# User manual | MA10192: Mathematics 1 - Department of Mathematical Sciences

```MA10192: Mathematics 1
Lecture notes: Chapters 1-9
Semester 1, 2005
Gunnar Traustason
1
1
Based on the notes of my predecessors André Léger and Alastair Spence
Contents
1 Exponentials and logarithms
1.1 The real numbers . . . . . . . . . . . . . . . . . . . . . . .
1.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 Inverses . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Exponentials . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Powers . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 The Exponential Function . . . . . . . . . . . . . .
1.4 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 The logarithm function . . . . . . . . . . . . . . .
1.5 Various examples and applications of the exponentials and
1.5.1 Equations with Exponentials and Logs . . . . . . .
1.5.2 Examples from Chemical Engineering . . . . . . .
1.5.3 Application of Logs - Parameter Estimation . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
the logarithms
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
2 Trigonometry
2.1 Angles and the unit circle . . . . . . . . . . . . . . . . . . . .
2.2 The right angled triangle . . . . . . . . . . . . . . . . . . . .
2.3 The trigonometric functions for arbitrary angles . . . . . . . .
2.4 Trigonometric equations . . . . . . . . . . . . . . . . . . . . .
2.4.1 Some trigonometric equations . . . . . . . . . . . . . .
2.4.2 To Convert a Sum to a Harmonic Form . . . . . . . .
2.5 The inverses to the trigonometric functions . . . . . . . . . .
2.5.1 The inverse sine function sin−1 x or arcsin x . . . . . .
2.5.2 The Inverse Cosine Function (cos−1 (x) or arccos(x)) .
2.5.3 The Inverse Tangent Function (tan−1 (x) or arctan(x))
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3
3
4
4
4
5
5
7
7
7
9
9
9
10
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
11
11
11
13
16
16
17
19
19
20
20
3 Inequalities
21
3.1 Solving inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 Sketching Reciprocal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4 Limits and Asymptotes
4.1 Definition and basic properties of limits . . . .
4.2 Calculating limits with help of standard limits
4.3 Asymptotes . . . . . . . . . . . . . . . . . . . .
4.4 Physical Example of Limits . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
25
25
26
28
31
5 Differentiation
5.1 Definition and geometric interpretation .
5.2 Common Derivatives . . . . . . . . . . .
5.3 Higher Order Derivatives . . . . . . . .
5.4 Rules for Differentiation . . . . . . . . .
5.5 Logarithmic Differentiation . . . . . . .
5.6 Implicit Differentiation . . . . . . . . . .
5.7 Derivatives of Inverse Functions . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
32
32
33
33
34
34
36
36
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
.
.
.
.
.
.
.
5.8
5.9
5.10
5.11
5.12
5.13
5.14
5.15
5.16
Parametric Differentiation . . . . . . . . . . . . . . .
Application to a real life situation . . . . . . . . . .
Application of the derivative for evaluation of limits:
Stationary points . . . . . . . . . . . . . . . . . . . .
Global max and min on a closed bounded interval .
Curve Sketching . . . . . . . . . . . . . . . . . . . .
Application to Optimisation . . . . . . . . . . . . . .
Differentials - Estimating Small Changes . . . . . . .
Leibniz Rule . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
L’Hopital’s Rule
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
37
37
38
39
41
41
42
43
43
6 Taylor and Maclaurin Series
6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . .
6.2 Calculating Maclaurin series and polynomials . . .
6.3 Estimating the error term . . . . . . . . . . . . . .
6.4 Manipulation of Maclaurin Series . . . . . . . . . .
6.5 Taylor Series . . . . . . . . . . . . . . . . . . . . .
6.6 Newton’s Method for solving non-linear equations .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
45
45
46
48
49
49
50
7 Integration
7.1 Primitives . . . . . . . . . . . . . . .
7.2 Integration by parts and substitution
7.3 Primitives and areas . . . . . . . . .
7.4 Properties of Definite Integrals . . .
7.5 Techniques of Integration . . . . . .
7.5.1 Substitution . . . . . . . . . .
7.5.2 Integration by Parts . . . . .
7.5.3 Integrals using Trig Formula
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
52
52
54
55
56
57
57
58
59
8 Ordinary Differential Equations (ODE’s)
8.1 ODE of lst order . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Separable equations . . . . . . . . . . . . . . . . . . . . . . . .
8.3 An application . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4 2nd order homogenous linear ODE’s with constant coefficients .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
60
60
66
67
68
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
9 Functions in several variables
70
9.1 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
9.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
9.3 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
2
Chapter 1
Exponentials and logarithms
You can read more in J/S sections §1.1 ,§1.7 and §1.10 - §1.12.
1.1
The real numbers
These have evolved through the times as follows:
I. The natural numbers,
N (for counting)
0, 1, 2, . . .
II. The integers,
Z (for debit-credit)
. . . , −2, −1, 0, 1, 2, . . .
III. The rationals,
Q (for dividing into parts)
m/n where m, n ∈ Z and n 6= 0.
These can be arranged on the “rational line”
−5/3
•
0 1/2
• •
1
•
2
•
→
Q
The rationals can also be represented as finite or infinite periodic decimals:
1/3
=
0.33 · · ·
3/2
=
1.5
With these we can perform the arithmetic operations +, −, • and /.
IV. The real numbers,
R
We want to use numbers for measuring and evaluating various terms within the natural sciences
length, area, mass, energy, pressure, · · ·
Unfortunately,
Q doesn’t suffice for this.
3
Example.
↑
Q
a2 = 12 + 12 = 2 (Pythagoras)
1•
•
·
a·
··
·•
• ◦
1 a
→
Q
But there is no rational a such that a2 = 2. This means that the rational line has “holes” and we can’t
even use the rationals to evaluate lengths of simple straight line segments. We need to extend the rational
numbers to deal with this handicap. This we do by filling in all these holes by new numbers. The√result are
the√real numbers. How can these be represented? For example, if we denote the number a above by 2. What
is 2?
Notice that
1.42 < 2 < 1.52
1.412 < 2 < 1.422
1.4142 < 2 < 1.4152
⇒
√
1.4 < √2 < 1.5
1.41 < √2 < 1.42
1.414 < 2 < 1.415
..
.
We can continue indefinitely and what we get is an infinite non-periodic decimal
√
2 = 1.41421356 · · ·
The real numbers: All decimals (finite, infinite, periodic, non-periodic).
1.2
Functions
This is another fundamental notion for natural sciences. Often various terms are related to each other through
a formula. For example a particle with a mass m that moves with speed v has kinetic energy E = 1/2 · mv 2 .
Here E is a function in v, namely E = f (v) where
f (v) = 1/2 · mv 2 .
1.2.1
Definition
A function f is determined by a domain Df and a “rule” that assigns, to every element x in the domain, a
value f (x). The set of all the values is denoted by Vf and sometimes called the range of the function.
Usually f (x) is determined from a formula.
Example. f (x) =
√
x − 1, x ≥ 1.
Here Df = [1, ∞) and Vf = [0, ∞)i.
1.2.2
Inverses
Definition. We say that f is injective if distinct elements in Df always have distinct values (i.e. a 6= b ⇒
f (a) 6= f (b)).
Examples. (a)The function
f (x) = x3 , x ∈ R
4
is injective as it is strictly increasing and thus no value occurs twice.
(b) The function
f (x) = x2 , x ∈ R
is on the other hand not injective as there are values that occurs more than once. For example f (1) = f (−1).
If f is injective then there is a unique source for each value (as no value occurs more than once). There
is thus a function F that reverses the process: has Vf as its domain, Df as its range and takes each value
back to its source.
f
→
y = f (x) iff x = F (y)
x
y
←
Df
F
Vf
Notice that
DF
=
Vf
VF
=
Df
F (f (x))
= x
f (F (y))
= y
Example. The function
f (x) = x3 , x ∈ R
has as inverse
F (x) =
1.3
1.3.1
√
3
x, x ∈ R.
Exponentials
Powers
For any real number a, a2 = a × a and subsequently an = a × a × · · · × a for any positive integer. We want
|
{z
}
n times
to extend the definition to the more general case when n is a real number.
A. an where n is an integer.
Example.
divide by 2 ↑
..
.
1
2−2 = 2·2
= 14
1
−1
2 =2
20 = 1
21 = 2
22 = 2 · 2 = 4
23 = 2 · 2 · 2 = 8
..
.
multiply by 2 ↓
Similarly for an for an arbitrary a. In particular
a−n =
As
1
a/b
1
.
an
= b/a
(explanation:
1
1 · b/a
b/a
=
=
= b/a)
a/b
a/b · b/a
1
5
we have in particular
(a/b)−n = (b/a)n .
The Exponential Laws
1. an × am = an+m .
2. an ÷ am = an−m .
3. (an )m = an×m .
4. (ab)n = an bn , similarly
a n
b
=
an
bn .
Notice also that
5. a0 = 1, for any a 6= 0.
6. 1n = 1.
7. a−n = 1/an .
8. If an = am then n = m.
The main point to make here is that all these laws are natural. Instead of memorising them, I suggest that
you recall instead each time you use them why they are natural.
Examples
1. 22 × 23 = 25 = 32
2. (22 )3 = 26 = 64
3. 22 ÷ 23 = 2−1 =
1
21
4. (3x)3 = 27x3
2
5. 52 = 25
4
2
2 −2
6. 3
= 32 =
9
4
=
1
2
B. an where n is a real number and a > 0.
What is the logical definition of 21/3 ?
If the power laws above are still to hold, we should have
1
So we should set 21/3 =
√
3
(21/3 )3 = 2 3 ·3 = 21 = 2.
2. Similarly
√
3
22/3 = (21/3 )2 = ( 2)2 .
So the following general definition is very natural.
√
Definition. am/n = ( n b)m .
√
Now what about irrational numbers. What is for example 2
2
?
This can be defined according to the following principle
√
1.4, 1.41, 1.414, . . . → 2√
21.4 , 21.41 , 21.414 , . . . → 2 2
√
What this means is that if the first sequence is a sequence of rationals√that has 2 as a limit then the second
sequence is going to tend to some limit. This limit we can define as 2 2 . Similarly for ax in general where x
is a real number.
6
More examples
√
1. (32)1/2 = (25 )1/2 = 25/2 = 22 .21/2 = 4 2
√
1/2
1/2 3 1/2
2. 4x3 = 22 x3
= 22
x
= 2x3/2
1/4×3
3. 163/4 = (16)
1.3.2
= 161/4
3
= 23 = 8
The Exponential Function
let a > 0, a 6= 1. The function
f (x) = ax , x ∈ R
is called the exponential function with base a.
10
9
Blue: y = (1.5)x
Green: y = 2x
Red: y = 3x
Light Blue: y = (0.5)x
8
7
6
5
4
3
2
1
1.5x
2x
3x
0.5x
0
!1
!3
!2
!1
0
1
2
3
Figure 1.1: The Exponential Functions
In general, we obtain an exponentially increasing function in x if a > 1, and exponentially decreasing if
0 < a < 1. Here the domain is Df = R and the range is Vf = (0, ∞).
Example. Suppose that we have a population of 10 millions. Supposing that we have the constant population
growth rate of 1% per year, what is the population going to be in 10 years time?
Solution. The answer is 10 · (1.01)10 ≈ 11.05 millions.
An important special case for the exponential functions is when the base a is the irrational number e =
d x
2.71828182 . . . . The function f (x) = ex is often written exp(x) it has the following key property that dx
e = ex
and crops up frequently (as we will see) when dealing with differential equations.
1.4
Logarithms
Example. We continue from last example. After how many years will the population double in size?
Solution. We are now interested in the inverse problem as we want to solve the equation (a = 1.01)
ax = 2.
That is we want to find the source for the exponential function f (x) = ax that gives the value 2. The answer
is denoted loga 2. So F (x) = loga x is the inverse function to f (x) = ax . With a calculator we see that the
answer is loga 2 ≈ 69.7 years.
1.4.1
The logarithm function
The function loga x is defined to be the inverse of the function ax . It thus has as domain the interval (0, ∞)
(the range of the function ax ) and range (−∞, ∞) (the domain of ax ). Also by the inverse property
y = loga x iff x = ay .
7
Natural Log graph
3
2
y
1
0
!1
!2
!3
!1
!0.5
0
0.5
1
x
1.5
2
2.5
3
Figure 1.2: The log Function y = loga x
In other words: to which power must a be raised to obtain x? Answer loga x.
If a = e then y = loge x = ln x. This is called the natural logarithm
Example.
log2 8
=
3
since 23 = 8
log3 (1/9) = −2 since 3−2 = 1/9
√
√
log5 ( 5) = 1/2 since 51/2 = 5.
Since loga x is strictly increasing or decreasing and since this is the inverse to ax we have the following
properties
If loga x = loga y then x = y
If ln x = ln y then x = y
aloga x = x
eln x = x
x
loga a = x
ln ex = x
Moreover we have the following laws:
The logarithm laws
1.
2.
3.
4.
5.
6.
loga (xy) = loga (x) + loga (y)
loga (x/y) = loga (x) − loga (y)
loga xn = n · loga x
loga 1/x = − loga x
loga 1 = 0
loga a = 1
ln xy = ln x + ln y
ln x/y = ln x − ln y
ln xn = n · ln x
ln 1/x = − ln x
ln 1 = 0
ln e = 1
Proof of 1 and 2 Suppose that x = an and y = am . Notice that it follows that n = loga x and m = loga y.
Then
(1) loga xy = loga an · am = loga an+m = n + m = loga x + loga y
(2) loga x/y = loga an /am = loga an−m = n − m = loga x − loga y.
In fact it is sufficient to use ln. All the other logarithms can be calculated from this one using the following
change-of-base formula:
loga x
logb x =
.
loga b
In particular if we choose a = e we have
logb x =
8
ln x
.
ln b
1.5
1.5.1
Various examples and applications of the exponentials and the
logarithms
Equations with Exponentials and Logs
Eq 1. ln 4x = 3.
Solution. Taking exponentials of both sides gives
eln 4x = e3
⇔ 4x = e3
1
⇔ x = e3
4
Eq 2. 2x = 5.
Solution. Taking natural logs of both sides gives
ln 2x = ln 5
⇔ x ln 2 = ln 5
ln 5
⇔ x=
.
ln 2
Eq 3. ln(x + 1) = ln x + 3.
Solution. Taking exponentials
eln(x+1) = eln x+3 = eln x e3 = xe3
⇔
x + 1 = xe3
1 = x(e3 − 1)
1
.
⇔ x= 3
e −1
⇔
Eq 4. ln(x + x2 ) = ln x + ln x2 .
Solution. The RHS is equivalent to ln x3 using property 3 of ln. Hence taking exponentials
x + x2 = x3 ⇔ x x2 − x − 1 = 0.
√
So that x = 0 or x = 1±2 5 . But the arguments of log must be strictly positive so x = 0 and x =
√
allowed.Therefore x = 1+2 5 is the only solution.
