all x in I .
Theorem 4.1. (1) If F is an antiderivative of f on an interval I, then so is F + C for any
constant C. Furthermore, any antiderivative of f on I is of the form F + C for some constant
C. This can be expressed as
Z
f (x) dx = F(x) + C.
R
f (x) dx is called an indefinite integral.
(2) Let ↵ and
be any constants. Then
Z
Z
Z
↵f (x) + g(x) dx = ↵ f (x) dx +
g(x) dx.
Example 4.1. An anti-derivative of xn is
1 n+1
,
n+1 x
Z
xn dx =
where n , 1 as
1 n+1
x
+ C.
n+1
Example 4.2. Find all the anti-derivatives of 2x cos 3x.
R
Solution. 2x cos 3x dx = x2
1
3 sin 3x + C.
67
⇣
d
1 n+1
dx n+1 x
⌘
= xn . Thus
4.2 Standard Integrals
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
(ax + b)n dx =
1
ax+b
(ax+b)n+1
(n+1)a
+C
(n , 1)
dx = 1a ln |ax + b| + C
eax+b dx = 1a e ax+b + C
sin(ax + b) dx =
1
a cos(ax + b) + C
cos(ax + b) dx = 1a sin(ax + b) + C
tan(ax + b) dx = 1a ln | sec(ax + b)| + C
sec(ax + b) dx = 1a ln | sec(ax + b) + tan(ax + b)| + C
csc(ax + b) dx =
1
a ln | csc(ax + b) + cot(ax + b)| + C
cot(ax + b) dx =
1
a ln | csc(ax + b)| + C
sec2 (ax + b) dx = 1a tan(ax + b) + C
csc2 (ax + b) dx =
1
a cot(ax + b) + C
sec(ax + b) tan(ax + b) dx = 1a sec(ax + b) + C
1
a csc(ax + b) + C
csc(ax + b) cot(ax + b) dx =
1
a2 +(x+b)2
dx = 1a tan 1 ( x+b
a )+C
p
1
a2 (x+b)2
dx = sin 1 ( x+b
a )+C
p
1
a2 (x+b)2
dx = cos 1 ( x+b
a )+C
1
a2 (x+b)2
dx =
1
x+b+a
2a ln x+b a
+C
1
(x+b)2 a2
dx =
1
x+b a
2a ln x+b+a
+C
p
1
(x+b)2 +a2
p
1
(x+b)2 a2
p
dx = ln (x + b) + (x + b)2 + a2 + C
p
dx = ln (x + b) + (x + b)2 a2 + C
21.
Rp
a2 x2 dx =
x
2
22.
Rp
x2 a2 dx =
x
2
p
a2 x2 + a2 sin
p
x 2 a2
2
1x
a
+C
p
a2
2 a2 | + C
2 ln |x + x
68
Example 4.3. Let a , 0. Show that
Solution. We have to show
b
a,
For x >
R
1
ax+b
dx = 1a ln |ax + b| + C, for x ,
b
a.
d 1
1
dx ( a ln |ax + b|) = ax+b .
we have |ax + b| = ax + b. Thus
d 1
d 1
1
( ln |ax + b|) =
( ln(ax + b)) =
.
dx a
dx a
ax + b
b
a,
For x <
we have |ax + b| = (ax + b). Thus
d 1
d 1
1
a
1
( ln |ax + b|) =
( ln( (ax + b))) = ·
=
.
dx a
dx a
a (ax + b) ax + b
⌅
Example 4.4. Find
(a)
(b)
R
R
1
p
dx
x2 4x+29
1
p
3+6x 9x2
Solution. (a)
R
dx
1
p
dx
x2 4x+29
=
R
p
1
dx
(x 2)2 +52 p
= ln (x 2) + (x 2)2 + 52 + C
p
= ln (x 2) + x2 4x + 29 + C.
(b)
R
1
p
3+6x 9x2
R
1
dx
(3x 1)2
1
dx
( 23 )✓2 (x ◆13 )2
1
1 x 3
p
R 22
= 13 p
dx =
= 13 sin
2
3
+C
= 13 sin 1 ( 3x2 1 ) + C.
⌅
Exercise 4.1. Find
(a)
(b)
R⇣
R
⌘
3x 1 2
2x+1
dx
(2e2x 1 e x )2
ex+1
dx
69
Ans: (a)
9x
4
15
4 ln |2x + 1|
25
8(2x+1)
+ C, (b) 43 e3x
3
4e 2 x
1 3x 1
+ C.
3e
Trigonometric Identities Useful for Integration
1. sec2 x 1 = tan2 x
2. csc2 x 1 = cot2 x
3. sin A cos A = 12 sin 2A
4. cos2 A = 12 (1 + cos 2A)
5. sin2 A = 12 (1 cos 2A)
6. sin A cos B = 12 (sin(A + B) + sin(A B))
7. cos A sin B = 12 (sin(A + B) sin(A B))
8. cos A cos B = 12 (cos(A + B) + cos(A B))
9. sin A sin B =
1
2 (cos(A + B)
cos(A B))
R
Example 4.5. Find cos 6x sin 3x dx
Solution.
R
cos 6x sin 3x dx =
=
R
1
2
1
x
2 (sin 2
R
sin (
x
6 )) dx
sin 2x + sin 6x dx
= 12 ( 2 cos 2x
= cos 2x
6 cos 6x ) + C
3 cos 6x + C.
⌅
R
1
Example 4.6. Show that cos4 x dx = 38 x + 14 sin 2x + 32
sin 4x + C.
Solution. First we have
cos4 x = ( 12 (1 + cos 2x))2 = 14 (1 + 2 cos 2x + cos2 2x)
= 14 (1 + 2 cos 2x + 12 (1 + cos 4x))
= 38 + 12 cos 2x + 18 cos 4x.
70
R
R
Thus cos4 x dx= 38 + 12 cos 2x + 18 cos 4x dx
1
= 38 x + 14 sin 2x + 32
sin 4x + C.
⌅
Exercise 4.2. Find
Ans:
1
2 tan 2x
R⇣
sin 4x
1+cos 4x
⌘2
dx
x + C.
4.3 Partial Fractions
Let P(x) and Q(x) be two polynomials. Suppose Q(x) is a product of linear or quadratic
P(x)
factors with real coefficients. Then, the rational function Q(x) can be expressed as a sum of
simple fractions whose denominators are factors of Q(x).
Factors of Q(x)
Partial fractions
ax + b
A
ax + b
(ax + b)2
A
B
+
ax + b (ax + b)2
ax2 + bx + c, b 2 4ac < 0
The rational function
P(x)
Q(x)
Ax + B
ax2 + bx + c
is said to be a proper fraction if the degree of P(x) is smaller
that the degree of Q(x). Otherwise, it is called an improper fraction. If
fraction, one can perform long division to write it as A(x) +
fraction.
Examples
(1)
5
1
2x + 4
2x + 4
3
3
=
=
+
.
x2 9 (x 3)(x + 3) x 3 x + 3
14
8
3x2 + x + 4
2x + 10
3
3
=
3
+
=
3
+
.
x+2 x 1
x2 + x 2
x2 + x 2
(2) is an example of an improper fraction.
Z
3x2 + x + 4
Example 4.7. Find
dx.
x2 + x 2
(2)
71
B(x)
,
Q(x)
P(x)
Q(x)
where
is an improper
B(x)
Q(x)
is a proper
Solution.
Z
3x2 + x + 4
dx =
x2 + x 2
Z
3
14
3
x+2
8
3
+
x 1
14
8
ln |x + 2| + ln |x 1| + C.
3
3
dx = 3x
⌅
4.4 Integration by Substitution
Theorem 4.2. Let u = g(x) be a di↵erentiable function whose range is some interval I and let
f be continuous on I . Then,
Z
Z
Example 4.8. Find
p
e 3x
2e 3x + 4
f (g(x))g 0 (x) dx =
Z
f (u) du.
dx.
du
= 6e3x , so that du = 6e3x dx. Then
dx
Z
Z
1 p 3x
e 3x
1
1p
dx =
u +C =
2e + 4 + C.
p
p du =
3
3
6 u
2e 3x + 4
Solution. Let u = 2e 3x + 4. Then
Z
Exercise 4.3. Find
⌅
(3 tan 4x)5
dx.
cos2 4x
1
24 (3
tan 4x)6 + C.
Z
8
Exercise 4.4. Find
dx.
p
x ln x
p
Ans: 16 ln x + C.
Ans:
Interchanging the roles of u and x in Theorem 4.2 and then renaming u and x by x and t
respectively, we get
Theorem 4.3. Let x = g(t) be a di↵erentiable function whose range is some interval I and let
f be continuous on I . Then
Z
Z
f (x) dx =
f (g(t))g 0 (t) dt.
72
Z
Example 4.9. Find
ln x
dx.
x
dx
= e t , so that dx = et dt. Then
dt
Z
Z
Z
ln(et ) t
(ln x)2
ln x
t2
dx =
e
dt
=
t
dt
=
+
C
=
+ C.
x
et
2
2
Solution. Let x = e t . Then
⌅
Trigonometric Substitution
qExpression
a2 (x + b)2
q
a2 + (x + b)2
q
(x + b)2 a2
Substitution
x + b = a sin ✓,
⇡
2
x + b = a tan ✓,
⇡
2
x + b = a sec ✓, 0 < ✓ <
⇡
2
✓
<✓<
Identity involved
⇡
2
1 sin2 ✓ = cos2 ✓
⇡
2
1 + tan2 ✓ = sec2 ✓
or ⇡  ✓ <
3⇡
2
sec2 ✓ 1 = tan2 ✓
Z p
25 4x2
Example 4.10. Find
dx.
x2
p
5
2x
Solution. Let x = sin ✓. That is ✓ = sin 1 . Then 2x = 5 sin ✓, so that 25 4x2 =
2
5
p
dx 5
5
2
25 25 sin ✓ = 5 cos ✓. Also
= cos ✓, so that dx = cos ✓d✓. Then
dt 2
2
Z p
Z
25 4x2
5 cos ✓ 5
dx =
· cos ✓ d✓
2
5
2
x
( sin ✓)2
2
Z
= 2 cot2 ✓ d✓
Z
= 2 csc2 ✓ 1 d✓
= 2 cot ✓ 2✓ + C
1p
2x
=
25 4x2 2 sin 1 ( ) + C.
x
5
⌅
Z
Example 4.11. Find
p
1
x 9x2 + 1
dx.
73
p
p
1
Solution. Let x = tan ✓. That is ✓ = tan 1 (3x). Then 1 + 9x2 = 1 + tan2 ✓ = sec ✓. Also,
3
dx 1
1
2
= sec ✓, so that dx = sec2 ✓d✓. Then
d✓ 3
3
Z
Z
1
1
1
dx
=
· sec2 ✓ d✓
p
1
2
3
x 9x + 1
tan ✓ sec ✓
Z 3
= csc ✓ d✓
= ln | csc ✓ + cot ✓| + C
p
9x2 + 1 1
= ln
+
+C
3x
3x
p
9x2 + 1 + 1
= ln
+ C.
3x
Exercise 4.5. Find
Ans:
Z p
9
1
1 x 3
2 sin ( 3 ) + 2 (x
⌅
6x x2 dx.
p
3) 6x x2 + C
4.5 Integration by Parts
Recall the product rule of di↵erentiation:
d
(f (x)g(x)) = f 0 (x)g(x) + f (x)g 0 (x).
dx
Integrating both sides of the above equation with respect to x gives
Z
f (x)g(x) =
Z
0
f (x)g(x) dx +
f (x)g 0 (x) dx,
or
Z
Z
0
f (x)g(x) dx = f (x)g(x)
f (x)g 0 (x) dx.
This suggests the following way of performing integration by parts:
74
Z
integrate
f0
keep g
keep f
di↵erentiate
g
z}|{ z}|{ Z z}|{ z}|{
0
f (x)g(x) dx = f (x) · g(x)
f (x) g 0 (x) dx.
Z
Example 4.12. Find
x sin 3x dx.
Solution.
Z
Z
x sin 3x dx =
(sin 3x) · x, dx
integrate sin 3x
keep integral of sin 3x
z }| { keep x Z z }| {
cos 3x
cos 3x z}|{
=
x
3
3
x cos 3x sin 3x
=
+
+ C.
3
9
di↵erentiate x
z}|{
1
dx
⌅
Z
Example 4.13. Find
x ln x dx.
integrate x
Z
Solution.
z}|{ keep ln x
z}|{ Z
x2
x ln x dx =
ln x
2
=
x2 ln x
2
keep integral of
x
di↵erentiate ln x
z}|{
x2
2
z}|{
1
dx
x
x2
+ C.
4
⌅
Basic rules to determine which function to integrate and which function to di↵erentiate.
The choice of integration follows the reverse order of the following:
75
Types of functions
Examples
Logarithmic Function
ln(ax + b) or its higher powers
di↵erentiate it
Inverse Trigonometric Functions
sin 1 (ax + b), cos 1 (ax + b), tan 1 (ax + b)
di↵erentiate it
Algebraic Functions
Power functions xa , polynomials
di↵erentiate it
sin(ax + b), cos(ax + b),
tan(ax + b),
csc(ax + b), sec(ax + b), cot(ax + b), or
a combinations of these
di↵erentiate it
Trigonometric Functions
Exponential Functions
e ax+b
Z
Exercise 4.6. Find
(2x + 1) ln(2x 3) dx.
x2
2
Ans: (x2 + x) ln(2x 3)
Z
Exercise 4.7. Find
2
Ans: ( 18 + x2 ) tan
1 2x
Z
Exercise 4.8. Find
Ans:
5x
2
15
4 ln(2x
3) + C.
x tan 1 (2x) dx.
x
4
+ C.
sin 2x
dx.
e 2x
1 2x
(sin 2x + cos 2x) + C.
4e
Z
Exercise 4.9. Find
sec3 x dx.
Ans: 12 (sec x tan x + ln | sec x + tan x|) + C.
4.6 Riemann Sums and Definite Integrals
Let f be continuous on [a, b].
76
Remark
integrate it
integrate it
The following limit exists and is known as the definite integral of f from x = a to x = b.
8
!
!9
n
>
>
X
>
>
b
a
b
a
<
=
lim >
f
a
+
k(
)
.
>
>
>
n!1 :
;
n
n
k=1
It is denoted by
Z
b
f (x) dx.
a
Geometrically the definite integral gives the area of the region under the graph of f from
x = a to x = b (at least in the case when f (x) 0 and a < b). The numbers a and b are
respectively called the lower and upper limits of the definite integral. The function f (x) is
called the integrand of the definite integral.
!
!
n
X
b a
b a
The finite series
f a + k(
) is known as a Riemann sum of f .
n
n
k=1
Z
b
Summing up, we have the following definition of
f (x) dx.
a
Z
b
a
8
!
!9
n
>
>
>
b a >
<X b a
=
f (x) dx = lim >
f a + k(
) >
.
>
>
n!1 :
;
n
n
k=1
77
Approximation. For sufficiently large n,
Z
b
a
!
!
n
X
b a
b a
f (x) dx ⇡
f a + k(
) .
n
n
k=1
Z
Example 4.14. Use Riemann sum to compute
The summation formula
n
X
k=1
3
0
x2 dx.
1
k 2 = n(n + 1)(2n + 1) is needed to compute the sum.
6
Ans: 9.
Solution. Here f (x) = x2 , a = 0, b = 3. Thus for each n, the corresponding Riemann sum is
given by
n
n
X
X
3 0
3 0
3 3k 2
· f (0 + k ·
)=
( ) .
n
n
n n
k=1
k=1
Using the given summation formula, we get
n
n
X
3 3k 2 27 X 2 27 n(n + 1)(2n + 1) 9(n + 1)(2n + 1)
( ) = 3
k = 3·
=
.
n n
6
n
n
2n2
k=1
k=1
Thus
Z
3
0
n
X
9(n + 1)(2n + 1)
3 3k 2
( ) = lim
n!1
n!1
n n
2n2
x2 dx = lim
k=1
9(1 + n1 )(2 + n1 ) 9(1 + 0)(2 + 0)
=
= 9.
n!1
2
2
= lim
4.7 Fundamental Theorem of Calculus (FTC)
The method of using Riemann sums to evaluate definite integrals is tedious. The following
result, known as the First Fundamental Theorem of Calculus (FTC 1), provides us with a
simpler way of calculating definite integrals when the anti-derivatives of the integrand can
be found.
Theorem 4.4 (FTC 1). Let f be continuous on [a, b] and let F be an anti-derivative of f .
Then,
Zb
f (x) dx = F(b) F(a).
a
78
That is,
Z
b
F 0 (x) dx = F(b) F(a).
a

Notation: Often we denote F(x)
Z
b
b
a
= F(b) F(a). Then the FTC 1 takes the form

F (x) dx = F(x)
0
a
b
a
= F(b) F(a).
Remark. We will see later that FTC 1 follows from Theorem 4.5 (FTC 2).
Z
Example 4.15. Evaluate
e
1
1
(ln x) 3
dx.
x
Ans: 34 .
1
Solution: We need to find an anti-derivative of f (x) =
tution u = ln x to get
Z
1
(ln x) 3
dx =
x
Z
(ln x) 3
x
. For this, we make the substi-
1
4
3
(ln x) 3 d(ln x) = (ln x) 3 + C.
4
4
3
Pick C = 0, and we get an F(x) = (ln x) 3 . Then by FTC 1, we have
4
Z
e
1
1

4
(ln x) 3
3
dx = (ln x) 3
x
4
e
4
3
= (ln e) 3
4
1
4
3
3
(ln 1) 3 =
4
4
3
0= .
4
Remark. We often present our solution more compactly as follows:
Z
e
1
1
(ln x) 3
dx =
x
Z

e
1
1
(ln x) 3 d(ln x)
4
3
= (ln x) 3
4
e
4
3
= (ln e) 3
4
1
4
3
3
(ln 1) 3 =
4
4
3
0= .
4
⌅
Z
Example 4.16. Evaluate
⇡
2
0
x cos x dx.
79
Ans:
⇡
2
1.
Solution. Using integration by parts, we have
Z ⇡

2
x cos x dx = x sin x
0
Z
⇡
2
0
⇡
2
0

⇡
=(
0) + cos x
2
⇡
= + (0 1)
2
⇡
=
1.
2
sin x dx
⇡
2
0
⌅
Example 4.17. Use a Riemann sum to show that lim
n!1
n
X
k=1
1
1
= ln 4.
3k + n 3
Solution. Recall that
!
! Zb
n
X
b a
b a
lim
f a + k(
) =
f (x) dx.
n!1
n
n
a
k=1
To express lim
n!1
n
X
k=1
1
as a Riemann sum, we need to identify the function f (x) and the
3k + n
interval [a, b]. First we have
n
X
k=1
n
n
k=1
k=1
X
1
1
1 1X
1
4 1
=
=
.
3
4
1
3k + n
k( n ) + 1 n 3
1 + k( n ) n
1
From this we see that a = 1, b = 4 and f (x) = . Therefore,
x
n
n
X 1
1X
1
4 1
lim
= lim
4
1
n!1
n!1
3k + n
3
1 + k( n ) n
k=1
k=1
Z 4

4
1
1
1
1
=
dx = ln x = ln 4.
3 1 x
3
3
1
Next we have the Second Fundamental Theorem of Calculus (FTC 2) as follows:
Theorem 4.5 (FTC 2). Let f be continuous on [a, b]. The function g defined by
Zx
g(x) =
f (t) dt, a  x  b
a
80
is continuous and di↵erentiable on (a, b), and g 0 (x) = f (x). That is
Zx
d
f (t) dt = f (x).
dx a
Remark. FTC 2 essentially says that the area function
integrand) f .
Rx
a
f (t) dt is an anti-derivative of (the
Sketch of Proof of Theorem 4.5 (FTC 2). For small h,
Z
x+h
g(x + h) g(x) =
Z
Z
a
f (t) dt
a
x+h
=
x
f (t) dt
f (t) dt
x
= area under the graph of f over [x, x + h]
⇡ f (x) · h.
y
.
...
....
........
...
...
...
....
.....
..
.
.
.
..
...
.....
....
...
.....
...
.......
.......................................
.
...
.
.
.
... ..........
...
...... ........................
...... ........................................
...
..........................
......
...
.......
..............................
.........
.
.
.
.
.
.
...
.
.
.
..............................
.
.
.
.
.
.
.......
..............................
...
..............................
...
..............................
...
..............................
...
..............................
...
..............................
.
..................................................................................................................................................................................
....
y = f (t)
•
f(x)
x x+h
0
Thus
Letting h ! 0, we get
t
g(x + h) g(x)
⇡ f (x).
h
g(x + h) g(x)
= f (x).
h
h!0
g 0 (x) = lim
⌅
As mentioned earlier, one can deduce FTC 1 from FTC 2 as follows:
Proof of Theorem 4.4 (FTC 1). Let F be an antiderivative
a continuous function f over
Z of
x
an interval I , and let a, b be in I . By Theorem 4.5 (FTC 2),
f (t) dt is also an antiderivative
a
of f . Thus by Theorem 4.1, we have
Z
x
F(x) =
f (t) dt + C
a
81
for some constant C. Then
F(b) F(a) =
⇣Z
b
⇣Z
f (t) dt + C)
a
a
Z
b
f (t) dt + C) =
a
f (t) dt.
a
⌅
Remark. If g(x) is di↵erentiable, then using the Chain Rule, one has
d
dx
Z
g(x)
f (t) dt = f (g(x))g 0 (x).
a
Proof. We let
Z
u
F(u) =
f (t) dt
and u = g(x).
a
Thus
Z
g(x)
F(g(x)) =
f (t) dt.
a
By Theorem 4.5 (FTC 2), one has
dF
= f (u).
du
Thus,
d
dx
Z
g(x)
d
F(g(x))
dx
dF du
=
·
(by Chain Rule)
du dx
= f (g(x))g 0 (x).
f (t) dt =
a
⌅
Example 4.18. Find
d
dx
Z
sin x p
2
1 + t 6 dt.
p
Ans: cos x 1 + sin6 x.
Solution. We apply the formula
d
dx
Z
g(x)
f (t) dt = f (g(x))g 0 (x)
a
p
(with f (t) = 1 + t 6 and g(x) = sin x). Note that g 0 (x) = cos x. Thus
d
dx
Z
sin x p
2
1 + t 6 dt =
q
p
1 + (sin x)6 · cos x = cos x 1 + sin6 x.
82
⌅
Properties of Definite Integrals
Let c 2 [a, b] and ↵, 2 R.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Rb
a
Rc
c
↵ dx = ↵(b a)
f (x) dx = 0
Rb
(↵f (x) + g(x)) dx =
a
Rb
a
Rb
a
Rb
a
Rb
a
Rb
a
Rb
a
f (x) dx =
12.
a
f (x) dx
a
f (x) dx +
Ra
f (x) dx =
b
↵f (x) dx +
Rb
c
Rb
g(x) dx
a
f (x) dx
f (x) dx
0 if f (x)
0 for a  x  b
f (x) dx  0 if f (x)  0 for a  x  b
Rb
f (x) dx
a
f (x) dx 
10. m(b a) 
11.
Rc
Rb
Ra
a
Ra
a
Rb
Rb
a
a
g(x) dx if f (x)
g(x) for a  x  b
g(x) dx if f (x)  g(x) for a  x  b
f (x) dx  M(b a) if m  f (x)  M for a  x  b
f (x) dx = 0 if f is an odd function defined on [ a, a]
f (x) dx = 2
Ra
0
f (x) dx if f is an even function defined on [ a, a]
4.8 Miscellaneous Examples
Z
Example 4.19. (Integrals of the type
Z
Evaluate
1
2
px + q
ax2 + bx + c
3x + 7
dx.
x2 + 4x + 5
Solution.
83
dx)
Z
1
3x + 7
x2 + 4x + 5
2
Z
=
1
=
=
3 2x + 4
1
+ 2
dx
2
2 x + 4x + 5 x + 4x + 5
2
Z
1
3 2x + 4
1
+
dx
2
2 x + 4x + 5 1 + (x + 2)2
2

dx
1
3
ln |x2 + 4x + 5|
2
2
h
i
+ tan 1 (x + 2)
1
2
3
= (ln 2 ln 1) + tan 1 (1) tan 1 (0)
2
3
⇡
= ln 2 + .
2
4
⌅
Z
Example 4.20. (Integrals of the type
Z
Show that
Solution:
Z 1
2
Z
=
2
Z
=
=
1
1
2

2
1
p
p
2x + 3
7⇡
dx = 2 3 +
p
6
4x x2
3x + 7
x2 + 4x + 5
px + q
ax2 + bx + c
dx)
4.
dx
3 2x + 4
1
+
dx
2 x2 + 4x + 5 x2 + 4x + 5
3 2x + 4
1
+
dx
2
2 x + 4x + 5 1 + (x + 2)2
3
ln |x2 + 4x + 5|
2
1
2

