THE UNIVERSITY OF MELBOURNE

SCHOOL OF MATHEMATICS AND STATISTICS

MAST10006 Calculus 2

Lecture Notes





STUDENT NAME: EMAIL:



This compilation has been made in accordance with the provisions of Part VB of the copyright act for the teaching purposes of the University.
This booklet is for the use of students of the University of Melbourne enrolled in the subject MAST10006 Calculus 2.































© School of Mathematics and Statistics, University of Melbourne, January 2019.

These notes have been written by Christine Mangelsdorf, Anthony Morphett and Binzhou Xia. Reproduction of any part of this work other than that authorised by Australian
Copyright Law without permission of the copyright owners is unlawful.


Edition 11, July 2019.
Table of Contents
Section 0 - Notation used in MAST10006 Calculus 2 2
Section 1 - Limits, Continuity, Sequences and Series 9
Section 2 - Hyperbolic Functions 85
Section 3 - Complex Numbers 117
Section 4 - Integral Calculus 141
Section 5 - First Order Ordinary Differential Equations 183
Section 6 - Second Order Ordinary Differential Equations 249
Section 7 - Functions of Two Variables 309
1 / 391
Section 0 - Notation used in MAST10006 Calculus 2
Standard Abbreviations
1. such that or given that: |
2. therefore: ∴
3. for all: ∀
4. there exists: ∃
5. equivalent to: ≡
6. that is: i.e
7. approximate: ≈
8. much smaller than:
2 / 391
Standard Notation for Sets of Numbers
1. natural numbers: N = {1, 2, 3, . . . }
2. integers: Z = {0,±1,±2, . . . }
3. rational numbers: Q = {mn | m,n ∈ Z,n , 0}
4. real numbers: R (rational numbers plus irrational numbers)
5. complex numbers: C = {x + iy | x, y ∈ R, i2 = −1}
6. R2 = {(x, y) | x, y ∈ R} (xy plane)
7. R3 = {(x, y, z) | x, y, z ∈ R} (3 dimensional space)
3 / 391
Standard Notation for Intervals
1. element of: ∈
so a ∈ X means “a is an element of the set X ”
2. open interval: (a, b)
so x ∈ (0, 1) means “0 < x < 1”
3. closed interval: [a, b]
so x ∈ [0, 1] means “0 ≤ x ≤ 1”
4. partial open and closed interval: (a, b] or [a, b)
so x ∈ [0, 1) means “0 ≤ x < 1”
5. not including: \
so x ∈ R \ {0} means “x is any real number excluding 0”.
Alternatively, we could write (−∞, 0) ∪ (0,∞) where ∪
means the ”union of the two intervals”.
4 / 391
More Standard Notation
1. natural logarithm: log x
base 10 logarithm: log10 x
Alternative notations for natural logarithms used in
textbooks: loge x, ln x
2. inverse trigonometric functions: arcsin x, arctan x etc
Alternative notations used in textbooks: sin−1 x, tan−1 x etc
3. implies: ⇒
so p⇒ q means “p implies q”
4. if and only if (iff): ⇔ (means both ⇐ and ⇒)
so p⇔ q means “p implies q” AND “q implies p”
5. approaches: →
so f (x)→ 1 as x→ 0 means “f (x) approaches 1 as x
approaches 0”
5 / 391
Greek Alphabet
α alpha ν nu
β beta ξ xi
γ gamma o omicron
δ delta pi pi
or ε epsilon ρ rho
ζ zeta σ sigma
η eta τ tau
θ theta υ upsilon
ι iota φ phi
κ kappa χ chi
λ lambda ψ psi
µ mu ω omega
6 / 391
7 / 391 8 / 391
Section 1: Limits, Continuity, Sequences, Series
Limits
Let f be a real-valued function.
We say that f has the limit L as x approaches a,
lim
x→a f (x) = L,
if f (x) gets arbitrarily close to L whenever x is close enough to a
but x , a.
Note:
If it exists, the limit L must be a unique finite real number.