1.5.2
√
1− 5
2
aren’t
Examples from Chemical Engineering
pH for acids
pH is a measure of the strength of an acid (Broented-Larry theory of acids and bases - acids are proton
donors), where the formula
pH = − log10 [H + ]
gives the concentration in mol/dm−3 . Hence, acids with high [H + ] have low pH and conversely, bases have
low [H + ] and thus high pH.
9
Chemical Decay
A chemical, c, decays exponentially according to the rule c = 10e−kt where t is time and k is unknown.
After 30 hours c = 2.5. Find k.
Solution. Concentration at t = 30 is c = 2.5. Therefore we must solve 2.5 = 10e−30k . This gives
1
1
= e−30k ⇔ ln
= ln e−30k = −30k
4
4
1
1
1
⇔ k = − ln
= −
. (−2 ln 2)
30
4
30
1
⇔ k=
ln 2.
15
Carbon Dating
A constant proportion of the carbon atoms in any living creature is made up of the radioactive isotope
14
C of carbon. This proportion remains constant in any living creature. However, when a living organism
dies it ceases its metabolism of carbon and the process of radioactive decay into normal carbon begins. This
full fills the equation (a solution to a certain differential equation that arises from experiment)
N (t) = N (0)e−kt
where N (0) is the amount of 14 C when the animal died and N (t) is the amount t years later. This can be
used to estimate how long time ago the animal died.
Example. Given the decay constant k ≈ 0.0001216 find how long ago an animal died if it has only 88.55% of
its 14 C left.
Solution. If 88.55% of 14 C is remaining then the ratio of the initial amount of
amount N (t) found at our unknown time t is such that
0.8855 =
14
C denoted N (0) to the
N (t)
= e−kt .
N (0)
Taking natural logs on both sides gives
ln(0.8855) = −kt = −(0.0001216)t ⇔ t = −
ln(0.8855)
= 1000.
0.0001216
So we can conclude that the animal died 1000 years ago.
1.5.3
Application of Logs - Parameter Estimation
Consider an equation of the form y = kxn where k and n are unknown constants. Question: Is it possible to
find values of k and n given certain values of x and y from the graph of y = kxn ? (Recall the equation of a
straight line y = mx + c for y-intercept c and slope m).
Method: Take ln and the result is
ln y
ln y
|{z}
=
ln(kxn )
=
ln k + ln xn
= n |{z}
ln x + |{z}
ln k
X
Y
c
which looks like the equation of a straight line except we have ln y and ln x instead of y and x. If we plot ln y
against ln x then gradient = n and intercept on ln y axis is ln k.
10
Chapter 2
Trigonometry
You can read more in J/S: 1.5, 1.6 and 1.8.
2.1
Angles and the unit circle
S
B
θ
A
R
Figure 2.1: Circle of radius R.
Let S equal the length of the arc from A to B. Then
θ=
S
R
Here we consider a oriented angle. It is positive if we move counterclockwise and negative if we move clockwise.
For example, for a quarter circle S =
1. 60o =
π
3
2. 45o =
π
4
3. 30o =
π
6
πR
2 ,
hence θ = π/2 radians or 90o . Remember
Generally we convert from Degreees to Radians by θ =
θo π
180
Recall. The area of the sector that the angle θ cuts out of a circle of radius R is given by A =
2.2
The right angled triangle
11
θR2
2 .
Consider a right angled triangle.
hypotenuse c
opposite b
Pythagoras: a2 + b2 = c2
x
Figure 2.2: Right Angled Triangle
From the triangle we define the value of the trigonometric functions cos, sin and tan for angles 0 ≤ x ≤ π/2.
• sin x =
b
c
• cos x =
a
c
• tan x =
b
a
Remarks. 1) We have that tan(x) = b/a =
(2) cos2 x + sin2 x = (a/c)2 + (b/c)2 =
a2 +b2
c2
b/c
a/c
=
=
c2
c2
sin x
cos x .
= 1.
Two triangles to remember
From the following two triangles (fig. 2.3 and 2.4) one can read off the exact values of the trigonometric
functions in π/6, π/4 and π/3.
• sin π4 = cos π4 = √12 and tan π4 = 1.
• sin
• sin
π
3
π
6
√
=
3
2 ,
π
3
cos
= 12 , cos
π
6
1
2
and tan
π
3
=
3
2
and tan
π
6
=
=
√
=
√
3.
√1 .
3
√
2
45o =
1
π
4
1
Figure 2.3: Isoceles Triangle
12
30 =
√
2
60 =
π
6
3
π
3
1
Figure 2.4: Half an Equilateral Triangle
Laws 1.
As the sum of the acute angles in a right angled triangle is π/2, the other acute angle is π/2 − x. From
this one can thus read directly from the triangle
c
b
x
a
Figure 2.5: Right Angled Triangle
that:
cos(π/2 − x) = b/c = sin x
sin(π/2 − x) = a/c = cos x
tan(π/2 − x) = a/b = 1/ tan x = cot x.
2.3
The trigonometric functions for arbitrary angles
13
Consider a circle of radius r with the centre at the origin.
ŷ
r
y
θ
x̂
x
Figure 2.6: Cartesian (x̂, ŷ) and polar r, θ coordinates
Let θ be as in diagram. Then we define
cos(θ)
=
sin(θ)
=
tan(θ)
=
x
r
y
r
y
.
x
In particular if r = 1 then cos(θ) = x and sin(θ) = y.
Notice that
Dcos = R
Vcos = [−1, 1]
Dsin = R
Vsin = [−1, 1].
By looking at the unitcircle one can read off the following table:
x
cos
sin
tan
0
1
0
0
π/4
√
1/√2
1/ 2
1
π/2
0
1
undef.
3π/4√
−1/
√ 2
1/ 2
−1
π
−1
0
0
5π/4√
−1/√2
−1/ 2
1
3π/2
0
−1
undef.
7π/4
√
1/ √
2
−1/ 2
−1
2π
1
0
0
Notice also that
cos(x + 2π) = cos x, sin(x + 2π) = sin x, tan(x + π) = tan x.
14
Geometric presentation of tan:
ŷ
·· t
1 ·
θ y
x
x̂
Figure 2.7: Geometric presentation of tan
The gradient of the line that forms the angle θ is y/x = tan(θ). But this gradient is also equal to t/1. Hence
t = tan(θ). From this presentation it is clear that tan(θ) → ∞ when θ → π/2− and tan(θ) → −∞ when
θ → −π/2+ .
Notice that
Dtan = R \ {· · · , −π/2, π/2, 3π/2, · · · },
Vtan = R
For the graphs of the functions cos, sin and tan, see either your written notes or look at section 1.6 in J/S.
The signs of sin, cos, sin can be read easily from the unit circle. We get the following scheme
+
+
−
+
−
+
−
−
−
+
+
−
sin
cos
tan
Laws 2. From the unit circle
··
··
1
·· x
··
·· ···
·
··
··
··
··
Figure 2.8: Laws 2
15
one can read off directly the following laws:
cos(−x) = cos(x),
cos(π − x) = − cos(x),
sin(−x) = − sin(x), sin(π − x) = sin(x),
tan(−x) = − tan(x), tan(π − x) = − tan(x),
2.4
2.4.1
cos(π + x) = − cos(x),
sin(π + x) = − sin(x),
tan(π + x) = tan(x).
Trigonometric equations
Some trigonometric equations
Let u be a given angle. From the unit circle (see figure 2.8) we see that
cos v = cos u ⇔ v = u + n · 2π or v = −u + n · 2π
sin v = sin u ⇔ v = u + n · 2π or v = (π − u) + n · 2π
tan v = tan u ⇔ v = u + n · π
where n can be any integer.
Examples. (1) cos x =
√
3/2
Solution. We know from one of the triangles that we should remember that one solution is x = π/6. According
to the general scheme above the general solution is therefore
x = π/6 + n · 2π or x = −π/6 + n · 2π, n ∈ Z.
(2) sin 3x = sinx.
Solution. If v = 3x and u = x then according to general situation we have
3x = x + n · 2π
or
3x = π − x + n · 2π
m
2x = n · 2π
or
4x = π + n · 2π
m
x=n·π
or x = π/4 + n · (π/2)
where n ∈ Z.
(3) sin 5x = cos 3x.
Solution. First use sin 5x = cos(π/2 − 5x). Then we want to solve
cos(π/2 − 5x) = cos 3x.
Now we can use the general method
π/2 − 5x = 3x + n · 2π
or π/2 − 5x = −3x + n · 2π
m
8x = π/2 − n · 2π
or
2x = π/2 − n · 2π
m
x = π/16 + n · (π/4)
or x = π/4 + n · π
where n ∈ Z.
(4) tan 2x = − tan x.
16
Solution. We use tan(−x) = − tan x. Thus we need to solve
tan 2x = tan(−x).
The solution according to the general method is
2x = −x + n · π ⇔ 3x = nπ ⇔ x = n · (π/3)
where n ∈ Z.
Laws 3. Here are some more formulas that one meets frequently. (For more formulas see J/S page 26)
cos(x + y)
=
cos x · cos y − sin x · sin y
sin(x + y)
=
sin x · cos y + cos x · sin y
cos 2x
=
cos2 x − sin2 x
=
2 cos2 x − 1
=
1 − 2 sin2 x
=
2 sin x · cos x.
sin 2x
Examples. (1) Calculate the exact value of sin 75o .
Solution. We have
sin 75o
sin(45o + 30o )
=
sin 45o · sin 30o + cos 30o · sin 45o
√
3 1
1 1
·√
= √ · +
2
2 2
2
√
3+1
√ .
=
2 2
=
(2) Find the exact value of cos π/8.
Solution. Solving the equation cos 2x = 2 · cos2 x − 1 for cos x, gives
cos2 x =
In particular,
1 + cos 2x
.
2
√
√
1 + cos π/4
1 + 1/ 2
2+1
√ .
cos π/8 =
=
=
2
2
2 2
2
It follows that
p√
cos π/8 =
2.4.2
2+1
.
23/4
To Convert a Sum to a Harmonic Form
(Read more in J/S page 15/16). Given a, b and u find c and φ so that
cos (x + φ)
a cos(x) + b sin(x) = c
|
{z
}
| {z }
sum of cos and sin
harmonic function
for amplitude c and phase shift φ.
• Step 1: Find P = (a, −b) on x, y plane
• Step 2: Draw line from P to origin.
√
Length OP is given by c = a2 + b2 and φ is the polar angle of P.
17
y
(a, −b)
φ
x
Figure 2.9: Harmonic Sum Description
Sketching Harmonic Solution
To sketch for general harmonic function y = c cos (ωx + φ) note the following.
1. Observe one wavelength of the cosine function has length 2π/ω.
2. The beginning of our cosine wave now starts from x = − ωφ and one wavelength ends at x = − ωφ +
It is out of phase by φ units in x.
2π
ω .
3. Also note our wave is magnified by constant c, so the range of y is between y ∈ [−c, c], rather than
y ∈ [−1, 1].
Harmonic Function: y=c*cos(wx+phi))
3
c
2
y
1
0
!!/"
(!/2 !#)"
(2!!#)/"
!1
!2
!3
!2
0
2
4
6
8
10
12
x
Figure 2.10: Harmonic Sketch
Examples of converting a sum to a harmonic form
√
1. cos(x) − sin(x) = c cos(x + φ). Now a = 1, b = −1 so that P = (1, 1). Therefore c = 2 and φ = π4 .
√
Hence cos(x) − sin(x) = 2 cos x + π4 .
√
√
√
2. cos(x) − 3 sin(x) = c cos(x
+ φ). Now a = 1, b= − 3 so that P = (1, + 3). Therefore c = 2 and
√
φ = − π3 . Hence cos(x) + 3 sin(x) = 2 cos x + π3 .
√
3. − sin(x) + 4 cos(x) = c cos(x + φ). Now a = 4, b = −1 so that P = (4, 1). Therefore c = 17, but finding
φ is not so straightforward. Looking at the triangle φ satisfies tan(φ) = 14 or sin(φ) = √117 . So φ is the
angle whose sin is √117 , such that φ = sin−1 √117 . This leads us on to inverse trig functions.
18
y
√
17
1
φ
x
4
Figure 2.11: Harmonic Example 3
2.5
2.5.1
The inverses to the trigonometric functions
The inverse sine function sin−1 x or arcsin x
Consider y = sin(x) restricted to the interval − π2 ≤ x ≤
π
2
and −1 ≤ y ≤ 1.
sin(x)
1
0.8
0.6
0.4
y
0.2
0
!!/2
!/6
!/2
!0.2
!0.4
!0.6
!0.8
!1
!1.5
!1
!0.5
0
x
0.5
1
1.5
Figure 2.12: sin(x)
The inverse sine function sin−1 (x) or (arcsin(x)) is defined as follows. For −1 ≤ x ≤ 1, y = sin−1 (x) if
x = sin(y) and −π/2 ≤ y ≤ π/2. In words ’y is the angle whose sine is x’ and lies between −π/2 and π/2.
sin(x)
!/2
1.5
1
!/6
y
0.5
0
!0.5
!1
!1.5
!1
!0.8
!0.6
!0.4
!0.2
!!/2
0
x
0.2
0.4
0.6
0.8
1
Figure 2.13: y = sin−1 (x)
Since arcsin x is the inverse of sin x, −π/2 ≤ x ≤ π/2, we have
Darcsin = [−1, 1],
Varcsin = [−π/2, π/2].
19
Also
arcsin(sin x) = x if
−π/2 ≤ x ≤ π/2
sin(arcsin x) = x if
−1 ≤ x ≤ 1.
Warning. arcsin(sin π) = arcsin(0) = 0 but not π.
Examples
1. sin−1
2. sin−1
2.5.2
1
2
= π6 since sin
− 12 = − π6
π
6
=
1
2
The Inverse Cosine Function (cos−1 (x) or arccos(x))
The function cos x, 0 ≤ x ≤ π is injective. The inverse is called cos−1 or arccos. (For the graph see your
handwritten notes or J/S page 20). So
arccos(x) = y ⇔ x = cos(y), and 0 ≤ y ≤ π.
Note that
Darccos = [−1, 1],
Varccos = [0, π].
Also
0≤x≤π
arccos(cos x) = x
if
cos(arccos x) = x
if −1 ≤ x ≤ 1.
Examples
√ 1. cos−1 − 23 =
2. cos−1
2.5.3
1
2
=
2π
3
π
3
The Inverse Tangent Function (tan−1 (x) or arctan(x))
The function tan x, −π/2 < x < π/2 is injective. The inverse is called tan−1 or arctan. (For the graph see
your handwritten notes or J/S page 20). So
arctan(x) = y ⇔ x = tan(y), and − π/2 < y < π/2.
Note that
Darctan = R,
Varctan = [−π/2, π/2].
Also
arctan(tan x) = x if −π/2 < x < π/2
tan(arctan x) = x if x ∈ R.
Examples
1. tan−1 (1) = π/4
√
2. tan−1 (−1/ 3) = −π/6.