+ tan 1 (x + 2)
=
3
(ln 2 ln 1) + tan 1 (1) tan 1 (0)
2
=
3
⇡
ln 2 + .
2
4
1
2
⌅
84
x
x
Exercise 4.10. Using the identity 1 + cos x + sin x = 2 cos2 (1 + tan ), evaluate
2
2
Z ⇡
2
1
dx.
0 1 + cos x + sin x
Ans: ln 2.
Exercise 4.11. Using the substitution x =
Z
⇡
2
0
⇡
2
y, show that
sin2 x
dx =
1 + cos x + sin x
Z
⇡
2
0
cos2 x
dx.
1 + cos x + sin x
Exercise 4.12. Using the results in exercise 4.10 and 4.11 and the identity sin2 x + cos2 x = 1,
show that
Z
⇡
2
0
sin2 x
ln 2
dx =
.
1 + cos x + sin x
2
4.9 Improper Integrals
The definite integrals that we have studied so far have the following characteristics: (i) the
domain of integration is a finite closed interval [a, b], (ii) the function or the integrand has
finite values on the domain of integration. It is possible that we would encounter problems
that do not meet these conditions.
The integral for the area under the curve y = xe
integration which is infinite.
x
from x = 0 to x = 1 has domain of
The graph of y = xe x .
The integral for the area under the curve y =
the function
p1
x
p1
x
whose value at x = 0 is infinite.
85
from x = 0 to x = 10 requires us to integrate
The graph of y =
p1 .
x
In either case, the integrals are called improper integrals and are calculated as limits.
Definition 4.2. Integrals with infinite limits of integration are improper integrals of Type I.
1. If f (x) is continuous on [a, 1), then
Z
1
Z
b!1
a
2. If f (x) is continuous on ( 1, b], then
Z
b
1
Z
a! 1
c
1
f (x) dx.
a
b
f (x) dx = lim
3. If f (x) is continuous on ( 1, 1), then
Z1
Z
f (x) dx =
1
b
f (x) dx = lim
f (x) dx.
a
Z
1
f (x) dx +
f (x) dx,
c
where c is any real number.
In each case, if the limit is finite, we say that the improper integral converges and that the limit
is the value of the improper integral. If the limit fails to exist, then we say that the improper
integral diverges.
Example 4.21. Evaluate the Type I improper integral
Z1
ln x
dx.
x2
1
Solution. We first use integration by parts to compute the indefinite integral.
Z
Z
(1 + ln x)
1
1
1 1
ln x 1
ln x · 2 dx = ln x ·
·
dx =
+C =
+ C.
x
x x
x
x
x
x
86
Thus
Z
1
1
ln x
dx = lim
x2
b!1
Z
b
1
ln x
dx = lim
x2
b!1
"
(1 + ln x)
x
#b
= lim 1
1
b!1
1 + ln b
= 1.
b
ln b
= 0 by L’Hôpital’s rule.
b!1 b
Here lim
⌅
Example 4.22. Evaluate the Type I improper integral
Z1
1
dx.
2
1 1+x
Z
Z1
1
1
dx
and
dx.
2
1
+
x
1
+
x2
1
0
Z0
Z0
1
1
⇡
⇡
dx = lim
dx = lim (0 tan 1 b) = ( ) = .
2
2
2
2
1
+
x
1
+
x
b!
1
b!
1
1
b
0
Solution. We need to evaluate
Similarly,
Z
1
0
Therefore,
Z
1
⇡
dx = .
2
2
1+x
1
1
⇡ ⇡
dx = + = ⇡.
2
2 2
1 1+x
Definition 4.3. Integrals of functions that become infinite at a point within the interval of
integration are improper integrals of Type II.
1. If f (x) is continuous on (a, b] and is discontinuous at a, then
Z
b
Z
f (x) dx = lim+
c!a
a
b
f (x) dx.
c
2. If f (x) is continuous on [a, b) and is discontinuous at b, then
Z
b
Z
c
f (x) dx = lim
c!b
a
f (x) dx.
a
3. If f (x) is discontinuous at c with a < c < b, then
Z
b
a
Z
c
f (x) dx =
Z
b
f (x) dx +
a
c
87
f (x) dx.
⌅
In each case, if the limit is finite, we say that the improper integral converges and that the limit
is the value of the improper integral. If the limit fails to exist, the improper integral diverges.
Example 4.23. Evaluate the Type II improper integral
Z
1
0
Solution.
Z1
0
Z
1
(x 1)
2
3
dx = lim
c!1
c
0
1
(x 1)
2
3
1
2
(x 1) 3
dx.
h
1 ic
1
dx = lim 3(x 1) 3 = lim 3(c 1) 3 + 3 = 3.
c!1
0
c!1
⌅
Exercise 4.13. Evaluate the Type I improper integral
Z1
tan 1 x
dx.
1 + x2
0
Ans:
⇡2
8 .
Exercise 4.14. Evaluate the Type II improper integral
Z
1
0
Ans:
1
dx.
2 x(1 + x)
p
⇡
4.
Exercise 4.15. Evaluate the Type I improper integral
Z1
1
dx.
x
x
1 e +e
Ans:
⇡
2.
88
Chapter 5
Applications of Integration
Read Thomas’ Calculus, Chapter 6.
5.1 Area Between Curves
Theorem 5.1. Let f and g be continuous on [a, b] with f (x) g(x) for all a  x  b. The area
of the region bounded by the curves y = f (x), y = g(x), and the lines x = a and x = b is given
by
Zb
A=
(f (x) g(x)) dx.
a
In particular, if g(x) = 0, we obtain the following result.
89
Theorem 5.2. Let f be continuous on [a, b] with f (x) 0 for all a  x  b. The area of the
region bounded by the curve y = f (x), the x-axis y = 0, and the lines x = a and x = b is given
by
Zb
A=
f (x) dx.
a
In general we have the following result.
Theorem 5.3. Let f and g be continuous on [a, b] (not necessarily with f (x) g(x) for all
a  x  b). The area of the region bounded by the curves y = f (x), y = g(x), and the lines x = a
and x = b is given by
Zb
A=
|f (x) g(x)| dx.
a
To evaluate the above integral, we split it into two or more integrals, each corresponding to
the region where either f (x) g(x) 0 or f (x) g(x)  0.
In particular, if g(x) = 0, then we obtain the following result.
Theorem 5.4. Let f be continuous on [a, b]. The area of the region bounded by the curve
y = f (x), and the lines x = a and x = b is given by
Z
b
A=
a
|f (x)| dx.
To evaluate the above integral, we split it into two or more integrals, each corresponding to
the region where either f (x) 0 or f (x)  0.
Example 5.1. Find the area of the region bounded the curve y = x(9 x2 ), ( 2  x  2), the x-axis,
the line x = 2 and the line x = 2.
90
Solution. Let f (x) = x(9 x2 ). Notice that f (x)
Z
2
Z
2
0
Area =
=
"
=
2
0 for 0  x  2 and f (x)  0 for 2  x  0.
|x(9 x2 )| dx
2
Z
x(9 x ) dx +
9x2 x4
+
2
4
#0
+
2
2
0
" 2
9x
2
x(9 x2 ) dx
x4
4
#2
0
= 0 ( 18 + 4) + (18 4) 0 = 28.
⌅
Example 5.2. Find the area of the region bounded by the curves y = e 2x
10 ex , (x 0), and
(a) the x-axis,
(b) the y-axis.
7
2
Ans: (a)
12 ln 3 + 10 ln 10 + ln 2,
(b) 12 ln 3 6.
Solution.
1. e 2x
2=0,x=
ln 2
.
2
91
2, (x
0), and y =
2.
e 2x
2 = 10 ex () e2x + ex
() (ex + 4)(ex
12 = 0
3) = 0 , x = ln 3.
3. 10 e x = 0 () x = ln 10.
Z
(a) Green Area =

ln 3
ln 2
2
(e
2x
Z
2) dx +
ln 10
ln 3
(10 e x ) dx
ln 3 
ln 10
1 2x
= e
2x
+ 10x e x
ln 2
2
ln 3
2
9
=(
2 ln 3) (1 ln 2) + (10 ln 10 10) (10 ln 3 3)
2
7
=
12 ln 3 + 10 ln 10 + ln 2.
2
Z
(b) Yellow Area =
=
ln 3
Z 0ln 3
(10 e x ) (e2x
12 ex
0
= 12x e x
2) dx
e 2x dx
1 2x ln 3
e
2
0
9
= (12 ln 3 3
) ( 1
2
= 12 ln 3 6.
1
)
2
Theorem 5.5. Let f and g be continuous on [c, d] with f (y) g(y) for all c  y  d. The area
of the region bounded by the curves x = f (y), x = g(y), and the lines y = c and y = d is given
by
92
Z
d
A=
(f (y) g(y)) dy.
c
In general, we have the following result.
Theorem 5.6. Let f and g be continuous on [c, d] (not necessarily with f (y) g(y) for all
c  y  d). The area of the region bounded by the curves x = f (y), x = g(y), and the lines y = c
and y = d is given by
Zd
A=
|f (y) g(y)| dy.
c
To evaluate the above integral, we split it into two or more integrals, each corresponding to
the region where either f (y) g(y) 0 or f (y) g(y)  0.
Example 5.3. Find the area of the region bounded the curve y = ln(x + 2), y = 2 ln x, the x-axis.
Ans: 4 ln 2 1.
Solution.
93
Solving y = ln(x + 2) and y = 2 ln x gives (x, y) = (2, ln 2). Also y = ln(x + 2) , x = ey
y
y = 2 ln x , x = e 2 .
h y
i2 ln 2
R 2 ln 2 y
Thus the area is A = 0
e 2 ey + 2 dy = 2e 2 ey + 2y
= 4 ln 2 1.
2 and
0
⌅
5.2 Volume of Solid of Revolution by Disk Method
Theorem 5.7. When the plane region bounded by the curve y = f (x) and the lines x = a and
x = b is revolved completely about the x-axis, the volume of the solid formed is
Z
b
V =⇡
f (x)2 dx.
a
The above formula is known as the disk method.
Remark. Informally we can visualize the above formula as follows: Think of the expression
⇡(f (x)2 dx as the volume of a thin vertical circular disk of radius f (x) and thickness dx (so
that its volume is ⇡ · (radius)2 · thickness = ⇡(f (x)2 dx). Also we think of the definite integral
Zb
sign
as the process of ‘taking the limit of a Riemann sum’. Then the formula essentially
a
says that the volume V of solid of revolution is equal to the limit of the Riemann sum (i.e.
Zb
the definite integral
) of the volumes of the thin circular disks (given by ⇡(f (x)2 dx)
a
94
Theorem 5.8. Let f and g be continuous on [a, b] with f (x) g(x) 0 for all a  x  b. When
the region bounded by the curves y = f (x) and y = g(x) for a  x  b is revolved completely
about the x-axis, the volume of the solid formed is
Z
b
V =⇡
f (x)2 dx ⇡
a
Z
b
g(x)2 dx.
a
Theorem 5.9. Let f be continuous on [c, d]. When the plane region bounded by the curve
x = f (y) and the lines y = c and y = d is revolved completely about the y-axis, the volume of
95
the solid formed is
Z
b
V =⇡
f (y)2 dy.
a
Remark. Informally we can visualize the above formula as follows: The volume V of the
Zd
solid of revolution is equal to the limit of a Riemann sum (i.e.
) of volumes of thin
c
horizontal circular disks (given by ⇡ · (f (y))2 · dy).
Theorem 5.10. Let f and g be continuous on [c, d] with f (y) g(y) 0 for all c  y  d.
When the region bounded by the curves x = f (y) and x = g(y) for c  y  d is revolved
completely about the y-axis, the volume of the solid formed is
Z
d
V =⇡
Z
2
f (y) dy
c
d
⇡
g(y)2 dy.
c
Example 5.4. Find the volume of the solid generated by rotating completely the region bounded
by the curve y 2 = 4x, the line y = 2x 4 and the y-axis about the
(a) x-axis,
(b) y-axis.
Ans: (a)
22⇡
3 ,
(b)
16⇡
15 .
Solution. The two curves intersect at (1, 2), (4, 4). The required region is the one shaded in
purple in the figure.
96
(a) (Rotating about the x-axis) Using the disk method, the volume is
Z
V =⇡
1
0
Z
= 4⇡
(2x 4)2 (
4x)2 dx
"
1
0
p
x
2
x3
5x + 4 dx = 4⇡
3
5x2
+ 4x
2
#1
=
0
22⇡
.
3
(b) (Rotating about the y-axis) Using the disk method, the volume is
Z
Z 2
1 22
1
V =⇡
( y ) dy + ⇡
( (y + 4))2 dy
2 4
4 2
" 5 #0
"
# 2
3
y
(y + 4)
2 2
16⇡
=⇡
+⇡
= ⇡( + ) =
.
80 2
12
5 3
15
4
0
⌅
5.3 Cylindrical Shell Method
Consider the solid of revolution obtained by revolving the region R about the y-axis. If we
apply the disk method, it is necessary to express the equation y = f (x) in the form x = g(y).
Quite often, it is difficult or even impossible to do so.
97
The following result, known as the method of cylindrical shell, provides a solution to this
problem.
Theorem 5.11. When the plane region bounded by the curve y = f (x) and the lines x = a and
x = b, where 0  a < b, is revolved completely about the y-axis, the volume of the solid formed
is
Z
b
V = 2⇡
x|f (x)| dx.
a
Remark. Informally we can visualize the above formula as follows: Think of the expression
2⇡x · |f (x)| · dx as the volume of a thin vertical cylindrical shell of base radius x and of
height |f (x)| and of thickness dx (so that its volume is given by (surface area of the cylinder)·
thickness = (2⇡x · |f (x)|) · dx). Then the formula says that the volume V of the solid of
Zb
revolution is equal to the limit of a Riemann sum (i.e.
) of volumes of thin vertical
a
cylindrical shells (given by 2⇡x · |f (x)| · dx).
98
Theorem 5.12. When the plane region bounded by the curve y = f (x), y = g(x) and the lines
x = a and x = b, where 0  a < b, is revolved completely about the y-axis, the volume of the
solid formed is
Zb
V = 2⇡
x|f (x) g(x)| dx.
a
Theorem 5.13. When the plane region bounded by the curve x = f (y) and the lines y = c and
y = d, where 0  c < d, is revolved completely about the x-axis, the volume of the solid formed
99
is
Z
d
V = 2⇡
y|f (y)| dy.
c
Remark. Informally we can visualize the above formula as follows: The volume V of the
Zd
solid of revolution is equal to the limit of a Riemann sum (i.e.
) of volumes of thin
horizontal cylindrical shells (given by 2⇡y · |f (y)| · dy).
c
Theorem 5.14. When the plane region bounded by the curve x = f (y), x = g(y) and the lines
y = c and y = d, where 0  c < d, is revolved completely about the x-axis, the volume of the
solid formed is
Zd
V = 2⇡
y|f (y) g(y)| dy.
c
Example 5.5. The regions bounded by the curve y = 2x x2 (1  x  3), the line x = 1, the line
x = 3 and the x-axis is revolved completely about the y-axis. Calculate the volume of the solid
generated.
Ans: 9⇡.
Solution. We use the shell method. The volume is given by
100
Z
V = 2⇡
Z
3
1
Z
= 2⇡
"
x|f (x)| dx = 2⇡
2
2
1
Z
2
x(2x x ) dx +
1
2x3
= 2⇡
3
x4
4
#2
"
x|2x x2 | dx
3
2
!
2
x( 2x + x ) dx
x4
+ 2⇡
4
1
2x3
3
#3
= 2⇡(
2
11 43
+ ) = 9⇡.
12 12
Example 5.6. Sketch the curve whose equation is y = ln(2x 1) for x > 12 .
The region bounded by this curve, the axes and the line y = ln 3 is rotated completely about the
x-axis. Calculate the volume of the solid generated.
Ans: ⇡(3 ln 3 + 12 (ln 3)2 2).
Solution. We use the shell method. First
1
y = ln(2x 1) , x = (e y + 1).
2
The volume is given by
Z
V = 2⇡
Z
= 2⇡
"
ln 3
0
ln 3
0
= ⇡ ye y
Z
y f (y) dy = 2⇡
ln 3
0
y
1 y
(e + 1) dy
2
Z ln 3
1 y
y (e + 1) dy = ⇡
ye y + y dy
2
0
#ln 3
y2
1
y
e +
= ⇡(3 ln 3 + (ln 3)2 2).
2 0
2
101
⌅
5.4 Arc Length of a curve
Let f be continuous on [a, b]. Using Riemann sums, we can prove:
The length of the curve y = f (x), a  x  b is
Z bq
1 + f 0 (x)2 dx.
a
Remark. To visualize the above expression, we can think of a curve as consisting of many
small slanted line segments, and each slanted line segment is the hypotenuse of a rightangled triangle of base length dx and of height dy. Then
Theorem, the length
qby Pythagoras’ p
p
dy
of a slanted line segment is given by (dx)2 + (dy)2 = 1 + ( dx )2 · dx = 1 + f 0 (x)2 dx. Then
Zb
the length of the curve is the limit of a Riemann sum (i.e.
) of lengths of small slanted
a
p
line segments (given by 1 + f 0 (x)2 dx).
Example 5.7. Calculate the length of the curve
y=
Solution. Given y =
x4 +3
6x
x4 + 3
, 1  x  2.
6x
= 16 (x3 + 3x ), we have
1
3
1
1
y 0 = (3x2
) = (x2
).
2
6
2
x
x2
⇣
⌘2
p
Thus 1 + y 02 = 1 + 14 (x2 x12 )2 = 12 (x2 + x12 ) so that 1 + y 02 = 12 (x2 + x12 ).
h 3
i2
R 2p
R2
The arclength is 1 1 + y 02 dx = 1 12 (x2 + x12 ) dx = 12 x3 1x = 17
12 .
1
⌅
Similarly, if the curve is given as the graph of a function of y, we have the following formula.
102
The length of the curve x = g(y), p  y  q is
Z qq
1 + g 0 (y)2 dy.
p
Exercise 5.1. Calculate the length of the curve
y = 1+
3
Ans: 23 (5 2
✓
3x
2
◆ 23
, 0x
16
.
3
1).
Exercise 5.2. Sketch the curves y = 2 e x 1 and y = 4x2 3 for x 0 on a single diagram. It is
given that the two curves meet at a point where x = 1. Calculate the area of the region bounded by
the two curves and the y-axis.
Ans:
8
3
+ 1e .
Exercise 5.3. Let In =
(a) Show that for n
R1
0
(2x + 1)n e
x dx,
where n is a non-negative integer.
1,
In = (1
3n
) + 2nIn 1 .
e
(b) The region bounded by the curve y = (2x + 1)2 e x , the axes and the line x = 1 is rotated
completely about the y-axis. Use the result in (a) to find the value of the solid generated.
Ans: (b) ⇡(66
172
e ).
Exercise 5.4. Consider the region R bounded by y = 2x2 , the line x = 2 and the x-axis. For
0 < p < 2, the vertical line x = p divides R into two parts R1 and R2 , where R1 denotes the part
on the right of x = p and R2 denotes the part on the left of x = p. Let V1 be the volume of the
solid generated by revolving R1 about the x-axis, and V2 be the volume of the solid generated
by revolving R2 about the y-axis. Find R1 and R2 in terms of p. Find also the value of p that
maximizes the total volume given by V = V1 + V2 .
Ans: p = 1 gives the maximum V .
103
104
Chapter 6
Sequences and Series
Read Thomas’ Calculus, Chapter 10.
6.1 Sequences
An infinite sequence of numbers is an infinite ordered list of numbers
a1 , a2 , a3 , · · · , an , · · ·
For each n, the nth number in the list is called the nth term of the sequence.
We usually denote a sequence by {an }1
n=1 (or simply {an } when the reference to n is clear).
Formally,
Definition 6.1. An infinite sequence of numbers is a function whose domain is the set of
positive integers.
In this formal definition, an is the value of the function evaluated at n.
Example 6.1. The sequence of arithmetic progression is given by {a + (n 1)d}1
n=1 , where a is the
first term of the sequence and d is called the common di↵erence.
The function that defines this sequence is: f (n) = a + (n 1)d.
The sequence of geometric progression is given by {ar n 1 }1
n=1 , where a is the first term of the
sequence and r is known as the common ratio of the geometric progression.
The function that defines this sequence is: f (n) = ar n 1 .
For example, 1, 3, 5, 7, 9, . . ., is an arithmetic progression (with a = 1, d = 2).
1 1 1 1
1
1, , 2 , 3 , 4 , . . ., is a geometric progression (with a = 1, r = ).
2 2 2 2
2
105
n
We now consider the sequence {an }1
n=1 , where an = n+1 , and determine what happens to an
when n is getting large. The following graph shows how the terms approach 1.
n
We say that the sequence { n+1
} approaches 1 as n increases and write
n
lim
n!1 n + 1
= 1.
We now state formally the meaning of a limit of a sequence.
Let {an }1
n=1 be a sequence of real numbers.
Then lim an is the value an approaches as n tends to positive infinity.
n!1
If lim an exists as a real (finite) number L, then we say that the sequence {an } converges
n!1
(or more detailedly, {an } converges to L). Sometimes we simply write an ! L.
We say that the sequence {an } diverges if lim an does not exist as a real (finite) number.
n!1
Example 6.2. (i) As we have seen, we have lim
n
n!1 n + 1
Thus the sequence
⇢
= 1.
n
n
converges (to 1), and we also write
! 1.
n+1
n+1
(ii) Note that the sequence
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, · · ·
does not converge to any real number. Thus the sequence {( 1)n } diverges.
(iii) Clearly we have lim 2n = 1. But 1 is not a real number, so the sequence {2n} diverges.
n!1
106
6.2 Finding the Limit of a Sequence
This following theorem gives a shortcut to evaluate the limit of some sequences using the
limit of functions.
Theorem 6.1. Let f (x) be a function, and {an } be a sequence such that f (n) = an for all n. If
lim f (x) = L, then lim an = L.
x!1
n!1
Example 6.3. Find the following limits:
(i)
1
.
n!1 n
lim
(ii)
ln n
.
n!1 n
lim
1
1
Solution. (i) Consider the function f (x) = . Note that = f (n) for all n.
x
n
Since lim
1
x!1 x
= 0, it follows from Theorem 6.1 that lim
1
n!1 n
= 0.
(ii) By L’Hôpital’s rule, we have
lim
x!1
Thus we also have lim
n!1
ln x
1/x
1
= lim
= lim = 0.
x!1 1
x!1 x
x
ln n
= 0.
n
⌅
6.3 Limit Laws for Sequences
If {an } and {bn } are convergent sequences and c is a constant, then we have
• lim can = c lim an .
n!1
n!1
• lim (an ± bn ) = lim an ± lim bn .
n!1
n!1
n!1
• lim an bn = lim an · lim bn .
n!1
n!1
n!1
lim an
an n!1
=
, if lim bn , 0.
n!1 bn
n!1
lim bn
• lim
n!1
107
Example 6.4. From the previous example, we know that
lim
1
n!1 n
=0
and
lim
n!1
ln n
= 0.
n
It follows that we have
2n2 + 3 ln n
3 ln n
= lim (2 + 2 )
2
n!1
n!1
n
n
ln n 1
= 2 + 3 lim
·
n!1 n
n
= 2 + 3 · 0 · 0 = 0.
lim
Theorem 6.2. (Squeeze Theorem for Sequence) If an  bn  cn for all n and lim an = lim cn = L,
n!1
then lim bn = L.
n!1
n!1
Example 6.5. Show that if lim |an | = 0, then lim an = 0.
n!1
n!1
Solution. Note that |an |  an  |an | for all n,
and it is given that lim |an | = 0, so that we also have lim |an | =
n!1
n!1
follows from Squeeze Theorem that lim an = 0.
lim |an | = 0 = 0. Thus it
n!1
n!1
⌅
Example 6.6. Show that lim
n!
= 0.
n!1 nn
Solution. Since 0 
n!
nn
=
n
n
⇥
n 1
n
follows from Squeeze Theorem.
⇥
n 2
n
⇥ ··· ⇥
2
n
⇥
1
n

1
n
1
= 0, the result
n!1 n
for all n and lim
⌅
6.4 Series
An expression of the form
1
X
n=1
an = a1 + a2 + a3 + a4 + · · ·
is called an infinite series or simply a series. To compute the value (called the sum) of this
infinite series, we construct a new sequence {Sn } defined by
Sn =
n
X
i=1
ai = a1 + a2 + · · · + an .
108
{Sn } is called the sequence of partial sums of the given series. Then the sum of the infinite
series is defined as the limit of the sequence {Sn }. In other words, we have
1
X
an = lim Sn .
n!1
n=1
We say that the series
1
X
n=1
an converges if the sequence {Sn } converges (which means that
1
X
lim Sn exists as a real number). Thus the series
n!1
an is convergent means that it has a
n=1
finite sum.
We say that the series
1
X
n=1
an diverges if the sequence {Sn } diverges (which means that lim Sn
n!1
does not exist as a real number). Thus the series
1
X
an is divergent means that it does not
n=1
have a finite sum.
Example 6.7. The constant sequence {1} is a convergent sequence, since lim 1 = 1, which is a
n!1
real number.
Now consider the series
1
X
n=1
1 = 1 + 1 + 1 + 1 + ···.
For each n, the nth partial sum is given by Sn =
n
X
1 = n. Thus, we have
1 = lim Sn =
n=1
i=1
lim n = 1.
1
X
n!1
n!1
But 1 is not a real number, and thus the series
Example 6.8.
1
X
1 is divergent.
n=1
(a) An important example of an infinite series is the geometric series
1
X
ar n 1 , (a , 0).
n=1
The geometric series is convergent with its sum equal to
when |r|
1.
(b) Is the series
1
X
22n 31
n
convergent or divergent?
n=1
109
a
1 r
when |r| < 1, and it is divergent
(c) Find the sum of the series
1
X
n=1
xn , for |x| < 1.
Solution. (a) Note that for each n,
8
>
a(1 r n )
>
>
<
Sn =
ar i 1 = >
1 r
>
>
:an
i=1
n
X
if r , 1,
if r = 1.
8
>
>
if a > 0,
<1
Note that a , 0. Then lim an = >
>
n!1
: 1 if a < 0.
8
>
1
if r > 1,
>
>
>
<
n
lim r = >
0
if 1 < r < 1,
>
n!1
>
>
:does not exist if r  1.
Then it follows readily that
8
>
1
if a > 0 and r 1,
>
>
>
>
1
>
X
if a < 0 and r 1,
>
< 1
ar n 1 = lim Sn = >
a
>
n!1
>
if 1 < r < 1,
>
>
n=1
>
>1 r
:
does not exist if r  1.
In summary,
1
X
ar
n 1
n=1
divergent when |r|
(b)
1
X
22n 31
n
n=1
(c)
1
X
n=1
=3
x =
1
X
n=1
1 r
) when |r| < 1, and
1.
1 ✓ ◆n
X
4
n=1
n
is convergent (with its sum =
a
x · xn
1
3
is a geometric series with r =
1
X
ar n
1
is
n=1
4
> 1, and thus it is divergent.
3
is a geometric series with a = x and r = x. When |x| < 1, it follows from
(a) that the series is convergent and
1
X
n=1
xn =
x
1 x
.
⌅
Example 6.9. Show that the series
1
X
n=1
1
is convergent, and find its sum.
n(n + 1)
110
Solution. For each n,
Sn =
n
X
i=1
1
=(
1
=1
1
X
Therefore,
1
X
n=1
n=1
n ✓
X
1
1
=
i(i + 1)
i
1
1
)+(
2
2
1
.
1+n
⇣
1
= lim 1
n!1
n(n + 1)
1
i +1
i=1
1
1
)+(
3
3
◆
1
1
) + ··· + (
4
n
1
)
n+1
1 ⌘
= 1, which is a real number. Thus the series
1+n
1
converges.
n(n + 1)
⌅
Theorem 6.3. If
1
X
an and
n=1
constant) and
(a)
(b)
1
X
can = c
n=1
1
X
1
X
bn are convergent series, so are the series
n=1
an , and
n=1
(an + bn ) =
n=1
1
X
an +
n=1
1
X
bn .
n=1
Example 6.10. From Example 6.9, we know that
1
X
n=1
1
. It follows that
n(n + 1)
1
1
1
2
X
X
X
2
4
2
1
3
( n+
)=
+
4
=
3
n(n + 1)
3n
n(n + 1) 1
n=1
n=1
n=1
(Note that
can (where c is a
n=1
1
X
(an + bn ). Moreover,
n=1
1
X
1
X
1
3
+ 4 · 1 = 1 + 4 = 5.
1
X
2
2
1
is a geometric series with a = and r = .)
n
3
3
3
n=1
⌅
Lemma 6.4. If the series
1
X
an is convergent, then lim an = 0.
n!1
n=1
111
Proof. Suppose
all n
2. Then
1
X
n=1
an = L. Let Sn = a1 + · · · + an . Then lim Sn = L. Note that an = Sn Sn
n!1
1
for
lim an = lim (Sn Sn 1 ) = L L = 0.
n!1
n!1
⌅
Theorem 6.5. (The nth Term Test for Divergence)
If lim an does not exist or if lim an , 0, then the series
n!1
n!1
1
X
an is divergent.
n=1
This is also known simply as the nth term test.
Example 6.11. Is the series
1
X
n=1
n2
convergent or divergent?
7n2 + 3
n2
1
1
1
= lim
=
= , 0. Thus by the nth term test, the series
n!1 7n2 + 3
n!1 7 + 3
7+0 7
2
Solution. lim
1
X
n=1
n
n2
is divergent.
7n2 + 3
⌅
Warning. The nth term test is inconclusive if lim an = 0.
n!1
1
X
1
1
X
1
. Note that both series satisfy the condin
n2
n=1
n=1
1
1
X
X
1
1
tion lim an = 0. But we will later show that
is divergent, while
is convergent.)
n!1
n
n2
(To see this, consider the two series
and
n=1
n=1
⌅
For series of nonnegative terms (i.e., each an
Theorem 6.6. A series
1
X
0), we have the following fundamental result:
an of nonnegative terms converges if and only if its partial sums
n=1
are bounded from above (i.e., there exists a constant K such that Sn < K for all n.)
Remark. Basically, this theorem means that if each an
0, then the sum of the series
1
X
n=1
is either a finite number or 1.
112
an
1
X
1
Example 6.12. The series
is convergent.
n2
n=1
Solution. Let Sn =
n
X
1
. Then for each n,
k2
k=1
1
1
1
1
+ 2 + ··· +
+ 2
2
2
2
3
(n 1)
n
1
1
1
1
 1+
+
+ ··· +
+
1⇥2 2⇥3
(n 2) ⇥ (n 1) (n 1) ⇥ n
1
1 1
1
1
1
1
= 1 + (1
)+(
) + ··· + (
)+(
)
2
2 3
n 2 n 1
n 1 n
1
=2
< 2.
n
Sn = 1 +
Thus the sequence of partial sums {Sn } is bounded above (by 2). Therefore by the above
1
X
1
theorem,
converges.
⌅
n2
n=1
In general, it is difficult to determine if a given series in convergent. In the next few sections,
we will discuss some methods that will enable us to test the convergence of certain series.
6.5 Integral Test
Example 6.13. The harmonic series
1
X
1
n=1
n
is divergent.
Solution. Observe that
✓
◆ ✓
◆ ✓
◆
1
1 1
1 1 1 1
1 1
1
1+ + + + + + + + +
+ ··· +
+···
2
3 4
5 6 7 8
9 10
16
| {z } |
{z
} |
{z
}
2 1
8 1
4 1
=
=
=
4 2
16 2
8 2
From this bracketing of the terms, one can see that the partial sums {Sn } satisfy
S2
1
, S4
2
1
2 · , S8
2
1
3 · , S16
2
and more generally,
S 2k
k
2
for all k.
113
1
4 · ,··· ,
2
So the sequence of partial sums is not bounded from above. By the above theorem, the
harmonic series diverges.
⌅
For n = 1, 2, . . ., the rectangle with height n1 erected on the unit interval [n, n + 1] has area n1 .
The harmonic series may be viewed as the sum of the areas of these rectangles. Graphically
we see that the union of all these rectangles contains the green region R bounded by the
graph of y = 1x andR the x-axis for x 1. However, the region R has an infinite area as the
1
improper integral 1 1x dx diverges. This implies the harmonic series diverges.
Theorem 6.7. (Integral Test) Let {an } be a sequence of positive terms. Suppose that an = f (n),
1
X
where f is a continuous, positive, decreasing function of x for all x 1. Then the series
an
n=1
Z1
is convergent if and only if the improper integral
f (x) dx is convergent. In other words:
1
Z
(i) If
1
1
Z
(ii) If
f (x) dx is convergent, then
1
1
f (x) dx is divergent, then
1
X
an is convergent.
n=1
1
X
an is divergent.
n=1
Example 6.14. Test the convergence of the series
1
X
n=1
Z
1
1
1
n2 + 1
.
⇡ ⇡ ⇡
= . Thus the im2 4 4
b!1
b!1
1
Z1
1
X
1
1
proper integral
dx
converges.
By
the
integral
test,
converges. (But note
2+1
2+1
x
n
1
n=1
⇡
that is not the sum of the series.)
4
⌅
Solution.
x2 + 1
dx = lim [tan
1
x]b1 = lim (tan
114
1
b tan
1
1) =
Theorem 6.8. (The p-series) The p-series
1
X
1
is convergent if and only if p > 1.
np
n=1
" p+1 #b
✓
1
x
1
1
Proof. Let p > 1.
dx
=
lim
=
lim
p 1
p
p + 1 1 b!1 1 p b
b!1
1 x
converges by the integral test.
Z
1
◆
1 =
1
p 1
. Thus the series
If p = 1, then the series is the harmonic series which is divergent.
Z1
⌘
1
1 ⇣ 1 p
Let 0 < p < 1. Then 1 p > 0.
dx
=
lim
b
1
= 1. Thus the series diverges
p
b!1 1 p
1 x
by the integral test.
Let p  0. Then lim
1
n!1 np
, 0. By Theorem 6.5 (the nth term test), the series diverges.
⌅
Example 6.15. The series
The series
1
X
1
1
p is divergent, since it is a p-series with p =  1.
2
n
n=1
1
X
1
3
p is convergent, since it is a p-series with p = > 1.
2
n n
n=1
6.6 The Comparison Test
Theorem 6.9. (Comparison Test)
Suppose
1
X
n=1
(i) If
1
X
(ii) If
an and
1
X
n=1
bn are series with nonnegative terms such that 0  an  bn for all n.
bn is convergent, then
n=1
1
X
n=1
an is divergent, then
1
X
an is convergent.
n=1
1
X
bn is divergent.
n=1
Remark. Roughly speaking, (i) means that if the bigger series has a finite sum, then the
smaller series also has a finite sum.
Also, (ii) means that if the sum of the smaller series is 1, then the sum of the bigger series
is also 1.
115
Remark. When applying the Comparison Test, we often compare a given series with an
appropriate p-series or geometric series.
Example 6.16. Determine the convergence of the series
1
X
n=1
7
2n2 + 4n + 3
.
1
1
X
X
7
1
=
7
converges
n2
n2
n=1
n=1
1
X
7
as it is a p-series with p = 2 > 1. Thus by the comparison test,
converges.
2n2 + 4n + 3
Solution. Note that 0 
7
7