9 / 391
Example 1.1: If f (x) =
{
2x x , 1
4 x = 1 , evaluate limx→1
f (x).
-3 -2 -1 1 2 3 x
-1
1
2
3
4
f HxL
Solution:
Note:
The limit of f as x approaches a does not depend on f (a). The
limit can exist even if f is undefined at x = a.
10 / 391
Example 1.2: If f (x) =
1
x2
, evaluate lim
x→0 f (x).
x
f HxL
Solution:
11 / 391
Example 1.3: If f (x) =
{
1 x < 0
2 x ≥ 0 , evaluate limx→0 f (x).
x
f HxL
2
1
Solution:
12 / 391
We can describe this behaviour in terms of one-sided limits.
We write
Theorem:
lim
x→a f (x) = L if and only if limx→a− f (x) = L and limx→a+ f (x) = L.
Thus the limit exists if and only if the left and right hand limits
exist and are equal.
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Limit Laws
Let f and g be real-valued functions and let c ∈ R be a constant.
If lim
x→a f (x) and limx→a g(x) exist, then
1. lim
x→a [f (x) + g(x)] = limx→a f (x) + limx→a g(x).
2. lim
x→a [cf (x)] = c limx→a f (x).
3. lim
x→a [f (x)g(x)] = limx→a f (x) · limx→a g(x).
4. lim
x→a
[
f (x)
g(x)
]
=
lim
x→a f (x)
lim
x→a g(x)
provided lim
x→a g(x) , 0.
5. lim
x→a c = c.
6. lim
x→a x = a.
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Example 1.4: Use the limit laws to evaluate lim
x→2
x3 + 2x2 − 1
5 − 3x .
Solution:
15 / 391
Limits as x Approaches Infinity
We say that f has the limit L as x approaches positive infinity,
lim
x→∞ f (x) = L,
if f (x) gets arbitrarily close to L whenever x is sufficiently large
and positive.
We say that f has the limit M as x approaches negative infinity:
lim
x→−∞ f (x) = M
if f (x) gets arbitrarily close to M whenever x is sufficiently large
and negative.
Note:
1. L and M must be finite.
2. Limit laws (1)-(5) apply.
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Example 1.5: Evaluate lim
x→∞ e
−x.
x
H L
H0,1L
Solution:
17 / 391
Evaluating Limits with Indeterminate Forms
We say a function
f (x)
g(x)
has indeterminate form
0
0
as x→ a if
lim
x→a f (x) = limx→a g(x) = 0.
Example 1.6: Evaluate lim
x→2
x2 − 4
x − 2 .
Solution:
18 / 391
We say a function
f (x)
g(x)
has indeterminate form
∞
∞ as x→ a if
f (x)→∞ and g(x)→∞.
Example 1.7: Evaluate lim
x→∞
3x2 − 2x + 3
x2 + 4x + 4
.
Solution:
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We say a function f (x) − g(x) has indeterminate form ∞−∞ as
x→ a if f (x)→∞ and g(x)→∞.
Example 1.8: Evaluate lim
x→∞
(√
x2 + 1 − x
)
.
Solution:
20 / 391
Sandwich Theorem:
If g(x) ≤ f (x) ≤ h(x) when x is near a but x , a, and
lim
x→a g(x) = limx→a h(x) = L
then
lim
x→a f (x) = L.
x
y
hHxL
gHxL
f HxL
a
L
21 / 391
Note:
1. “x is near a but x , a” means that x lies in (b, a) ∪ (a, c) for
some b < a < c.
2. The validity of Sandwich Theorem is based on the fact that
g(x) ≤ f (x) ≤ h(x)
⇒|f (x) − L| ≤ |g(x) − L| + |h(x) − L| for all L.
Can you prove this inequality or even the stronger
conclusion that |f (x) − L| ≤ max{|g(x) − L|, |h(x) − L|}?
3. Sandwich Theorem also works for limits as x→∞.
For example, if g(x) ≤ f (x) ≤ h(x) when x ∈ (c,∞) for some
real number c, and lim
x→∞ g(x) = limx→∞ h(x) = L, then
lim
x→∞ f (x) = L.
The similar theorem holds for x→ −∞.