20
Chapter 3
Inequalities
3.1
Solving inequalities
We will often have to deal with inequalities later on in the course. For example in order to sketch a graph to
a function we need to solve
f 0 (x) > 0
to find out where the function is increasing.
Example 1. Solve −1 < 1 − 4x ≤ 3.
Solution. We deal with the left hand side (LHS) and right hand side (RHS) of the inequality seperately:
−1 < 1 − 4x
0 < 2 − 4x
4x < 2
divide both sides by 4
1
x<
2
or
1 − 4x ≤ 3
subtract 1 from both sides
−4x ≤ 2
divide both sides by 4
1
−x ≤
2
multiply each side by -1 and so reflect the inequality
1
x≥−
2
Hence combining the solutions from the LHS and RHS we have − 21 ≤ x < 21 . 2
One has to be extra careful when dealing with inequalities.
Example. Solve
1
= 1.
x−2
Solution. We multiply through by x − 2 one both sides. This gives
1 = x − 2 ⇔ x = 3.
Example 2. Solve the inequality
1
>1
x−2
False attempt. We multiply again through by x − 2. This gives
1 > x − 2 ⇔ x < 3.
21
However if we check the inequality on x = 0 (which is less than 0) we should have
1
> 1.
−2
But this is absurd as the LHS is negative. 2
Where did we cheat? The problem is that when multiplying with x − 2 we might (for some values of x)
be multiplying with negative number. But then the inequality should change. For example
−1 < 2
but
(−1) · (−1) > (−1) · 2.
There is a way around this:
Solution to Example 1. Subtract 1 from both sides. This gives
1
1
1
>1 ⇔
− >0
x−2
x−2 1
(x − 2)
1
−
>0
⇔
x−2
x−2
1−x+2
>0
⇔
x−2
3−x
⇔
>0
x−2
What have we gained by doing this? Now we can write a sign table for this fraction. Notice that the numerator
is zero when x = 3 and the denominator is zero when x = 2.
x
3−x
x−2
3−x
x−2
+
-
2
+
0
o
+
+
+
3
0
+
0
+
-
From the table we read that the fraction is strictly positve when 2 < x < 3.
Answer: 2 < x < 3.
Remark. Instead of solving this inequality algebraically one could have done it grahphically. Look at
1
the graph of x−2
(see figure 3.3 (b)). It is clear that the value of this function is strictly greater than 1 if and
only if x is strictly between 2 and 3.
Example 3. Solve x2 > 4.
Solution. Here again one has to be careful. One cannot simply argue that x2 > 22 ⇒ x > 2. There
are two solutions to the equation x2 = 4, namely −2 and 2. When x < −2 or x > 2 we have that the
inequality holds wheras it does not hold when −2 ≤ x ≤ 2.
Answer: x < −2 or x > 2.
Example 4. Solve
x>
1
.
x
Solution. Firstly
x 1
x2 − 1
− >0⇔
> 0.
1 x
x
22
Next we write a sign table.
x
x2 − 1
x
x2 −1
x
+
-
−1
0
0
0
0
o
+
1
0
+
0
+
-
+
+
+
From the table we read when the fraction is strictly greater than zero.
Answer. −1 < x < 0 or x > 1.
Graphical Solution
Graphs of y=1/x and y=x
3
2
y
1
0
!1
!2
!3
!3
!2
!1
0
x
1
2
3
Figure 3.1: Plot of y = 1/x and y = x
Example 5. Solve
x+3
≤ 2.
x−1
Solution. We have
x+3
≤2
x−1
⇔
⇔
⇔
⇔
x+3 2
− ≤0
x−1 1
x + 3 2(x − 1)
−
≤0
x−1
x−1
x + 3 − 2(x − 1)
≤0
x−1
5−x
≤0
x−1
The numerator is zero when x = 5 and the denominator is zero when x = 1.
x
5−x
x−1
3−x
x−2
+
-
1
+
0
o
+
+
+
5
0
+
0
+
-
Answer: x < 1 or x ≥ 5.
23
Graphical Solution
Graphs of y=(x+3)/(x!1) and y=2
10
y
5
0
!5
!5
0
5
10
x
Figure 3.2: Plot of y =
3.2
x+3
x−1
and y = 2
Sketching Reciprocal Functions
How does one sketch in general a reciprocal function. For example if we want to sketch the graph for
could do this in few steps as follows:
Sketch: i) y = x1 , ii) y =
1
x−2
and iii) y =
x−1
x−2 .
Note y =
=
x−2+1
x−2
=1+
(x!1)/(x!2)
3
2
2
2
1
1
1
0
0
0
y
3
!1
!1
!1
!2
!2
!2
!3
!3
!2
!1
0
x
(a) y =
1
x−2
1/(x!2)
3
y
y
1/x
x−1
x−2
1
1
x
2
3
!3
!3
!2
!1
0
x
(b) y =
1
2
3
1
x−2
Figure 3.3: Reciprocal Functions
24
x−1
x−2 ,
!3
!3
!2
!1
0
x
(c) y =
1
x−1
x−2
2
3
we
Chapter 4
Limits and Asymptotes
4.1
Definition and basic properties of limits
What do we mean by
1
=0
x
(“the limit for 1/x is zero as x tends to infinity”) alternatively expressed
lim
x→∞
1
→ 0 as x → ∞
x
(“1/x tends to zero as x tends to infinity”)?
Loosely speaking, 1/x can be made as close to zero as we please by choosing x big enough. How big x
needs to be chosen depends ofcourse on how close we want to be to zero:
|1/x − 0| < 0.1
if
x > 10
|1/x − 0| < 0.01
if
x > 100
|1/x − 0| < if
x > ω = 1/
(Recall If x is a real number, the modulus of x, written |x| is defined as
x x≥0
|x| =
−x x < 0
Examples: |3| = 3, | − 2| = −(−2) = 2).
Generally. We write
lim f (x) = A
x→a
or
f (x) → A as
x→a
to mean that f (x) can be made as close to A as we please by choosing x close enough to a.
Rules of limits. Suppose
lim f (x) = A,
x→a
and
where A, B are some real numbers. Then
(1) f (x) + g(x) → A + B
25
lim g(x) = B
x→a
(2) f (x) · g(x) → A · B
(3) f (x)/g(x) → A/B (if B 6= 0).
Also if f (x) → b as x → a and g(x) → c as x → b. Then
(4) g(f (x)) → c as x → a.
Examples. (1) We have limx→0 sin x = 0 we have, using (1)-(3) that
sin2 x + 1
02 + 1
=
= −1.
x→0 sin x − 1
0−1
lim
(2) Since e−x → 0 as x → ∞ and sin x → 0 as x → 0 we have from (4) that
sin(e−x ) → 0
as x → ∞.
Warning In (1)-(3) we are assuming that A, B are real numbers. One should never apply “false formulas”
like 0 · ∞ = 0, ∞/∞ = 1, 0/0 = 0. This leads to a catastrophe!
Example 3. What is
x2 − 3x + 2
?
x→1
x−1
Solution. We have that x2 − 3x + 2 and x − 1 both tend to 0 as x → 1. We can’t however conclude that the
2
fraction x −3x+2
tends to 0/0 = 1. Instead use factorisation
x−1
lim
x2 − 3x + 2
(x − 1)(x − 2)
=
= x − 2 → 1 − 2 = −1
x−1
x−1
as x → 1.
Definition. By
x → a+
we mean that x tends to a from above/right. Similarly
x → a−
means that x tends to a from below/left.
Example 4.
limx→0+
limx→0−
Since these are different the limit limx→0
4.2
|x|
x
|x|
x
|x|
x
= limx→0+
= limx→0−
x
x =1
−x
x = −1.
does not exist on the other hand.
Calculating limits with help of standard limits
I. Comparison of ln x, xα and ax as x → ∞.
Let α > 0 and a > 1. One can show that the growth of ln x is slower than the growth of xα which then
is slower than the growth of ax . (Compare the graphs of these three functions). In other words
ln x
→0
xα
α
x
→0
ax
as
x→∞
as
x→∞
26
or equivalently
xα
→∞
ln x
x
a
→∞
xα
as
x→∞
as
x→∞
Example 5. Calculate
ex + x2 + ln x
x→∞
3ex + 2x
Solution. We shorten with the stongest term in the denominator.
lim
ex + x2 + ln x
3ex + 2x
2
=
→
1 + xex + lnexx
3 + (2/e)x
1+0+0
1
=
3+0
3
as x → ∞. (Notice that since 2/e < 1 we have that (2/e)x tends to zero).
II. Some standard limits of type “0/0” when x → 0
You should know the following standard limits (important for differentiation of ln x, sin x and ex ). As we
will see we can also calculate other limits from these standard ones.
Standard limits.
sin(x)
x
ex − 1
lim
x→0
x
ln(1 + x)
lim
x→0
x
lim
x→0
=
1
(4.1)
=
1
(4.2)
=
1.
(4.3)
Example 6. Calculate
lim
x→0
Solution. We have
sin 5x
3x
sin 5x 5x
sin 5x
=
·
→ 1 · 5/3 = 5/3.
sin 3x
5x
3x
Example 7. Calculate
2
ex − 1
lim
.
x→0 ln 1 + 7x
Solution. We have
2
2
2
ex − 1
ex − 1
7x
x2
ex − 1
7x
x
=
·
·
=
·
· →1·1·0=0
2
2
ln 1 + 7x
x
ln 1 + 7x 7x
x
ln 1 + 7x 7
as x → 0.
27
4.3
Asymptotes
Definition: An asymptote to a curve is a straight line to which the curve approaches as the distance from the
origin increases.
Look at the graphs in figure 4.1. It is clear from the graphs that the horizontal line y = 0 is an assympot for y = e−x , and y = 1/x as x → ∞. Also y = 1 is an asympote for y = 1 − e−x when x → ∞. In
fact
lim e−x
x→∞
lim
1
x→∞ x
−x
lim 1 − e
x→∞
=
0
=
0
=
1.
Also y = 0 is an horizontal asympote as x → −∞. Which also follows from
lim
x→−∞
1
= 0.
x
On the other hand there is no horizontal asympote when x → ∞ for y = ex and indeed limx→∞ ex = ∞.
For (c) the situation is different. We don’t have any horizontal asympote but x = 0 is a vertical asymptote. As x approaches 0 from the right we have that y → −∞.
Graph of y=exp(!x) and y=exp(!2x)
Graph of y=exp(x)
1.5
Graph of y=ln(x)
5
5
4
4
1
3
2
3
2
0
y
y
1
y
0.5
0
!1
1
!2
!0.5
!3
0
!4
0.5
1
1.5
2
(a) Ex1: y =
2.5
x
3
e−x ,
3.5
4
y=
4.5
!1
5
0
0.2
0.4
e−2x
0.6
0.8
1
x
1.2
1.4
1.6
1.8
2
!5
0
0.2
0.4
0.6
0.8
(b) Ex2: y =
ex
Graph of y=1!exp(!x)
1.5
4
3
1
2
1
0.5
!1
!2
0
!3
!4
!5
!5
!4
!3
!2
!1
0
x
1
(d) Ex4: y =
2
3
4
5
1.2
1.4
1.6
(c) Ex3: y = ln(x)
Graph of y=1/x
5
0
1
x
y
0
y
!1
!0.5
1
x
0
0.5
1
1.5
2
2.5
x
3
3.5
(e) Ex5: y = 1 −
Figure 4.1: Some asymptotes
28
4
e−x
4.5
5
1.8
2
I. How to find horizontal asymptotes analytically.
We have that y = a is a horizontal asymptote to y = f (x) if and only if
lim f (x) = a or
x→∞
lim f (x) = a.
x→−∞
II. How to find vertical asymptotes analytically.
We have that x = a is a vertical asymptote if and only if
lim f (x) = ±∞ or
x→a+
lim f (x) = ±∞.
x→a−
(A candidate for x = a we normally find where f (a) is not defined).
Example 1. Find all vertical and horizontal asympotes to
y=
x−2
.
2x + 3
Then sketch the graph.
Solution.
Step 1. Horizontal asymptotes. We have
x−2
2x + 3
=
→
1 − x2
2 + x3
1/2.
As x → ∞ and also as x → −∞. So y = 0 is a horizontal asymptote as x → ±∞. (You can write this line as
a dotted line in your coordinate system).
x−2
is not defined when 2x + 3 = 0 ⇔ x = −3/2. So
Step2. Vertical asymptotes. We have that the fraction 2x+3
x = −3/2 is a candidate for a vertical asymptot. When x is little bigger than x = −3/2 we have that
x−2
−7/2
≈
= −big
2x + 3
+small
So
lim
x→−3/2+
x−2
= −∞.
2x + 3
When x is little less than −3/2, we have that
x−2
−7/2
≈
= +big
2x + 3
−small
and
lim
x→−3/2−
x−2
= ∞.
2x + 3
Thus x = −3/2 is a vertical asymptot both when we approach −3/2 from the left and from the right.
Step 3. Sketch the graph. This you can do by first making a table. (It is a good practise also to check
where the curve meets the x-axis and the y-axis). See figure 4.2.
29
(x!2)/(2x+3)
3
2
y
1
0
3/2
!2/3
!1
!2
!3
!3
!2
!1
0
x
1
2
Figure 4.2: Plot of y =
3
x−2
2x+3
Example 2. Find all the horizontal and vertical aympotes to
y=
2x + 1
x
and sketch the graph.
Solution.
Step 1. Horizontal asymptotes. We have
2x + 1
x
=
2+
1
x
→ 1/2.
As x → ∞ and also as x → −∞. So y = 0 is a horizontal asymptote as x → ±∞. (You can write this line as
a dotted line in your coordinate system).
Step2. Vertical asymptotes. We have that the fraction 2 + 1/x is not defined when x = 0. So x = 0 is a
candidate for a vertical asymptot. When x is little bigger than 0 we have that
2 + 1/x ≈ 2 +
1
= 2 + big = big
+small
So
lim 2 + 1/x = ∞.
x→0+
When x is little less than 0, we have that
2 + 1/x ≈ 2 +
1
= 2 − big = −big
−small
and
lim− 2 +
x→0
1
= −∞.
x
Thus x = 0 is a vertical asymptote both when we approach 0 from the left and from the right. (You can add
this line as a dotted line to your coordinate system).
Step 3. Sketch the graph. See figure 4.3.
30
(2x+1)/(x)
8
7
6
5
4
y
3
2
1
0
!1
!2
!3
!3
!2
!1
0
x
1
Figure 4.3: Plot of y =
4.4
2
3
2x+1
x
Physical Example of Limits
B
The recommended correlation for viscosity of liquids is ln µ = A + T +C
where µ is the viscosity, T is temperature, with A, B constants. The constant c is given by C = 17.71 − 0.19Tb where Tb is the normal boiling
point in Kelvin. Hence for n-Propane c = −26.18
Question 1: Plot ln µ for A = −1 and B = 2.