for all n
2n2 + 4n + 3 n2
1. We know
n=1
⌅
Example 6.17. Determine the convergence of the series
1
X
n=1
Solution. Note that for all n
2n 1
1
2n
1
.
1,
1 n
· 2 = 2n
2
1
=) 0 
1
1

.
2n 1 2n 1
1
X
1
1
is convergent, since it is a geometric series with |r| = < 1. Thus by the
n
1
2
2
n=1
1
X 1
comparison test,
is convergent.
2n 1
The series
n=1
⌅
6.7 The Ratio Test and Root Test
Theorem 6.10. (The Ratio Test) Suppose
(ii) If L > 1, then
1
X
an is a series such that lim
n!1
n=1
finite number or 1).
(i) If 0  L < 1, then
1
X
1
X
an is absolutely convergent. That is
n=1
1
X
n=1
an is divergent.
n=1
(iii) If L = 1, then the ratio test is inconclusive.
116
an+1
= L (L is a
an
|an | is convergent.
Proof. (Optional Reading Exercise) See Section 6.13.
Remark. Theorem 6.13 (later) will tell us:
1
X
n=1
Thus in (i) (when 0  L < 1), the series
Remark. For a geometric series
1
X
n=1
L = lim
n!1
1
X
⌅
|an | is convergent =)
1
X
an is convergent .
n=1
an itself is also convergent.
n=1
an =
1
X
ar n 1 , one has
n=1
an+1
ar n
= lim
= lim |r| = |r|.
n!1 ar n 1
n!1
an
Thus roughly speaking, the Ratio Test says that if a series resembles a geometric series, then
its convergence/divergence property also resembles that of the geometric series (in the cases
of (i) and (ii)).
Remark. (iii) can be illustrated by the convergent series
1
X
1
n=1
n
1
X
1
and the divergent series
n2
n=1
, as both series have L = 1.
Example 6.18. Test the series
1
X
n3
( 1)n n
3
n=1
for absolute convergence.
Solution.
lim
n!1
(n+1)3
3n+1
3
n
1)n 3n
( 1)n+1
(
✓
◆
✓
◆
1 n+1 3
1
1 3 1
= lim
1+
= < 1.
n!1 3
n!1 3
n
n
3
= lim
1
X
n3
Thus by the ratio test,
( 1)n n converges absolutely.
3
n=1
⌅
Example 6.19. Test the series
1
X
n!
n=1
nn
for absolute convergence.
117
Solution.
(n + 1)!
(n + 1)n+1
nn
1
L = lim
= lim
=
.
n
1 n
n!
n!1
n!1 (n + 1)
lim
(1
+
)
n!1
n
nn
By L’Hôpital’s rule,
1
1
lim
lim ln(1 + )x
lim x ln(1 + )
x!1
1 x
x!1
x!1
x
x
lim (1 + ) = e
=e
=e
x!1
x
1
· ( x12 )
(1+ 1x )
1
lim
lim
1
x!1
1
x!1 1 + 1
x2
x = e 1+0 = e.
=e
=e
ln(1 + 1x )
1
x
1
1
Thus we have lim (1 + )n = e, and it follows that L = < 1.
n!1
n
e
Hence by the Ratio Test, the series
1
X
n!
n=1
nn
is absolutely convergent.
⌅
Exercise 6.1. Test for convergence of the following series.
(a)
1
X
2n + 5
n=1
(b)
3n
1
X
(2n)!
n=1
(c)
n!n!
1
X
4n (n!)2
n=1
(2n)!
Ans. (a) convergent, (b) divergent, (c) divergent.
Our next test resembles the Ratio Test closely:
Theorem 6.11. (The Root Test) Suppose
(ii) If L > 1, then
1
X
an is a series such that lim
n=1
finite number or 1).
(i) If 0  L < 1, then
1
X
1
X
an is absolutely convergent.
n=1
an is divergent.
n=1
(iii) If L = 1, then the root test is inconclusive.
118
n!1
p
n
|an | = L (L is a
1
X
Remark. For a geometric series
an =
n=1
L = lim
n!1
1
X
ar n 1 , one has
n=1
p
p
1
n 1
n
n
|an | = lim |ar n 1 || = lim |a| n |r| n = |a|0 · |r|1 = 1 · |r| = |r|.
n!1
n!1
As such, the conclusions in (i) and (ii) in the Root Test are consistent with what we know
about convergence/divergence property of a geometric series.
1
X
1
Remark. Again (iii) can be illustrated by the convergent series
and the divergent
n2
n=1
1
X
1
series
, as both series have L = 1.
n
n=1
Example 6.20. Test the series
1 ✓
◆
X
2n + 3 n
n=1
3n + 2
for convergence.
Solution.
lim
r✓
n
n!1
By the root test,
1 ✓
◆
X
2n + 3 n
n=1
3n + 2
2n + 3
3n + 2
◆n
= lim
2n + 3
n!1 3n + 2
=
2
< 1.
3
is convergent.
⌅
Example 6.21. Test the series
1
X
nn
33n+1
n=1
for convergence.
r
Solution. lim
n!1
n
nn
33n+1
= lim
n!1
n
1
33+ n
= 1. By the root test,
1
X
nn
is divergent.
33n+1
n=1
⌅
Exercise 6.2. Test for convergence of the following series.
(a)
1
X
n2
n=1
2n
(b)
1 ✓
◆
X
1 n
(c)
1+n
1
X
2n
n=1
n3
n=1
Ans. (a) convergent, (b) divergent, (c) convergent.
119
6.8 Alternating Series
An alternating series is a series whose terms are alternatively positive and negative.
Example 6.22. An example of an alternating series is given by
1
X
( 1)n
11
n
n=1
=
1
1
1 1
+
2 3
1 1
+
4 5
1
+ ··· ,
6
which is known as the alternating harmonic series. The name reflects the fact that
1
X
1
n=1
n
1
X
n=1
( 1)n
11
n
=
is the harmonic series.
Remark. The harmonic series
ing harmonic series
1
X
( 1)n
n=1
1
X
1
n=1
11
n
n
is divergent. Nonetheless, it turns out that the alternat-
is convergent.
This can be seen as follows:
1
X
( 1)n
n=1
11
n
=(
1
1
1
1
)+(
2
3
1
1
)+(
4
5
1
) + ···
6
1
1
1
+
+
+ ···
1·2 3·4 5·6
1
X
1
=
.
(2n 1) · 2n
=
n=1
For each n
1, we have
2n 1
n =) 0 
1
1
 2.
(2n 1) · 2n 2n
1
1
X
1
1 X 1
The series
= ·
is convergent, since it is a p-series with p = 2 > 1. Thus
2
2n2
n2
n=1
n=1
1
1
X
X
1
1
by the Comparison Test,
is convergent. Thus the series
( 1)n 1 is also
(2n 1) · 2n
n
n=1
n=1
convergent.
The above conclusion also follows from the following result:
120
Theorem 6.12. (The Alternating Series Test) If bn is a sequence of positive numbers such that
(i) bn is decreasing (that is, bn
bn+1 for all n), and
(ii) lim bn = 0,
n!1
then the alternating series
1
X
( 1)n 1 bn = b1 b2 + b3 b4 + · · ·
n=1
is convergent.
Example 6.23. Use the Alternating Series Test to show that the series
1
X
( 1)n
n
1
is convergent.
n=1
1
X
( 1)n 1
1
Proof.
is an alternating series with each bn = > 0. For each n 1, we have
n
n
n=0
⇢
1
1
1
1
1
=
> 0, and thus the sequence
is decreasing. Also we have lim = 0.
n!1 n
n n + 1 n(n + 1)
n
1
X
n
1
( 1)
Thus by the alternating series test, the series
is convergent.
n
n=0
Example 6.24. (i) Show that the series
1
X
( 1)n 1 n2
is convergent.
n3 + 1
n=1
(ii) Deduce that the series
1
X
( 1)n n2
is convergent.
n3 + 1
n=1
1
X
( 1)n 1 n2
n2
is
an
alternating
series
with
each
b
=
> 0. For each n
n
n3 + 1
n3 + 1
n=1
we have, by direct calculation,
Solution. (i)
n2
n3 + 1
Thus the sequence
n
n2
n3 +1
o
(n + 1)2
n2 ((n + 1)3 + 1) (n + 1)2 (n3 + 1)
=
(n + 1)3 + 1
(n3 + 1)((n + 1)3 + 1)
4
n + 2n3 + n2 2n 1
= 3
(n + 1)((n + 1)3 + 1)
n4 + 2n2 (n 1) + (n2 1)
=
0.
(n3 + 1)((n + 1)3 + 1)
is decreasing.
121
1,
1
n2
n
=
lim
n!1 n3 + 1
n!1 1 +
Also, lim
1
n3
=
0
= 0.
1+0
Thus by the alternating series test,
1
X
( 1)n 1 n2
is convergent.
n3 + 1
n=1
(ii) From (i), we know that
1
X
( 1)n 1 n2
is convergent.
n3 + 1
n=1
Thus by Theorem 6.3,
1
1
X
X
( 1)n n2
( 1)n 1 n2
=
1
·
is also convergent.
n3 + 1
n3 + 1
n=1
n=1
⌅
6.9 Absolute Convergence
Given a series
1
X
an , we can construct a new series
n=1
n=1
values of the terms of the original series.
Theorem 6.13. If
1
X
n=1
1
X
|an | is convergent, then
1
X
|an |, whose terms are the absolute
an is convergent.
n=1
Proof. Note that 0  (an + |an |)  2|an | for all n. Since
1
X
ison test,
(an + |an |) converges too. Therefore,
1
X
n=1
2|an | converges, we have by compar-
n=1
1
X
n=1
1
1
X
X
an =
(an + |an |)
|an |
n=1
n=1
converges.
⌅
Definition 6.2. A series
1
X
an is said to be absolutely convergent if
n=1
The series
1
X
1
X
n=1
|an | is convergent.
an is said to be conditionally convergent if it is convergent but not absolutely
n=1
122
convergent.
Theorem 6.13 states that every absolutely convergent series is convergent.
1
X
( 1)n
Example 6.25. Show that the series
n2
1
is absolutely convergent.
n=1
Solution. The series
1
X
( 1)n
n2
1
=
n=0
Therefore,
1
X
( 1)n
n2
1
1
X
1
is convergent, since it is a p-series with p = 2 > 1.
n2
n=0
converges absolutely.
n=0
1
X
( 1)n
Example 6.26. Show that the series
n
1
is conditionally convergent.
n=1
Solution. From Example 6.23, we know, by the alternating series test, that
1
=
n=0
Therefore,
1
is
n=0
convergent.
1
X
( 1)n
Also,
n
1
X
( 1)n
n
1
X
( 1)n
n
1
X
1
n=0
1
n
is the harmonic series which is divergent.
converges conditionally.
n=0
⌅
Example 6.27. Show that the series
1
X
sin n
n=1
n2
is absolutely convergent.
6.10 Power Series
A power series is a series of the form
1
X
n=0
cn x n = c0 + c1 x + c2 x 2 + c3 x 3 + · · · ,
where x is a variable, and the cn ’s are constants called coefficients of the series. For each fixed
x, the power series is a series of numbers that we can test for convergence or divergence.
123
More generally, a series of the form
1
X
n=0
cn (x a)n = c0 + c1 (x a) + c2 (x a)2 + c3 (x a)3 + · · · ,
is called a power series centred at a or a power series about a.
Note that the power series
1
X
cn (x a)n always converges at x = a.
n=0
Example 6.28. For what values of x is the series
1
X
n!xn convergent?
n=0
Solution. If x , 0, then lim
n!1
Therefore,
1
X
1
X
(n + 1)!xn+1
=
lim
(n
+
1)|x|
=
1.
By
ratio
test,
n!xn diverges.
n!1
n!xn
n=0
n!xn converges if and only if x = 0.
n=0
⌅
Example 6.29. For what values of x is the series
1
X
(x 7)n
convergent?
n
n=0
Solution.
(x 7)n+1
n+1
lim
n!1 (x 7)n
n
n
|x 7| = |x 7|. By ratio test, we have the following conn!1 n + 1
= lim
clusions (i) and (ii).
1
X
(x 7)n
(i) If |x 7| < 1, then
is absolutely convergent.
n
n=0
1
X
(x 7)n
(ii) If |x 7| > 1, then
is divergent.
n
n=0
1
X
( 1)n
(iii) If x = 6, the series becomes
, which is an alternating series. By the alternating
n
n=0
series test, it is convergent.
1
X
1
( iv) If x = 8, the series becomes
, which is the harmonic series which is divergent.
n
n=0
Summarizing, the series
1
X
n=0
(x 7)n
is convergent if and only if x 2 [6, 8).
n
⌅
124
Theorem 6.14. For a given power series
1
X
cn (x a)n , exactly one of the following possibili-
n=0
ties holds:
(i) The series converges at x = a only.
(ii) The series converges for all x.
(iii) There is a positive number R such that the series converges absolutely if |x a| < R and
diverges if |x a| > R.
The number R in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is R = 0 in case (i) and R = 1 in case (ii). The interval of
convergence of a power series is the interval consisting of all values of x for which the series
converges.
The interval of convergence can be (a R, a + R), [a R, a + R), (a R, a + R], [a R, a + R].
In some cases, we can compute R by the following method.
Theorem 6.15. Consider the power series
or lim
p
n
n!1
1
X
cn (x a)n , where cn , 0 for all n. If lim
n=0
n!1
cn+1
=L
cn
|cn | = L, where L is a real number or 1, then R = L1 .
By convention, if L = 0, then R = 1, and if L = 1, then R = 0.
Proof. Suppose lim
n!1
Thus lim
n!1
cn+1
= L.
cn
cn+1 (x a)n+1
c
= lim n+1 |x a| = L|x a|.
n!1 cn
cn (x a)n
By ratio test, the series converges absolutely for L|x a| < 1, that is |x a| < L1 ; and the series
diverges for L|x a| > 1, that is |x a| > L1 . Therefore the radius of convergence is L1 .
p
The second case where lim n |cn | = L follows similarly by root test.
n!1
⌅
125
Example 6.30. Find the radius of convergence and interval of convergence of the power series
1
X
( 3)n xn
.
p
n
+
1
n=0
Solution.
( 3)n+1
p
n+2
lim ( 3)n
n!1 p
n+1
r
= lim 3
n!1
n+1
= 3. Thus the radius of convergence is 13 .
n+2
1
X
( 1)n
When x = the series becomes
, which is an alternating series. By the alternating
p
n=0 n + 1
series test, it is convergent.
1
X
1
When x = 13 , the series becomes
, which is divergent by integral test.
p
n
+
1
n=0
1
3,
Consequently, the interval of convergence is (
1 1
3 , 3 ].
⌅
Example 6.31. Find the radius of convergence and interval of convergence of the power series
1
X
n(x + 2)n
.
3n+1
n=0
n+1
3n+2
n
n!1 n+1
3
n+1 1
= . Thus the radius of convergence is 3. The centre of the
n!1 3n
3
power series is at x = 2. Next we consider the two endpoints x = 2 ± 3, that is, x = 5, 1.
1
1
X
X
n( 3)n
n( 1)n
When x = 5, the series becomes
,
that
is
, which is divergent by the nth
3
3n+1
n=0
n=0
term test .
1
1
X
X
n(3)n
n
When x = 1, the series becomes
, that is
, which is also divergent by the nth
n+1
3
3
n=0
n=0
term test .
Solution. lim
= lim
Consequently, the interval of convergence is ( 5, 1).
⌅
Example 6.32. Find the radius of convergence and interval of convergence of the power series
1
X
(4x 3)2n+1
.
4n n4/3
n=1
Solution. Note that we cannot apply Theorem 6.15 directly since all the coefficients of the
even powers of x are zero. Let un =
(4x 3)2n+1
.
4n n4/3
126
(4x 3)2n+3
un+1
4n n4/3
= lim n+1
·
n!1 un
n!1 4
(n + 1)4/3 (4x 3)2n+1
✓
◆4
n 3 |4x 3|2 |4x 3|2
= lim
=
.
n!1 n + 1
4
4
lim
By ratio test, the power series converges absolutely for
and diverges for |x 34 | > 12 .
|4x 3|2
4
<1,
|4x 3|
2
< 1 , |x
3
4|
< 12 ,
1
Therefore, the radius of convergence is .
2
1
1
X 22n+1
X 2
At x = 54 , the series is
=
, which is convergent since it is a p-series with
n n4/3
4/3
4
n
n=1
n=1
p=
4
3
> 1.
At x = 14 , the series is
p=
4
3
> 1.
1
1
X
( 2)2n+1 X 2
=
, which is convergent since it is a p-series with
4n n4/3
n4/3
n=1
n=1
Therefore, the interval of convergence is [ 14 , 54 ].
⌅
6.11 Power Series Representation
Recall that for |x| < 1, the geometric series
1
X
n=0
1
X
xn = 1 + x + x2 + x3 + · · · + xn + · · · =
xn is called a power series representation of the function
1
.
1 x
1
1 x
about x = 0.
n=0
Example 6.33. (i) Find a power series representation of
(ii) Find a power series representation of
1
about x = 0.
1+x
x3
about x = 0.
x+2
Solution. (i) We make use of the above geometric series.
1
X
1
1
=
=
( x)n = 1 x + x2 x3 + x4 + · · · ,
1 + x 1 ( x)
n=0
127
which is valid for | x| < 1. That is,
1
X
1
=
( 1)n xn
1+x
for |x| < 1.
n=0
(ii) Using the power series representation in (i), we get
1
⇣ ⌘
x3
x3
1
x3 X
n x n
=
·
=
·
(
1)
x+2
2 1 + 2x
2
2
n=0
That is,
1
X ( 1)n
x3
=
xn+3
x+2
2n+1
n=0
x
for | | < 1.
2
for |x| < 2.
⌅
Example 6.34. Find a power series representation of
1
about x = 0.
x2 + 3x + 2
Solution.
1
1
=
2
x
+
1
x + 3x + 2
1
1
=
x+2 1+x
which is valid when both |x| < 1 and
|x| < 1 and
Thus
1
X
1 1
( 1)n xn
x =
21+ 2
n=0
1
1 X
x
·
( 1)n ( )n ,
2
2
n=0
x
< 1. Note that
2
x
< 1 () |x| < 1 and |x| < 2 () |x| < 1.
2
1
X
1
=
( 1)n (1
x2 + 3x + 2
n=0
1
)xn
2n+1
for |x| < 1.
⌅
Theorem 6.16. If the power series
function f defined by
1
X
cn (x a)n has radius of convergence R > 0, then the
n=0
f (x) =
1
X
cn (x a)n
n=0
is di↵erentiable on the interval |x a| < R and
(i) f 0 (x) =
1
X
n=1
ncn (x a)n 1 , for |x a| < R.
128
Z
(ii)
f (x) dx =
1
X
cn
n=0
(x a)n+1
+ C, for |x a| < R.
n+1
Example 6.35. Find a power series representation of ln(1 x) and its radius of convergence.
Solution. For |x| < 1,
1
1 x
=
1
X
xn . Thus by Theorem 6.16 (ii),
n=0
Z
1
1 x
=)
dx =
ln(1 x) =
1
X
xn+1
n=0
1
X
n=0
n+1
+C
xn+1
+ C,
n+1
for |x| < 1.
When x = 0, we have 0 = 0 + C so that C = 0. Therefore,
1
X
xn+1
ln(1 x) =
n=0
n+1
=
1
X
n=0
1 n+1
x ,
n+1
for |x| < 1.
The radius of convergence is 1 by ratio test.
⌅
Example 6.36. Find a power series representation of tan
Solution. For |x| < 1,
1
1
n=0
n=0
1 x.
X
X
1
2 n
=
(
x
)
=
( 1)n x2n . Thus by Theorem 6.16 (ii),
1 + x2
Z
1
X
2n+1
1
n x
dx
=
(
1)
+C
2n + 1
1 + x2
n=0
1
X
x2n+1
=) tan 1 x =
( 1)n
+ C,
2n + 1
n=0
for |x| < 1.
When x = 0, we have 0 = 0 + C so that C = 0. Therefore,
tan
1
x=
1
X
x2n+1
( 1)n
,
2n + 1
n=0
for |x| < 1.
⌅
6.12 Taylor and Maclaurin Series
By repeated use of Theorem 6.16 (i), we deduce the following.
129
Theorem 6.17. If f has a power series representation at x = a, that is
f (x) =
1
X
n=0
cn (x a)n , |x a| < R, for some R > 0,
then its coefficients are given by the formula
cn =
f (n) (a)
.
n!
If f has a power series representation at x = a, then it is unique and has the form
f (x) =
1
X
f (n) (a)
n!
n=0
(x a)n .
This is called the Taylor series of f at x = a.
The Maclaurin series of f is the special case of Taylor series when a = 0:
f (x) =
1
X
f (n) (0)
n=0
n!
xn .
Exercise 6.3. Assume that each of the functions ex , sin x and cos x has a Maclaurin series representation. Find the Maclaurin series and its radius of convergence for (a) ex , (b) sin x, (c) cos x.
x
Ans: (a) e =
1 n
X
x
n=0
1
1
X
X
2n+1
x2n
n x
, (b) sin x =
( 1)
, (c) cos x =
( 1)n
.
n!
(2n + 1)!
(2n)!
n=0
n=0
All three Maclaurin series converge for all x in R.
d n ex
Solution. (a) Let’s work out the Maclaurin series of e x . Since
= e x for all n
n
dx
d n ex
(0) = 1 for all n 0.
dxn
0. Thus
Therefore, the Maclaurin series of ex is
1+
x x2
xn
+
+ ··· +
+ ··· .
1! 2!
n!
1
(n + 1)!
1
Since lim
= lim
= 0, the radius of convergence is 1 and so the interval of
1
n!1
n!1 n + 1
n!
convergence is R.
130
Assuming
ex
x
has a Maclaurin series representation, we thus have e =
1 n
X
x
n=0
n!
for all x 2 R.
(b), (c): Exercise.
⌅
6.13 Appendix: The Precise Definition of the Limit of a Sequence and some Proofs
Remark: Section 6.13 will be excluded from the assessments (quizzes and the Final Exam).
Remark: The precise definition of the limit of a sequence and the proofs given in this section
will will be studied in detail in MA2108 Mathematical Analysis I.
Definition 6.3. The sequence {an } converges to the number L if for every positive number ✏,
there corresponds an integer N such that for all n,
n > N implies that |an L| < ✏.
If no such number exists, we say that {an } diverges. If {an } converges to L, we write
lim an = L
n!1
or simply an ! L and call L the limit of the sequence.
We write lim an = 1 if an is arbitrarily large by taking n sufficiently large. Formally, it
n!1
means for every M > 0, there exists a number N such that
Example 6.37. Show that lim
1
n!1 n
n > N ) an > M.
= 0.
Solution. Given ✏ > 0, choose a positive integer N such that N1 < ✏. Then for any integer
n > N , we have | n1 0| = n1 < N1 < ✏. This verifies the definition of limit. Therefore we have
1
shown that lim = 0.
n!1 n
⌅
Now we give the proofs of the Ratio Test (Theorem 6.10) and the Alternating Series Test
(Theorem 6.12).
131
Proof of Theorem 6.10 (Ratio Test). Notation and setting as in Theorem 6.10. Suppose
a
lim n+1 = L < 1. Choose r such that L < r < 1. Then there exists a positive integer N such
n!1 an
that
a
n > N ) n+1 < r.
an
P
n N 1 is converThen |an |  |aN +1 |r n N 1 for
P all n > N . Since the geometric series |aN +1 |r
gent, by comparison test, |an | is convergent.
a
Suppose lim n+1 = L > 1. Choose r such that 1 < r < L. Then there exists a positive integer
n!1 an
N such that
a
n > N ) n+1 > r.
an
for all n > N . Since lim |aN +1 |r n N 1 = 1 as r > 1, we have lim |an | = 1.
n!1
n!1
1
X
Thus lim an , 0. Therefore,
an is divergent by the test for divergence (Theorem 6.5).
|aN +1 |r n
Then |an |
n!1
N 1
n=1
⌅
Proof of Theorem 6.12 (Alternating Series Test). Notation and setting as in Theorem 6.12.
n
X
Let Sn =
( 1)i 1 bi . Then S2n = (b1
i=1
b2 ) + · · · + (b2n
1
b2n ) = b1
(b2
b3 )
···
(b2n
2
b2n 1 ) b2n . The first equality shows that S2n 0 and S2(n+1) S2n , The second equality
shows that S2n  b1 for all n. The sequence {S2n } is non-decreasing and bounded above, so it
have a limit, say
lim S2n = L.
n!1
As S2n+1 = S2n + b2n+1 , we have
lim S2n+1 = lim S2n + lim b2n+1 = L + 0 = L.
n!1
n!1
n!1
Lastly, for any term Sn with n 2, there exists a positive integer m such that S2m  Sn 
S2m+1 . [For example, S2  S2  S3 , S2  S3  S3 , S4  S4  S5 , S4  S5  S5 , etc.] Also, as n
tends to infinity, m tends to infinity. Thus
lim S2m  lim Sn  lim S2m+1 .
m!1
n!1
m!1
Since lim S2m = lim S2m+1 = L, we have by Squeeze Theorem that lim Sn = L. This means
m!1
m!1
m!1
1
X
the series
( 1)n 1 bn is convergent.
n=1
⌅
132
Chapter 7
Vectors and Geometry of Space
Read Thomas’ Calculus, Chapter 12.
In this chapter,
• we introduce the coordinate systems for three-dimensional space R3 . This provides the
setting for our study of calculus of functions of two and three variables. The simplest geometric notion is the distance between two points in space.
• we define vectors geometrically followed by the study their algebraic properties. We emphasize the power of algebraic manipulation of vectors. In particular, we look at dot product
and cross product of vectors and their applications.
• we define lines and planes in R3 .
7.1 The 3D-Coordinate System
We set up the 3D coordinate system by fixing a point O in space (called the origin) and take
three lines passing through O that are perpendicular to each other. These lines are labeled
as x-axis, y-axis and z-axis respectively.
A point P in space can be represented by an ordered triple (a, b, c) where a , b and c are
projections of the point P onto the x-, y- and z-axis respectively. The three dimensional
space is also called the xyz-space.
The direction of the z-axis is determined by the right-hand rule:
133
Any two of the axes determine a plane:
Example 7.1. Describe and sketch the surface in R3 represented by the equation x + y = 2.
Solution. Note that x +y = 2 represents a line on the xy-plane. However, in R3 , it represents
the plane containing all points whose x- and y-coordinate sum to 2. This is a vertical plane.
⌅
Theorem 7.1 (Distance Formula).
The distance |P1 P2 | between the points P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) is
q
|P1 P2 | = (x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2 .
Consequently, we have the following equation of a sphere:
134
Theorem 7.2 (Equation of Sphere).
An equation of a sphere with center C(h, k, l) and radius r is
(x h)2 + (y
k)2 + (z l)2 = r 2 .
7.2 Vectors
It can be complicated (messy) to describe an object in space directly using the coordinates
x, y and z. It turns out to be easier by using vectors.
A vector is often represented by an arrow.
• The length of the arrow represents the magnitude of the vector.
• The arrow points in the direction of the vector.
For instance, suppose a particle moves along a line segment from point A to point B.
The vector v has initial point A (the tail) and terminal point B (the tip). We indicate this by
!
writing v = AB . Call this the displacement vector of a particle from A to B.
We denote a vector by either:
• Printing a letter in boldface v, or
• Putting an arrow above the letter v~.
The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.
Note that a vector does not depend on its initial point.
135
Definition 7.1 (Adding Vectors – The Triangle Law).
Let u and v be two vectors. Then their sum u + v is the vector from the initial point of u to the
terminal point of v when we position the vectors so that the initial point of v coincide with the
terminal point of u.
Definition 7.2 (Scalar Multiplication).
Let c 2 R and u be a vector. The scalar multiple cu is the vector whose length is |c| times the
length of u and whose direction is the same as u if c > 0 and is opposite to u if c < 0. If c = 0
or u = 0, then cu = 0.
Notice two nonzero vectors are parallel if they are scalar multiple of each other. By the
di↵erence u v, we mean
u v = u + ( v).
To treat vectors systematically (algebraically), we place the initial point of a vector u at the
origin O. In doing so, the terminal point has the coordinate (u1 , u2 ) or (u1 , u2 , u3 ) for some
u1 , u2 , u3 2 R, depending on whether u is a vector in R2 or R3 .
Denote u by
u = hu1 , u2 , u3 i.
u1 , u2 , u3 are called the components of u.
hu1 , u2 , u3 i is also called the position vector of the point (u1 , u2 , u3 ).
In other words, position vectors are vectors whose initial point is the origin.
Any vector can be represented by a position vector.
What is so nice about position vectors? The main advantage of representing vectors using
position vectors is that we can simplify calculations by algebraic manipulations!
• To add position vectors, we can just add their corresponding components. If a = ha1 , a2 , a3 i,
b = hb1 , b2 , b3 i then
a + b = ha1 + b1 , a2 + b2 , a3 + b3 i.
• To multiply a by a scalar c, we multiply each component by that scalar.
ca = hca1 , ca2 , ca3 i.
136
Theorem 7.3.
!
Given the points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ), the vector a representing AB is
a = hx2 x1 , y2 y1 , z2 z1 i.
Proof.
!
AB =
=
=
=
!
!
AO + OB
!
!
OB OA
hx2 , y2 , z2 i hx1 , y1 , z1 i
hx2 x1 , y2 y1 , z2 z1 i.
⌅
Theorem 7.4 (Properties of Vectors).
Suppose a, b and c are vectors, and c, d 2 R are scalars. Then
(1) a + b = b + a
(3) a + 0 = a
(5) c(a + b) = ca + cb
(7) (cd)a = c(da)
(2) a + (b + c) = (a + b) + c
(4) a + ( a) = 0
(6) (c + d)a = ca + da
(8) 1a = a
These properties are readily verified geometrically or algebraically.
Three vectors in V3 play a special role. They are
i = h1, 0, 0i, j = h0, 1, 0i, k = h0, 0, 1i.
These vectors are called the standard basis vectors. They have length 1 and point in the
direction of the positive x-, y- and z-axes respectively. Thus, any vector a = ha1 , a2 , a3 i can be
written as
a = ha1 , a2 , a3 i
= a1 h1, 0, 0i + a2 h0, 1, 0i + a3 h0, 0, 1i
= a1 i + a2 j + a3 k.
The length of a vector u is the length of any of its representation, and is denoted by | u|| (in
the textbook, the notation |u| is used). Using the distance formula, we have the following
The length of the vector u = hu1 , u2 , u3 i is
| u|| =
q
u12 + u22 + u32 .
A unit vector is a vector whose length is 1. For example, i, j and k are unit vectors.
137
Theorem 7.5.
If a , 0, then a unit vector in the same direction as a is given by
u = a/ | a|| .
Notice 1/ | a|| is a positive scalar, so u is in the same direction as a. Now,
| u|| =
a
1
=
| a|| = 1.
| a||
| a||
So u is the unit vector in the same direction as a.
⌅
7.3 The Dot Product
The dot product of two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i is defined to be
a · b = a1 b1 + a2 b2 + a3 b3 .
The dot product satisfies the following properties
Theorem 7.6 (Properties of Dot Product).
For vectors a, b and c and any scalar d,
(i) a · b = b · a (commutativity)
(ii) a · (b + c) = a · b + a · c (distributive law)
(iii) (da) · b = d(a · b) = a · (db)
(iv) 0 · a = 0
(v) a · a = | a||2 .
Notice a · b = 0 does not imply that a = 0 or b = 0.
For two nonzero vectors a and b in V3 , we define the angle ✓ between them to be the smaller
angle between a and b, formed by placing their initial points at the origin.
138
• a and b have the same direction i↵ ✓ = 0.
• a and b have opposite direction i↵ ✓ = ⇡.
• a and b are orthogonal (perpendicular) i↵ ✓ = ⇡2 .
Theorem 7.7.
Let ✓ be the angle between nonzero vectors a and b. Then
a · b = | a|| | b|| cos ✓.
Proof. Recall that the Law of Cosines says that
| a b||2 = | a||2 + | b||2 2 | a|| | b|| cos ✓.
| a b||2 = | ha1 b1 , a2 b2 , a3 b3 i||2
= (a1 b1 )2 + (a2 b2 )2 + (a3 b3 )2
= (a21 2a1 b1 + b12 ) + (a22 2a2 b2 + b22 )
+(a23 2a3 b3 + b32 )
= (a21 + a22 + a23 ) + (b12 + b22 + b32 ) 2(a1 b1 + a2 b2 + a3 b3 )
= | a||2 + | b||2 2a · b.
Rearranging,
2a · b = | a||2 + | b||2 | a b||2
= 2 | a|| | b|| cos ✓,
the last equality follows from the Law of Cosines.
So
a · b = | a|| | b|| cos ✓.
⌅
Example 7.2. Find the angle between the vectors a = h2, 1, 3i and b = h1, 5, 6i.
Solution. We have
cos ✓ =
It follows that
✓ = cos
1
a·b
11
=p p .
a|
b|
| || |
14 62
!
11
⇡ 1.953radian.
p p
14 62
⌅
139
Theorem 7.8.
Two vectors a and b are orthogonal if and only if a · b = 0.
Proof. If either a or b is 0 then a · b = 0 and a and b are orthogonal as 0 is considered to be
orthogonal to every vector.
We may assume that a and b are nonzero. Then
| a|| | b|| cos ✓ = a · b = 0
if and only if cos ✓ = 0, if and only if ✓ = ⇡2 , which is equivalent to saying that a and b are
orthogonal.
⌅
7.4 Projections
!
!
The figure shows two vectors a and b with the same initial point representing PQ and PR .
!
Let S be the foot of the perpendicular line from R to the line containing PQ .
!
The vector PS is called the vector projection of b onto a, denoted by
proja b.
The scalar projection of b onto a (also called the component of b along a) is defined to be
the signed magnitude of the vector projection, and is denoted by
compa b.
140
Notice
compa b = | b|| cos ✓ =
This value is negative if
Therefore,
⇡
2
a·b
.
| a||
< ✓  ⇡, where ✓ is the angle between a and b.
!
a
a·b a
a·b
a·b
proja b = compa b ⇥
=
=
a=
a.
2
a·a
| a||
| a|| | a|| | a||
Example 7.3. Let a = h 2, 3, 1i and b = h1, 1, 2i. Find the scalar projection and vector projection
of b onto a.
p
p
Solution. Notice | a|| = ( 2)2 + 32 + 12 = 14. So
compa b =
It follows that
a · b ( 2)(1) + 3(1) + 1(2)
3
=
=p .
p
| a||
14
14
3 a
3
3 9 3
proja b = p
=
a = h , , i.
7 14 14
14 | a|| 14
⌅
Theorem 7.9 (Distance from a point to a plane).
The (shortest) distance from a point P(x0 , y0 , z0 ) to the plane ax + by + cz = d is given by
ax0 + by0 + cz0 d
.
p
a2 + b 2 + c 2
Proof. A normal vector to the plane is n = ha, b, ci. (See Section 7.7.) Pick any point Q(x1 , y1 , z1 )
on the plane so that ax1 + by1 + cz1 = d. Then the shortest distance from P to the plane is
!
projn QP
=
=
=
=
!
compn QP
!
QP · n
| n||
hx0 x1 , y0 y1 , z0 z1 i · ha, b, ci
| n||
ax0 + by0 + cz0 d
.
p
a2 + b 2 + c 2
⌅
Example 7.4. Find the distance from the point (2, 3, 4) to the plane x + 2y + 3z = 13.
141
Solution. We have (x0 , y0 , z0 ) = (2, 3, 4) and a = 1, b = 2, c = 3, d = 13. Using the formula, the
distance is
5
|2(1) + ( 3)(2) + 4(3) 13|
=p .
p
14
12 + 22 + 32
⌅
7.5 The Cross Product
We now define a second type of product of vectors, called the cross product or vector product. While the dot product is a scalar, the cross product is a vector.
For two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i, define the cross product of a and b to be
a⇥b =
i j k
a1 a2 a3
b1 b2 b3
a1 a3
a a
a2 a3
i
j+ 1 2 k
b2 b3
b1 b3
b1 b2