22 / 391
Example 1.9: Evaluate lim
x→0
[
x2 sin
(1
x
)]
.
Solution:
23 / 391
Example 1.10: Evaluate lim
x→0
[
x sin
(1
x
)]
.
Solution:
24 / 391
Continuity
Let f be a real-valued function.
The function f is continuous at x = a if
lim
x→a f (x) = f (a).
25 / 391
Example 1.11: Let
f (x) =
{
2x x , 1
4 x = 1.
Is f continuous at x = 1?
-3 -2 -1 1 2 3 x
-1
1
2
3
4
f HxL
Solution:
26 / 391
Example 1.12: Let f (x) =

x2 − 4
x − 2 x , 2
4 x = 2.
Is f continuous at x = 2?
Solution:
27 / 391 28 / 391
At the endpoints of a domain, we cannot take both left and right
hand limits, so we use the appropriate limit to test continuity.
1. A function f is left continuous (continuous from the left) at
x = a if lim
x→a− f (x) = f (a).
2. A function f is right continuous (continuous from the right)
at x = a if lim
x→a+ f (x) = f (a).
29 / 391
Example 1.13: Is f (x) =
√
x continuous in its domain?
x
y
1 2 3 4
1
2
y =
√
x
Solution:
30 / 391
Let f and g be real-valued functions and let c ∈ R be a constant.
Continuity Theorem 1:
If the functions f and g are continuous at x = a, then the
following functions are continuous at x = a:
1. f + g,
2. cf ,
3. fg,
4.
f
g
if g(a) , 0.
Note:
The theorem follows from limit laws.
31 / 391
Recall that ( g ◦ f )(x) = g( f (x)).
Continuity Theorem 2:
If f is continuous at x = a and g is continuous at x = f (a), then
g ◦ f is continuous at x = a.
Continuity Theorem 3:
The following function types are continuous at every point in
their domains: polynomials, trigonometric functions,
exponentials, logarithms, nth root functions, hyperbolic
functions.
32 / 391
Example 1.14: Let f (x) =
log x + sin x√
x2 − 1
.
For which values of x is f continuous?
Solution:
33 / 391 34 / 391
Example 1.15: f (x) =

x3 − cx + 8, x ≤ 1
x2 + 2cx + 2, x > 1.
For which values of c is f continuous? Justify your answer.
Solution:
35 / 391 36 / 391
Theorem:
If f is continuous at b and lim
x→a g(x) = b then
lim
x→a f [g(x)] = f
[
lim
x→a g(x)
]
= f (b).
Note:
This theorem also holds for limits as x→∞, as long as b ∈ R is
finite.
37 / 391
Example 1.16: Evaluate lim
x→∞ sin (e
−x) .
Solution:
38 / 391
Differentiability
Let f : R→ R be a real-valued function. The derivative of f at
x = a is defined by
f ′(a) = lim
h→0
f (a + h) − f (a)
h
.
The function f is differentiable at x = a if this limit exists.
Geometrically, f is differentiable at x = a if the graph y = f (x)
has a tangent line at x = a given by
y − f (a) = f ′(a)(x − a)
which gives a good approximation to the graph near x = a.
Note:
We can also define left differentiable and right differentiable.
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x
f HxL
a
Differentiable at x=a
x
f HxL
Not differentiable
at x=0
If f is differentiable at x = a, the linear approximation of f near
x = a is given by
f (x) ≈ f (a) + f ′(a)(x − a)
Theorem:
If f is differentiable at x = a, then f is continuous at x = a.
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L’Hoˆpital’s Rule
Let f and g be differentiable functions near x = a, and g′(x) , 0
at all points x near a with x , a. If
lim
x→a
f (x)
g(x)
has the indeterminate form
0
0
or
∞
∞ then
lim
x→a
f (x)
g(x)
= lim
x→a
f ′(x)
g′(x)
if the limit involving the derivatives exists.
Note:
L’Hoˆpital’s Rule also holds when x approaches infinity.
41 / 391
Example 1.17: Evaluate lim
x→0
sin x
x
.
Solution:
42 / 391
Example 1.18: Evaluate lim
x→∞
3x2 − 2x + 3
x2 + 4x + 4
.