Question 2: What is the viscosity as T → ∞?
Solution 1: For plot see below.
Solution 2: Taking exponentials µ = exp −1 +
2
e−1 limT →∞ exp T −26
= e−1 .
2
T −26
= exp(−1) × exp
!1+2/(x!26)
8
7
6
5
4
y
3
2
1
0
!1
!2
!3
20
21
22
23
24
25
x
26
27
28
Figure 4.4: Plot of y = −1 +
31
29
30
2
T −26
2
T −26
. Hence limT →∞ µ =
Chapter 5
Differentiation
5.1
Definition and geometric interpretation
Let us consider some examples.
Example 1
Suppose a car is moving on a straight road (x-axis). Its displacement (distance) from the origin at time t is
given by x = s(t). After time t + ∆t the displacement is s(t + ∆t). Then the instantaneous velocity v(t) at
time t is defined by
s(t + ∆t) − s(t)
change in distance travelled
v(t) = lim
=
∆t→0
∆t
change in time
For notation purposes we define
lim
∆t→0
s(t + ∆t) − s(t)
ds(t)
=
= rate of change of distance with respect to time
∆t
dt
Acceleration a(t) of the car at time t is then given by
d
dv(t)
=
a(t) =
dt
dt
ds(t)
dt
=
d2 s(t)
dt2
Example 2
Consider the change in surface area of a sphere (e.g. an expanding balloon). Then we define
• Surface area Sa with radius r is given by Sa = 4πr2
2
• After some time the surface area has changed such that Sa = 4π (r + ∆r)
4π r2 + 2r∆r + ∆r2 − 4πr2
4π 2r∆r + ∆r2
Change in Area
4π(r + ∆r)2 − 4πr2
=
=
=
= 4π (2r + ∆r)
∆r
∆r
∆r
Now using limit rules
lim
∆r→0
Change in Area
= 8πr = Rate of change of area with respect to radius
Now for the general mathematical defininition:
Let f (x) be a function defined on a real open interval and let x0 be any point in the interval, then
Definition
The limit lim∆x→0
f (x0 +∆x)−f (x0 )
∆x
is said to be the derivative of f (x) at x0 and it is denoted f 0 (x0 ) or
32
df (x0 )
dx .
So f 0 (x0 ) is the rate of change in the value at x = x0 with respect to the input.
Geometric Interpretation
We have that f 0 (x0 ) can be interpreted as the gradient of the tangent line to the curve y = f (x) at the
point (x0 , f (x0 ). (See your handwritten notes or J/S 2.1 for a figure).
5.2
Common Derivatives
f (x)
f (x)
f (x)
f (x)
f (x)
f (x)
f (x)
f (x)
=
=
=
=
=
=
=
=
f 0 (x)
f 0 (x)
f 0 (x)
f 0 (x)
f 0 (x)
f 0 (x)
f 0 (x)
f 0 (x)
xn
ex
ekx
ln(x) (x > 0)
ax
sin x
cos x
tan x
=
=
=
=
=
=
=
=
nxn−1
ex
kekx
1
x
(ln a) ax
cos x
− sin x
1
cos2 x
In the fifth case we prove the derivative by rearranging f (x) such that f (x) = ax = elna x = ex ln a , hence
differentiating we pull down ln(a) and then rearrange ex ln a as ax .
Example 1
Prove that (ex )0 = ex . We have
ex+h − ex
ex eh − ex
eh − 1
=
= ex
→ ex · 1
h
h
h
as
h→0
Here we have used one of the standard limits from last chapter.
The formula (xn )0 = nxn−1 can be useful when differentiating 1/xn or expressions involving roots.
Example 2. (1/x)0 = (x−1 )0 = (−1) · x−2 = −1/x2 .
√
Example 3. ( x)0 = (x1/2 )0 =
5.3
1
2
· x−1/2 =
1
√
.
2· x
Higher Order Derivatives
We denote the second derivative by
d2 y
dx2
=
d
dx
dy
dx
. Hence to differentiate sin(x) twice we have
d2 (sin(x))
d
(cos(x)) = − sin(x)
=
dx2
dx
2 d3 y
d y
d
Similarly for third order derivatives dx
3 = dx
dx2
Example
2
d y
Show that y = e2x satisifes dx
2 − 4y = 0.
If y = e2x then
d dy
d
d2 y
d 2x =
=
2e2x = 2
e
= 2.2e2x = 4e2x = 4y
2
dx
dx dx
dx
dx
33
5.4
Rules for Differentiation
You should know these rules. Here u and v are functions of x and k is a constant. Then
1. Sum Rule
d
dx (u
+ v) =
du
dx
+
2. Factor Rule For k constant
dv
dx
dk
dx
= 0, and
d
dx (ku)
= k du
dx
dv
= u dx
+ v du
dx
d
d
d
Example: dx
3x + 1 sin(x) = 3x2 + 1 dx
(sin(x))+sin(x) dx
3x2 + 1 = 3x2 + 1 cos(x)+6x sin(x)
1 du
d
1
du
dv
u dv
u
4. Quotient Rule dx
v = v 2 v dx − u dx = v dx − v 2 dx
2
2
x
1
1−x2
d
x2 + 1 .1 − x.2x = x(x+1−2x
= (1+x
Example: dx
2 +1)2
2 )2
x2 +1 = (x2 +1)2
3. Product Rule
d
dx (uv)
2
dy du
dy
= du
. dx
5. Chain Rule If y = y(u) and u = u(x) then dx
2
Example: If y = ln x + 1 . Let y(u) = ln(u) and u(x) = x2 + 1 then
dy
dy du
1
2x
=
.
= .2x = 2
dx
du dx
u
x +1
2
Example: If y = (sin(x) + x) , then let u = sin(x) + x, hence
dy
du
= 2u
= 2 (sin(x) + x) (cos(x) + 1)
dx
dx
Examples
Find
1.
d
dx
3.
d
dx
tan(x) =
4.
d
dx
(x ln x) = x. x1 + ln(x).1 = 1 + ln(x)
sin(x)
d
2. dx
= x12 (x cos(x) − sin(x))
x
2
ex
d
dx
sin(x)
cos(x)
=
1
cos2 (x)
cos2 (x) − sin(x) (− sin(x)) =
1
cos2 (x)
. Let y(u) = eu , u(x) = x2 . Then applying the chain rule:
dy
dx
=
dy du
du . dx
2
= eu .2x = 2xex
dy
dy du
ln 2x2 + 1 . Let y(u) = ln(u), u = 2x2 + 1. Then dx
= du
. dx = u1 · 4x = 2x4x
2 +1
d
6. dx
sin(cos(x2 )) Let f (x) = sin(u), u = cos(x2 ) = cos(v), v = x2 . Then
df
df du dv
2
2
dx = dx . dv . dx = cos(cos(x ). − sin(x ) .2x
2 2
2
2
dy
d
d
x2
2 x2
7. dx
y(x) = dx
sin xex
. Let y = sin(u), u = xex . Then du
and dx
= cos(xex )ex 1 + 2x2
dx = e +2x e
5.
d
dx
5.5
Logarithmic Differentiation
(Read more in J/S §3.7 page 57).
Generally suppose that we have some function y = y(x) and that we want to differentiate it. Let u = ln y then
du
du dy
1 dy
=
·
= ·
.
dx
dy dx
y dx
Hence
dy
d
=y·
ln y.
dx
dx
This can often be useful as sometimes it is much easier to differentiate ln y than y. Let us cnsider an example.
34
2
1
We want to differentiate y(x) = x 4 e−x cos(−3x). We take ln on both sides. This gives
1
2
ln(y(x)) = ln x 4 + ln e−x − ln cos3 (x)
1
ln(x) − x2 − 3 ln(cos(x))
4
=
Now using chain rule on ln(y(x)) we find that
1 dy
y dx
11
1
− 2x − 3
(− sin(x))
4x
cos(x)
1
− 2x + 3 tan(x)
4x
=
=
⇓
dy
dx
1
y
− 2x + 3 tan(x)
4x
2 1
1
x 4 e−x
−
2x
+
3
tan(x)
cos3 (x) 4x
=
=
Examples
1. If y = x3x find
dy
dx .
Now ln(y) = ln x3x = 3x ln(x), hence differentiating
1 dy
y dx
dy
dx
3
2. If y =
ex
x sin2 (x)
find
dy
dx .
1
+ 3 ln(x) = 3 (ln(x) + 1)
x
=
3x.
=
3x3x (ln(x) + 1)
=
3
ln ex − ln(x) − ln sin2 (x)
=
x3 − ln(x) − 2 ln (sin(x))
Now
ln(y)
Differentiating with respect to x gives:
1 dy
y dx
=
1
3x − − 2
x
2
1
sin(x)
(cos(x))
⇓
dy
dx
3. If y =
sin(x)
1
(1+x2 ) 2 e−x
then find
dy
dx .
3
=
ex
x sin2 (x)
1
1
2
3x − − 2
(cos(x))
x
sin(x)
Now
ln(y)
=
ln(sin(x)) =
1
ln(1 + x2 ) − ln (e−x
2
Differentiating with respect to x gives:
1 dy
1
1
1
=
(cos(x)) −
(2x) + 1
y dx
sin(x)
2 1 + x2
x
= cot(x) −
+1
1 + x2
⇓
dy
sin(x)
x
= =
cot(x)
−
+
1
1
dx
1 + x2
(1 + x2 ) 2 e−x
35
5.6
Implicit Differentiation
You can read more in J/S §3.8.
Sometimes we don’t know the function y = y(x) explicitly but only implicitly from a formula. Let’s look
at an example.
If x2 + 4xy + y 2 = 100, find
x yields
dy
dx
in terms of x and y. We denote y 0 =
2x + 4y + 4xy 0 + 2yy 0
=
dy
dx ,
hence differentiating with respect to
0
⇓
0
= − (2x + 4y)
y (4x + 2y)
⇓
dy
dx
5.7
= −
x + 2y
2x + y
Derivatives of Inverse Functions
You can read more in J/S §3.9.
One can determine the derivative of the inverse function F (x) to f (x), using
1
dy
= dx
dx
dy
where the right hand side should be expressed in x. Let us look at an example.
Example 1. Find the derivative of ln x.
Solution. As ln x is the inverse of ex , we have
y = ln x ⇔ x = ey .
Thus
1
1
dy
1
= = y = .
dx
dx
e
x
dy
Example 2. Find the derivative of arcsin x.
Solution. We have
y = arcsin x ⇔ x = sin y and − π/2 ≤ y ≤ π/2
Thus
dy
1
1
= =
.
dx
dx
cos y
dy
We need to express this in x = sin y. But
cos y = ±
q
p
1 − sin2 y = ± 1 − x2
and as cos y is positive in the interval [−π/2, π/2], it follows that
(arcsin x)0 = √
36
1
.
1 − x2
5.8
Parametric Differentiation
You can read more in J/S §3.10.
Sometimes we have x and y as functions of a parameter t (often time), so that x = x(t) and y = y(t).
Consider the circle prescribed by x = 2 cos(t), y = 2 sin(t) for 0 ≤ t < 2π. Then x2 + y 2 = 4.
ŷ
2
y
t
x
x̂
Figure 5.1: Paramaterised Circle
To find
dy
dx
we use
dy
dx
=
dy
dt
dx
dt
in terms of t. Hence for a circle
dy
dt
dx
dt
Therefore
dy
=
dx
At (x, y) = (0, 1) we have
circle.
dy
dx
dy
dt
dx
dt
=
2 cos(t)
= −2 sin(t).
=−
x
cos(t)
=− .
sin(t)
y
= −0/1 = 0 corresponding to a zero rate of change of y in x at the top of the
Example
If x = t2 and y = t3 find
dy
dx .
dy
=
dx
5.9
dy
dt
dx
dt
=
3t2
3
3√
= t=
x
2t
2
2
Application to a real life situation
Example. Sand falls from a chute and forms a conical pile in such way that the vertical angle θ emains
constant. (See figure from your handwritten notes. Or draw one yourself).
Suppose that the radius of the base of the cone is r = r(t) and that the height is h = h(t). Show that
if r increases at a rate α cm/s then the volume increases at a rate πrhα cm3 /s.
Solution. Step 1. Mathematical formulation.
We have already drawn a figure to understand what is going on.
given: r0 (t) = α
need: v 0 (t) where v(t) is the volume at the timepoint t sec.
37
Step 2. We need a formula for v that we can differentiate.
πr2 h
.
3
Thinking ahead we don’t know h0 (t). However (look at the figure) tan(θ) = r/h. Thus h = r/ tan(θ) and
v=
v=
π
πr2 · r/ tan(θ)
=
· r3 .
3
3 tan(θ)
Step 3. We differentiate v with respect to time.
We use the chain rule
dv
dv dr
π
π
r
=
·
=
· 3r2 · α = · r · · α = π · rhα.
dt
dr dt
3 tan(θ)
3
θ
5.10
Application of the derivative for evaluation of limits: L’Hopital’s
Rule
Suppose f, g are differentiable functions defined on an open interval. If either
lim f (x) = lim g(x) = 0
x→x0
x→x0
or
lim f (x) = lim g(x) = ±∞
x→x0
x→x0
then
lim
x→x0
f (x)
f 0 (x)
= lim 0
g(x) x→x0 g (x)
Examples
1. Find limx→0
cos(x)−1
.
x
In this case f (x) = cos(x) − 1 and g(x) = x. We have f (0) = g(0) = 0. Thus
lim
x→0
2. Similarly find limx→0
cos(x) − 1
− sin(x)
− sin(0)
= lim
=
=0
x→0
x
1
1
sin(x)
x
0
sin(x)
(sin(x))
cos(x)
= lim
= lim
= cos(0) = 1
0
x→0
x→0
x→0
x
1
(x)
lim
3. Calculate limx→0
sin(x)−x
.
x3
First we try to apply L’Hopitals rule
lim
x→0
sin(x) − x
cos(x) − 1
0
= lim
=
x→0
x3
3x2
0
We still have the indeterminate form. 00 . So let’s try to apply L’Hopitals rule again.
cos(x) − 1
− sin(x)
1
= lim
=−
2
x→0
x→0
3x
6x
6
lim
and we conclude that
1
sin(x) − x
=−
3
x→0
x
6
lim
38
4. Find limx→∞ ln(x)
x2 . Here we deal with the indeterminate form ∞/∞. Differentiating the numerator and
denominator we find
ln(x)
x→∞ x2
lim
5. Find limx→∞
=
lim
1
x
Using L’Hopital’s rule once
2x
1
= 0 Using L’Hopital again
= lim
x→∞ 2x2
x→∞
x2
ex .
We use L’Hopitals rule twice.
lim f (x) = lim
x→∞
5.11
x→∞
2
2x
= lim x = 0
x→∞ e
ex
Stationary points
Let y = f (x) be differentiable. The points a for which f 0 (a) = 0 are called stationary points.
We consider four cases.