= (a2 b3 a3 b2 )i (a1 b3 a3 b1 )j + (a1 b2 a2 b1 )k.
=
To compute a ⇥ b, we must write the components of a in the second row of the determinant,
and the components of b in the third row. The order is important!
One of the most important properties of the cross product is the following theorem.
Theorem 7.10.
The vector a ⇥ b is orthogonal to both a and b.
Proof. To show a ⇥ b is orthogonal to a, we compute their dot product as follows:
a2 a3
a1 a3
a a
a
a + 1 2 a3
b2 b3 1
b1 b3 2
b1 b2
= (a2 b3 a3 b2 )a1 (a1 b3 a3 b1 )a2 + (a1 b2 a2 b1 )a3
= 0.
(a ⇥ b) · a =
A similar computation shows that (a ⇥ b) · b = 0.
⌅
The vector a ⇥ b points in a direction perpendicular to a and b. This can be given by the
right-hand rule as follows:
142
What is the geometric meaning of the length | a ⇥ b||? This is given by the following theorem.
Theorem 7.11.
If ✓ is the angle between a and b, then
| a ⇥ b|| = | a|| | b|| sin ✓.
We can use cross product
• to find the area of a parallelogram
• to find the distance from a point to a line in R3 .
If a and b are represented by directed line segments with the same initial point then they
determine a parallelogram with base | a||, altitude | b|| sin ✓. Therefore, the area of the parallelogram is given by
| a|| | b|| sin ✓ = | a ⇥ b|| .
The distance from Q to the line through P and R is
!
!
PQ ⇥ PR
!
.
PQ sin ✓ =
!
PR
143
Some of the usual laws of algebra hold for cross products.
Theorem 7.12.
If a, b and c are vectors and d is a scalar, then
(i) a ⇥ b = b ⇥ a
(ii) (da) ⇥ b = d(a ⇥ b) = a ⇥ (db)
(iii) a ⇥ (b + c) = a ⇥ b + a ⇥ c
(iv) (a + b) ⇥ c = a ⇥ c + b ⇥ c
7.6 Lines
How do we write down an equation of a line in space? To do it, we must describe the
behavior of a general point on the line. We shall see that vectors can help us to achieve this
goal with minimal e↵ort.
So let P(x, y, z) denote an arbitrary point on the line L.
Let r and r0 denote the position vectors of P and P0 respectively, where P0 is a point on L
which we have fixed. Our aim is to describe r, the position vector of an arbitrary point on L.
!
!
Let v be a vector parallel to L, so P0 P = tv for some scalar t since P0 P and v are parallel.
Then
!
r = r0 + P0 P ,
so that
r = r0 + tv
which is a vector equation of L.
Each parameter t gives the position vector r of a point on L. As t varies, the line is traced
out by the tip of the vector r.
144
We can write the vector equation in the component form:
v = ha, b, ci, r0 = hx0 , y0 , z0 i, r = hx, y, zi.
Two vectors are equal if and only if the corresponding components are equal. Therefore, we
have
r = r0 + tv
hx, y, zi = hx0 , y0 , z0 i + tha, b, ci.
Theorem 7.13 (Parametric Equation of Line).
x = x0 + at, y = y0 + bt, z = z0 + ct.
Usually the parameter t (in the parametric equation of line) takes values on the entire R or
an interval I .
The numbers a, b and c are called direction numbers of the line L.
The vector equation and parametric equations of a line are not unique.
If we change the point r0 or the parameter t or choose a di↵erent parallel vector v, then the
equations change. Therefore, direction numbers are not unique.
Example 7.5. Find an equation of the line passing through P(1, 2, 1) and Q(5, 3, 4).
Solution. A vector parallel to the line is
!
PQ = h5 1, 3 2, 4 ( 1)i = h4, 5, 5i.
Pick a point on the line, say (1, 2, 1). Then the parametric equations for the line are
x = 1 + 4t, y = 2 5t, z = 1 + 5t.
⌅
Let L1 and L2 be two lines in R3 , with parallel vectors a and b, respectively, and let ✓ be the
angle between a and b.
• The lines L1 and L2 are parallel whenever a and b are parallel.
• If L1 and L2 intersect then ✓ is an angle between L1 and L2 . Notice ⇡
between the lines.
✓ is also an angle
In 2-D, two lines are either parallel or intersect. This is not true in 3-D. Nonparallel, nonintersecting lines are called skew lines.
Example 7.6. Show that the lines
L1 : x 2 = t, y
1 = 2t, z 5 = 2t,
L2 : x 1 = s, y
2 = s, z 1 = 3s,
are skew.
145
Solution. The lines are not parallel since a vector parallel to L1 is a = h 1, 2, 2i and a vector
parallel to L2 is b = h1, 1, 3i. Since a is not a scalar multiple of b, these vectors are not
parallel.
Assume for a contradiction that L1 and L2 intersect. Then there must exist a choice of the
parameter t and s such that the values of x, y and z are the same. In particular, for the
x-coordinate,
2 t = 1 + s,
so that s = 1 t.
On the other hand, the y-coordinate must satisfy
y = 1 + 2t = 2 s.
Substituting s = 1 t into the last equation, we have t = 0 and so s = 1.
Now, the z-coordinate must satisfy
z = 5 + 2t = 5,
z = 1 + 3s = 4,
which is absurd!
Hence our assumption that L1 and L2 intersects was wrong. So the lines are skew, as desired.
⌅
7.7 Planes
To get an equation for the plane, we need to describe an arbitrary point P(x, y, z) on the
plane. Again, we use vectors to help us do that.
Let r and r0 denote the position vectors of P and P0 respectively, where P0 is a fixed point on
the given plane.
!
Then r r0 is represented by P0 P .
146
The normal vector n (which is orthogonal to the plane) is always orthogonal to r r0 . Therefore we have
Theorem 7.14 (Vector Equation of Plane).
which can be written as
n · (r r0 ) = 0
n · r = n · r0 .
To obtain a scalar equation for the plane, write the vectors in component form and equate
corresponding components:
n = ha, b, ci, r = hx, y, zi, r0 = hx0 , y0 , z0 i.
Then n · r = n · r0 becomes
ha, b, ci · hx, y, zi = ha, b, ci · hx0 , y0 , z0 i
ax + by + cz = ax0 + by0 + cz0 .
Theorem 7.15 (Linear Equation of Plane).
ax + by + cz + d = 0,
where
d = (ax0 + by0 + cz0 ).
Example 7.7. Find an equation of the plane that passes through the points P(1, 3, 2), Q(3, 1, 6),
R(5, 2, 0).
Solution. First, we need a vector n orthogonal to the plane. This can be given by
!
!
n = PQ ⇥ PR
147
Notice
So
!
!
PQ = h2, 4, 4i, PR = h4, 1, 2i.
i j
k
2 4 4
n =
4 1 2
= 12i + 20j + 14k.
With the point P(1, 3, 2) and the normal vector n, an equation of the plane is:
12(x 1) + 20(y
3) + 14(z 2) = 0
or after simplifications,
6x + 10y + 7z = 50.
⌅
Two planes are parallel if their normal vectors are parallel.
If two planes are not parallel, then
• They intersect in a straight line.
• The angle between the two planes is defined as the acute angle between their normal
vectors
Example 7.8. (a) Find the angle between the planes x + 2y + z = 3 and x 4y + 3z = 5.
(b) Find the line of intersection of these two planes.
Solution. (a) The normal vectors of these planes are
n1 = h1, 2, 1i, n2 = h1, 4, 3i.
So, if ✓ is the angle between them, then
n1 · n2
| n1 | | n2 |
1(1) + 2( 4) + 1(3)
= cos 1 p
p
1 + 4 + 1 1 + 16 + 9
4
= cos 1 p
⇡ 108.7
156
✓ = cos
1
148
Therefore, the angle between the planes is 71.3 .
(b) Solving both equations for x,
x = 3 2y
z and x = 5 + 4y
3z.
Setting them to be equal gives
3 2y
Solving for z gives
z = 5 + 4y
3z.
z = 3y + 1.
Substituting this into the first equation,
x = 3 2y
(3y + 1) = 5y + 2.
Let y = t be the parameter, we obtain a parametric equation for the line of intersection
x = 5t + 2, y = t, z = 3t + 1.
⌅
Example 7.9. Consider the planes ↵ : 3x + 4y + 12z = 26 and
: 3x + 4y + 12z = 39.
(a) Find the distance between ↵ and .
(b) Let ` be the line with parametric equations
x = 2 + 3t, y = 2 + 4t, z = 1 + 12t.
Find the intersection point Q of ` with .
30 25
Ans: (a) 1, (b) ( 29
13 , 13 , 13 ).
Solution. (a) Note that P(2, 2, 1) is a point on ↵, and the two planes ↵ and
Then the distance between ↵ and
is given by
|3(2) + 4(2) + 12(1) 39|
13
=
= 1.
p
13
32 + 42 + 122
149
are parallel.
(b) Substituting the parametric equations of ` into the equation of , we have
3(2 + 3t) + 4(2 + 4t) + 12(1 + 12t) = 39 , 26 + 169t = 39 , t =
1
13 .
Therefore, the intersection point Q is
(2 +
3
4
12
29 30 25
, 2 + , 1 + ) = ( , , ).
13
13
13
13 13 13
Note that PQ is perpendicular to ↵ and , and kPQk = 1.
⌅
Exercise 7.1. Let a and b be vectors in R3 . Prove that (a · b)2 + |a ⇥ b|2 = |a|2 |b|2 .
Exercise 7.2. Find the distance from the point P(3, 3, 3) to the line ` : x = t, y = 2t , z = t.
Ans.
p
2.
Exercise 7.3. Find the point on the surface z = x2 +y 2 +10 that is nearest to the plane x+2y z = 0.
Ans. ( 12 , 1, 45
4 ).
150
Chapter 8
Functions of Several Variables
Read Thomas’ Calculus, Chapter 14.
8.1 Vector Functions of One Variable
Recall the vector equation of line:
r(t) = r0 + tv.
We have seen that the tip of the vector r(t) traces out a line as t varies.
We can rewrite the above as follows:
r(t) = hx0 , y0 , z0 i + tha, b, ci = hx0 + ta, y0 + tb, z0 + tci.
Notice that each component of r(t) is a scalar function of t.
In general, a vector-valued function is
r(t) = hf (t), g(t), h(t)i
where f (t), g(t) and h(t) are scalar functions of t.
Formally,
A vector-valued function r(t) is a mapping from its domain D ✓ R to its range R ✓ V3 , so
that for each t 2 D, r(t) = v for exactly one vector v 2 V3 .
We write a vector-valued function as
r(t) = f (t)i + g(t)j + h(t)k
or
r(t) = hf (t), g(t), h(t)i
for some scalar function f , g and h (called the component functions of r).
151
Suppose r(t) traces out the curve C, we say that r(t) is a parametrization of C.
A curve C can have more than one parametrizations.
For example, both
r(t) = ht, t 2 i, t 2 R
r(t) = ht 3 , t 6 i, t 2 R
parameterize the parabola f (x) = x2 on the xy-plane.
Example 8.1. Sketch the curve traced out by the vector-valued function r(t) = sin ti 3 cos tj+2tk.
Solution. There is a relationship between x and y here:
✓ ◆2
y
x +
= sin2 t + cos2 t = 1
3
2
which is the equation of an ellipse in 2-D. In 3-D, since the equation does not involve z, it
becomes the equation of an elliptic cylinder whose axis is the z-axis.
The curve will wind its way up the cylinder anticlockwise as t increases.
We call this curve an elliptical helix.
152
⌅
8.2 Calculus of Vector Functions
To extend di↵erentiation and integration to vector-valued functions, just think
‘component-wise’!!!
Definition 8.1. The derivative r0 (t) of the vector-valued function r(t) is defined by
r(t + 4t) r(t)
4t!0
4t
for any values of t for which the limit exists.
r0 (t) = lim
When the limit exists for t = a, we say that r is di↵erentiable at t = a.
The derivative of a vector-valued function can be found directly from the derivatives of the
components.
Theorem 8.1 (Derivative of Vector-valued Function).
Let r(t) = hf (t), g(t), h(t)i and suppose that the components f , g and h are all di↵erentiable at
t = a. Then r is di↵erentiable at t = a and its derivative is given by
r0 (a) = hf 0 (a), g 0 (a), h0 (a)i.
153
Example 8.2. Let r(t) = h3t 2 + 2t + 1, sin(⇡t), e 2t i. Find r0 (t), and compute r0 (1).
Solution.
d
d
d
(3t 2 + 2t + 1),
sin(⇡t), e 2t i
dt
dt
dt
2t
= h6t + 2, ⇡ cos(⇡t), 2e i.
r0 (t) = h
Hence
r0 (1) = h6 + 2, ⇡ cos ⇡, 2e2 i = h8, ⇡, 2e 2 i.
⌅
Theorem 8.2 (Derivative Rules).
Suppose r(t) and s(t) are di↵erentiable vector-valued functions, f (t) is a di↵erentiable scalar
function and c is a scalar constant. Then
(i)
(ii)
d
0
0
dt (r(t) + s(t)) = r (t) + s (t)
d
0
dt (cr(t)) = cr (t)
(t)r(t) = f 0 (t)r(t) + f (t)r0 (t)
(iii)
d
dt f
(iv)
d
0
0
dt r(t) · s(t) = r (t) · s(t) + r(t) · s (t)
(v)
d
0
0
dt (r(t) ⇥ s(t)) = r (t) ⇥ s(t) + r(t) ⇥ s (t).
8.3 Tangent Vector and Tangent Line to a Curve
Recall that one interpretation of the derivative of a scalar function is that the value of the
derivative at a point gives the slope of the tangent line to the curve at that point.
There is a similar interpretation for the derivative of vector-valued functions.
Recall
r0 (a) = lim
4t!0
r(a + 4t) r(a)
.
4t
154
Notice that for 4t > 0, the vector
As 4t ! 0,
r(a+4t) r(a)
4t
r(a+4t) r(a)
4t
points in the same direction as r(a + 4t) r(a).
approaches r0 (a).
This is a vector tangent to the curve at r(a). We call r0 (a) a tangent vector to the curve at
t = a.
Example 8.3. Find the tangent line L to the curve r(t) = hcos t, sin t, ti at (0, 1, ⇡/2).
Solution. At point (0, 1, ⇡/2), t = ⇡/2. Since r0 (t) = h sin t, cos t, 1i, a direction of the tangent
line L is
r0 (⇡/2) = h sin(⇡/2), cos(⇡/2), 1i = h 1, 0, 1i
So a parametric equation of L is
x = 0 + ( 1)t, y = 1 + (0)t, z = ⇡/2 + (1)t,
that is
x = t, y = 1, z = ⇡/2 + t.
⌅
8.4 Arc Length of a Space Curve
Suppose that a smooth curve C is traced out by the vector-valued function r(t) = hf (t), g(t), h(t)i,
where f , f 0 , g, g 0 , h, h0 are all continuous for t 2 [a, b], and the curve is traversed exactly once
as t increases from a to b. Then the arc length of C is given by the following result.
155
Theorem 8.3 (Arc Length Formula).
Let C be the curve given by
r(t) = hf (t), g(t), h(t)i, a  t  b
where f 0 , g 0 and h0 are continuous. If C is traversed exactly once as t increases from a to b,
then its length is
Z bq
s =
f 0 (t)2 + g 0 (t)2 + h0 (t)2 dt
Z
a
b
=
r0 (t) dt
a
Example 8.4. Find the arclength of the curve traced out by the endpoint of the vector-valued
function r(t) = h2t, ln t, t 2 i for 1  t  e.
Solution.
s =
=
=
=
=
Z er
✓ ◆2
1
22 +
+ (2t)2 dt
t
1
Z er
1 + 4t 2 + 4t 4
dt
t2
1
Z er
(1 + 2t 2 )2
dt
t2
1
Ze
Z e✓
◆
1 + 2t 2
1
dt =
+ 2t dt
t
1 t
⇣1
⌘
2 e
ln |t| + t 1
= (ln e + e2 ) (ln 1 + 1) = e2 .
156
⌅
So far we have seen functions of one variable, i.e. the domain is a subset of R
function
(scalar) f (t)
(vector) r(t)
Domain D
D✓R
D✓R
Range R
R✓R
R ✓ R2 or R3
However, in the real world, physical quantities often depend on two or more variables.
8.5 Functions of Two Variables
Definition 8.2.
A function f of two variables is a rule that assigns to each ordered pair of real numbers (x, y)
in a set D ✓ R2 = R ⇥ R a unique real number denoted by f (x, y).
If a function f is given by a formula and no domain is specified, then the domain of f is
understood to be:
the set of all pairs (x, y) for which the given expression is a well-defined real number.
Example 8.5. Find the domain of
f (x, y) = x ln(y 2 x).
Solution. ln(y 2 x) is defined only when y 2 x > 0, that is, x < y 2 .
So the domain of f is
D = {(x, y) : x < y 2 }.
157
⌅
One way of visualizing z = f (x, y) is to draw its graph.
If f is a function of two variables with domain D, then the graph of f is the set of all points
(x, y, z) 2 R3 such that z = f (x, y) and (x, y) 2 D.
The graph of a function f of two variables is also called the surface S with equation z =
f (x, y).
We can visualize the graph S of f as lying directly above or below its domain D in the
xy-plane.
Graphing functions of more than one variable is not simple!
For most functions of two variables, to identify the surface, we must
(1) take hints from the expressions z = f (x, y)
(2) think through the traces and piece together the clues
Example 8.6. Match the functions f (x, y) = ln(x2 + y 2 ) and g(x, y) = cos(x2 + y 2 ) to the surfaces
shown below:
Solution. Notice both functions contain the expression x2 +y 2 . This is significant: given any
value r and any point (x, y) on the circle x2 + y 2 = r 2 , the height of the surface at the point
(x, y) is a constant. That is, the surface has circular cross sections parallel; to the xy-plane.
However, both surfaces shown have this property, so we cannot yet tell which surface is
matched by which function.
Notice the cosine of any angle lies between 1 and 1. So the second graph is g(x, y).
158
An important property of f (x, y) is that the logarithm tends to 1 as its argument x2 + y 2
approaches 0. This appears in the first graph.
⌅
Another way to visualize functions of several variables is to use the contour plot which
provide the same information condensed into a 2-D picture.
Definition 8.3 (Level Curve).
A level curve of f (x, y) is the two-dimensional graph of the equation f (x, y) = k for some
constant k.
Definition 8.4 (Contour Plot).
A contour plot of f (x, y) is a graph of numerous level curves f (x, y) = k, for representative
values of k.
To sketch contour plots, we use values of k that are equally spaced. The surface is:
• steep where the level curves are close together.
• flatter where the level curves are farther apart.
Example 8.7. Sketch some level curves of h(x, y) = 4x2 + y 2 .
Solution. If k < 0, then 4x2 + y 2 = k has no solution, so there is no level curves for k < 0.
If k = 0, then there is only one solution (0, 0), so the level curve is a single point (0, 0).
159
If k > 0, then 4x2 + y 2 = k is an ellipse:
y2
+ p 2 = 1.
( k/2)2
k
p
x2
Thus, larger k gives rise to an ellipse with longer major and minor axes.
⌅
8.6 Cylinders and Quadric Surfaces
We have seen that the graph of functions of two variables are surfaces in space R3 . To
appreciate the calculus of functions of two variables, we need more examples of surfaces in
space other than planes (ax + by + cz = d)and spheres ((x a)2 + (y b)2 + (z c)2 = d 2 ).
It is not easy to draw surfaces in R3 . Our goal here is to identify and sketch some special
type surfaces, namely the cylinders and some quadric surfaces.
8.6.1 Cylinders
When we mention the word cylinder, we probably think of the following object
160
Actually, the term cylinder is used to refer to a surface more general that the one we saw.
Definition 8.5.
A surface is a cylinder if there is a plane P such that all the planes parallel to P intersect the
surface in the same curve (when viewed in 2-dimension).
Example 8.8. Show that the surface given by
y 2 + z2 = 1
is a cylinder.
Solution. Notice x is missing in the equation. When x = 0, y 2 + z2 = 1 is a circle with radius
1 in the yz-plane, which is the intersection of the surface and the yz-plane.
Generally, x = k represent a plane parallel to the yz-plane, and the intersection the surface
and this plane is always the circle y 2 + z2 = 1.
Therefore, the surface y 2 + z 2 = 1 is a cylinder.
In fact, any equation in x, y and z where one of the variable is missing is a cylinder.
⌅
Example 8.9. Sketch the graph of the surface z = x2 .
Solution. Notice that the equation of the graph, z = x2 , does not involve y.
This means that any vertical plane with equation y = k (parallel to the xz-plane) intersects
the graph in a curve with equation z = x2 . So the surface z = x2 is a cylinder.
161
⌅
8.6.2 Quadric Surface
Definition 8.6 (Quadric Surface).
A quadric surface is the graph of a second-degree equation in three variables x, y and z:
Ax2 + By 2 + Cz2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0
where A, B, . . ., J are constants.
There are six basic quadric surfaces. But we shall focus on two of them:
• Elliptic paraboloid
• Ellipsoid
Definition 8.7 (Elliptic paraboloid – symmetric about the z-axis).
x2 y 2 z
+
=
a2 b 2 c
The graph of the elliptic paraboloid
x2
a2
y2
+ b2 =
z
c
when c > 0.
162
Example 8.10. Identify and sketch the surface
x2 + 2z 2 6x y + 10 = 0.
Solution. By completing squares, we rewrite the equation as
(y
1) = (x 3)2 +
z2
1/2
It represents an elliptic paraboloid. However, it has been shifted so that its vertex is the
point (3, 1, 0), and is symmetric about the line which is parallel to the y-axis and passes
through (3, 1, 0).
⌅
163
Definition 8.8 (Ellipsoid).
x2 y 2 z2
+
+ =1
a2 b 2 c 2
If a = b = c, then the ellipsoid is a sphere.
Example 8.11. Sketch the quadric surface with equation
x2 +
y 2 z2
+ = 1.
9
4
Solution. The surface intersects the xy-plane (z = 0) in the ellipse
x2 +
y2
= 1.
9
In general, the surface intersects the plane z = k in the ellipse
x2 +
provided 1
k2
4
y2
=1
9
k2
,
4
> 0, that is 2 < k < 2.
The surface also intersects every plane x = k (which is parallel to the yz-plane (x = 0)) in the
ellipse
y 2 z2
+ = 1 k2,
9
4
provided 1 k 2 > 0, that is 1 < k < 1.
164
The surface also intersects every planes y = k which is parallel to the xz-plane (y = 0) in
ellipse
z2
k2
x2 + = 1
,
4
9
2
provided 1 k9 > 0, that is 3 < k < 3.
⌅
8.7 Functions of Three Variables
Definition 8.9.
A function f of three variables is a rule that assigns to each ordered triple of real numbers
(x, y, z) in a set D ✓ R3 = R ⇥ R ⇥ R a unique real number denoted by f (x, y, z).
Unlike functions of two variables, it is very difficult to visualize a function f of three variables by its graph.
That would lie in a four-dimensional space!!!
However, we do gain some insight into f by examining its level surfaces (counterparts of
level curves in two-variable case).
165
Definition 8.10 (Level Surface).
A level surface of f (x, y, z) is the three-dimensional graph of the equation f (x, y, z) = k for
some constant k.
If the point (x, y, z) moves along a level surface, the value of f (x, y, z) remains fixed.
Example 8.12. Find the level surfaces of the function
f (x, y, z) = x2 + y 2 + z2 .
Solution. The level surfaces are:
where k
x2 + y 2 + z2 = k
0.
p
These form a family of concentric spheres with radius k.
⌅
8.8 Partial Derivatives
Recall that for a function f of a single variable, we define the derivative function as
f (x + h) f (x)
h
h!0
f 0 (x) = lim
for any values of x for which the limit exists.
166
At a particular point x = a, we interpret f 0 (a) as the instantaneous rate of change of f with
respect to x at that point.
We want to generalize the notion of derivative to functions of more than one variable.
The idea is to ‘vary’ one variable and keep other variable(s) fixed.
Definition 8.11 (Partial Derivative).
If f is a function of two variables, its partial derivatives are the functions f x and f y defined by:
f (x + h, y) f (x, y)
,
h
h!0
f x (x, y) = lim
f (x, y + h) f (x, y)
.
h
h!0
f y (x, y) = lim
Example 8.13. Let f (x, y) = x2 y. Then
(x + h)2 y x2 y
(x2 y + 2xhy + h2 y) x2 y
= lim
= lim (2xy + hy) = 2xy.
h
h
h!0
h!0
h!0
2
2
x (y + h) x y
f y (x, y) = lim
lim x2 = x2 .
h
h!0
h!0
f x (x, y) = lim
There are many alternative notations for partial derivatives:
Instead of f x , we can write f 1 or D1 f (to indicate di↵erentiation with respect to the first
variable x) or
@f
.
@x
To compute the partial derivative f x , one may simply do the following: Treat (temporarily) the other variable y in f (x, y) as a constant, and di↵erentiate f (x, y) with respect to the
variable x.
Similarly, to compute f y , one may simply do this: Treat (temporarily) the other variable x in
f (x, y) as a constant, and di↵erentiate f (x, y) with respect to the variable y.
Example 8.14. For f (x, y) = exy + yx , compute f x and f y . Find also
Solution. Treating y as a constant, we have
1
f x (x, y) = ye xy + .
y
Treating x as a constant, we have
f y (x, y) = xexy
167
x
.
y2
@f
@f
(2, 1) and
(2, 1).
@x
@y
Hence at (x, y) = (2, 1), we have
@f
1
(2, 1) = f x (2, 1) = 1 · e 2·1 + = e 2 + 1,
@x
1
@f
2
(2, 1) = f y (2, 1) = 2 · e 2·1
= 2e 2 2.
@y
12
⌅
To give a geometric interpretation of partial derivatives, we recall that the equation z =
f (x, y) represents a surface S (the graph of f ).
If f (a, b) = c, then the point P(a, b, c) lies on S.
By fixing y = b, we are restricting our attention to the curve C1 in which the vertical plane
y = b intersects S. That is, C1 is the trace of S in the plane y = b.
Likewise, the vertical plane x = a intersects S in a curve C2 .
Both the curves C1 and C2 pass through P.
• The curve C1 is the graph of the function g(x) = f (x, b). So, the slope of its tangent T1 at P
is: g 0 (a) = f x (a, b).
• The curve C2 is the graph of the function h(x) = f (a, y). So, the slope of its tangent T2 at P
is: h0 (b) = f y (a, b).
Thus, the partial derivatives f x (a, b) and f y (a, b) can be interpreted geometrically as:
The slopes of the tangent lines at P(a, b, c) to the traces C1 and C2 of S in the planes y = b
and x = a.
For functions of more than two variables, such as w = f (x, y, z), we can similarly define
fx , fy , fz ,
Example 8.15. Find
x3 + y 3 + z 3 + 6xyz
@z
@x
and
@z
@y
@f @f @f
@w @w @w
,
,
or
,
,
.
@x @y @z
@x @y @z
if z is defined implicitly as a function of x and y by the equation
= 1.
168
Solution. Take partial derivative with respect to x on both sides:
3x2 + 3z2
Solving for
@z
,
@x
@z
@z
+ 6yz + 6yx
= 0.
@x
@x
we have
(3z 2 + 6yx)
Similarly, we have
x2 + 2yz
@z
@z
= (3x2 + 6yz) =)
= 2
.
@x
@x
z + 2xy
y 2 + 2xz
@z
= 2
.
@y
z + 2xy
⌅
8.9 Higher Order Partial Derivatives
If f is a function of two variables, then its partial derivatives f x and f y are also functions of
two variables.
So, we can consider their partial derivatives
(f x )x , (f x )y , (f y )x , (f y )y .
These are called the second partial derivatives of f .
If z = f (x, y), we use the following notation:
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
(f x )x = f xx = f 11 =
@ @f
@x @x
(f x )y = f xy = f 12 =
@ @f
@y @x
(f y )x = f yx = f 21 =
@ @f
@x @y
(f y )y = f yy = f 22 =
@ @f
@y @y
=
@2 f
@x2
=
@2 z
@x2
=
@2 f
@y@x
=
@2 z
@y@x
=
@2 f
@x@y
=
@2 z
@x@y
=
@2 f
@y 2
=
@2 z
@y 2
Thus, the notation f xy means that we first di↵erentiate with respect to x and then with
respect to y.
In computing f yx , the order is reversed.
Example 8.16. Find all second-order partial derivatives of f (x, y) = x2 y
Solution. First, we compute the first-order derivatives:
1
f x = 2xy + ,
x
169
y 3 + ln x.
f y = x2 3y 2 .
Then we have
@
@x
f xx =
f xy =
f yx
f yy
⇣
⌘
2xy + 1x = 2y
@
@y
⇣
2xy + 1x
⌘
=
⇣
⌘
@
= @x
x2 3y 2 =
⇣
⌘
@
= @y
x2 3y 2 =
1
,
x2
2x,
2x,
6y.
⌅
Notice f xy = f yx in the preceding example. This is not a coincidence.
It turns out that the mixed partial derivatives f xy and f yx are equal for most (not all) functions that one meets in practice.
The following theorem, discovered by the French mathematician Alexis Clairaut (1713–
1765), gives conditions under which we can assert that f xy = f yx .
Theorem 8.4 (Clairaut’s Theorem).
Suppose f is defined on a disk D that contains (a, b). If the functions f xy and f yx are both
continuous on D, then
f xy (a, b) = f yx (a, b).
Partial derivatives of order 3 and higher can also be defined. For example, f xyy = (f xy )y .
Using Clairaut’s Theorem, it can be shown that
f xyy = f yxy = f yyx
if these functions are continuous.
8.10 Tangent Planes
Recall that we use derivative f 0 (a) to get the tangent line to the curve y = f (x) at x = a:
y = f (a) + f 0 (a)(x a).
In the same spirit, we shall use partial derivatives to obtain the tangent plane to the surface
z = f (x, y) at a given point.
170
Consider the surface S which is the graph of z = f (x, y). Suppose f has continuous first
partial derivatives.
Let P(a, b, c) be a point on S. Notice c = f (a, b).
Let C1 and C2 be the curves obtained by intersecting the vertical planes y = b and x = a with
the surface S. Notice P lies on both C1 and C2 .
Let T1 and T2 be the tangent lines to the curves C1 and C2 at the point P.
Then, the tangent plane to the surface S at the point P is defined to be the plane that contains
both tangent lines T1 and T2 .
How to find an equation for the tangent plane?
Recall that any plane passing through P(a, b, c) has an equation of the form
n · hx a, y
b, z ci = 0
where n is a vector normal to the plane.
Notice the tangent line T1 lies on the plane y = b. Along T1 at x = a, a change of 1 unit in x
corresponds to a change of f x (a, b) in z (here we require f x to be continuous). The value of y
does not change along the line. A vector with the same direction as T1 is
h1, 0, f x (a, b)i.
Similarly, a vector with the same direction as T2 is
h0, 1, f y (a, b)i.
We have found two vectors parallel to the tangent plane:
h1, 0, f x (a, b)i, h0, 1, f y (a, b)i.
171
A vector normal to the plane is given by the cross product:
h0, 1, f y (a, b)i ⇥ h1, 0, f x (a, b)i = hf x (a, b), f y (a, b), 1i.
Theorem 8.5 (Equation of Tangent Plane).
Suppose f (x, y) has continuous first partial derivatives at (a, b). A normal vector to the tangent
plane at (a, b, f (a, b)) to the surface z = f (x, y) is
hf x (a, b), f y (a, b), 1i.
Further, an equation of the tangent plane is given by
f x (a, b)(x a) + f y (a, b)(y
b) (z f (a, b)) = 0
or
z = f (a, b) + f x (a, b)(x a) + f y (a, b)(y
b).
Example 8.17. Find the tangent plane to the elliptic paraboloid z = 2x2 + y 2 at the point (1, 1, 3).
Solution. Notice
f x (x, y) = 4x, f x (1, 1) = 4,
f y (x, y) = 2y, f y (1, 1) = 2.
The equation of the plane is
z = f (1, 1) + 4(x 1) + 2(y
z = 4x + 2y
1),
3.
The figure shows the elliptic paraboloid and its tangent plane at (1, 1, 3) that we found in the
preceding example
172
⌅
8.11 Di↵erentiability and Chain Rule
For single-variable function f (x), we say that f is di↵erentiable at a if and only if f 0 (a)
exists. For two-variable function f (x, y), it is tempting to say that f is di↵erentiable at (a, b)
if f x (a, b) and f y (a, b) exist. However, such definition would fail to capture the true nature of
‘di↵erentiability’!
Definition 8.12.
Informally, we say that f is di↵erentiable at (a, b) if the tangent plane at (a, b) is a good
approximation to f at points close to (a, b).
The above definition is not precise since we do not define what do we mean by ‘a good
approximation’. Do not worry! All the functions we encounter in this course will be di↵erentiable at points in its domain. Recall that the Chain Rule for functions of a single variable