Solution:
43 / 391
Example 1.19: Evaluate lim
x→∞
(
x−
1
3 log x
)
. (0 · ∞)
Solution:
44 / 391
Aside - What is a limit really?*
Recall our definition of limit:
lim
x→a f (x) = L if f (x) gets arbitrarily close to L whenever x is close
enough to a but x , a.
What do ‘arbitrarily close’ and ‘close enough’ mean?
More formally,
for any arbitrary positive real number ε, there is a positive real
number δ such that
|f (x) − L| < whenever |x − a| < δ.
This formal definition of limit is covered in MAST20026 Real
Analysis, but as a taster we do two examples.
* Slides 45-47 are not examinable in MAST10006.
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Aside - What is a limit really?*
Example 1.20: Using the definition, prove that
lim
x→1
2x = 2
Solution:
For an arbitrary positive real number ε,
|f (x) − 2| = 2|x − 1| < ε if and only if |x − 1| < 1
2
ε = δ.
This shows that |f (x) − 2| can be arbitrarily small whenever
|x − 1| is small enough but not equal to 0.
In other words, f (x) can be arbitrarily close to 2 whenever x is
close enough to 1 but x , 1.
Therefore,
lim
x→1 f (x) = 2.
46 / 391
Aside - What is a limit really?*
Example 1.21: Sketch a proof of the limit law
lim
x→a [f (x) + g(x)] = limx→a f (x) + limx→a g(x)
Solution:
Suppose lim
x→a f (x) = L and limx→a g(x) = M.
For an arbitrary positive real number ε, to make
|f (x) + g(x) − (L + M)| < ε
we only need to make |f (x) − L| < ε
2
and |g(x) −M| < ε
2
.
These will be satisfied whenever x is close enough to a but
x , a since lim
x→a f (x) = L and limx→a g(x) = M.
Hence f (x) + g(x) can be arbitrarily close to L +M whenever x is
close enough to a but x , a, which means that
lim
x→a [f (x) + g(x)] = L + M = limx→a f (x) + limx→a g(x).
47 / 391 48 / 391
Sequences
A sequence is a function f : N→ R.
It can be thought of as an ordered list of real numbers
a1, a2, a3, a4, . . . , an . . .
Thus, f (n) = an.
The sequence is denoted by {an}, where an is the nth term.
Example
1,
1
2
,
1
3
,
1
4
,
1
5
, . . . an =
1
n
Example
1,−1, 1,−1, 1,−1, . . . an = (−1)n−1
49 / 391
The graph of a sequence {an} can be plotted on a set of axes
with n on the x-axis and an on the y-axis.
Example: an = 1n Example: an = (−1)n−1
2 4
n
−1
1
an
1 3 5 2 4
n
−1
1
an
1 3 5
50 / 391
Limits of Sequences
A sequence {an} has the limit L if an approaches L as n
approaches infinity. Note, that L must be finite.
We write
lim
n→∞ an = L
or an → L as n→∞.
If the limit exists we say that the sequence converges.
Otherwise, we say that the sequence diverges.
51 / 391
Example 1.22: Determine whether the following sequences
converge or diverge: (a)
{1
n
}
(b)
{
(−1)n−1
}
(c) {n}
Solution:
52 / 391
The only difference between lim
n→∞ an = L and limx→∞ f (x) = L is
that n is a natural number whereas x is a real number.
Theorem:
Let f : R→ R be a real function and {an} be a sequence of real
numbers such that an = f (n). If
lim
x→∞ f (x) = L then limn→∞ an = L.
This means that we can use the techniques for evaluating limits
of functions to evaluate limits of sequences.
Note:
lim
n→∞ an = L 6=⇒ limx→∞ f (x) = L.
eg. an = sin(2pin), f (x) = sin(2pix).
53 / 391
Theorem:
Let {an} and {bn} be sequences of real numbers and c ∈ R a
constant.
If lim
n→∞ an and limn→∞ bn exist, then
1. lim
n→∞ [an + bn] = limn→∞ an + limn→∞ bn.
2. lim
n→∞ [can] = c limn→∞ an.