Case 1.
x
f 0 (x)
f (x)
-
&
-
x0
0
+
+
%
+
In this case we have local minimum at x0 . (see figure 5.3 (b))
39
Case 2.
x
f 0 (x)
f (x)
+
+
%
x0
0
+
-
&
-
In this case we have local maximum at x0 . (see figure 5.3 (a))
Case 3.
x
f 0 (x)
f (x)
+
+
%
x0
0
+
+
+
%
+
In this case we have neither a local maximum nor local minimum (what we have is a terrace point (see
figure 5.3 (c)).
Case 4.
x
f 0 (x)
f (x)
-
&
-
x0
0
-
&
-
Again we have a terrace point.
The second derivative test: Sometimes one can read from the second derivative whether we have a local
maximum or local minimum. Assume f (x) is twice differentiable at a stationary point x0 (so f 0 (x0 ) = 0) then
1. If f 00 (x0 ) > 0 then f has a local minimum at x0 .
2. If f 00 (x0 ) < 0 then f has a local maximum at x0 .
Consider f (x) = x2 and g(x) = −x2 then f has a local minimum since f 00 (x) = 2 > 0 and g has a local
maximum since g 00 (x) = −2 < 0. (See figure 5.2)
!x*x)
15
10
10
5
5
0
0
y
y
x*x)
15
!5
!5
!10
!10
!15
!4
!15
!3
!2
!1
0
x
(a) y =
1
2
3
4
x2
!4
!3
!2
!1
0
x
(b) y =
1
2
3
4
−x2
Figure 5.2: Local Minimum and Maximum
Example
Consider f (x) = x4 − 2x2 . Then f 0 (x) = 4x3 − 4x = 4x(x − 1)(x + 1), hence there are three turning points
at x = 0, −1, 1. Using f 00 (x) = 12x2 − 4, the turning points are such that at x = 0 we have a local maximum
(f 00 (0) = −4 < 0 ) and at x = −1, 1 we have local minima (f 00 (±1) = 8 > 0).
Local maxima and minima occur at critical points f 0 (x) = 0 where f 0 changes sign.
40
4
3.5
4
4
3.5
3.5
3
3
2.5
2.5
2
2
(1)
f =0
3
(1)
2.5
(1)
f >0
f <0
y
y
y
2
1.5
1.5
1
1
0.5
0.5
f(1)>0
f(1)<0
f(1)>0
1.5
1
f(1)=0
f(1)=0
0
0
0
!0.5
!0.5
!0.5
!0.5
!0.5
!0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
f(1)>0
0.5
0
0.5
1
1.5
2
x
2.5
3
3.5
4
0
0.5
(a) Local Maxima
1
1.5
2
2.5
3
3.5
4
x
x
(b) Local Minima
(c) No Max,Min
Figure 5.3: Local Minimum and Maximum, No max/min
5.12
Global max and min on a closed bounded interval
Fact. Every continuous function on [a, b] has a largest value (global max) and a smallest value (global min).
Furthermore if x1 , . . . , xm are the stationary points then
global max
=
max {f (a), f (b), f (x1 ), . . . , f (xm )}
global min
=
min {f (a), f (b), f (x1 ), . . . , f (xm )}.
4
2
y
0
!2
!4
!6
!8
0
1
2
3
4
5
6
7
x
Figure 5.4: Multiple Maxima and Minima
Example
Find the global maximum and minimum of f (x) =
3
1+x2
on [−2, 2].
6x
0
Clearly f is differentiable such that f 0 (x) = − (1+x
2 )2 . There is only one turning point (f (x) = 0) at x = 0.
Now for x < 0 it is clear that f 0 (x) > 0 and for x > 0 we see f 0 (x) < 0. Hence x = 0 is a local maximum with
value f (0) = 3.
Now we test the interval points: f (−2) =
We have global minimum
5.13
3
5
3
5
and f (2) = 35 .
occuring at x = ±2.
Curve Sketching
Read more in J/S p97 §4.4.
We went through an example in lectures that I gave out as handout. Here are some more examples.
41
fig:prog31
1. Sketch y = x ln(x) for 0 < x < ∞.
• At x = 1 we have y = 0 and as x → ∞, y → ∞.
• At x = 0 we use L’Hopitals rule such that limx→0 x ln(x) = limx→0
limx→0 −x = 0
• We seek stationary points,
−1
1
e.
−1
therefore x = e
=
y = (e−1 ) ln(e−1 ) = −e
dy
dx
= ln(x) + 1, hence at turning point
The stationary point is a minimum since
= − 1e
2
d y
dx2
ln(x)
x−1
dy
dx
=
= limx→0
1
x
−x−2
=
= 0 → ln(x) = −1,
1
x
> 0 and has value
1.5
y
1
0.5
0
!0.5
!0.5
0
0.5
1
1.5
2
x
Figure 5.5: y = x ln(x)
2. Sketch y = xe−x for 0 < x < ∞.
• As x → ∞, y → 0, and at x = 0 y(0) = 0
• Critical points when
dy
dx
= e−x − xe−x = 0, hence when (1 − x)e−x = 0, so when x = 1 and y = e−1
dy
dx
• Sign table shows that as x < 1,
> 0 and for x > 1,
dy
dx
< 0, so the stationary point is a maximum.
0.5
0.4
0.3
0.2
y
0.1
0
!0.1
!0.2
!0.3
!0.4
!0.5
!0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
x
Figure 5.6: y = xe−x
5.14
Application to Optimisation
Cut a square of side xm from each corner of a 5m by 2m rectangular sheet. Fold along dotted lines to make a
rectangular open box. Find values of x that makes volume V maximum. Now length of box is 5 − 2x, width
is 2 − 2x and height is x.
Hence volume is such that
V = x(5 − 2x)(2 − 2x) = x(10 − 14x + 4x2 )
Thus
dV
= 12x2 − 28x + 10.
dx
Turning point occurs for 12x2 − 28x + 10 = 0, hence using the standard formula for roots of a quadratic we
have x = 0.44 or 1.89. We can discount x = 1.89 since width can not be negative. We can then show V is
2
maximum since ddxV2 = 24x − 28 < 0.
42
5 − 2x
x
x
x
2 − 2x
2
x
5
5.15
Differentials - Estimating Small Changes
dy
dx
= f 0 (x), we write
dy = f 0 (x) dx
Can interprete this as small change dx in the input causes a small change dy in the output. Good approximation
if dx is small.
Example
How does the surface area of a spherical pellet change as radius changes from 1cm to 1.01cm. What is the
relative change in surface area in terms of the relative change in radius?
Surface area S = 4πr2 , hence dS = S 0 dr such that dS = (8πr)dr. If at r = 1cm, then dr = 0.01 then change
in surface area dS = 8π(1)(0.01) = 0.08π.
The relative change in area
dS
S
=
8πrdr
4πr 2
= 2r dr
The relative change in volume V = 43 πr3 is then
5.16
dV
V
=
4πr 2 dr
4
3
3 πr
= 3r dr
Leibniz Rule
By Binomial Theormem (a + b)2 = a2 + 2ab + b2 = a2 b− + 2ab + a0 b2 . Using notation f = f (0) ,
2
d f
dx2
= f (2) etc, then
d2
(f g)
dx2
d3
(f g)
dx3
=
f (2) g (0) + 2f (1) g (1) + f (0) g (2)
=
f (3) g (0) + 3f (2) g (1) + 3f (1) g (2) + f (0) g (3)
43
df
dx
= f (1) ,
Example
If y = x4 ex we set f (x) = x4 and g(x) = ex then
d3
(f g)
dx3
3 k
X
3
d f d3−k g(x)
=
k
dxk dx3−k
0
= f (3) g (0) + 3f (2) g (1) + 3f (1) g (2) + f (0) g (3)
= x4 ex + 3 4x3 ex + 3 4.3x2 ex + (4.3.2x) ex
= ex x4 + 12x3 + 36x2 + 24x
44
Chapter 6
Taylor and Maclaurin Series
Read more in J/S Chapter 5.
6.1
Definitions
We are going to see how we can approximate, as closely we want, with polynomials many of the functions that
we have encountered previously in this course. This is how sin x, cos x, ex , . . . are calculated by your calculator.
But how do we find the polynomial of degree n that best approximates a given function f (x). Let us first deal
with polynomials and see how we can decipher the polynomial from f (0), f 0 (0), f (2) (0), . . ..
Example. Take a polynomial f (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 of degree 4. Then
f 0 (x)
=
a1 + 2a2 x + 3a3 x2 + 4a4 x3
f (2) (x)
=
2a2 + 3 · 2 · a3 x + 4 · 3 · a4 x2
f
(3)
(x)
=
3 · 2 · a3 + 4 · 3 · 2 · a4 x
f
(4)
(x)
=
4 · 3 · 2 · a4 .
We therefore have that
f (0)
= a0
0
f (0)
= a1
f 00 (0)
=
2a2
f
(3)
(0)
=
3 · 2 · a3
f
(4)
(0)
=
4 · 3 · 2 · a4
which implies that
a0
=
f (0)
a1
=
f 0 (0)
a2
=
f (2) (0)/2!
a3
=
f (3) (0)/3!
a4
=
f (4) (0)/4!.
Hence
f (2) (0) 2 f (3) (0) 3 f (4) (0) 4
x +
x +
x .
2!
3!
4!
More generally: if f (x) = a0 + a1 x + · · · + an xn , then
f (x) = f (0) + f 0 (0)x +
f (x) = f (0) + f 0 (0)x +
f (2) (0) 2
f (n) (0) n
x + ··· +
x .
2!
n!
45
Definition. Let f (x) be any function that is n-times differentiable at 0. Then the polynomial
f (2) (0) 2
f (n) (0) n
x + ··· +
x
2!
n!
Pn (x) = f (0) + f 0 (0)x +
is called the Maclaurin polynomial of degree n for f (x).
So how well does this polynomial approximate f (x)?
Fact. If f (x) is an (n + 1)-times differentiable function defined on some open interval containing 0. Then
f (x) = f (0) + f 0 (0)x + · · · +
{z
|
Pn (x)
f (n+1) (0) n f (n+1) (z) n+1
x +
x
n!
} |(n + 1)!
{z
}
error term
where z = z(x) depends on x but is always between 0 and x.
Normally (for us) we have for given x that
error term → 0 as n → ∞,
or equivalently
Pn (x) → f (x) as n → ∞.
We often express the limit as an infinite sum and write
f (x) = f (0) + f 0 (0)x +
f (2) (0) 2 f (3) (0) 3
x +
x + ···
2!
3!
and refer to it as the Maclaurin series for f (x).
(Warning: The formula is not always valid for all x. Sometimes only in certain interval around 0 and sometimes
in fact only for x = 0 (which is rare for us)).
6.2
Calculating Maclaurin series and polynomials
Example 1. Find the Maclaurin Series for f (x) = ex .
f (x) = exp(x),
f (0) = 1
0
f 0 (0) = 1
00
f 00 (0) = 1
f (x) = exp(x),
f (x) = exp(x),
Hence the Maclaurin Expansion for exp(x) is given by
ex = 1 + x +
x2
x3
x4
+
+
+ ...
2!
3!
4!
(It is valid for all real numbers x).
The Maclaurin expansion becomes more accurate by adding successively higher order terms. (see figure
6.1)
46
2
1.5
1
0.5
y
0
!0.5
!1
!1.5
!2
!2
!1.5
!1
!0.5
0
x
0.5
1
1.5
2
Figure 6.1: Maclaurin Series y = ex
Example 2 Calculate p4 (x) for f (x) = sin(x).
f (x) = sin(x),
f (0) = 0
f 0 (x) = cos(x),
f 0 (0) = 1
f 00 (x) = − sin(x),
f 00 (0) = 0
f 000 (x) = − cos(x),
f 000 (0) = −1
f (iv) (x) = sin(x),
f (iv) (0) = 0
Hence
p4 (x)
0+x+0·
=
= x−
x3
x4
x2
+
+0·
2!
3!
4!
x3
6
More useful examples
1.
1
1−x
= 1 + x + x2 + x3 + x4 + · · · =
P∞
n=0
xn and is valid for |x| < 1.
(Recall that this is the formula for geometric series. Which can be proved as follows:
(1 + x + x2 + . . . + xn )(1 − x)
=
(1 + x + x2 + · · · + xn ) − (x + x2 + · · · + xn + xn+1 )
=
1 − xn+1
and thus
1 + x + x2 + · · · + xn =
1
1 − xn+1
→
1−x
1−x
as n → ∞.)
2. (1 + x)α = 1 + αx +
α(α−1) 2
x
2
+ . . . . For example (1 + x)1/2 = 1 −
x2
2!
+
x4
4!
−
x6
6!
4. ln(1 + x) = x −
x2
2
+
x3
3
− . . . is valid for |x| < 1
3. cos(x) = 1 −
x
2
+ 18 x2 + . . . .
+ ...
One way of deriving the last formula without going straight to the definitions is to notice first that
1
(ln(1 + x))0 =
1+x
1
=
1 − (−x)
=
1 + (−x) + (−x)2 + (−x)3 + · · ·
=
1 − x + x2 − x3 + · · ·
Then we think backwards. Which series would have this last expression as it’s derivative and the constant
term 0 (after all ln(1 + 0) = 0). The answer is
ln(1 + x) = x −
x2
x3
x4
+
−
+ ···
2
3
4
47
6.3
Estimating the error term
Recall.
f (x) = f (0) + f 0 (0)x + · · · +
|
{z
Pn (x)
f (n+1) (0) n f (n+1) (z) n+1
x +
x
n!
} |(n + 1)!
{z
}
error term
where z = z(x) depends on x but is always between 0 and x.
Example 1. Calculate e such that the error is at most 1/10000.
Solution.
xn
ez
x2
+ ··· +
+
xn+1
ex = 1 + x +
2!
n!
(n
+
1)!
|
{z
} |
{z
}
Pn (x)
error term
where z lies between 0 and x. In particular for x = 1 this gives
e=1+1+
1
ez
1
+ ··· +
+
2!
n!
(n + 1)!
| {z }
error term
where 0 ≤ z ≤ 1. We want n big enough so that the error term is at most 1/10000. We do some estimation.
The worst case would be if z = 1. Now
ez
e
3
3
1
≤
<
=
=
.
(n + 1)!
(n + 1)!
(n + 1)!
1 · 2 · 3 · · · (n + 1)
2 · 4 · 5 · · · (n + 1)
So we want 2 · 4 · 5 · · · (n + 1) to be at least 10000. We make an inspection
2·4·5·6
=
240
2·4·5·6·7
=
1680
2·4·5·6·7·8
=
13440
(not enough)
(not enough)
(Bingo!).
So we need to take n + 1 = 8, i.e. n = 7.
Conclusion. e ≈ 1 + 1 +
1
2!
+ ··· +
1
7!
(with error less than 1/10000).
Example 2. Estimate the error in approximating sin x by the Maclaurin polynomial P3 (x) of degree 3
in the interval −0.1 < x < 0.1.