gives the following rule for di↵erentiating a composite function.
If y = f (x) and x = g(t), where f and g are di↵erentiable functions, then y is indirectly a
di↵erentiable function of t, and
dy dy dx
=
.
dt dx dt
We shall extend the chain rule to functions of several variables. This takes several slightly
di↵erent forms, depending on the number of independent variables.
The first version deals with a function z = f (x, y) where x = g(t) and y = h(t) are both functions of a single variable t:
z = f (g(t), h(t)).
173
Theorem 8.6 (The Chain Rule - Case 1).
Suppose that z = f (x, y) is a di↵erentiable function of x and y, where x = g(t) and y = h(t) are
both di↵erentiable functions of t. Then, z is a di↵erentiable function of t, and
dz @f dx @f dy
=
+
.
dt @x dt @y dt
Example 8.18. For z = f (x, y) = x2 ey , x = g(t) = t 2 1 and y = h(t) = sin t, find the derivative
dz
dt .
Solution. First, compute the partial derivatives:
@z
@z
= 2xey ,
= x2 ey .
@x
@y
Next, compute the derivatives:
dy
dx
= 2t,
= cos t.
dt
dt
Therefore, using the chain rule,
dz
@z dx @z dy
=
+
dt
@x dt @y dt
= 2xe y (2t) + x2 ey cos t
= 2(t 2 1)esin t (2t) + (t 2 1)2 esin t cos t.
⌅
Notice, in the preceding example, you could have first substituted for x and y and then
compute the derivative of
f (g(t), h(t)) = (t 2 1)2 esin t
using the usual rules of di↵erentiation for functions of a single variable.
We can easily extend The Chain Rule to the case of a function f (x, y) where x and y now are
both functions of two independent variables s and t, x = g(s, t) and y = h(s, t).
Then, z is indirectly a function of s and t:
z = f (g(s, t), h(s, t)).
We wish to find
@z @z
,
.
@s @t
Recall that, in computing @z
, we hold s fixed and compute the ordinary derivative of z with
@t
respect to t. (This is the situation in The Chain Rule - Case 1)
Similarly, in computing @z
, we hold t fixed and compute the ordinary derivative of z with
@s
respect to s. (This is the situation in The Chain Rule - Case 1)
We have the following (for free!)
174
Theorem 8.7 (The Chain Rule - Case 2).
Suppose that z = f (x, y) is a di↵erentiable function of x and y, where x = g(s, t) and y = h(s, t)
are both di↵erentiable functions of s and t. Then,
@z @f @x @f @y
=
+
,
@s @x @s @y @s
@z @f @x @f @y
=
+
.
@t @x @t @y @t
Case 2 of the Chain Rule contains three types of variables:
• s and t are independent variables.
• x and y are called intermediate variables.
• z is the dependent variable.
Example 8.19. If z = e x sin y, where x = st 2 and y = s 2 t, find
@z
@s
and
@z
.
@t
Solution. Applying Case 2 of Chain Rule,
@z
@z @x @z @y
=
+
@s
@x @s @y @s
= (ex sin y)(t 2 ) + (ex cos y)(2st)
2
2
= t 2 est sin(s 2 t) + 2stest cos(s 2 t).
@z
@z @x @z @y
=
+
@t
@x @t @y @t
= (e x sin y)(2st) + (ex cos y)(s 2 )
2
2
= 2stest sin(s 2 t) + s 2 est cos(s 2 t).
⌅
To remember the Chain Rule, it’s helpful to draw a tree diagram, as follows:
We draw branches from the dependent variable z to the intermediate variables x and y to
indicate that z is a function of x and y.
175
Then, we draw branches from x and y to the independent variables s and t. On each branch,
we write the corresponding partial derivative.
To find @z
, we find the product of the partial derivatives along each path from z to s and
@s
then add these products:
@z @z @x @z @y
=
+
.
@s @x @s @y @s
Similarly, we find
@z
@t
by using the paths from z to t.
Theorem 8.8 (The Chain Rule - General Version).
Suppose that u is a di↵erentiable function of n variables x1 , . . . , xn , and each xj is a di↵erentiable function of m variables t1 , . . . , tm . Then u is a function of t1 , . . . , tm and
@u
@u @x1 @u @x2
@u @xn
=
+
+ ··· +
@ti @x1 @ti @x2 @ti
@xn @ti
for each i = 1, . . . , m.
Example 8.20. Write out the Chain Rule for the case where w = f (x, y, z, t) and x = x(u, v),
y = y(u, v), z = z(u, v), t = t(u, v).
Solution. The figure shows the tree diagram.
176
With the aid of the tree diagram, we can now write the required expressions:
@w @w @x @w @y @w @z @w @t
=
+
+
+
.
@u
@x @u @y @u @z @u @t @u
@w @w @x @w @y @w @z @w @t
=
+
+
+
.
@v
@x @v @y @v @z @v @t @v
⌅
Example 8.21. If w = f (x2 y 2 , y 2 x2 ) and f is di↵erentiable, show that
y
@w
@w
+x
= 0.
@x
@y
Solution. Introduce intermediate variables:
u = x2 y 2 , v = y 2 x2 .
Using Chain Rule,
@w @w @u @w @v @w
@w
=
+
=
(2x) +
( 2x)
@x @u @x @v @x @u
@v
and
@w @w @u @w @v @w
@w
=
+
=
( 2y) +
(2y)
@y
@u @y @v @y @u
@v
Therefore
y
@w
@w
+x
@x
@y
!
!
@w
@w
@w
@w
=
(2xy) +
( 2xy) +
( 2xy) +
(2xy) = 0.
@u
@v
@u
@v
⌅
177
8.12 Implicit Di↵erentiation
Consider a surface defined by an equation
F(x, y, z) = 0
where F(x, y, z) is di↵erentiable.
Suppose z is implicitly defined as a function of independent variables x and y, that is, for
every choice of x and y, there is a unique z such that F(x, y, z) = 0.
Suppose we are interested in
@z
.
@x
@z
If we can solve the above equation for z, say z = f (x, y), then we can compute @x
directly.
But life is complicated enough that we may not be able to solve for z. Using Chain Rule to
di↵erentiate the equation F(x, y, z) = 0 with respect to x:
@F @x @F @y @F @z
+
+
= 0.
@x @x @y @x @z @x
But
@y
@x
= 1,
=0
@x
@x
since x and y are independent variables. Therefore, if
@z
=
@x
@F
@x
@F
@z
=
@F
@z
, 0, then
Fx
.
Fz
Theorem 8.9 (Implicit Di↵erentiation: Two Independent Variables).
Suppose the equation F(x, y, z) = 0, where F is di↵erentiable, defines z implicitly as a di↵erentiable function of x and y. Then,
Fy (x, y, z)
Fx (x, y, z) @z
@z
=
,
=
@x
Fz (x, y, z) @y
Fz (x, y, z)
provided Fz (x, y, z) , 0.
Example 8.22. Find
@z
@x
if
x3 + y 3 + z 3 + 6xyz = 1.
Solution. Let F(x, y, z) = x3 + y 3 + z3 + 6xyz 1. Then
Fx = 3x2 + 6yz, Fz = 3z2 + 6xy.
Therefore, by the Implicit Di↵erentiation Theorem,
3x2 + 6yz
@z
Fx
=
=
.
@x
Fz
3z2 + 6xy
⌅
178
8.13 Increments and Di↵erentials
Definition 8.13.
Let z = f (x, y). Suppose 4x and 4y are increments in the independent variable x and y
respectively.
Then the increment in z is defined by
4z = f (x + 4x, y + 4y) f (x, y).
Definition 8.14.
Let z = f (x, y). Suppose 4x and 4y are increments in the independent variable x and y
respectively.
Then the di↵erentials of the independent variables x and y are
dx = 4x, dy = 4y.
The di↵erential (or total di↵erential) of the dependent variable z is
dz = f x (x, y)dx + f y (x, y)dy.
Notice that
• the increment 4z is the change in z as (x, y) changes from (a, b) to (a + 4x, b + 4b).
• the di↵erential dz is the change in the tangent plane as (x, y) changes from (a, b) to (a +
4x, b + 4b).
Example 8.23. Let z = 2x2
from (1, 1) to (0.98, 1.03).
xy. Find 4z. Use this result to find the change of z if (x, y) changes
179
Solution.
4z = f (x + 4x, y + 4y) f (x, y)
⇣
⌘
= 2(x + 4x)2 (x + 4x)(y + 4y) (2x2 xy)
= (4x y)4x x4y + 2(4x)2 4x4y.
As (x, y) changes from (1, 1) to (0.98, 1.03), we have 4x = 0.98 1 = 0.02 and 4y = 1.03 1 =
0.03. Substituting these values into the expression of 4z above, we obtain
4z = 0.0886.
⌅
From the previous example, it seems quite complicated to calculate 4z. Is there a way to
approximate 4z?
It turns out that dz gives a good approximation of 4z provided 4x and 4y are small and
f (x, y) is di↵erentiable.
Theorem 8.10.
Suppose f is di↵erentiable at (a, b). Let 4x and 4y be small increments in x and y respectively
from (a, b). Then
4z ⇡ dz = f x (a, b) dx + f y (a, b) dy = f x (a, b)4x + f y (a, b)4y.
Example 8.24. The base radius and height of a circular cone are measured as 10cm and 25cm
respectively, with a possible error in measurement of as much as 0.1cm in each. Use di↵erential to
estimate the maximum error in the calculated volume of the cone.
Solution. The volume of the cone is V = ⇡r 2 h/3. So
dV = Vr dr + Vh dh =
2⇡rh
⇡r 2
dr +
dh.
3
3
Since each error is at most 0.1cm, we can take dr = 0.1 and dh = 0.1 along with r = 10, h = 25
to give
dV =
500⇡
100⇡
(0.1) +
(0.1) = 20⇡.
3
3
The maximum error required is 20⇡cm3 .
⌅
180
8.14 Directional Derivatives and the Gradient Vector
Imagine you are hiking in the Grand Canyon. Let’s think of your altitude at the point given
by longitude x and latitude y as a function f (x, y).
Facing east (in the direction of positive x-axis), the slope is given by the partial derivative
@f
.
@x
Facing north (in the direction of positive y-axis), the slope is given by the partial derivative
@f
.
@y
How to compute the slope when you are facing any given direction, say north-east?
Definition 8.15 (Directional Derivative).
The directional derivative of f (x, y) at (x0 , y0 ) in the direction of unit vector u = ha, bi is
f (x0 + ha, y0 + hb) f (x0 , y0 )
h
h!0
Du f (x0 , y0 ) = lim
provided this limit exists.
By looking at the figure above, we can think of the directional derivative Du f (x0 , y0 ) as the
slope to the point P(x0 , y0 , z0 ) on the surface in the direction given by u.
Notice
• if u = i = h1, 0i then
Di f = f x .
181
• if u = j = h0, 1i then
Dj f = f y .
In other words, the partial derivatives of f with respect to x and y are just special cases of
the directional derivative.
In practice, we do not usually compute the directional derivative using the definition. Instead, we compute it using the dot product of the vector consisting of partial derivatives and
the unit direction vector u.
Theorem 8.11 (Computing Directional Derivative).
If f (x, y) is a di↵erentiable function, then f has a directional derivative in the direction of any
unit vector u = ha, bi and
Du f (x, y) = f x (x, y)a + f y (x, y)b.
We can rewrite it in terms of vectors:
Du f (x, y) = hf x , f y i · ha, bi = hf x , f y i · u.
Consider the vector hf x , f y i =
@f
@f
i + @y j.
@x
It turns out that this vector has much significance. So we give it a special name.
Definition 8.16 (Gradient).
The gradient of f (x, y) is the vector-valued function
rf (x, y) = hf x , f y i = f x i + f y j =
@f
@f
i+
j
@x
@y
provided both partial derivatives exist.
rf is read ‘del f ’.
With this notation, we have
Du f (x, y) = rf (x, y) · u
Example 8.25. Find the directional derivative of the function f (x, y) = x2 y 3
(2, 1) in the direction of the vector v = 2i + 5j.
Solution. First compute the gradient vector at (2, 1):
rf (x, y) = 2xy 3 i + (3x2 y 2 4)j
rf (2, 1) = 4i + 8j.
182
4y at the point
Notice v is NOT a unit vector, since
p
p
| v|| = 22 + 52 = 29.
The unit vector in the direction of v is
u=
v
2
5
= p i + p j.
| v||
29
29
Therefore
Du f (2, 1) = rf (2, 1) · u
2
5
= h 4, 8i · h p , p i
29 29
32
= p .
29
⌅
For functions of three variables, we can define directional derivative in a similar manner.
Definition 8.17 (3-D Directional Derivative).
The directional derivative of f (x, y, z) at (x0 , y0 , z0 ) in the direction of unit vector u = ha, b, ci
is
f (x0 + ha, y0 + hb, z0 + hc) f (x0 , y0 , z0 )
Du f (x0 , y0 , z0 ) = lim
h
h!0
provided this limit exists.
Just as with functions of two variables, we have
Theorem 8.12 (Computing 3-D Directional Derivative).
where
Du f (x0 , y0 , z0 ) = rf (x0 , y0 , z0 ) · u
rf = hf x , f y , f z i =
@f
@f
@f
i+
j+
k
@x
@y
@z
is the gradient vector.
What is so significant about rf ?
183
Theorem 8.13 (Level Curve vs rf ).
Suppose f (x, y) is di↵erentiable function of x and y at (x0 , y0 ).
Suppose rf (x0 , y0 ) , 0. Then rf (x0 , y0 ) is perpendicular/normal to the level curve f (x, y) = k
at the point (x0 , y0 ) where f (x0 , y0 ) = k.
Using a similar argument, we can prove that this phenomenon also holds for level surfaces
F(x, y, z) = k.
Theorem 8.14 (Level Surface vs rf ).
Suppose F(x, y, z) is di↵erentiable function of x, y and z at (x0 , y0 , z0 ). Suppose S is the level
surface F(x, y, z) = k containing (x0 , y0 , z0 ). Let C be any curve that lies on S and passes
through (x0 , y0 , z0 ). Let r(t) be a parametric equation of C such that r(t0 ) = hx0 , y0 , z0 i.
Suppose rF(x0 , y0 , z0 ) , 0. Then
rF(x0 , y0 , z0 ) · r0 (t0 ) = 0,
That is, the rF(x0 , y0 , z0 ) is perpendicular/normal to tangent vector r0 (t0 ) to any curve C on
the surface S that passes through (x0 , y0 , z0 ).
F(x0 , y0 , z0 ) = k, rF(x0 , y0 , z0 ) · r0 (t0 ) = 0.
Consequently, the tangent plane to the level surface F(x, y, z) = k at (x0 , y0 , z0 ) is given by
the equation
184
Theorem 8.15 (Tangent Plane to Level Surface).
rF(x0 , y0 , z0 ) · hx x0 , y
y0 , z z0 i = 0
or equivalently,
Fx (x0 , y0 , z0 )(x x0 ) + Fy (x0 , y0 , z0 )(y
y0 ) + Fz (x0 , y0 , z0 )(z z0 ) = 0.
Example 8.26. Find the equations of the tangent plane and normal line at the point ( 2, 1, 3) to
the ellipsoid
x2
z2
+ y 2 + = 3.
4
9
Solution. The ellipsoid is the level surface (with k = 3) of the function
F(x, y, z) =
x2
z2
+ y2 + .
4
9
Therefore,
x
2z
Fx (x, y, z) = , Fy (x, y, z) = 2y, Fz =
2
9
Fx ( 2, 1, 3) = 1, Fy ( 2, 1, 3) = 2, Fz ( 2, 1, 3) =
2
.
3
The equation of the tangent plane at ( 2, 1, 3) is
rF( 2, 1, 3) · hx ( 2), y
1(x + 2) + 2(y
1)
1, z ( 3)i
2
(z + 3) = 0,
3
which simplifies to
3x 6y + 2z + 18 = 0.
The normal vector to the plane is h3, 6, 2i. So the parametric equations of the normal line
are
x = 2 + 3t, y = 1 6t, z = 3 + 2t, t 2 R.
⌅
185
Remark. Theorem 8.5 is a special case of Theorem 8.15, and one may deduce Theorem 8.5
from Theorem 8.15 as follows:
Suppose S is the surface defined by z = f (x, y), and (a, b, f (a, b)) is a point on S. Now consider
the function F(x, y, z) = f (x, y) z. Notice that
z = f (x, y) () f (x, y) z = 0 () F(x, y, z) = 0.
Thus we may regard S as the level surface F(x, y, z) = 0. Note also that at a point (x0 , y0 , z0 ),
we have
Fx (x0 , y0 , z0 ) = f x (x0 , y0 ), Fy (x0 , y0 , z0 ) = f y (x0 , y0 ), Fz (x0 , y0 , z0 ) = 1.
Then from Theorem 8.15 (with (z0 , y0 , z0 ) = (a, b, f (a, b)), the equation of the tangent plane at
(a, b, f (a, b)) to S (as the level surface F(x, y, z) = 0) is given by
Fx (a, b, f (a, b))(x a) + Fy (a, b, f (a, b))(y
b) + Fz (a, b, f (a, b))(z f (a, b)) = 0
or equivalently,
f x (a, b)(x a) + f y (a, b)(y
b) (z f (a, b)) = 0,
which is the same equation as given in Theorem 8.5.
⌅
Let’s return to the directional derivative Du f (x, y, z) and ask some questions.
We know that Du f (x, y, z) = rf (x, y, z) · u is a scalar function of x, y and z (because it is a dot
product of two vectors). Geometrically, we think of Du f (x0 , y0 , z0 ) as the rate of change of f
at (x0 , y0 , z0 ) in the direction of u.
Question: at a given point (x0 , y0 , z0 ), in which direction does f change the fastest? In other
words, what is the maximum rate of change of f at (x0 , y0 , z0 )?
The answer lies in rf !
Let ✓ be the angle between rf and u. Then
Du f
= rf · u
= | rf | | u|| cos ✓
= | rf | cos ✓ since u is a unit vector
Du f = | rf | cos ✓.
The maximum value of cos ✓ is 1 and this happens when ✓ = 0.
So the maximum value of Du f is | rf | and it occurs when ✓ = 0, i.e. u points in the direction
of rf .
The minimum value of cos ✓ is 1 and this happens when ✓ = ⇡.
So the minimum value of Du f is | rf | and it occurs when ✓ = ⇡, i.e. u points in the direction
of rf .
186
Theorem 8.16 (Maximizing Rate of Increase/Decrease of f ).
Suppose f is a di↵erentiable function of two or three variables. Let P denote a given point.
Assume rf (P) , 0. Let u be a unit vector making an angle ✓ with rf . Then
Du f (P) = | rf (P)|| cos ✓.
Moreover,
• rf (P) points in the direction of maximum rate of increase of f at P (maximum value of
Du f (P) is | rf (P)||)
• rf (P) points in the direction of maximum rate of decrease of f at P (minimum value of
Du f (P) is | rf (P)||)
Example 8.27. Let f (x, y) = xe y . In what direction does f have the maximum rate of change at
the point P(2, 0)? What is this maximum rate of change.
Solution. Note that
rf (x, y) = hf x , f y i = he y , xe y i.
f increases fastest in the direction of the gradient vector
rf (2, 0) = h1, 2i.
The maximum rate of change is
p
p
| rf (2, 0)|| = 12 + 22 = 5.
⌅
8.15 Extrema of Functions of Two Variables
In the real world, we always seek to optimize our resources.
Given our constraints (our time, ability, finance, health, family background and what not),
getting an A in MA1521 itself can be seen as an optimization problem.
Similar to the study of extrema of functions of one variable, two key concepts are the local
maximum/minimum and the absolute maximum/minimum.
Definition 8.18 (Local and Absolute Maximum).
Let f (x, y) : D ! R. Then
• f has a local maximum at (a, b) if f (x, y)  f (a, b) for all points in some disk with center
(a, b). The number f (a, b) is called a local maximum value.
187
• f has an absolute maximum at (a, b) if f (x, y)  f (a, b) for all points in the domain D.
The number f (a, b) is called a absolute maximum value.
Definition 8.19 (Local and Absolute Minimum).
Let f (x, y) : D ! R. Then
• f has a local minimum at (a, b) if f (x, y) f (a, b) for all points in some disk with center
(a, b). The number f (a, b) is called a local minimum value.
• f has an absolute minimum at (a, b) if f (x, y) f (a, b) for all points in the domain D. The
number f (a, b) is called a absolute minimum value.
8.15.1 Local Extrema
A key observation that will be used repeatedly when finding local extrema of functions is
the following.
Theorem 8.17.
If f has a local maximum or minimum at (a, b) and the first-order derivatives of f exist there,
then
f x (a, b) = f y (a, b) = 0.
Proof. Let g(x) = f (x, b). Then g is a function of a single variable x. If f has a local maximum/minimum at (x, y) = (a, b) then g has a local maximim/minimim at x = a. So g 0 (a) = 0.
But g 0 (a) = f x (a, b). So f x (a, b) = 0.
Similarly, f y (a, b) = 0.
⌅
There is a geometric interpretation of the preceding theorem:
If f has a tangent plane at a local maximum/minimum (a, b), then the tangent plane has
equation
z = f (a, b) + f x (a, b)(x a) + f y (a, b)(y
b) = f (a, b),
that is the tangent plane is a horizontal plane parallel to the xy-plane.
If I stand on a local maximum/minimum then ......
188
In the following, I am definitely NOT standing on local maximum/minimum ......
Definition 8.20 (Critical or Stationary Point).
Let f (x, y) : D ! R. Then a point (a, b) is called a critical point of f if
• f x (a, b) = 0 and f y (a, b) = 0, OR
• one of the partial derivatives does not exist.
Clearly,
(a, b) local maximum/minimum point
=)
(a, b) critical point.
However, the converse IS NOT TRUE!
Example 8.28. Find the extreme (maximum/minimum) values of f (x, y) = y 2 x2 .
Solution. Extreme values can only occur at critical points. Since f x = 2x and f y = 2y, the
only critical point is (0, 0).
We still have to check whether f (0, 0) is a maximum/minimum value.
Note that f (0, 0) = 0.
If y = 0 and x , 0, then
189
f (x, y) = x2 < 0 = f (0, 0).
If x = 0 and y , 0, then
f (x, y) = y 2 > 0 = f (0, 0).
Therefore, f (0, 0) cannot be an extreme value for f . So f has no extreme values.
⌅
In the preceding example, we see that a critical point needs not be an extreme point. But
the behavior of the critical point in the preceding example is interesting.
If you look at the graph of f (x, y) = y 2 x2 , you will see that f (0, 0) = 0 is a maximum in the
direction of the x-axis but a minimum in the direction of the y-axis.
This motivates the following definition.
Definition 8.21 (Saddle Point).
Let f (x, y) : D ! R. Then
A point (a, b) is called a saddle point of f if
• it is a critical point of f , AND
• every open disk centered at (a, b) contains points (x, y) 2 D for which f (x, y) < f (a, b) and
points (x, y) 2 D for which f (x, y) > f (a, b).
190
Suppose you are standing on a surface and you are standing upright (parallel to
the z-axis). Moreover, when you begin walking, some directions take you uphill
while other directions take you downhill.
Then you are standing at a saddle point!
We cannot rely on our visualization of 3D-graphs to locate extreme points.
Luckily, we have the second derivative test to determine whether a given critical point is
local maximum/minimum, saddle point or neither.
Theorem 8.18 (Second Derivative Test).
Suppose f (x, y) has continuous second-order partial derivatives on some open disk centered at
(a, b). Suppose f x (a, b) = f y (a, b) = 0 (that is (a, b) is a critical point). Define the discriminant
D for the point (a, b) by
D = D(a, b) = f xx (a, b)f yy (a, b)
h
i2
f xy (a, b) .
(a) If D > 0 and f xx (a, b) > 0, then f (a, b) is a local minimum.
(b) If D > 0 and f xx (a, b) < 0, then f (a, b) is a local maximum.
(c) If D < 0, then (a, b) is a saddle point of f .
(d) If D = 0, then no conclusion can be drawn.
Example 8.29. Locate and classify all critical points for f (x, y) = x3 2y 2 2y 4 + 3x2 y.
Solution. We have
f x = 3x2 + 6xy,
f y = 4y
8y 3 + 3x2 .
Step 1. Locate critical points.
Let’s solve the system
If x = 0, then by (2) we have 4y
solution (0, 0).
If x , 0, then by (1) we have y =
3x2 + 6xy = 0
(1)
4y 8y 3 + 3x2 = 0 (2)
8y 3 = 0 , 4y(1 + 2y 2 ) = 0 , y = 0. Thus we obtain one
x
2.
Substituting this into (2), we have
191
4( 2x ) 8(
x , 0.
Using y =
x 3
2)
x
2,
+ 3x2 = 0 , 2x + x3 + 3x2 = 0 , x(x + 2)(x + 1) = 0 , x = 1, 2. Note that
we obtain the two solutions ( 1, 12 ), ( 2, 1).
Therefore, the critical points are (0, 0), ( 1, 12 ), ( 2, 1).
Step 2. Classify (if possible) these critical points using Second Derivative Test.
We need
f xx = 6x + 6y,
f xy = 6x,
f yy = 4 24y 2 .
At the critical point ( 2, 1), we have
D( 2, 1) = f xx ( 2, 1)f yy ( 2, 1) [f xy ( 2, 1)]2 = ( 6) · ( 28) ( 12)2 = 24 > 0.
Note also that f xx ( 2, 1) = 6 > 0. Thus by the Second Derivative Test, we know that f has a
local maximum point at ( 2, 1).
We can make similar computations at the other two critical points (0, 0) and ( 1, 12 ), and the
result is tabulated as follows:
critical point
(0, 0)
( 1, 12 )
( 2, 1)
D
0
6<0
24 > 0
f xx
6<0
2nd Derivative Test’s Conclusion
inconclusive
saddle point
local maximum
We need a di↵erent analysis to deal with the critical point (0, 0).
Notice in the plane y = 0, f (x, y) = f (x, 0) = x3 . We know from Calculus that this curve has
an inflection point at x = 0. So there is no local extremum at this point.
Moreover, when we start walking from (0, 0) in positive direction of x, we will be walking
uphill; in the negative direction of x, we will be walking downhill.
So (0, 0) is a saddle point.
⌅
Exercise 8.1. Find the critical points of f (x, y) = 3xe
Ans. Local minimum at
p1 ,
2
Local maximum at
x2 y 2
and classify them.
p1 .
2
Exercise 8.2. A delivery company only accepts rectangular boxes the sum of whose length and
girth (perimeter of a cross-section) does not exceed 270 cm. Find the dimensions of an acceptable
box of largest volume.
192
Ans. 90cm ⇥ 45cm ⇥ 45cm.
193
194
Chapter 9
Double Integrals
Read Thomas’ Calculus, Chapter 15.
9.1 Riemann Sum
Having studied derivatives for functions of several variables and their applications, we now
turn to introducing the idea of integral for functions of several variables. It turns out that
these ideas are useful in many practical problems.
Recall that in Calculus of Single Variable, our attempt to find the area under a curve led to
the definition of a definite integral.
We now seek to find volume under a surface and in the process we arrive at the definition of
a double integral.
We start by reviewing how we arrive at the definite integral of functions of a single variable:
Step 1. Suppose f (x) is defined for a  x  b. We divide the interval [a, b] into n subintervals
of equal size 4x = bn a .
Step 2. We choose sample points xi⇤ from these subintervals and form the Riemann Sum
n
X
f (xi⇤ )4x.
i=1
Step 3. Take the limit of such sum as n ! 1 to obtain the definite integral of f from a to b:
Z
b
a
In the special case where f (x)
f (x) from a to b.
f (x)dx = lim
n
X
n!1
0, the integral
Rb
a
195
f (xi⇤ )4x.
i=1
f (x)dx represents the area under the curve
9.2 Volume and Double Integral
Suppose f (x, y) is a function of two variables defined on a closed rectangle
R = [a, b] ⇥ [c, d] = {(x, y) 2 R2 : a  x  b, c  y  d}.
Suppose f (x, y)
0. The graph of f is a surface with z = f (x, y) above the region R.
Let S be the solid that lies above R and under the graph of f .
How can we find the volume of S?
We can estimate the volume of S as follows:
Step 1. divide the rectangle R into subrectangles. We do this by
• dividing the interval [a, b] into m subintervals [xi 1 , xi ] of equal length 4x =
• dividing the interval [c, d] into n subintervals [yj 1 , yj ] of equal length 4y =
Form subrectangles
Rij = [xi 1 , xi ] ⇥ [yj 1 , yj ],
for all 1  i  m, 1  j  n.
Each of these subrectangles has area 4A = 4x4y.
196
b a
m ,
d c
n .
and
⇤
Step 2. Choose a sample point (xij
, yij⇤ ) in each Rij . Then approximate the part of S lies
⇤
above Rij by a thin rectangle box with base Rij and height f (xij
, yij⇤ ). The volume of this box
is given by
⇤
f (xij
, yij⇤ )4A.
It follows that by adding the volumes of all these thin boxes, we get an approximation of the
total volume of S:
V⇡
m X
n
X
⇤
f (xij
, yij⇤ )4A.
i=1 j=1
Our intuition tells us that the approximation becomes better as m, n ! 1. So we would
expect
V = lim
m,n!1
m X
n
X
i=1 j=1
197
⇤
f (xij
, yij⇤ )4A.
We make the following definition:
Definition 9.1 (Double Integral).
The double integral of f over the rectangle R is
ZZ
f (x, y) dA = lim
m,n!1
R
m X
n
X
⇤
f (xij
, yij⇤ )4A
i=1 j=1
⇤
provided the limit exists and is the same for any choice of the sample points (xij
, yij⇤ ) in Rij , for
1  i  m, 1  j  n.
When this happens, we say that f is integrable over R.
Remark. It can be shown that all continuous functions are integrable.
By comparing our definition of integral and volume, we have
Theorem 9.1 (Volume as a Double Integral).
If f (x, y) 0, the volume V of the solid that lies above the rectangle R and below the surface
z = f (x, y) is
ZZ
V=
f (x, y) dA.
R
Some properties of double integral:
Assuming all the integrals exist, we have
ZZ
ZZ
ZZ
(f (x, y) + g(x, y)) dA =
1.
R
f (x, y) dA +
R
ZZ
ZZ
2.
cf (x, y) dA = c
R
g(x, y) dA.
R
f (x, y) dA.
R
198
3. If f (x, y)
g(x, y) for all (x, y) 2 R, then
ZZ
ZZ
f (x, y) dA
R
g(x, y) dA.
R
9.3 Iterated Double Integral
Recall that it is usually difficult to evaluate a single integral directly from definition, but the
Fundamental Theorem of Calculus provides a much easier method:
Z
b
f (x) dx = F(b) F(a)
a
where F(x) is an antiderivative of f (x).
For double integral, it is even more difficult to compute it from first principles. We now
see how to express a double integral as an iterated integral which can be evaluated by
calculating two single integrals.
Suppose f (x, y) is integrable over the rectangle R = [a, b] ⇥ [c, d].
Rd
We use the notation c f (x, y) dy to mean that x is held fixed and f (x, y) is integrated with
respect to y from c to d.
This procedure is called partial integration with respect to y. Notice the similarity to partial di↵erentiation.
Rd
So c f (x, y) dy is a function of x, as it depends on the value of x: set
Z
d
A(x) =
f (x, y) dy.
c
We now integrate A(x) from a to b:
Z
b
a
A(x) dx =
Z b "Z
a
d
#
f (x, y) dy dx.
c
The integral on the right-hand side is called an iterated integral.
Usually, we omit the brackets:
Z bZ
a
d
c
f (x, y) dy dx =
Z b "Z
a
199
d
c
#
f (x, y) dy dx.
Definition 9.2 (Iterated Integral).
Z bZ
a
d
f (x, y) dy dx
c
means we first integrate with respect to y from c to d (keeping x fixed) and then with respect
to x from a to b.
Z dZ
c
b
f (x, y) dx dy
a
means we first integrate with respect to x from a to b (keeping y fixed) and then with respect
to y from c to d.
Example 9.1. Evaluate the iterated integral
Z 2Z
1
3
0
x2 y dx dy.
Solution. We first integrate with respect to x and then with respect to y:
Z 2Z
1
3
0
2
x y dx dy =
=
Z 2 "Z
1
1
Z
=
2
1
"
=
=
0
Z 2"
#
3
x3 y
3
2
x y dx dy
#3
dy
0
9y dy
9y 2
2
#2
1
27
.
2
⌅
Example 9.2. Evaluate the iterated integral
Z 3Z
0
2
1
x2 y dy dx.
Solution. We first integrate with respect to y and then with respect to x:
200
Z 3Z
0
2
1
2
x y dy dx =
=
Z 3 "Z
0
Z 3"
0
Z
#
2
2
x y dy dx
1
x2 y 2
2
#2
dx
1
3
3 2
x dx
0 2
" 3 #3
x
=
2 0
=
=
27
.
2
⌅
Notice in both of the preceding examples, we obtained the same answer.
It seems that the order of integration (with respect to x or y first) does not matter. This is
similar to Clairaut’s Theorem for mixed partial derivatives.
Indeed, if f is continuous on R, this is always true. Moreover, the (iterated) integral is equal
to the corresponding double integral.
The following theorem gives a practical way for evaluating a double integral by expressing
it as an iterated integral (in either order):
Theorem 9.2 (Fubini’s Theorem).
If f is continuous on the rectangle R = [a, b] ⇥ [c, d], then
ZZ
f (x, y) dA =
R
Z bZ
a
d
f (x, y) dy dx =
c
Z dZ
c
b
f (x, y) dx dy.
a
More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a
finite number of smooth curves, and the iterated integrals exist.
Example 9.3. Evaluate
ZZ
y sin(xy) dA
R
where R = [1, 2] ⇥ [0, ⇡].
Solution 1. Using Fubini’s Theorem, let’s integrate first with respect to x.
201
ZZ
Z
y sin(xy) dA =
R
Z
=
Z
=
=
⇡Z 2
0
1
⇡
0
⇡
0