3. lim
n→∞ [anbn] = limn→∞ an · limn→∞ bn.
4. lim
n→∞
[an
bn
]
=
lim
n→∞ an
lim
n→∞ bn
provided lim
n→∞ bn , 0.
54 / 391
Sandwich Theorem:
Let {an}, {bn} and {cn} be sequences of real numbers.
If an ≤ cn ≤ bn for all n > N for some N, and
lim
n→∞ an = limn→∞ bn = L
then
lim
n→∞ cn = L.
55 / 391
The Factorial Function
The factorial function n! (n = 0, 1, 2, ...) is defined by
n! = n(n − 1)! , 0! = 1
or
n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1
Therefore
1! = 1
2! = 2 × 1 = 2
3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
Example
(2n + 2)! = (2n + 2) × (2n + 1) × (2n) × (2n − 1) × ... × 3 × 2 × 1
or
(2n + 2)! = (2n + 2) × (2n + 1) × (2n)!
56 / 391
Standard Limits
(1) lim
n→∞
1
np
= 0 (p > 0) (2) lim
n→∞ r
n = 0 (|r| < 1)
(3) lim
n→∞ a
1
n = 1 (a > 0) (4) lim
n→∞ n
1
n = 1
(5) lim
n→∞
an
n!
= 0 (a ∈ R) (6) lim
n→∞
logn
np
= 0 (p > 0)
(7) lim
n→∞
(
1 +
a
n
)n
= ea (a ∈ R) (8) lim
n→∞
np
an
= 0 (p ∈ R, a > 1)
Note:
Standard limits (1), (3), (4), (6), (7), (8) also hold for limits of
real-valued functions as x→∞. Standard limit (2) also holds
for x→∞ when 0 ≤ r < 1.
57 / 391
Example 1.23: Evaluate lim
n→∞
[(n − 2
n
)n
+
4n2
3n
]
.
Solution:
58 / 391
Example 1.24: Find the limit of the sequence
an =
3n + 2
4n + 2n
, n ≥ 1.
Solution:
59 / 391
Note:
The order hierarchy can be used to help identify the largest
term in an expression:
logn np an n!
60 / 391
Example 1.25: Prove Standard Limit 6:
lim
n→∞
logn
np
= 0 (p > 0)
Solution:
Note:
We must change to a continuous variable x ∈ R before applying
L’Hoˆpital’s rule.
61 / 391
Example 1.26: Evaluate lim
n→∞
[
log(3n − 2) − logn] .
Solution:
Note:
We must change to a continuous variable x ∈ R before applying
the continuity theorem. 62 / 391
Example 1.27: Evaluate lim
n→∞
1 + sin2
(
npi
3
)
√
n
.
Solution:
63 / 391 64 / 391
Adding Infinitely Many Numbers
Starting with any sequence {an}, adding the an’s together in
order gives a sequence {sn}:
s1 = a1,
s2 = a1 + a2,
s3 = a1 + a2 + a3,
...
...
...
...
The sequence of partial sums {sn} may or may not converge. If
it does converge, we call
S = lim
n→∞ sn = limn→∞(a1 + a2 + . . . + an)
the sum of the an’s.
65 / 391
Example 1.28: Find the sum S of an =
(1
2
)n
,n ≥ 1.
Solution:
66 / 391
Series
The series with terms an is denoted by the sum
∞∑
n=1
an.
If lim
n→∞ sn exists, we say that the series converges. Otherwise we
say that the series diverges.
Example
The sequence {n} = 1, 2, 3, 4, . . .
The series
∞∑
n=1
n = 1 + 2 + 3 + 4 + . . .
The sequence and series both diverge to infinity, so the sum
does not exist.
67 / 391
Application: Decimals
The decimal representation of a number is actually a series.
Example
The sequence
{ 1
10n
}
= 0.1, 0.01, 0.001, . . ..
The series
∞∑
n=1
1
10n
= 0.1 + 0.01 + 0.001 + . . . = 0.11111111 . . .
The sequence converges to 0 while the series converges to
1
9
.