Solution. Let f (x) = sin x. We have
sin x
x3
f (5) (z) 5
= x−
+ 0 · x4 +
x
3!{z
5!
|
}
P3 (x)=P4 (x)
= x−
sin z 3
x3
+
x
6
120
where z lies between 0 and x. We now estimate the error term
sin z 5 1
1
5
−5
−8
120 · x ≤ 120 (0.1) ≤ 120 · 10 ≈ 8 · 10 .
Conclusion. The error is at most 8 · 10−8 (very small indeed).
48
6.4
Manipulation of Maclaurin Series
Sometimes we can calculate the Maclaurin Series without going straight to the definiton using some clever
Example 1. Calculate the Maclaurin Series for sin x2 .
Solution. We use the fact that we already know the Maclaurin Series for sin x. We have
y3
y5
y7
+
−
+ ··· .
3!
5!
7!
sin y = y −
Putting y = x2 then gives
sin x2 = x2 −
x10
x12
x6
+
−
+ ···
3!
5!
7!
Example 2. Calulate the Maclaurin polynomial P3 (x) of degree 3 for ex sin x.
Solution. We know that
ex = 1 + x +
x3
x2
+
+ (higher terms)
2!
3!
and
sin x = x −
x3
+ (higher terms).
3!
Therefore
ex sin x
x2
x3
x3
+
+ h.t.) · (x −
+ h.t)
2
6
6
x3
x3
−
+ h.t.
= x + x2 +
2
6
1
= x + x2 + x3 + h.t.
3
=
(1 + x +
Answer. P3 (x) = x + x2 + x3 /3.
6.5
Taylor Series
We might want to approximate f (x) near a general point a (not neccessarily 0). This we can achieve using
the work for Maclaurin series as follows: let x = a + h and
g(h) = f (a + h).
Using the Maclaurin expansion we get
g(h) = g(0) + g 0 (0)h +
g 00 (0) 2
· h + ···
2!
That is
f (a + h) = f (a) + f 0 (a)h +
f 00 (a) 2
· h + ···
2!
That is (since x = a + h ⇔ h = x − a)
f (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
(x − a)2 + . . .
2!
This last series is called the Taylor series for f (x) around a.
49
Example Find the first 3 terms in the Taylor series of sin(x) about c = π/4.
Solution. We have
P3 (x) = f (π/4) + f 0 (π/4) · (x − π/4) +
f 00 (π/4)
f (3) (π/4)
· (x − π/4)2 +
(x − π/4)3 .
2!
3!
We calculate next the coefficents:
1
f (π/4) = √
2
1
f 0 (π/4) = √
2
f (x) = sin(x),
f 0 (x) = cos(x),
1
f 00 (π/4) = − √
2
1
f 000 (π/4) = − √
2
f 00 (x) = − sin(x),
f 000 (x) = − cos(x),
Hence,
6.6
1
1
−1
−1
P3 (x) = √ + √ · (x − π/4) + √ · (x − π/4)2 + √ · (x − π/4)3 .
2
2
2 2
6 2
Newton’s Method for solving non-linear equations
Read more in Evans p534 and J/S §4.6 page 104
Our aim is to find the solution of f (x) = 0.
Numerical method using Taylor Series.
We start with a intial guess x1 (rough approximation) for the solution. This could be arrived at by considering a graph for the function for example.
Next we want to improve the approximation. In reality the solution is
x1 + h
where h is the error (hopefully small but we want to make it even smaller). We estimate h using Taylor series
0
=
f (x1 + h)
=
f (x1 ) + f 0 (x1 ) · h +
≈
f (x1 ) + f 0 (x1 ) · h
f 00 (x1 ) 2
· h + ···
2!
This implies that (if things have gone well)
h≈−
f (x1 )
.
f 0 (x1 )
So an improved approximation should be
x2 = x1 −
f (x1 )
.
f 0 (x1 )
We can iterate this progress in order to get better and better approximations
x3
= x2 −
f (x1 )
f 0 (x1 )
..
.
xn+1
= xn −
50
f (xn )
.
f 0 (xn )
If the method works (not always the case) the sequence
x1 , x2 , x3 , . . .
tends to the solution of the equation f (x) = 0.
Example. Let f (x) = ex − 2x − 1 = 0.
Solution. In this case
f 0 (x) = ex − 2.
We start with an initial guess x1 = 1.3 for the root of the equation. Using Newton’s method
x2
=
x3
+
e1.3 − 2(1.3) − 1
= 1.259
e1.3 − 2
e1.259 − 2(1.259) − 1
1.259 −
= 1.257
e1.259 − 2
1.3 −
51
Chapter 7
Integration
Read more in Evans Chapter 15, J/S §14
Integration has many applications. For example we can use it to
1. calulate areas, volumes, mass, energi, . . .
2. solve differential equations.
We start with the reverse of differentiation. We will see that this is closely related to the problem of finding
an area under the curve of a given function.
7.1
Primitives
A. Definitions.
Definition. Let I be an open interval. The function
F (x), x ∈ I
is said to be a primitive to the function
f (x), x ∈ I,
0
if F (x) = f (x).
Remark. Then also
(F (x) + c)0 = F 0 (x) + c0 = f (x) + 0 = f (x)
for any constant c. Conversely if F (x) and G(x) are two primitives to f (x), then
(G(x) − F (x))0 = G0 (x) − F 0 (x) = f (x) − f (x) = 0
and G(x) − F (x) is some constant c. Thus if F (x) is one primitive to f (x) then any primitive is of the form
F (x) + c
with c constant. We use the notation
Z
f (x) dx = F (x) + c
(sometimes referred to as the indefinite integral of f (x)).
Some general rules.
Z
Z
(f (x) + g(x)) dx
Z
af (x) dx
=
f (x) dx +
Z
= a f (x) dx
52
Z
g(x) dx
where a is a constant. In general the problem of finding primitives is more difficult than differentiation. For
example there is no simple product rule that gives the primitive to f (x)g(x) when we know the primitives to
f (x) and g(x).
B. Some primitives.
Z
a dx = ax + c
Z
Z
Z
(a constant)
xα dx
=
xα+1
+c
α+1
1
dx
x
=
ln |x| + c
ex dx
= ex + c
(α 6= −1 a constant)
Z
cos x dx =
sin x + c
Z
sin x dx = − cos x + c
Z
1
dx
cos2 x
Z
1
dx
2
Z sin x
1
dx
1 + x2
Z
1
√
dx
1 − x2
=
tan x + c
= − cot x + c
=
arctan x + c
=
arcsin x + c
In fact all these follow from formulas we have got for differentiation. Just differentiate the RHS and check
that you get what is under the intergral on the LHS.
C. Two useful rules.
Generally we also have:
Z
and
f 0 (x)
dx = ln |f (x)| + c
f (x)
Z
(7.-12)
1
· F (ax) + c
a
f (ax) dx =
(7.-12)
where a is a constant and F (x) is the primitive to f (x).
Proof of 7.1. If y = ln |u| and u = f (x), then
Proof of 7.2. If y =
1
a
dy
dx
· F (u) and u = ax, then
1
a
=
·
dy
du
dy
dx
·
du
dx
=
=
1
a
dy
du
·
1
u
· f 0 (x) =
f 0 (x)
f (x) .
2
· a = F 0 (u) = f (ax). 2
Example 1. We have
Z
cos 5x dx =
1
· sin 5x + c
5
Warning. One can not generalise this and take the Primitve of the outer function divided by the inner
derivative. For example
Z
sin x2
cos x2 dx 6=
+ c.
2x
53
Example 2
Z
cos x
dx = ln | sin x| + c
sin x
A part from these methods there are basically two methods that one uses when calculating primitives: “integration by parts” and “substitution”.
7.2
Integration by parts and substitution
A. Integration by parts.
Theorem If F (x) is a primitive to f (x) then
Z
Z
f (x) · g(x) dx = F (x) · g(x) − F (x) · g 0 (x) dx
Proof Must show that the RHS is a primitive to f (x) · g(x). But
Z
(F (x)g(x) − F (x)g 0 (x) dx)0 = F 0 (x)g(x) + F (x)g 0 (x) − F (x)g 0 (x)
=
F 0 (x)g(x)
=
f (x)g(x) 2.
Remark. The way this result can be useful is when the integral on the RHS is simpler than the one of the
LHS. Notice that the new product F (x) · g 0 (x) that we integrate on the RHS, comes from integrating one
of f (x), g(x) and differentating the other. Sometimes it may be better to change the order and work with
g(x) · f (x) instead if that leads to a simpler integral on the RHS.
Example 1. Calculate
R
x · ex dx.
Solution. Integrating or differenting ex makes no difference. However, it is better to differentate x than
integrating it. So we change the order (here f (x) = ex , g(x) = x, F (x) = ex , g 0 (x) = 1)
Z
Z
x
x
e · x dx = e · x − ex · 1 dx
= xex − ex + c.
Example 2. Calculate
R
ln x dx.
Solution. We use a common trick. We write ln x = 1 · ln x and then apply integration by parts (f (x) =
1, g(x) = ln x, F (x) = x, g 0 (x) = 1).
Z
Z
1
1 · ln x = x · ln x − x ·
x
Z
= x · ln x − 1 dx
=
x · ln x − x + c.
B. Substitution. Suppose that x is a differentable function in t. Then
Z
Z
dx
f (x) dx = f (x) ·
· dt.
dt
Here we should think of the LHS as expressed in x and the RHS as expressed in t.
54
Method. (i) Replace f (x) by an expression in t. (ii) Replace dx by
We now have an integral of a function in t.
dx
dt
· dt (an expression in t). (iii)
This method is often useful as the expression in t can be much simpler.
Example 1. Calculate
Solution. Let t =
√
R
x·
√
x − 1 dx.
x − 1. Soving for x gives x = t2 + 1. We then have
√
t =
x−1
x =
dx
=
dt
t2 + 1
2t
and
Z
x·
√
Z
x − 1 dx
=
Z
=
=
=
Example 2. Calculate
R
cos x
1+sin2 x
(t2 + 1)t · 2t dt
(2t4 + 2t2 ) dt
2 5 2 3
·t + ·t
5
3
2 √
2 √
· ( x − 1)5 + ( x − 1)3 + c
5
3
dx.
Solution. We work here a bit differently. We observe that the derivative of sin x is cos x and therefore
dt
if we put t = sin x then dt = dx
· dx = cos x · dx. Having though ahead like this we now begin the calculations
t
=
sin x
dt
=
cos x · dx
and
Z
7.3
cos x
dx
1 + sin2 x
=
1
· dt
1 + t2
arctan(t)
=
arctan(sin x).
=
Primitives and areas
Consider a function f (x) that is integrable (has primitive) and with positive values on an open interval I.
Consider the graph of this function (draw a figure or look at your handwritten notes).
Let A(x) be the area under the curve between a and x (where x ≥ a). If we take a small ∆x then the
region between x and x + ∆x is approximately a rectangle with height f (x) and width ∆x. Therefore
A(x + ∆x) − A(x)
f (x) · ∆x
≈
= f (x)
∆x
∆x
Hence
A(x + ∆x) − A(x)
= f (x).
∆x
So A(x) is a primitive to f (x). Let then F (x) be any primitive to f (x). Then
A0 (x) = lim
∆x→0
A(x) = F (x) + c
55
(7.-32)
for some constant c. As the area between a and a is 0, we have A(a) = 0 and (7.3) gives c = −F (a). The area
between a and b is therefore
A(b) = F (b) + c = F (b) − F (a)
.
Definition.
Rb
If a f (x) is finite, f (x) is said to be integrable and
Z
a
b
b
f (x)dx = F (b) − F (a) = [F (x)]a
or F (x)|ba
where F (x) is an indefinite integral (antiderivative) of f (x). See Handout Sheet 15: The Definite Integral
Examples
h 3 i2
R2
1. 1 x2 dx = x3
=
1
2.
R 1/2
3.
R2
0
1
dx
1 x
4.
R1
6.
R1
3
(2x + 1) dx =
23
3
1
8
−
1
3
=
8
3
−
1
3
4 1/2
(2x + 1) |0
=
7
3
=2−
1
8
=
15
8
2
= [ln(x)]1 = ln(2) − ln(1) = ln(2)
1
dx
0 x
2
= [ln(x)]1 = ln(1) − ln(0) = ∞ The integrand does not exist.
R
2
2
5. x22x
+4 dx = ln(x + 4) + k = ln k(x + 4)
2x
dx
0 x2 +4
7.4
1.
2.
3.
4.
5.
6.
1
= ln(x2 + 4) 0 = ln(5) − ln(4) = ln
5
4
Properties of Definite Integrals
Rb
Rb
Rb
(f (x) ± g(x)) dx = a f (x)dx ± a g(x)dx
Rb
Rb
cf (x)dx = c a f (x)dx
a
Rb
Rc
Rb
If c ∈ [a, b] then a f (x)dx = a f (x)dx + c f (x)dx
Rb
Ra
f (x)dx = − b f (x)dx
a
Ra
f (x)dx = 0
a
R
R
b
b
a f (x)dx ≤ a |f (x)|dx
a
7. Mean Value Theorem for integrals. If f (x) is continuous, then there is a point c ∈ (a, b) such that
Rb
f (x)dx = (b − a)f (c). Geometrically, the is area under y = f (x) is equal to the area of rectangle
a
height f (c).
56
x2!2
abs(x2!2)
16
14
14
12
12
10
10
8
8
1
0.5
y
y
y
16
6
6
4
4
2
Area 1
Area 3
0
Area 3
Area 2
0
Area 2
!2
!4
0
Area 1
2
!2
!3
!2
!1
0
x
1
2
3
4
!4
!3
!2
!1
0
x
1
2
3
!0.5
4
fig:prog36
7.5
0
0.2
0.4
0.6
0.8
1
1.2
x
R
R
(a) Rule 6 (x2 − 2) = A1 − A2 + A3 (b) Rule 6 |x2 − 2| = A1 + A2 + A3
(c) Rule 7: Mean Value Theorem
Techniques of Integration
7.5.1
Substitution
A change of variables often makes the integration easier. In theory x = g(u) then
Hence for definite integrals:
Z
b
Z
x=b
f (x)dx =
a
Z
= g 0 (u) and dx = g 0 (u)du.
u=β
f (g(u)) g 0 (u)du where
f (x)dx =
x=a
dx
du
a = g(α), b = g(β)
u=α
Examples - Indefinite Integral Substitution
R
2
1
1. Solve (4x − 1) dx. Set u = 4x − 1, then du
dx = 4 and dx = 4 du.
Z
2. Solve
R
Z
2
(4x − 1) dx =
∴
√ 1
dx.
1−x2
Z
∴
u2
1
du =
4
4
Set x = sin(u) then
√
1
dx =
1 − x2
dx
du
Z
Z
1
4
1 3
u
3
+c=
1
3
(4x − 1) + c
12
= cos(u) and dx = cos(u)du.