y sin(xy) dx dy
[ cos(xy)]21 dy
( cos 2y + cos y) dy
1
sin 2y + sin y
2
⇡
0
= 0.
Solution 2. By Fubini’s Theorem, we should get the same answer if we first integrate with
respect to y:
ZZ
y sin(xy) dA =
R
Z 2Z
1
⇡
0
y sin(xy) dy dx.
We first need to compute
Z
⇡
0
y sin(xy) dy.
Using integration by parts,
Z
⇡
0
"
y sin(xy) dy =
y cos(xy)
x
#⇡
Z
⇡
0
0
cos(xy)
dy
x
=
⇡ cos ⇡x 1
+ 2 [sin xy]⇡0
x
x
=
⇡ cos ⇡x sin ⇡x
+
.
x
x2
Now, integrating the first term by parts, we have
Z ✓
◆
⇡ cos ⇡x
sin ⇡x
dx =
x
x
Z
sin ⇡x
dx
x2
So
Z ✓
◆
⇡ cos ⇡x sin ⇡x
sin ⇡x
+
dx =
.
2
x
x
x
Hence
202
Z 2Z
1
⇡
0
y sin(xy) dy dx =
=

sin ⇡x
x
2
1
sin 2⇡
+ sin ⇡ = 0.
2
⌅
Now, we make some observation about the last example.
• Though both solutions give the same answer, the first solution is much easier than the
second one. Therefore, when we evaluate double integrals, it is wise to choose the right
order of integration that yields simpler calculations.
• Consider the surface z = y sin(xy).
This function takes both positive and negative values on R = [1, 2] ⇥ [0, ⇡].
RR
For such a function, R f (x, y) dA is a di↵erence of volumes: V1 V2 where V1 is the volume
above R and below the graph of f and V2 is the volume below R and above the graph.
The fact that the integral is 0 in the preceding example means that these two volumes V1
and V2 are equal.
203
9.4 A Special Case
Sometimes f (x, y) can be factored as the product of a function of x only and a function of y
only. That is
f (x, y) = g(x)h(y).
Then Fubini’s Theorem gives
ZZ
Z dZ
f (x, y) dA =
b
g(x)h(y) dx dy
R
c
a
Z d "Z
=
#
b
g(x)h(y) dx dy
c
a
In the inner integral, y is a constant, so h(y) is a constant and we can write
ZZ
f (x, y) dA =
Z d"
Z
h(y)
R
c
#
b
g(x) dx dy
a
Z
! Z
b
=
g(x) dx
a
since
Rb
a
!
d
h(y) dy
c
g(x) dx is a constant.
To summarize:
Theorem 9.3 (A Special Case).
ZZ
Z
g(x)h(y) dA =
R
! Z
b
g(x) dx
a
!
d
h(y) dy
c
where R = [a, b] ⇥ [c, d].
Example 9.4. Evaluate
ZZ
2x
dA
R y
where R = [3, 4] ⇥ [1, 2].
Solution: By Theorem 9.3,
ZZ
ZZ
⇣Z 4
⌘ ⇣ Z 2 1 ⌘ h i4 h
i2
2x
1
dA =
2x· dA =
2x dx ·
dy = x2 · ln |y| = (42 32 )·(ln 4 ln 1) = 7 ln 4.
3
1
y
R y
R
3
1 y
⌅
204
9.5 Double Integral over General Region
So far we have defined double integrals over domains which are rectangles. In this section,
we shall define double integrals over domains which are more general than rectangles. In
particular, they are regions which are bounded between two continuous curves. They are
called Type I and Type II regions respectively.
Definition 9.3. Type I Region
A plane region D is said to be of Type I if it lies between the graphs of two continuous functions
of x, that is,
D = {(x, y) : a  x  b, g1 (x)  y  g2 (x)}
where g1 (x) and g2 (x) are continuous on [a, b].
Some examples of Type I region:
Definition 9.4. Type II Region
A plane region D is said to be of Type II if it lies between the graphs of two continuous
functions of y, that is,
D = {(x, y) : c  y  d, h1 (y)  x  h2 (y)}
where h1 (y) and h2 (y) are continuous on [c, d].
Some examples of Type II region:
205
How do we compute the integral of f (x, y) over Type I region D?
Theorem 9.4. Double Integral over Type I Domain
If f is continuous on a Type I domain D such that
D = {(x, y) : a  x  b, g1 (x)  y  g2 (x)}
then
ZZ
f (x, y) dA =
D
Z bZ
a
g2 (x)
g1 (x)
f (x, y) dy dx.
Observe that the expression on the right-hand side of
ZZ
f (x, y) dA =
D
Z bZ
a
g2 (x)
g1 (x)
f (x, y) dy dx
is an iterated integral similar to the ones we have for rectangle region, except that in the
inner integral, we regard x as being constant not only in f (x, y) but also in the limits of the
integration, g1 (x) and g2 (x).
Similarly, we have
Theorem 9.5. Double Integral over Type II Domain
If f is continuous on a Type II domain D such that
D = {(x, y) : c  y  d, h1 (y)  x  h2 (y)}
206
then
ZZ
f (x, y) dA =
Z dZ
D
Example 9.5. Evaluate
and y = 1 + x2 .
RR
D
h2 (y)
h1 (y)
c
f (x, y) dx dy.
(x + 2y) dA where D is the region bounded by the parabolas y = 2x2
Solution.
Step 1. Identify the region.
Notice the parabolas intersect when 2x2 = 1 + x2 , that is, x = ±1.
We note that D is a Type I region:
D = {(x, y) : 1  x  1, 2x2  y  1 + x2 }.
Step 2. Set up the iterated integral.
Therefore,
ZZ
Z
(x + 2y) dA =
D
1
Z
1+x2
2x2
1
(x + 2y) dy dx.
Step 3. Evaluate the inner integral.
Z
1+x2
2x2
(x + 2y) dy =
h
xy + y 2
iy=1+x2
y=2x2
= x(1 + x2 ) + (1 + x2 )2 x(2x2 ) (2x2 )2
= 3x4 x3 + 2x2 + x + 1.
Step 4. Complete the computation.
207
ZZ
Z
(x + 2y) dA =
D
"
1
1
( 3x4 x3 + 2x2 + x + 1) dx
x5
=
3
5
32
=
.
15
x4
x3 x2
+2 +
+x
4
3
2
#1
1
⌅
When we set up a double integral, it is helpful to draw a diagram.
For Type I region, it is helpful to draw a vertical arrow which starts at the lower
boundary y = g1 (x) and ends at the upper boundary y = g2 (x). This corresponds to
the inner integral.
For Type II region, it is helpful to draw a horizontal arrow which starts at the left
boundary x = h1 (x) and ends at the right boundary x = h2 (x). This corresponds to
the inner integral.
RR
Example 9.6. Evaluate D xy dA where D is the region bounded by the line y = x 1 and the
parabola y 2 = 2x + 6.
Solution. The region D can be of Type I or II:
But we prefer D as Type II because as a Type I region, the lower boundary of D is more
complicated, in particular, it consists of two parts: one for 3  x  1 and another for
1  x  5.
Therefore, we let
1
D = {(x, y) : 2  y  4, y 2 3  x  y + 1}.
2
208
So
ZZ
Z
xy dA =
D
Z
4
y+1
1 2
2y
2
3
xy dx dy.
Lets first compute the inner integral:
Z
y+1
y2
2
3
"
xy dx =
x2
y·
2
#x=y+1
x=
y2
2
3
y2
1
=
y(y + 1)2 y(
2
2
1
2
=
y5
+ 4y 3 + 2y 2
4
!
3 + 1)
!
8y .
2
Therefore,
ZZ
!
1 y5
3
2
xy dA =
+ 4y + 2y 8y dy
4
D
22
"
#4
y3
1 y6
4
2
=
+y +2
4y
2 24
3
2
= 36.
Z
4
⌅
Example 9.7. Find the volume of the tetrahedron T bounded by the planes x + 2y + z = 2, x = 2y,
x = 0 and z = 0.
Solution. For question like this, it is wise to draw two diagrams:
• one for the solid (tetrahedron) T ,
• another for the domain D.
How do we start drawing?
209
There is no general rule, it depends on the problem in the question.
Notice the plane x + 2y + z = 2 intersects the xy-plane in the line x + 2y = 2. (Set z = 0 in the
equation of the plane).
Together with the restrictions that the solid is bounded by z = 0 (above the xy-plane), x = 0
(the yz-plane) and the plane x = 2y, we see that T lies above the region D in the xy-plane
bounded by the lines:
• x = 2y,
• x + 2y = 2 (the intersection of the plane x + 2y + z = 2 and the plane z = 0),
• x = 0.
Notice that (1, 12 , 0) and (0, 1, 0) are two points on the plane x + 2y + z = 2.
There is another point on this plane: (0, 0, 2).
We can now draw the tetrahedron T as follows:
So the required volume V lies under the graph z = 2 x 2y and above
210
x
y1
2
D = {(x, y) : 0  x  1,
x
}.
2
Therefore
ZZ
V =
(2 x 2y) dA
D
=
=
=
Z 1Z
0
x/2
Z 1h
0
Z1
"
=
1 x/2
0
x3
3
2y
(2 x 2y) dy dx
y2
xy
iy=1
x/2
y=x/2
dx
(x2 2x + 1) dx
#1
2
x +x
0
1
=
.
3
⌅
Remark. (Interchanging the order of integration) When computing an iterated integral
over a domain which is of both type I and type II, it is sometimes easier to compute the
integral by interchanging the order of integration.
Example 9.8. Evaluate the iterated integral I =
Z 1Z
der of integration.
0
Z
[Remark. It is difficult to compute the inner integral
1
sin(y 2 ) dydx by interchanging the or-
x
1
sin(y 2 ) dy.]
x
Ans:
1
(1 cos 1).
2
Solution. The region of integration D is the triangular region bounded by the lines y = x, x =
0 and y = 1. Notice that D is a domain of both type I and type II.
211
Z 1Z
0
1
Z
2
2
sin(y ) dydx =
x
sin(y ) dA =
Z 1Z
D
=
=
Z 1h