In General
For a number x ∈ (0, 1) with decimal digits d1, d2, d3, d4, . . .
x = 0.d1d2d3d4 . . . =
∞∑
n=1
dn
10n
68 / 391
Properties of Series
Let
∞∑
n=1
an and
∞∑
n=1
bn be series, and c ∈ R\{0} a constant.
If
∞∑
n=1
an and
∞∑
n=1
bn converge then
1.
∞∑
n=1
(an + bn) =
∞∑
n=1
an +
∞∑
n=1
bn converges.
2.
∞∑
n=1
(can) = c
∞∑
n=1
an converges.
If
∞∑
n=1
an diverges then
∞∑
n=1
(can) diverges.
Note:
These follow from the properties of the limits of sequences. 69 / 391
Geometric Series
A geometric series has the form
∞∑
n=0
arn =
∞∑
n=1
arn−1 = a + ar + ar2 + ar3 + . . .
where a ∈ R\{0} and r ∈ R.
The series converges if |r| < 1 and diverges if |r| ≥ 1.
If |r| < 1, we have ∞∑
n=0
arn =
a
1 − r .
Note:
This follows from the fact that
n∑
k=0
ark =
a(1 − rn+1)
1 − r for r , 1.
70 / 391
Example 1.29: What does the series
∞∑
n=0
(1
2
)n
= 1 +
1
2
+
(1
2
)2
+
(1
2
)3
+ . . .
converge to?
Solution:
71 / 391
Harmonic p Series
A harmonic p series has the form
∞∑
n=1
1
np
.
The series converges if p > 1 and diverges if p ≤ 1.
Example
∞∑
n=1
1
n2
converges BUT
∞∑
n=1
1
n
diverges.
72 / 391
Divergence Test
If lim
n→∞ an , 0 then
∞∑
n=1
an diverges.
Note:
If lim
n→∞ an = 0 then
1.
∞∑
n=1
an may converge or diverge.
2. The Divergence Test is not relevant, so we need to use
another test to determine if
∞∑
n=1
an converges or diverges.
73 / 391
Example 1.30: Does the series
∞∑
n=1
n + 1
n
converge?
Solution:
74 / 391
Comparison Test
Let
∞∑
n=1
an and
∞∑
n=1
bn be positive term series.
1. If an ≤ bn for all n and
∞∑
n=1
bn converges, then
∞∑
n=1
an
converges.
2. If an ≥ bn for all n and
∞∑
n=1
bn diverges, then
∞∑
n=1
an diverges.
To apply the comparison test we compare a given series to a
harmonic p series or geometric series.
75 / 391
Example 1.31: Does
∞∑
n=1
7
2n2 + 4n + 3
converge or diverge?
Solution:
76 / 391
Example 1.32: Does
∞∑
n=1
n2 + 4
n3 + 1
converge or diverge?
Solution:
77 / 391 78 / 391
Ratio Test
Let
∞∑
n=1
an be a positive term series and
L = lim
n→∞
an+1
an
.
1. If L < 1,
∞∑
n=1
an converges.
2. If L > 1,
∞∑
n=1
an diverges.
3. If L = 1, the ratio test is inconclusive.
The ratio test is useful if an contains an exponential or factorial
function of n.
79 / 391
Example 1.33: Does
∞∑
n=1
10n
n!
converge or diverge?
Solution:
80 / 391
Example 1.34: Does
∞∑
n=1
(2n)!
n! n!
converge or diverge?
Solution:
81 / 391 82 / 391
83 / 391 84 / 391
Section 2: Hyperbolic Functions
Even Functions
A function f is an even function if
f (x) = f (−x)
Example
f (x) = cos x and f (x) = x2
Odd Functions
A function f is an odd function if
f (x) = −f (−x)
Example
f (x) = sin x and f (x) = x3
x
y
−2 −1 1 2
1
2
3
4
y = x2
x
y
−2 −1 1 2
−8
−4
4
8
y = x3
85 / 391
We define the hyperbolic cosine
function:
cosh x =
1
2
(
ex + e−x
)
, x ∈ R
x
y
−3 −2 −1 1 2 3
−1
1
2
3
4
y = cosh(x)
Properties
86 / 391
We define the hyperbolic sine
function:
sinh x =
1
2
(
ex − e−x) , x ∈ R x
y
−3 −2 −1 1 2 3
−4
−3
−2
−1
1
2
3
4
y = sinh(x)
Properties
87 / 391
We define the hyperbolic
tangent function:
tanh x =
sinh x
cosh x
=
1
2 (e
x − e−x)
1
2 (e
x + e−x)
=
ex − e−x
ex + e−x , x ∈ R.
x
y
−3 −2 −1 1 2 3
−2
−1
1
2
y = tanh(x)
Properties
88 / 391
Why call them hyperbolic functions?