Z
1
q
u2 du =
cos(u)du =
du = u + c = sin−1 (x) + c
2
1 − sin (u)
tan2 (x)dx. Set u = tan(x), then du = sec2 (x)dx = (1 + u2 )dx
Z
Z
Z Z
Z
u2
1
1
tan2 (x)dx =
du
=
1
−
du
=
1du−
du = u−arctan(u)+c = tan(x)−x+c
1 + u2
1 + u2
1 + u2
3. Solve
R
4. Solve
R
cos(2x + 5)dx. Set u = 2x + 5, then du = 2dx
Z
Z
1
1
1
cos(2x + 5)dx =
cos(u)du = sin(u) + c = sin(2x + 5) + c
2
2
2
5. Solve
R
x2
x3 −5 dx.
Set u = x3 − 5, then du = 3x2 dx
Z
x2
dx
3
x −5
=
1
3
Z
1
3
Z
1
1
1
du = ln(u) + c = ln(x3 − 5) + c
u
3
3
or
Z
x2
dx
3
x −5
=
3x2
1
dx = ln(x3 − 5) + c
−5
3
x3
57
Examples - Definite Integral Substitution
R2 2
dx
1. Solve 1 ln x(x) dx. Set x = eu , then du
= eu and dx = eu du, then
ln2 (x)
dx =
x
2
Z
∴
1
2. Solve
R1
0
2
√x
dx.
x3 +1
Z
x=2
x=1
ln2 (x)
dx =
x
Set u = x3 + 1 then
Z
1
∴
0
3. Solve
Z
1
0
7.5.2
R1
x
dx.
0 4x+2
x
dx =
4x + 2
2
6
u=ln(2)
u=0
du
dx
du
dx
u2 u
e du =
eu
ln(2)
Z
u2 du =
0
Z
2
1 3
u
3
ln(2)
=
0
1
3
(ln(2))
3
2
2 1 2 √
= u2 =
2−1
3
3
1
du
1
1
u2
= 4 and dx = 41 du with
3
1 1
3
(u − 2) . . du =
4
u 4
16
= 3x2 and x2 dx = 31 du.
x2
1
√
dx =
3
3
x +1
Set u = 4x + 2 then
Z
Z
Z
6
2
u−2
3
du =
u
16
Z
1
4
(u − 2) = x.
6
1−
2
2
3
3
6
=
[u − 2 ln(u)]2 =
(4 − 2 ln(3))
u
16
16
Integration by Parts
Read J/S page 325, Evans page 447 The formula for integrating by parts is obtained by integrating the product
rule for differentiation.
d
dv
du
(uv) = u
+v
dx
dx
dx
d
du
dv
=
(uv) − v
∴u
dx
dx
dx Z
Z
Z d
du
dv
dx =
(uv)dx − v dx
∴
u
dx
dx
dx
Z Z
dv
du
∴ u
dx = uv − v dx + c
dx
dx
Rb
R
b
b
dv
For definite integrals a u dx
dx = [uv]a − a du
dx vdx
Why do we do this? Well hopefully
parts.
R
v du
dx is easier to integrate than
Examples-Definite Integral Integration by Parts
R1
dv
1. Find 0 x ln(x)dx. Set u = ln(x) and dx
= x then v =
Z
0
2. Find
Z
0
1
R1
0
1
x2
2
R
dv
u dx
dx. So if in doubt integrate by
and du = x1 , hence integrating by parts
1 Z 1 2
1
Z
x2
x 1
1 1
1 1 2
1
x ln(x)dx =
ln(x) −
. dx = [0 − 0] −
xdx = −
x
=−
2
2 0
2 2
4
0 2 x
0
0
xe3x dx. Set u = x and
dv
dx
= e3x , then
du
dx
= 1 and v = 13 e3x .
1 Z 1
1
Z
1
1 3x
1
1 1 3x
1
1 1 3x
1
1
2
1
xe3x dx = x e3x −
e .1dx = e3 −
e dx = e3 −
e
= e3 − e3 +19 = e3 +
3
3
3
3
3
3
3
3
9
9
9
0
0
0
0
Rπ
dv
3. Find 0 x cos xdx. Set u = x and dx
= cos(x), then du
dx = 1 and v = sin(x).
Z π
Z π
Z π
π
π
x cos xdx = [x sin(x)]0 −
1. sin(x)dx = 0 −
sin(x)dx = [cos(x)]0 = cos(π) − cos(0) = −2
0
4. Find
0
R1
0
0
dv
1
ln(x)dx. Set u = ln(x) and dx
= 1 then du
dx = x and v = x.
Z 1
Z 1
1
1
1
ln(x)dx = [x ln(x)]0 −
x dx = [0 − 0] − [x]0 = −1
x
0
0
58
7.5.3
Integrals using Trig Formula
R
Rπ
Recall that cos(2x) = 12 sin(2x) + c then how do we integrate 0 sin2 (x)dx. We can use the trig formula
cos(2x) = 1 − 2 sin2 (x) and rearranging we find sin2 (x) = 21 (1 − cos(2x)), then
Z
0
If you think
R
π
1
sin (x)dx =
2
sin2 (x)dx =
2
1
3
Z
0
π
π
1
1
π
(1 − cos(2x)) dx =
x − sin(2x) =
2
2
2
0
sin3 (x) think again!!!
59
Chapter 8
Ordinary Differential Equations
(ODE’s)
You can read more in J/S Chapter 18,19 and Evans Chapter 19.
To motivate our study we start with an example from real life.
Example. A radioactive element disintegrates at a rate proportional to the amount that remains at any
given time (A known fact).
Mathematical formulation. Let y(t) be the amount that remains at time t. Then
dy
= −kt (∗)
dt
Problem. Find y(t) from (*).
This is an example of an ODE. For the solution to be unique we need also an intial condition (IC).
dy
dt
= −ky
y(0) = y0 (IC).
An ODE with a initial condition is often referred to as a “initial value problem”.
8.1
ODE of lst order
We now turn to the mathematics. In general, an ODE of 1st order is an equation of the form
dy
= f (x, y) (**).
dx
(To be of first order indicates that only first derivatives occur. To be ordinary means that only functions in
one variable are involved).
A solution to (**) is then a function
y(x), x ∈ I
0
for some open interval I such that y (x) = f (x, y(x)).
Example. The function y = e−3t is a solution to
dy
= −3y
dx
60
since
y 0 = −3e−3x = −3y.
We now move on to the problem of finding the solutions for ODE’s of particular form.
Definition. An equation of the form
dy
+ g(x)y = h(x)
dx
is called a linear ODE of lst order.
If h(x) = 0, it is said to be homogenous (or unforced).
If h(x) =
6 0, it is said to be nonhomogenous (or forced).
Solution method (using integrating factor)
Step 1. Take a primitive G(x) to g(x) and form
eG(x) (the integrating factor, IF)
Step 2. Multiply with IF on both sides
y 0 · eG(x) + y · g(x) · eG(x)
= h(x) · eG(x)
m
G(x) 0
(y · e
)
= h(x) · eG(x) .
Step 3. Integrate
y · eG(x)
Z
=
h(x)eG(x) dx
m
y
=
e−G(x) ·
Z
h(x) · eG(x) dx
A. Homogenous equations.
Example 1. Solve
dy
− 2y = 0.
dx
Solution. We go through the three steps.
I As −2x is a primitive to −2, we have IF = e−2x .
II (ye−2x )0 = 0 · e−2x = 0.
III ye−2x = c ⇔ y = c · e2x .
Answer. y = c · e2x , where c is constant.
Example 2. Solve
dy
− x · y = 0.
dx
Solution.
61
2
I. IF = e−1/2·x .
2
2
II. (y · e−1/2·x )0 = 0 · e−1/2·x = 0.
2
III. ye−1/2·x = c.
2
Answer. y = c · e1/2·x , c constant.
B. Nonhomogenous equations
We can still use the integrating factor method. Here we often need to do more work in terms of integration however.
Example 3. Solve
dy
− x · y = x.
dx
Solution. Here we have
2
I. IF = e−1/2·x .
2
2
II. (y · e−1/2·x )0 = x · e−1/2·x .
III. We therefore have
y · e−1/2·x
Z
2
=
t = −1/2 · x2
dt = −x dx
2
x · e−1/2·x dx
Z
=
−et dt
=
−et + c
=
−e−1/2·x + c.
2
This implies that
2
2
2
2
y = −e1/2·x · e−1/2·x + c · e1/2·x = −1 + c · e1/2·x . 2
Advantage. The method is quite general and works at least in theory.
Disadvantage. The integration can be difficult.
We’ll now describe an alternative method that is not so general but easier and works in fact very often.
Alternative method for non-homogenous equations.
We want to solve
dy
+ g(x) · y = 0.
dx
Step 1. We find one particular solution yp . (Using guess work)
Step 2. We find the general solution yh to
dy
+ g(x) · y = 0
dx
62
Then y = yp + yh is the general solution to our original equation. This is because
dy
+ g(x) · y
dx
dyp
dyh
+
+ g(x) · yp + g(x) · yh
dx
dx
dyp
dyh
= (
+ g(x)yp ) + (
+ g(x)yh )
dx
dx
= h(x) + 0
=
= h(x).
Example 3 (again)
dy
− xy = x.
dx
Solution. Step 1. Find yp .
(The general idea is to try to imitate the RHS, so y should be a polynomial. Since xy should be of degree 1, y should be a constant). Clearly y = −1 works.
yp = −1.
(8.-15)
Step 2. Find yh .
We need to solve
dy
− xy = 0.
dx
But this we solved in Example 2 and we know from this solution that
2
yh = c · e−1/2·x .
(8.-15)
2
Answer. y = yp + yh = −1 + c · e1/2·x .
C. Guessing a particular solution
If we want to use the second method it is good to have some general idea where to look for a particular
solution and the general idea is that we try to choose y similar to the RHS. Here are some examples.
Example 4. Find a particular solution to
dy
+ x · y = x3 .
dx
Solution. We try a polynomial and since x · y should be of degree 3, we try to choose y as a polynomial of
degree 2. So
y
dy
dx
= ax2 + bx + c
=
2ax + b.
So we must have
x3
=
dy
+x·y
dx
2ax + b + x(ax2 + bx + c)
=
ax3 + bx2 + (2a + c)x + b.
=
Comparing the LHS with the RHS gives that a = 1, b = 0 and 2a + c = 0. Which implies that a = 1, b = 0
and c = −2a = −2. Hence
63
Answer. yp = x2 − 2.
Example 5.
dy
dt
+ 2y = 4 · e−3t .
Solution. Try y = Ae−3t . Then
dy
dt
= −3Ae−t and the equation becomes
4e−3t = −3Ae−3t + 2Ae−3t = −Ae−3t .
Hence A = −4.
Answer. yp = −4 · e−3t .
Example 6. Solve
dy
+ y = 2 · e−t .
dt
Solution. If we try the same method here it fails: y = Ae−t gives
dy
dt
= −Ae−t and the equation becomes
2e−t = −Ae−t + Ae−t = 0
which can never work! If one tries here instead y = Ate−t then
dy
= Ae−t − Ate−t .
dt
2 · e−t
= A(1 − t)e−t + At · e−t
= Ae−t .
Which works for A = 2. Hence
Answer. y = 2t · e−t .
Generally:
dy
dt
+ cy = e−ct . Try y = At · e−ct .
D. Initial value problems
This time we add some intial condition that determines the solution fully. So we want to solve
dy
dt − y = cos(2t)
y(0) = y0
Method. First we find a general solution to the differential equation and then we use the initial condition to
determine it fully.
Example 7. Solve
dy
dt
− y = cos(2t)
y(0) = 0
Solution (using guess work).
Step 1. Particular solution yp .
We try
y
dy
dt
= A · sin 2t + B · cos 2t
=
2A · cos 2t − 2B · sin 2t.
64
Then our equation becomes
dy
− y = (2A − B) · cos 2t + (−2B − A) · sin 2t
dt
cos 2t =
which gives us the equation system
2A − B
=
1 (i)
−A − 2B
=
0. (ii)
Then (i)+2(ii) gives that −5B = 1 ⇔ B = −1/5. Then (ii) gives A = −2B = 2/5. Hence we have a particular
solution
yp = 2/5 · sin 2t − 1/5 · cos 2t.
Step 2. The general homogenous part yh .
We want to solve
Here the integrating factor is e−t
dy
− y = 0.
dt
and the equation can be rewritten as
(ye−t )0 = 0 ⇔ ye−t = c ⇔ y = cet .
Thus
yh = cet , c constant.
Step 3. Determine y fully using the initial condition.
We have
y = yp + yh = 2/5 · sin 2t − 1/5 · cos 2t + c · et .
The initial condition then gives
0 = y(0) = 2/5 · 0 − 1/5 + c ⇔ c = 1/5.
Answer y = 2/5 · sin 2t − 1/5 · cos 2t + 1/5 · et .
Example 8. Solve
dy
x · dx
− y = x3 , x > 0
y(1) = 1
Solution. (Using the IF-method).
First we transform the equation on standard form. In this case
1
dy
− · y = x2 .
dx x
Step 1. Solve the differential equation using IF-method.
The integrating factor is e− ln x = (eln x )−1 = x−1 = 1/x. According to the IF-method the equation can
now be rewritten as
(y ·
1 0
)
x
= x2 ·
1
=x
x
⇓
1
y·
x
=
y
=
1/2 · x2 + c
⇓
1/2 · x3 + cx.
65
Step 2. Use the initial condition to determine c.
1 = y(1) = 1/2 + c ⇔ c = 1/2.
Answer: y = 1/2 · x3 + 1/2 · x.
8.2
Separable equations
Sometimes an equation
dy
= f (x, y)
dx
can be written of the form
g(y) dy = f (x) dx.
It is then called separable.
Solution method. Interate both sides
Z
Z
g(y) dy =
f (x) dx
and the solve for y.
Example 1. Solve
dy
dx
= y2
y(0) = 1
Solution. After separation of variables then equation becomes
1
= dx ⇔
y2
1
=x+c
y
−1
⇔ y=
.
x+c
−
The initial condition then gives
y(0) = 1 ⇔ 1 = y(0) =
−1
⇔ c = −1.
c
−1
x−1 .
Example 2. Solve
√
=x y
y(0) = 0
dy
dx
Solution. After separation of variables, we have
1
√
√ dy = x dx ⇔ 2 y = x2 /2 + c ⇔ 4y = (x2 /2 + c)2 .
y
The initial condition then gives
0 = y(0) ⇔ c2 = 0 ⇔ c = 0.
x2
16 .
2
66
8.3
An application
A vessel has initially 200 liters of a salt solution with salt concentration 10 grams per liter. A salt solution is
now added that has a lower salt concentration of 2 grams per liter. This happens at rate 8 liters per hour. At
the same time liquid is removed at the rate 8 liters per hour (we assume that the salt solution is well mixed
during the whole procedure).
We want to know how the amount of salt in the container changes with time.
Mathematical formulation. Let y(t) be the amount of salt (in grams) in the container after t hours.