0
x sin(y 2 )
1
cos(y 2 )
2
ix=y
x=0
y
sin(y 2 ) dxdy
0
0
Z1
dy =
y sin(y 2 ) dy
0
1
1
= (1 cos 1).
2
0
9.6 Decomposing Domain into Smaller Domains
Double integrals are additive with respect to the domain: if D is the union of domains
D1 , . . . , Dn that do not overlap except possibly on boundary curves, then
Theorem 9.6. Additivity With Respect to Domain
ZZ
ZZ
ZZ
f (x, y) dA =
D
D1
f (x, y) dA + · · · +
f (x, y) dA.
Dn
Additivity may be used to evaluate double integrals over domain D which is neither of Type
I nor II but can be decomposed into finitely many domains of Type I or II.
"
Example 9.9. Suppose we want to compute the double integral
xy dA, where D is the shaded
region bounded by the curve y 2 = 2x, and the lines 2x 3y
below.
D
4 = 0 and 3x + 8y
y
.....
.......
....
..
...
...
...............
y2
...
......................
............... . .....
...
2................................................
.
...
.
.
.
.
3y
.
.
.
.
... ........................................
2
... ............ ...1............
.
.
.................. ..............
......................................................................................................................................................................................................................................
............... ......................
... ............ ..........................
... .............. ........................................
8y
........ . . . . . . . . ..........
...
........ . . . . . . . . ...........
........... . .........................
...
3
...............
................2................................
...
............... . . . . . . . .........
...
.............. . .......... ..........
............... . . . . ...........
...
................. . . . .........
.................. ..............
...
.................. ..........
...
..........................
1
2
...
....
..
.
• (8, 4)
x=
0
D
•
2
x=
D
+2
x=
D = D [D
x
+2
• (18, 6)
212
6 = 0 as shown
Notice that the region D is not a domain of type I or type II. Nonetheless, D is a union of two
domains of type I I as follows:
where
D = D1 [ D2 ,
y2
3y
x
+ 2},
2
2
y2
8y
D2 = {(x, y) : 6  y  0,
x
+ 2}.
2
3
D1 = {(x, y) : 0  y  4,
Then by Theorem 9.6, we have
"
"
xy dA =
D
"
xy dA +
xy dA
D1
=
Z 4Z
0
D2
3y
2 +2
y2
2
Z
xy dxdy +
0
6
Z
8y
3 +2
y2
2
xy dxdy,
and then the two iterated itegrals in the last line can be computed readily.
⌅
9.7 Properties of Double Integral
The following properties for double integral over D follow from the corresponding properties for double integrals over a rectangle region R:
Theorem 9.7.
ZZ
ZZ
ZZ
[f (x, y) + g(x, y)] dA =
f (x, y) dA +
D
D
Theorem 9.8.
ZZ
ZZ
cf (x, y) dA = c
D
f (x, y) dA
D
Theorem 9.9.
If f (x, y) g(x, y) for all (x, y) 2 D then
ZZ
f (x, y) dA
ZZ
D
g(x, y) dA.
D
213
g(x, y) dA
D
Example 9.10. (i) Let D be a domain in R2 . Then Theorem 9.7 and Theorem 9.8, we have
"
"
Z
2
2
(x + 2xy) dA =
x dA + 2
xy dA.
D
(ii) Show that
D
"
2
D
Z
2
(x + y ) dA
2xy dA.
D
D
Proof of (ii). For all (x, y) in D, we have
(x y)2
0 =) x2 2xy + y 2 0 =) x2 + y 2
"
"
2
2
Thus by Theorem 9.9, we have
(x + y ) dA
2xy dA.
D
2xy.
D
⌅
9.8 An Application – Finding Area
We can use double integral to compute area of a region D on the plane:
Theorem 9.10. Area of plane region
Let f (x, y) = 1 over a given region D. Then the area of D is
ZZ
A(D) =
1 dA.
D
Proof. By considering the constant function f (x, y) = 1 for all (x, y) in D, we see that
is the volume of the solid which is a cylinder whose base is A(D) and height 1.
Another way of computing the volume of a cylinder is
area of base ⇥ height
which is
A(D) · 1
214
RR
D
1 dA
in this case. So
ZZ
A(D) =
1 dA,
D
as required.
⌅
9.9 Double Integrals in Polar Coordinates
We have learned how to evaluate double integral over D where D is of the following type:
• rectangle;
• region of Type I or Type II.
Sometimes, the region D is not so easily described in terms of x and y coordinates.
Sometimes, such regions can be conveniently described using polar coordinates (r, ✓). The
following figure which shows the relationship between polar coordinates and the rectangle
coordinates:
Instead of using (x, y), we note that any point on the xy-plane can be represented by an
ordered pair (r, ✓) where
• r is the distance from the origin to the point
• ✓ is the angle from the positive x-axis to the straight line joining the origin and the point.
For example consider the following region given in terms of its polar coordinates:
215
Polar coordinates (r, ✓) of a point are related to the rectangle coordinate (x, y) by the equations
Theorem 9.11. Polar Coordinates Versus Rectangle Coordinates
r 2 = x2 + y 2 ,
x = r cos ✓,
y = r sin ✓.
Definition 9.5 (Polar Rectangle).
A polar rectangle is a region
R = {(r, ✓) : a  r  b, ↵  ✓  }.
How do we compute
ZZ
f (x, y) dA
R
where R is a polar rectangle?
RR
Recall that for R f (x, y) dA over the usual rectangle R, we can think of dA = dxdy as the
area of the ‘little rectangle’ 4A = 4x4y:
RR
To compute R f (x, y) dA over the polar rectangle R given by
R = {(r, ✓) : a  r  b, ↵  ✓  }.
we partition R as follows:
216
Then we can think of dA = r dr d✓ as the area of the ‘little polar rectangle’ 4A ⇡ 4r · r4✓:
Note that the arc of the polar rectangle is r 4✓ which depends on r.
Theorem 9.12. Change to Polar Coordinates in Double Integral
If f is continuous on a polar rectangle R given by
R = {(r, ✓) : 0  a  r  b, ↵  ✓  }
where 0 
↵  2⇡, then
ZZ
Z
Z
b
f (x, y) dA =
R
↵
f (r cos ✓, r sin ✓)r dr d✓.
a
The formula says that we convert from rectangle to polar coordinates in a double integral
by:
• writing x = r cos ✓, y = r sin ✓
• using the appropriate limits of integration for r and ✓
• replacing dA by r dr d✓ (do not forget the additional r in r dr d✓)
RR
Example 9.11. Evaluate R (3x + 4y 2 ) dA where R is the region in the upper half-plane bounded
by the circles x2 + y 2 = 1 and x2 + y 2 = 4.
Solution. The region R is shown below:
217
So
R = {(r, ✓) : 1  r  2, 0  ✓  ⇡}.
Changing to polar coordinates for the double integral, we have
ZZ
Z
2
(3x + 4y ) dA =
0
R
Z
=
Z
=
=
1
⇡Z
0
0
2
1
⇡h
Z 0⇡
Z
=
⇡Z 2
(3r cos ✓ + 4r 2 sin2 ✓)r dr d✓
(3r 2 cos ✓ + 4r 3 sin2 ✓) dr d✓
r 3 cos ✓ + r 4 sin2 ✓
7 cos ✓ +

15✓
= 7 sin ✓ +
2
=
r=1
d✓
(7 cos ✓ + 15 sin2 ✓) d✓
⇡
0
ir=2
15
(1 cos 2✓) d✓
2
15
sin 2✓
4
⇡
0
15⇡
.
2
⌅
Example 9.12. Find the volume of the solid bounded by the plane z = 0 and the paraboloid
z = 1 x2 y 2 .
Solution. Notice the plane and the paraboloid intersect in the circle x2 + y 2 = 1.
So the solid lies under the paraboloid and above the circular disk D given by x2 + y 2  1.
In polar coordinates, D is given by
218
D = {(r, ✓) : 0  r  1, 0  ✓  2⇡}.
Since 1 x2 y 2 = 1 r 2 , we have
ZZ
(1 x2 y 2 ) dA
D
Z 2⇡ Z 1
=
(1 r 2 )r dr d✓
0
0
! Z1
!
Z 2⇡
3
=
d✓
(r r ) dr
Volume =
0
"
= 2⇡
r2
0
#1
4
r
2
4
⇡
=
.
2
0
⌅
Exercise 9.1. Let R be the circular region bounded by the circle x2 + (y 1)2 = 1. It is known that
"
dA
⇡
= ,
2
2
2
a
R (1 + 2x + 2y )
where a is a positive integer. Determine the value of a.
[Hint: Use polar coordinates and evaluate the resulting integral by means of the substitution
t = tan ✓].
Ans. 6.
9.10 Surface Area
Let f be a di↵erentiable function of 2 variables defined on a domain
" D. We wish to find
the surface area of the graph of f over D. It is simply equal to
dS, where dS is the
D
di↵erential of the surface area of the graph of f . Therefore we need to express dS in terms
of the di↵erential dA of the domain. To do so, take any point P 0 (x, y) in D and let P be the
corresponding point on the graph of f . Consider an increment dx along the x-direction and
an increment dy along the y-direction at the point P 0 . Thus dA = |dxdy|. These increments
sweep out an increment of surface area on the surface at P. The di↵erential dS of this area
at P is given by the corresponding area on the tangent plane to the surface at P.
219
!
!
Let PQ be the vector on the tangent plane at P with x-component dx, and PR the vector
!
!
with y-component dy. Thus, PQ = hdx, 0, f x (x, y)dxi and PR = h0, dy, f y (x, y)dyi. The
!
!
area of the parallelogram spanned by PQ and PR is the magnitude of the cross product
!
!
PQ ⇥ PR .
i
j
k
!
!
PQ ⇥ PR = dx 0 f x dx = h f x , f y , 1idxdy.
0 dy f y dy
Therefore, dS = |h f x , f y , 1idxdy| =
q
f x2 + f y2 + 1 dA. Consequently,
"
Surface area =
D
" q
dS =
f x2 + f y2 + 1 dA.
D
Example 9.13. Find the area of the part of the paraboloid z = x2 + y 2 that lies under the plane
z = 9.
Solution. The paraboloid lies above the circular disk
D = {(r, ✓) | 0  ✓  2⇡, 0  r  3}.
220
The paraboloid is defined by z = f (x, y), where f (x, y) = x2 +y 2 . Thus we have f x = 2x, f y = 2y.
Then
" q
Surface area =
f x2 + f y2 + 1 dA
D
" q
=
1 + 4(x2 + y 2 ) dA
D
=
Z 2⇡Z 3 p
0
= 2⇡
=
0
h
1 + 4r 2 rdrd✓(change to polar coordinates)
i3
1
2 ) 32
(1
+
4r
12
0
p
⇡
(37 37 1).
6
⌅
Exercise 9.2. The surface area of the portion on the cylinder
y 2 + z2 = 1
bounded by the planes y = x + 2 and y = x 2 is equal to ⇡a. Determine the value of a.
Ans. 8
221
222
Chapter 10
Ordinary Di↵erential Equations
Read Thomas’ Calculus, Chapter 9.
10.1 First Order Ordinary Di↵erential Equations
Let y be a function of x. An equation involving x, y and at least one derivative of y is called
an ordinary di↵erential equation (ODE). The order of an ODE is the order of the highest
derivative that occurs in the equation. We consider only first order ordinary di↵erential
equations.
Separable ODE
A separable first order ODE is of the form
dy
= f (x)g(y).
dx
Separating the variables,
1
dy = f (x)dx.
g(y)
Integrating both sides,
Z
1
dy =
g(y)
Z
f (x) dx + C.
Example 10.1. Solve y 0 = (1 + y 2 )e x .
223
dy
Solution. First we note that the di↵erential equation
= (1+y 2 )ex is separable. We separate
dx
R 1
R
1
x dx. Thus
the variables to obtain
dy
=
e
dy
=
ex dx. That is tan 1 y = ex + C,
1 + y2
1 + y2
or y = tan(ex + C).
⌅
Example 10.2. Experiments show that a radioactive substance decomposes at a rate proportional
to the amount present. Starting with 2 mg at certain time, say t = 0, what can be said about the
amount available at a later time?
dy
Solution. Let y be the amount of substance in mg at time t in years. Then dt = ky, y(0) = 2,
R dy R
dy
where k is a positive constant. Thus y = kdt. Integrating, y =
kdt. That is ln |y| =
kt + C, or equivalently, |y| = e kt+C = eC e kt . Therefore, y = e C e kt or y = eC e kt . In other
words, y = Ae kt , where A is a constant. As y(0) = 2, we have 2 = Ae k⇥0 = A. Consequently,
y = 2e kt .
⌅
Remark. How to find k? The value of k depends on the radioactive substance. Usually we
can calculate k by looking up the half-life of the substance in a chemistry table.
For example, the half-life of the substance is T years. From the above solution, we know
y = Ae kt . Thus A2 = Ae kT . That is ln 2 = kT . From this we obtain k = lnT2 .
In the report ‘Stemming the tide 2020: The reality of the Fukushima radioactive water crisis’,
Greenpeace claimed that the contaminated water contained “dangerous levels of carbon14”, a radioactive substance that has the “potential to damage human DNA”.
Carbon-14 is unstable and has a half-life of 5730 ± 40 years.
Example 10.3. A copper ball is heated to 100 C. At time t = 0, it is placed in water which is
maintained at 30 C. At the end of 3 mins, the temperature of the ball is reduced to 70 C. Find
the time at which the temperature of the ball is 31 C.
[ Physical information: Experiments show that the rate of change of the temperature T of the
ball with respect to time t is proportional to the di↵erence between T and the temperature of the
surrounding medium.
Also heat flows so rapidly in copper that at any time the temperature is practically the same at all
points of the ball.]
Solution. Let T be Rthe temperature
of the ball at time t. Then dT
30), T (0) =
dt = k(T
R
dT
100, T (3) = 70. Thus T 30 = kdt. That is ln |T 30| = kt + C, or equivalently, T 30 = Ae kt .
T (0) = 100 ) 100 30 = Ae k⇥0 ) A = 70. Therefore, T = 30 + 70ekt .
224
T (3) = 70 ) 70 = 30 + 70e3k ) 4 = 7e3k ) k =
t
70e 3 (ln 4 ln 7) . Then
1
4
3 ln 7
= 13 (ln 4
ln 7). Therefore, T = 30 +
t
T = 31 =) 31 = 30 + 70e 3 (ln 4 ln 7)
t
1
=) (ln 4 ln 7) = ln
= ln 70
3
70
3 ln 70
=) t =
= 22.78 min.
ln 7 ln 4
⌅
Example 10.4. A skydiver together with his equipment has a combined weight of m kg. After
he jumps and the parachute opens at time t = 0, he falls freely and is descending with velocity
v m/s at the moment when the time is t s. The air resistance against his descending motion is
known to be bv 2 N, where b is a positive constant, and v is his
q velocity at that moment. Show that
the skydiver eventually approaches a terminal speed of k ⌘
acceleration due to gravity.
Solution. By Newton’s second law, we have m
r
Note that k ⌘
dv
= mg
dt
mg
b
m/s, where g = 9.81 m/s2 is the
bv 2 .
mg
=) mg = bk 2 , and the equation can thus be rewritten as
b
m
dv
dv
b 2
= bk 2 bv 2 =)
=
(v k 2 ).
dt
dt
m
Separating the variables, we get
dv
b
=
dt
2
m
k
1
1
1
b
=) (
)dv =
dt
2k v k v + k
m
1
1
2kb
=) (
)dv =
dt.
v k v +k
m
v2
225
Integrating, we get
Z
=)
=)
=)
=)
Z
1
2kb
(
)dv =
dt
v k v +k
m
2kb
ln |v k| ln |v + k| =
t +C
m
v k
2kb
ln
=
t +C
v +k
m
2kb
v k
= eC · e m t
v +k
2kb
v k
= Ae m t ,
v +k
1
where A is a constant (= e C or eC ). Solving for v, we obtain
0
2kb 1
BB 1 + Ae m t CC
B
CC k.
v = B@
2kb A
t
1 Ae m
From this, we see that
lim v =
t!1
1+0
· k = k.
1 0
⌅
Exercise 10.1. Solve the di↵erential equation
dy
= xe3x
dx
Ans: 12 e2y = 13 xe 3x
1 3x
9e
2y
.
+ C.
Exercise 10.2. A curve C that passes through the point (2, 1) is such that at any point (x, y) on
the curve,
x2
dy
= y(x3 + 4).
dx
Find the equation of the curve.
Ans: y = e
x2
2
4
x
.
10.2 Reduction to Separable Form
Certain first order di↵erential equations are not separable, but they can be made separable
by a simple change of variables.
226
This holds for equations of the form
y
y 0 = g( ),
x
y
y
where g is any function of x . Let v = . Then y = vx and y 0 = v + xv 0 . Then the equation
x
y
g(v) v
y 0 = g( ) can be written as v +xv 0 = g(v) or equivalently v 0 =
, which is separable.
x
x
We can now solve for v, and then solve for y.
Example 10.5. Solve 2xyy 0
y 2 + x2 = 0.
Solution. We may rewrite the equation as
y0
y 2 x2
=
or equivalently
2xy
y
y0
=
1 + ( x )2
y
2( x )
,
y
y
where the right hand side is a function of . Let v = , so that y 0 = (xv)0 = v + xv 0 . Then the
x
x
equation can be written as
v + xv 0 =
1 + v2
dv
1 + v2
() x
=
2v
dx
2v
Separating the variables, we get
v=
1 + v2
.
2v
2vdv
dx
=
.
2
x
1+v
Integrating, we get
Z
Z
2vdv
dx
=
x
1 + v2
2
=) ln |1 + v | = ln |x| + C
=) ln |x(1 + v 2 )| = C
=) |x(1 + v 2 )| = eC
=) x(1 + v 2 ) = A,
where A is a constant (= e C or
x(1 +
eC ). Therefore, we have
y2
) = A, or equivalently, x2 + y 2 = Ax.
x2
227
⌅
A di↵erential equation of the form y 0 = f (ax + by), where f is continuous and b , 0, can
be solved by setting u = ax + by. (If b = 0, then the equation itself is separable.)
Example 10.6. Solve (2x 4y + 5)y 0 + x 2y + 3 = 0. qquad
(*)
Solution. Note that the equation can be rewritten as
(x 2y + 3)
(x 2y) 3
=
,
2x 4y + 5
2(x 2y) + 5
y0 =
where the right hand side is a function of x 2y.
1 u0
. Thus the equation becomes
2
Let u = x 2y. Then u 0 = 1 2y 0 , and thus y 0 =
1 u0
u 3
=
,
2
2u + 5
which gives
u0 =
4u + 11
du 4u + 11
()
=
.
2u + 5
dx
2u + 5
Separating the variables, we get
which gives
1
2 (4u
+ 11)
4u + 11
2u + 5
du = dx,
4u + 11
1
2
du = dx () (1
1
)du = 2dx.
4u + 11
Integrating, we get
Z
(1
1
)du =
4u + 11
Z
2dx
1
ln |4u + 11| = 2x + C1
4
=) 4u ln |4u + 11| = 8x + 4C1
=) 4x 8y ln |4x 8y + 11| = 8x + 4C1 (since u = x 2y)
=) 4x + 8y + ln |4x 8y + 11| + C = 0, where C = 4C1 .
=) u
In the calculations above, the two expressions
2u + 5 = 2x 4y + 5
and
4u + 11 = 4x 8y + 11
have appeared in the denominators, and thus the equations
2x 4y + 5 = 0
and
228
4x 8y + 11 = 0
are possible solutions to the DE (*) that we may have missed out in our calculations.
2x + 5
1
Case 1: 2x 4y + 5 = 0. Then y =
and y 0 = . Substituting this into the left hand side
4
2
of (*), we get
(2x 4y + 5)y 0 + x 2y + 3 = 0 · y 0 + x 2 ·
2x + 5
1
+3 =
, 0.
4
2
Thus 2x 4y + 5 = 0 does not satisfy (*).
4x + 11
1
Case 2: 4x 8y + 11 = 0. Then y =
and y 0 = . Substituting this into the left hand
8
2
side of (*), we get
(2x 4y+5)y 0 +x 2y+3 = (2x 4·
4x + 11
1
4x + 11
11
1
11
+5)· +x 2·
+3 = (
+5)· +x x
+3 = 0.
8
2
8
2
2
4
Thus 4x 8y + 11 = 0 satisfies (*).
Thus the solutions to (*) are
4x + 8y + ln |4x 8y + 11| + C = 0
and
4x 8y + 11 = 0.
⌅
y
Exercise 10.3. Solve the initial value problem y 0 = x +
2x3 cos(x2 )
,
y
p
y( ⇡) = 0.
p
Ans: y = ±x 2 sin(x2 ).
Solution: Note that we may rewrite the di↵erential equation as
y0 =
y 2x2 cos(x2 )
+
,
y
x
x
y
y
where the right hand side is a function of and x. Let v = . Then y = xv and y 0 = v + xv 0 .
x
x
Thus the equation becomes
v + xv 0 = v +
2x2 cos(x2 )
v
2x cos(x2 )
v
dv 2x cos(x2 )
=)
=
.
dx
v
=) v 0 =
Separating the variables, we get
vdv = 2x cos(x2 )dx.
229
Integrating, we get
Z
Z
vdv =
2x cos(x2 )dx
1
=) v 2 = sin(x2 ) + C
2
1 y2
=) · 2 = sin(x2 ) + C
2 x
=) y 2 = 2x2 (sin(x2 ) + C).
Now
p
y( ⇡) = 0 =) 0 = 2⇡(sin ⇡ + C) =) C = 0.
p
Consequently, the solution is y 2 = 2x2 sin(x2 ), or y = ±x 2 sin(x2 ).
Exercise 10.4. Solve (x + 2y
⌅
1) + 3(x + 2y)y 0 = 0.
Ans: x + 3y + C = 3 ln |x + 2y + 2|, x + 2y + 2 = 0.
10.3 Linear First Order ODE
A linear first order ODE is of the form
dy
+ P(x)y = Q(x),
dx
where P(x), Q(x) are continuous functions. Note that the above ODE is separable if P(x) is
identically equal to Q(x). This is the standard form of a linear first order ODE.
R
Let I (x) = e P(x) dx . We call I (x) an integrating factor. Multiplying both sides of the above
ODE by I (x), we get
R
R
dy R P(x) dx
e
+ P(x)e P(x) dx y = Q(x)e P(x) dx .
dx
But
✓ R
◆
R
dy R P(x) dx
d
P(x) dx
P(x) dx
e
+ P(x)e
y=
ye
,
dx
dx
which can be shown by applying the product rule and the Fundamental Theorem of Calculus. Hence,
✓ R
◆
R
d
P(x) dx
ye
= Q(x)e P(x) dx .
dx
230
Thus we have shown that
d
(y · I (x)) = Q(x) · I (x).
dx
Integrating both sides gives
Z
y · I (x) =
Q(x) · I (x) dx
from which the solution for y can be obtained.
Example 10.7. Solve xy 0
3y = x2 , x > 0.
Solution.
First we rewrite the DE in the standard form y 0
R
3
x
dx
1
.
x3
= x. An integrating factor is
=
Multiplying the DE (the standard form) by this integrating factor, we
3
1
1
y
y
have (y 0
y) · 3 = x · 3 , which gives ( x3 )0 = x12 . Integrating, we have x3 = 1x + C. That is
x
x
x
y = x2 + Cx3 .
e
=e
3 ln x
3
xy
⌅
Example 10.8. Solve y 0
y = e 2x .
R
Solution. An integrating factor is e 1 dx = e x . Multiplying the DE by this integrating factor,
we have (y 0 y) · e x = e 2x · e x , which gives (ye x )0 = e x . Integrating, we obtain ye x = ex + C.
That is y = e2x + Cex .
⌅
Exercise 10.5. Solve the di↵erential equation
(x + 1)2
Ans: y =
1
x+1
dy
dx
(x + 1)y = 2, x > 1.
+ C(x + 1).
Exercise 10.6. Solve the di↵erential equation
dy 4 + y sin x
=
,
dx
cos x
⇡
⇡
<x< ,
2
2
given that y = 6 when x = 0.
Ans: y =
4x+6
cos x .
Exercise 10.7. An object of mass m dropped from rest in a medium that o↵ers a resistance proportional to the magnitude of the instantaneous velocity of the object. Let x(t) be the displacement
of the object measured vertically downward at time t so that x(0) = 0. Show that
k
mg
m2 g
t + 2 (e m t 1),
k
k
where k is the proportional (positive) constant of the force of resistance of the medium.
x(t) =
[Set up the DE for the velocity first: m dv
dt = mg
231
kv.]
10.4 The Bernoulli Equation.
An ODE in the form
y 0 + p(x)y = q(x)y n ,
where n , 0, 1, is called the Bernoulli equation. The functions p(x) and q(x) are continuous
functions on an interval J.
Let u = y 1 n . Substituting into the Bernoulli equation we get
u 0 + (1 n)p(x)u = (1 n)q(x).
This is a first order linear ODE.
Remark. (i) When n = 0 or 1, the Bernoulli equation itself is a first order linear ODE.
(ii) When n > 0, the constant zero function y(x) = 0 is automatically a solution of the
Bernoulli equation.
Example 10.9. Solve y 0 + y = x2 y 2 .
Solution. This is a Bernoulli equation with n = 2. Let z = y 1 2 = y 1 . Then z 0 = y 2 y 0 , so
that y 0 = y 2 z0 . Thus the given Bernoulli equation can be written as
y 2 z0 + y = x2 y 2 , z0
y
1
= x2 , z0
z = x2 .
This is a first order linear equation. Multiplying by the integrating factor e
have
(z 0 z)e x = x2 e x ,
which gives
Integrating, we get ze
x
=
R
Therefore,
x
+ 2xe
x
dx
= e x , we
(ze x )0 = x2 e x .
x2 e
x dx.
Using integration by parts, we have
Z
x2 e x dx = x2 e
Thus z = e x (x2 e
R
+ 2e
x
x
+ 2xe
x
+ 2e
x
+ C.
+ C) = x2 + 2x + 2 + Cex .
1
1
= x2 + 2x + 2 + Cex , and thus y = 2
.
y
x + 2x + 2 + Cex
Since n = 2 > 0, y = 0 is also a solution.
⌅
Exercise 10.8. Solve xy 0 + y = x4 y 3 .
Ans:
1
y2
= x4 + cx2 , or y = 0.
232
10.5 Applications of ODE
Example 10.10. At time t = 0, a tank contains 20 kg of salt dissolved in 100 litres of water.
Assume that water containing 14 kg of salt per litre is entering the tank at the rate of 3 litre per
min, and the well-stirred solution is leaving the tank at the same rate. Find the amount of salt at
any time t.
Solution. First note that the volume of the solution remains constant which is 100 litres. Let
Q be the amount of salt in kg at time t. The concentration of salt in the solution is Q/100 kg
per litre. Suppose at time t + dt, the amount of salt is Q + dQ. Then
dQ = salt input
Thus
salt output = 3 ⇥
dQ 3
=
dt
4
That is
1
Q
⇥ dt 3 ⇥
⇥ dt.
4
100
3Q
.
100
dQ
3
=
(Q 25).
dt
100
The general solution to this first order linear DE is Q = 25 + Ce
3t
20 = 25 + C so that C = 5. Consequently, Q = 25 5e 100 .
3t
100
. Since Q(0) = 20, we have
Note that lim Q(t) = 25. Thus after sufficiently long time, the salt concentration will apt!1
proach 25 kg per 100 litres.
⌅
Example 10.11. A body was found at a crime scene. You are a member of the CSI team and you
arrived at the crime scene at 8AM. Immediately upon arrival, you took the temperature of the
victim and found that it was 26 C. At 9AM, you took the temperature of the victim again and
found that it was 24 C. You estimate that the victim’s temperature was 37 C just before death
233
and that the temperature at the crime scene stayed approximately constant at 21 C. What is your
estimate on the time of death?
Remark: Newton’s law of cooling states that the rate of cooling of an object is proportional to the
di↵erence in temperature between the object and its surroundings.
Solution. Set time t = 0 at 8AM, where t is measured in hours. Let T be the temperature
of the body at time t. By Newton’s law of cooling, we have dT
21), where k is a
dt = k(T
kt
constant. The general solution is T = 21 + Ae . As T (0) = 26, we have 26 = 21 + A so that
A = 5. Therefore, T = 21 + 5ekt . At 9AM, that is 1 hour later, T (1) = 24. Thus 24 = 21 + 5ek so
that k = ln( 35 ). Hence
✓ ◆t
3
3 t
3
T = 21 + 5e t·ln( 5 ) = 21 + 5eln(( 5 ) ) = 21 + 5
.
5
If ⌧ is the time of death, then T (⌧) = 37. Therefore,
✓ ◆⌧
3
37 = 21 + 5
.
5
⇣ ⌘⌧
3
3
16
3
That is 16
=
, ln( 16
5
5
5 ) = ⌧ ln( 5 ) , ⌧ = ln( 5 )/ ln( 5 ) = 2.277 hours, (or equivalently
negative 2 hour 17 mins). Thus time of death is about 5 : 43AM.
⌅
Exercise 10.9. The Jurong lake has a volume of 700000 m3 . At time t = 0, the government starts
a water cleaning process so that only fresh clean water flows into the lake. After 5 years, it is found
that the pollution in the lake is reduced by 50%. If fresh water flows into the lake at a rate of r
cubic metres per year and lake water flows out to the sea at the same rate, what is the value of r
correct to the nearest thousands?
Ans: 97000.
Exercise 10.10. Newton’s law of cooling states that the rate of cooling of an object is proportional
to the di↵erence in temperature between the object and its surroundings. If an object is kept in an
environment whose temperature is kept constant at 15 C and the object takes 20 minutes to cool
from 95 C to 55 C, determine how much longer it will take for the object to cool down to 25 C.
Ans: 40 mins more.
Exercise 10.11. In a chemical reaction, the rate at which the mass, m (in grams) of a chemical
compound at time t (in seconds) is proportional to m2 9m + 18, (0 < m < 3). Initially (t = 0),
we assume m = 0. After 1 second, the mass of the chemical compound has increased to 2g. Write
down a di↵erential equation in m and t, and show that
m 6
= 2t+1 .
m 3
Find the exact mass of the chemical compound after 2 seconds.
Ans: 18/7 grams.
234
Chapter 11
More on ODE
Read Thomas’ Calculus, Chapter 9.
Remark: Chapter 11 will be excluded from the Final Exam.
11.1 Euler’s Method
Remark: Section 11.1 will be excluded from the Final Exam.
Not all the first order ODE’s can be solved explicitly in closed form. In that case, we have
to rely on numerical solutions. In this section, we introduce Euler’s method which is a
dy
numerical method in approximating a first order ODE. Given a di↵erential equation dx =
f (x, y) and an initial condition y(x0 ) = y0 , we can approximate the solution y = y(x) by its
linearization
L(x) = y(x0 ) + y 0 (x0 )(x x0 ) or L(x) = y0 + f (x0 , y0 )(x x0 ).
The linear function L(x) gives a good approximation to the solution in a short interval about
x0 . The idea of Euler’s method is to put together a sequence of such linearizations in a
successive manner to approximate the solution curve over a longer interval.
First we know that the point (x0 , y0 ) lies on the solution curve. Consider a small increment
from x0 to x1 ⌘ x0 +h. The graph of L(x) is the tangent line with slope f (x0 , y0 ) to the solution
curve y = y(x) at the point (x0 , y0 ). So if h is small, y1 ⌘ L(x1 ) is a good approximation to y(x1 ).
In other word, the point (x1 , y1 ) is close to the solution curve y = y(x).
235
Using the point (x1 , y1 ) and the slope f (x1 , y1 ) of the solution curve through (x1 , y1 ), we take
a second step. Setting x2 = x1 + h, we use the linearization of the solution curve through
(x1 , y1 ) to calculate y2 = y1 + f (x1 , y1 )h.
This gives the next approximation (x2 , y2 ) to the value along the solution curve y = y(x).
Continuing in this way, we take a third step from the point (x2 , y2 ) with slope f (x2 , y2 ) to
obtain the third approximation y3 = y2 + f (x2 , y2 )h, and so on.
In other words, we are building an approximation to one of the solution by following the
direction of the slope field of the di↵erential equation.
236
The following steps summarize Euler’s method. Suppose we wish to approximate the solution over the interval [a, b]. Choose an integer n as the number of steps. Let h = bn a . Let
x0 = a
x1 = x0 + h
x2 = x1 + h
..
.
b = xn = xn
1 + h.
Then calculate the approximations to the solution as follows.
y1 = y0 + f (x0 , y0 )h
y2 = y1 + f (x1 , y1 )h
..
.
yn = yn
1+f
(xn 1 , yn 1 )h.
The polygonal curve joining the points (x0 , y0 ), (x1 , y1 ), . . . , (xn , yn ) successively is an approximation to the solution curve of the DE y 0 = f (x, y) through the point (x0 , y0 ).
Example 11.1. Use Euler’s method to solve
y 0 = 1 + y, y(0) = 1,
on the interval [0, 1] starting at x0 = 0 by taking h = 0.1. Find the approximate value of y(1) and
compare it with the exact value.
Solution. Taking n = 10 and h = 0.1, the result is tabulated in the following table. The exact
solution to the DE is y = 2e x 1. Thus the exact value at x = 1 is y(1) = 2e 1 = 4.4366. The
approximate value is 4.1875.
Euler solution of y 0 = 1 + y, y(0) = 1, h = 0.1
x
y(Euler)
y(Exact)
Error
0
1
1
0
0.1 1.2
1.2103
0.0103
0.2 1.42
1.4428
0.0228
0.3 1.662
1.6997
0.0377
0.4 1.9282
1.9836
0.0554
0.5 2.221
2.2974
0.0764
0.6 2.5431
2.6442
0.1011
0.7 2.8974
3.0275
0.1301
0.8 3.2872
3.4511
0.1639
0.9 3.7159
3.9192
0.2033
1
4.1875
4.4366
0.2491
237
Exercise 11.1. Use Euler’s method to calculate the first three approximations to the initial value
problem:
y 0 = 2xy + 2y, y(0) = 3,
by taking h = 0.2.
Ans: y1 = 4.2, y2 = 6.216, y3 = 9.697.
Remark. In the film Hidden Figures, Katherine Goble resorts to Euler’s method in calculating the re-entry of astronaut John Glenn from Earth orbit.
11.2 2nd Order Linear Equations with Constant Coefficients
Remark: Section 11.2 will be excluded from the Final Exam.
Let us begin with second order homogenous linear equation with constant coefficients
y 00 + ay 0 + by = 0,
(11.1)
where a and b are real constants. We look for a solution of the form y = e x . Plugging into
(11.1) we find that, e x is a solution of (11.1) if and only if
2
+ a + b = 0.
238
(11.2)
(11.2) is called the auxiliary equation or characteristic equation of (11.1). The roots of (11.2)
are called characteristic values (or eigenvalues):
p
1
2 4b),
1 = ( a+ a
2
p
1
a2 4b).
2= ( a
2
1. If a2
is
4b > 0, (11.2) has two distinct real roots
y = c1 e
1x
2. If a2 4b = 0, (11.2) has one real root
1 = 2 ). The general solution of (11.1) is
1,
+ c2 e
2x
2,
and the general solutions of (11.1)
.
(we may say that (11.2) has two equal roots
x
y = c1 e
+ c2 xe x .
3. If a2 4b < 0, (11.2) has a pair of complex conjugate roots
1
= ↵ +i ,
2
=↵ i .
The general solution of (11.1) is
y = c1 e↵x cos( x) + c2 e ↵x sin( x).
Example 11.2. Solve y 00 + y 0
2y = 0, y(0) = 4, y 0 (0) = 5.
Solution. The characteristic values are 1, 2. Thus the solution is
y = ex + 3e
2x
.
⌅
Example 11.3. Solve y 00
4y 0 + 4y = 0, y(0) = 3, y 0 (0) = 1.
Solution. The characteristic values are 2 with multiplicity 2. Thus the solution is
y = (3 5x)e 2x .
⌅
Example 11.4. Solve y 00
2y 0 + 10y = 0.
Solution. The characteristic values are 1 + 3i, 1 3i. Thus the solution is
y = ex (c1 cos 3x + c2 sin 3x).
⌅
239
11.3 Method of Undetermined Coefficients
Remark: Section 11.3 will be excluded from the Final Exam.
Consider the equation y 00 + ay 0 + by = f (x), where a and b are real constants. To solve this
non-homogeneous linear DE, we look for a particular solution yp of y 00 + ay 0 + by = f (x).
Then the general solution is the sum of the general solution yc of the associated homogeneous linear DE: y 00 + ay 0 + by = 0 and this particular solution yp . That is
y = yc + yp .
Case 1. f (x) = Pn (x)e↵x , where Pn (x) is a polynomial of degree n
0.
We look for a particular solution in the form
y = Q(x)e ↵x ,
where Q(x) is a polynomial. Plugging it into y 00 + ay 0 + by = f (x) we find
Q00 + (2↵ + a)Q0 + (↵ 2 + a↵ + b)Q = Pn (x).
(11.3)
Subcase 1.1. If ↵ 2 + a↵ + b , 0, namely, ↵ is not a root of the characteristic equation, we
choose Q = Rn , a polynomial of degree n, and
y = Rn (x)e ↵x .
The coefficients of Rn can be determined by comparing the terms of same power in the two
sides of (11.3). Note that in this case both sides of (11.3) are polynomials of degree n.
Subcase 1.2. If ↵ 2 + a↵ + b = 0 but 2↵ + a , 0, namely, ↵ is a simple root of the characteristic
equation, then (11.3) is reduced to
Q00 + (2↵ + a)Q 0 = Pn .
(11.4)
We choose Q to be a polynomial of degree n+1. Since the constant term of Q does not appear
in (11.4), we may choose Q(x) = xRn (x), where Rn (x) is a polynomial of degree n.
y = xRn (x)e ↵x .
Subcase 1.3 If ↵ 2 +a↵ +b = 0 and 2↵ +a = 0, namely, ↵ is a root of the characteristic equation
with multiplicity 2, then (11.3) is reduced to
Q00 = Pn .
We choose Q(x) = x2 Rn (x), where Rn (x) is a polynomial of degree n.
y = x2 Rn (x)e↵x .
240
(11.5)
Example 11.5. Find the general solution of y 00
Solution. The homogeneous equation has
roots = 2, 1.
y0
2
2y = 4x2 .
2 = 0 as its characteristic equation with
Therefore the general solution of the associated homogeneous equation is y = c1 e2x + c2 e x .
Note that 4x2 = 4x2 e0x and 0 is not a root of the characteristic equation. We can try a
particular solution of the form
yp = A + Bx + Cx2 .
Substituting this into the equation, we have
2C
(B + 2Cx) 2(A + Bx + Cx2 ) = 4x2 .
Equating coefficients, we have
2C
B 2A = 0
2C 2B = 0
2C = 4
Thus A = 3, B = 2, C = 2, and y = 3 + 2x 2x2 .
The general solution is
y = c1 e 2x + c2 e
x
3 + 2x 2x2 .
⌅
Example 11.6. Solve y 00
2y 0 + y = xe x .
Solution. The general solution of the associated homogeneous DE is C1 ex + C2 xex .
Here ↵ = 1 is a double root of the characteristic equation
particular solution of the form y = x2 (A + Bx)e x .
2
2 + 1 = 0. Therefore, we try a
We have y 0 = (Bx3 + (A + 3B)x2 + 2Ax)ex and y 00 = (Bx3 + (A + 6B)x2 + (4A + 6B)x + 2A)ex .
Substituting these into the DE, we have (2A + 6Bx)ex = xe x . Thus A = 0 and B = 16 .
Consequently, the general solution is y = C1 ex + C2 xex + 16 x3 e x .
⌅
Case 2. f (x) = Pn (x)e↵x cos( x) or f (x) = Pn (x)e↵x sin( x), where Pn (x) is a polynomial of
degree n 0.
We first look for a solution of
y 00 + ay 0 + by = Pn (x)e (↵+i
)x
.
Using the method in Case 1 we obtain a complex-valued solution
z(x) = u(x) + iv(x),
241
(11.6)
where u(x) = <(z(x)), v(x) = =(z(x)). Substituting z(x) = u(x) + iv(x) into (11.6) and taking
the real and imaginary parts, we can show that u(x) = <(z(x)) is a solution of
y 00 + ay 0 + by = Pn (x)e↵x cos( x),
(11.7)
and v(x) = =(z(x)) is a solution of
y 00 + ay 0 + by = Pn (x)e↵x sin( x).
Example 11.7. Solve y 00
(11.8)
2y 0 + 2y = e x cos x.
Solution. The characteristic equation is
2
2 + 2 = 0 with roots 1 + i and 1 i.
The general solution of the associated homogeneous DE is
y = c1 e x cos x + c2 ex sin x.
2y 0 + 2y = e(1+i)x . Let’s find a particular solution.
Now consider the DE y 00
Since (1 + i) is a root of the characteristic equation, we should try a particular solution of the
form y = Axe (1+i)x .
Thus y 0 = (A + A(1 + i)x)e(1+i)x , y 00 = (2A(1 + i) + A(1 + i)2 x)e (1+i)x .
Therefore,
yh 00 2y 0 + 2
i
= (2A(1 + i) + A(1 + i)2 x) 2(A + A(1 + i)x) + 2Ax e(1+i)x
= 2Aie(1+i)x .
From this, 1 = 2Ai or A =
i
2.
Thus a particular solution is given by y =
i
(1+i)x , or equivalently, y
2 xe
= 12 xe x sin x
i
x
2 xe cos x.
Taking the real part, yp = 12 xe x sin x is a particular solution of the given DE.
Consequently, the general solution is
1
y = c1 e x cos x + c2 e x sin x + xex sin x.
2
⌅
Remark. Alternatively to solve (11.7) or (11.8), one can try a solution of the form
Qn (x)e ↵x cos( x) + Rn (x)e ↵x sin( x)
if ↵ + i is not a root of
2+a
+ b = 0, and
xQn (x)e ↵x cos( x) + xRn (x)e ↵x sin( x)
if ↵ + i is a root of
2+a
+ b = 0, where Qn and Rn are polynomials of degree n.
242
11.4 Appendix: Malthus Model of Population
Remark: Section 11.4 will be excluded from the Final Exam.
The total population N (t) of a country or a colony is clearly a function of time. N (t) though
should be integer valued and is great than 0, is considered as a continuous and in fact di↵erentiable
function of time, especially its value is usually very huge.
Given the population now, can one predict the future population?
Suppose B is a function giving the ‘ per capita birth rate” in a given society, i.e. B is the
number of babies born per second, divided by the total population N of the country at
that moment. Note that B could be small in a big country and large in a small country - it
depends on whether there is a strong social pressure on people to get married and have kids.
Now B could depend on time (people might gradually come to realise that large families are
no fun, etc..) and it could depend on N . But suppose you don’t believe these things. Instead
suppose people will always have as many kids as they can, no matter what. Then B is a
constant. Thus
number of babies born in time interval dt = BN dt.
Similarly, let D be the death rate per capita. Again it could be a function of t (in case
the society has better medicine, fewer smokers etc) or N (overcrowding leads to famine or
disease). But if we assume that it is constant, then
number of deaths in time interval dt = DN dt.
So the change in N , denoted by dN within the time interval dt is
dN = number of births
number of deaths,
provided there is no emigration or immigration. Thus
dN = (B D)N dt.
That is
dN
= (B D)N = kN ,
dt
(1)
where k = B D.
This model of society was put forward by Thomas Malthus in 1798. Clearly Malthus was
assuming a socially static society in which human reproductive behaviour never changes
with time or overcrowding, poverty etc.. What does Malthus’ model predict?
Suppose that the population now is N̂ and let t = 0 now.
R
R
dN
kt
From (1), dN
N = k dt = kt + C ) ln(N ) = kt + C ) N (t) = Ae .
dt = kN )
243
Since N̂ = N (0) = A, we get
N (t) = N̂ e kt .
(2)
(1) is the logistic equation and (2) is the solution of the standard Malthus’s model.
The population collapses if k < 0 (more deaths than births per capita), remain stable if (and
only if) k = 0, and it explodes if k > 0 (more births than deaths). Malthus observed that the
population of Europe was increasing, so he predicted a catastrophic population explosion;
since the food supply could not be expanded so fast, this would be disastrous.
In fact, this didn’t happen in Europe. So Malthus’ model is not quite correct; as many
millions went to the US, and many millions died in wars.
Malthus’ model can be improved. Note that Malthus’ model is interesting because it shows
that static behaviour patterns can lead to disaster. But precisely because the term e kt grows
so quickly, Malthus’ assumptions must eventually go wrong - obviously there is a limit to
the possible population. Eventually, if we don’t control B, then D will have to increase. So
we have to assume that D is a function of N .
Clearly, D must be an increasing function of N , but which function? The simplest possible
choice is
D = sN ,
(3)
where s is a constant.
Now we want to solve
Rewrite the equation as
dN
= BN
dt
dN
dt
sN 2 , N (0) = N̂ .
BN = sN 2 .
244
This is a Bernoulli equation.
2
1 dN
N dz
Let z = N 1 2 = N1 . Then dz
BN = sN 2 . That is
dt = N 2 dt . Thus
dt
linear equation in z. An integrating factor is eBt . Thus
dz
dt
+ Bz = s ,
Let N1 =
B
s
d(ze Bt )
dt
= se Bt , ze Bt = Bs eBt + C , z =
which is the carrying capacity. Then
Now N (0) = N̂ )
Thus
1
N̂
=
1
N1
+C ) C =
1
N̂
1
N
1
N1 .
1
1
1
=
+
N N1
N̂
s
B
+ Ce
=
1
N1
!
1
e
N1
Bt .
+ Ce
Bt
That is
1
N
dz
dt
+ Bz = s which is a
=
s
B
+ Ce
Bt .
Bt .
.
Rearranging,
N=
1+
⇣N
N1
1
N̂
⌘
1 e
Bt
.
(4)
Note that lim N = N1 , as B > 0. Also N (t) is increasing if N1 > N̂ , and N (t) is decreasing if
t!1
N1 < N̂ . Thus
dN
dt
, 0 if N1 , N̂ . (4) is the solution of the improved Malthus’ model.
Example 11.8. The growth of rabbits in your rabbit farm followed a logistic population model
with a birth rate per capita of 10 rabbits per rabbit per year. You observed that their number had
approached to a logistic equilibrium population of 2500 rabbits. One day your friend Dr. Good
visited your farm and suggested that you try to mix some of his latest invention of Vitamin X
into your rabbit feed to boost the reproduction rate. You followed his suggestion and after a long
period of time, observed that the rabbit population had reached a new logistic equilibrium of 3000
rabbits. If the new rabbit birth rate per capita after Vitamin X was introduced was B rabbits per
rabbit per year, what is the value of B?
245
Solution. We have
10
s
= 2500 so that s =
1
250 .
Thus
B
s
= 3000 ) B =
3000
250
= 12.
⌅
Exercise 11.2. Suppose N1 > 2N̂ . Show that there is a point of inflection on the graph of N at
t > 0.
246