Initially the concentration is 10 grams per liter and as we have 200 liters in the container, it follows that the
initial amount of salt in the container is
y(0) = 10 · 200 = 2000
grams. What is the rate of changes in the amount of salt w.r.t. time? The mixture that comes in has
concentration 2 grams per liter and as there are 8 liters that come in per hour. We have that the amount of
salt that comes in per hour is
0
yin
= 2 · 8 = 16
grams. How much goes out depends on the concentration at the given time t. After t hours the amount of
salt in the container is y(t) grams and as there are 200 grams in the container we have that the concentration
is y(t)/200 grams per liter. There are 8 liter that flow out per hour so the amount of salt that goes out per
hour is
0
yout
= 8 · y(t)/200 = y(t)/25
grams per hour. The total rate of change is therefore
1
dy
= 16 −
· y.
dt
25
We therefore need to solve the initial value problem
dy
dt + 1/25 · y = 16
y(0) = 2000.
Solution. Step 1. A particular solution yp .
As the RHS is a contstand we try y = c, constant. The derivative of y is then 0 and the equation becomes
1
· c ⇔ c = 25 · 16 = 400.
25
yp = 400.
Step 2. The homogenous part.
We want to solve
dy
1
+
· y = 0.
dt
25
1
The integrating factor here is e 25 ·t and the equation becomes
1
(y · e 25 ·t )0 = 0 ⇔ y · e1/25·t = c ⇔ y = ce−1/25·t .
yh = ce−1/25·t .
Step 3. Determine y fully.
67
y = yp + yh = 400 + c · e−1/25·t .
The initial condition gives
2000 = y(0) = 400 + c ⇔ c = 1600.
Answer. y(t) = 400 + 1600 · e−1/25·t .
Remark. Notice that y(t) → 400 as t → ∞. This fits as the concentration should eventually become
close to 2 grams per liter (as the incoming mixture) which would give 2 · 200 = 400 grams in the container.
8.4
2nd order homogenous linear ODE’s with constant coefficients
We consider here equations of the form
a·
dy
d2 y
+ cy = 0.
+b·
dt2
dt
where a, b and c are constants.
Solution method
Step 1. Solve the polynomial equation
ar2 + br + c = 0.
Suppose the solutions are r1 and r2 .
Step 2. We consider three cases.
Case 1. r1 6= r2 and both are real. In this case the general solution is
y = Aer1 t + Ber2 t .
Case 2. r1 = r2 . (Here the root is always real). The general solution is
y = (A + Bt)er1 t .
Case 3. r1 = α + iβ and r2 = α − iβ (β 6= 0).Here the general solution is
y = eαt (A cos βt + B sin βt).
Example 1. y 00 − 4y = 0.
Solution. Here r2 − 4 = (r − 2)(r + 2) = 0 ⇔ r = −2, or r = 2. So the answer is
y = A · e2t + B · e−2t ,
where A, B are constants.
Example 2. y 00 + 4y.
Solution. Here r2 + 4 = 0 ⇔ r4 = −4 ⇔ r = ±2i and the solution becomes
y = A cos 2t + B sin 2t
68
where A, B are constants.
Example 3. y 00 + 2y + 1 = 0.
Solution. Here r2 + 2r + 1 = (r + 1)2 = 0 ⇔ r = −1. The general solution is therefore
y = (A + Bt)e−t .
69
Chapter 9
Functions in several variables
You can read more in J/S §28.1-28.4, 28.6,29.1-29.2.
Physical terms often depend on more than one factor.
Example. (The ideal gas law)
p·V
R
where T is temperature, V volume and R is the gas constant. Here T is a function in two variables.
T = T (p, V ) =
We may be interested in the rate of change of T with respect to p and V . This leads to two partial derivatives
∂T
∂p
∂T
∂V
9.1
=
=
V
we differentiate w.r.t. p keeping V constant
R
p
we differentiate w.r.t. V keeping p constant.
R
Partial derivatives
In general if f (x, y) is a function in two variables then we define the two partial derivatives as before:
∂f
∂x
∂f
∂y
“differentiate w.r.t. x keeping y fixed”
“differentiate w.r.t. y keeping x fixed”.
Let’s compare the situation here with the one variable case.
Functions in one variable. y = f (x). We have seen before that if we change x to x + ∆x then the
resulting change ∆y in the output is circa
∆y ≈ f 0 (x) · ∆x.
Functions in several variables. z = f (x, y). Suppose that we change x to x + ∆x and y to y + ∆y. Then
the resulting change in the output is
∆z
= f (x + ∆x, y + ∆y) − f (x, y)
[f (x + ∆x, y + ∆y) − f (x, y + ∆y)] + [f (x, y + ∆y) − f (x, y)]
f (x + ∆x, y + ∆y) − f (x, y + ∆y)
f (x, y + ∆y) − f (x, y)
=
· ∆x +
· ∆y
∆x
∆y
∂f
∂f
≈
· ∆x +
· ∆y.
∂x
∂y
=
70
Example 1
Let z = f (x, y) = x3 cos(2y) + 2x2 + y.
• Find
∂f
∂x
and evaluate at (x, y) = (2, 0).
∂f
= 3x2 cos(2y) + 4x
∂x
• Find
∂f
∂y
∂f
(2, 0) = 12 + 8 = 20
∂x
and evaluate at (x, y) = (2, 0).
∂f
= −2x3 sin(2y) + 1
∂y
∂f
(2, 0) = 0 + 1 = 1
∂x
Higher derivatives. We can repeat differentating w.r.t. either x or y.
∂2f
∂x2
∂2f
∂y 2
∂2f
∂x∂y
∂2f
∂y∂x
“differentiate twice w.r.t. x
“differentiate twice w.r.t. y
“differentiate first w.r.t y and then w.r.t. x
“differentiate first w.r.t x and then w.r.t. y.
Example 2 (Continuation of Example 1). We have
∂2f
∂x2
=
=
=
2
∂ f
∂y∂x
=
=
=
∂2f
∂x∂y
=
=
=
∂2f
∂y 2
=
=
=
2
∂ f
Remark. The equation ∂x∂y
=
we work with in this course).
∂2f
∂y∂x
∂ ∂f
( )
∂x ∂x
∂
(3x2 · cos 2y + 4x)
∂x
6x · cos 2y + 4
∂ ∂f
( )
∂y ∂x
∂
(3x2 · cos 2y + 4x)
∂y
−6x · sin 2y
∂ ∂f
( )
∂x ∂y
∂
(−2x3 · sin 2y + 1)
∂x
−6x · sin 2y
∂ ∂f
( )
∂y ∂y
∂
(−2x3 · sin 2y + 1)
∂y
−4x3 · cos 2y.
holds for all reasonably well behaved functions (for example everything
71
Example 3 - Hemisphere
The hemisphere of radius 1 is given by the formula x2 +y 2 +z 2 = 1 for z ≥ 0. Then z = f (x, y) =
∂f
We wish to differentiate to find ∂f
∂x and ∂y . Recall the chain rule
1
√
f (x) = 1 − x2 2 = u
p
1 − x2 − y 2 .
where u = 1 − x2 , hence applying the chain rule
df
df du
1 1
x
=
= u− 2 (−2x) = −
1
dx
du dx
2
(1 − x2 ) 2
Hence for z = f (x, y) = 1 − x2 − y 2
12
let f (x, y) =
√
u for u = 1 − x2 − y 2 and then
∂z
df ∂u
−x
=
=
1
∂x
du ∂x
(1 − x2 − y 2 ) 2
∂z
df ∂u
−y
=
=
1
∂y
du ∂y
(1 − x2 − y 2 ) 2
Example 4
2
If f (x, t) = x2 − t2 , find
∂f ∂f
∂x , ∂t
∂f
and show that t ∂f
∂x + x ∂t = 0.
∂f
∂x
∂f
∂t
=
2 x2 − t2 .2x = 4x x2 − t2
=
2 x2 − t2 .(−2t) = −4t x2 − t2
∂f
Hence t ∂f
∂x + x ∂t = 0.
Example 5
The volume of a cuboid is v = xyz hence
∂V
∂x
∂V
∂y
∂V
∂z
= yz
= xz
= xy
Example 6
If u(x, t) = sin(x − 2t) show that
∂u
∂t
+ 2 ∂u
∂x = 0.
∂u
∂t
∂u
∂x
Hence
∂u
∂t
= −2 cos(x − 2t)
=
cos(x − 2t)
+ 2 ∂u
∂x = 0.
Example 7
2
∂u
∂t
∂u
∂x
∂2u
∂x2
1 − x4t
√
e
t
1
x2
2
∂ u
= t− 2 e− 4t satisfies ∂u
∂t = ∂x2 the Heat equation.
2
2
1
x2
1
x2
x −2
1 3 x2
x
1
= t− 2 e− 4t
t
− t− 2 e− 4t = t− 2 e− 4t
−
4
2
4t2
2t
x2
1
x2
2x
x
= t− 2 e− 4t −
= − 3 e− 4t2
4t
2t 2
2
2
2
x
x
x2
x2
1
x2
1
2x
1
1
x
x2
x
= − 3 e− 4t2 − 3 e− 4t2 −
= − 3 e− 4t2 + 5 e− 4t2 = t− 2 e− 4t
−
4t
4t2
2t
2t 2
2t 2
2t 2
4t 2
Show that u(x, t) =
72
9.2
Applications
A. Local max and min
It is useful here to compare the situation with functions in one variable.
Functions in one variable. If we are looking for a local max or local min x0 then a neccessary condition is that
f 0 (x0 ) = 0.
That is x0 , needs to be a stationary point. Then furthermore
a. If f 00 (x0 ) > 0 then x0 is a local min.
b. If f 00 (x0 ) < 0 then x0 is a local max.
Remark. If f 00 (x0 ) = 0 then anything can happen.
Functions in two variables. A neccessary condition here is that
∂f
∂f
(x0 , y0 ) = 0, and
(x0 , y0 ) = 0.
∂x
∂y
Such a point (x0 , y0 ) is called a stationary point. Before saying more we need some notation:
Notation. fx , fxy , . . . are shorthand notations for
0
, . . . are shorthand notations for
fx0 , fxy
∂f ∂ 2 f
∂x , ∂x∂y , . . ..
∂f
∂2f
∂x (x0 , y0 ), ∂x∂y (x0 , y0 ), . . ..
Now suppose that (x0 , y0 ) is a stationary point. Then furthermore
0 0
0 2
0
0
a. If fxx
fyy − (fxy
) > 0 and both fxx
> 0 and fyy
> 0 then (x0 , y0 ) is a local min.
0
0
0 2
0 0
< 0 then (x0 , y0 ) is a local max.
< 0 and fyy
) > 0 and both fxx
fyy − (fxy
b. If fxx
0 2
0 0
fyy − (fxy
) < 0 then (x0 , y0 ) is a saddle point. (See your handwritten notes for a drawing).
c. If fxx
0 0
0 2
Remark. If If fxx
fyy − (fxy
) = 0 then anything can happen.
We refer oftern to these cases as the four possible characteristics of a stationary point (x0 , y0 ).
B. Optimisation.
Example Consider a closed rectangular box of volume V fixed and of dimensions x, y, z. We will show
that the box with minimal surface area is a cube.
Solution. It is intuitively clear that there is some rectangular box of volume V that has smallest surface area. Well volume is given by V = xyz and the surface area by S = 2xy + 2yz + 2xz. Substituting in
V
z = xy
into the surface area equation we obtain
S = 2xy +
2V
2V
+
x
y
73
Then the critical points in the x and y direction are given by
∂S
2V
= 0 ⇒ 2y − 2 = 0
∂x
x
2V
∂S
= 0 ⇒ 2x − 2 = 0
∂y
y
⇒ x2 y = V
⇒ xy 2 = V
Hence V = xyz = x2 y = xy 2 implies x = y = z, a cube. All that is left to do is show that the critical point is
a minimum.
4V
∂2S
=
= 4 since V = x3
∂x2
x3
∂2S
4V
=
=4>0
∂y 2
y3
∂2S
= 2>0
∂y∂x
2 2 2 ∂ S
∂ S
∂ S
−
= 4 × 4 − 22 = 12 > 0
∂x2
∂y 2
∂y∂x
Hence the cube is a box with the minimal surface area for a given volume.
C. Differentials (small changes)
The Total Derivative dz is defined as
∂f
∂f
dx
+
dy
dz
=
|{z}
|{z}
|{z}
∂x
∂y
small change in x
estimate of change in z
small change in y
So dz estimates the change in z due to the change dx in x and dy in y.
Example
If z = 2x2 y 3 find the total derivative dz at x = 1, y = 2.
dz
=
=
∂f
∂f
dx +
dy
∂x
∂y
4xy 3 dx + 6x2 y 2 dy
Hence at x = 1, y = 2 we obtain dz = 32dx + 24dy. Hence the relative change is given by
dz
z
dz
z
=
=
4xy 3
6x2 y 2
dx
+
dy
2x2 y 3
2x2 y 3
dx
dy
2
+3
x
y
Alternatively when asking for relative changes take logs of z = 2x2 y 3 so that
ln(z) = ln(2) + 2 ln(x) + 3 ln(y)
Now we differentiate to obtain
dz
dx
dy
=2
+3
z
x
y
Example. In the example above if dx = 0.1 and dy = 0.1 then the change in the output would be approximately
dz = 32 · 0.1 + 24 · 0.1 = 5.6
and the relative change would be
dz
0.1
0.1
=2·
+3·
= 0.35
z
1
2
(i.e. 35%).
74
9.3
The Chain Rule
If z = z(x, y) with x = x(t), y = y(t) then
dz
=
dt
∂z
∂x
dx
+
dt
∂z
∂y
dy
dt
Example
2
A. If z = ln(x2 + y 2 ) with x = e−t and y = et then
dz
∂z dx
∂z dy
=
+
dt
∂x dt
∂y dt
2x
2y
−t
t2
=
−e
+
e
2t
x2 + y 2
x2 + y 2
1
−t
t2
−2xe
=
+
4yte
x2 + y 2
B. If z = z(x, y) with x = x(s, t), y = y(s, t) then
∂z
∂s
∂z
∂t
=
=
∂z
∂x
∂z
∂x
∂x ∂z ∂y
+
·
∂s
∂y ∂s
∂x ∂z ∂y
·
+
·
∂t
∂y ∂t
·
Example.(Polar coordinates). See figure from your handwritten lecturenotes. If z = z(x, y) and
x = r cos θ and y = r sin θ
then
∂z
∂r
=
=
∂z
∂θ
∂z ∂x ∂z ∂y
·
+
·
∂x ∂r
∂y ∂r
∂z
∂z
+ sin θ ·
.
cos θ ·
∂x
∂y
∂z ∂x ∂z ∂y
·
+
·
∂x ∂θ
∂y ∂θ
∂z
∂z
= −r sin θ ·
+ r cos θ ·
.
∂x
∂y
=
75
```