Let x = cosh t and y = sinh t then
89 / 391
So (x, y) = (cosh t, sinh t) denotes a point on the hyperbola
x2 − y2 = 1. Since x ≥ 1, the right hand branch of the hyperbola
can be parametrised by x = cosh t, y = sinh t, t ∈ R.
-4 -2 2 4
x
-4
-2
2
4
y
y=sinh t
x=cosh t
t
t=0
t>0
t<0
x2-y2=1
90 / 391
Application: Catenary
A flexible, heavy cable of uniform mass per length ρ and
tension T at its lowest point has shape
y =
T
ρg
cosh
(ρgx
T
)
where g is the acceleration due to gravity.
x
y
Support Support
91 / 391
Example 2.1: If cosh x = 1312 and x < 0 find sinh x and tanh x.
Solution:
92 / 391
93 / 391
Addition Formulae
sinh(x + y) = sinh x cosh y + cosh x sinh y
cosh(x + y) = cosh x cosh y + sinh x sinh y
sinh(x − y) = sinh x cosh y − cosh x sinh y
cosh(x − y) = cosh x cosh y − sinh x sinh y
94 / 391
Example 2.2: Prove the sinh(x + y) addition formula.
Solution:
95 / 391
Double Angle Formulae
sinh(2x) = 2 sinh x cosh x
cosh(2x) = cosh2 x + sinh2 x
cosh(2x) = 2cosh2 x − 1
cosh(2x) = 2sinh2 x + 1
These can be proved using the addition formulae.
96 / 391
Reciprocal Hyperbolic Functions
We define the three reciprocal hyperbolic functions:
sech x =
1
cosh x
, x ∈ R
x
y
−3 −2 −1 1 2 3
−0.5
0.5
1.0
y = sech(x)
cosech x =
1
sinh x
, x ∈ R \ {0}
x
y
−3 −2 −1 1 2 3
−2
−1
1
2
y = cosech(x)
97 / 391
Reciprocal Hyperbolic Functions
coth x =
1
tanh x
=
cosh x
sinh x
,
x ∈ R \ {0}
x
y
−3 −2 −1 1 2 3
−3
−2
−1
1
2
3
y = coth(x)
98 / 391
Basic Identities
cosh2 x − sinh2 x = 1
coth2 x − 1 = cosech2 x
1 − tanh2 x = sech2 x
99 / 391
Derivatives of Hyperbolic Functions
d
dx
(cosh x) = sinh x, x ∈ R
d
dx
(sinh x) = cosh x, x ∈ R
d
dx
(tanh x) = sech2 x, x ∈ R
d
dx
(sech x) = − sech x tanh x, x ∈ R
d
dx
(cosech x) = − cosech x coth x, x ∈ R \ {0}
d
dx
(coth x) = − cosech2 x, x ∈ R \ {0}
100 / 391
Example 2.3: Prove that
d (cosh x)
dx
= sinh x.
Solution:
101 / 391
Example 2.4: Let y =
√
sinh(6x), x > 0. Find
dy
dx
.
Solution:
102 / 391
Inverses of Hyperbolic Functions
We define three inverse hyperbolic functions.
1. Inverse hyperbolic sine function: arcsinh x
Since sinh x is a 1-1 function
domain arcsinh x = range sinh x = R.
range arcsinh x = domain sinh x = R.
arcsinh(sinh x) = x, x ∈ R.
sinh(arcsinh x) = x, x ∈ R.
x
y
−4 −3 −2 −1 1 2 3 4
−3
−2
−1
1
2
3
y = arcsinh(x)
y = sinh(x)
103 / 391
2. Inverse hyperbolic cosine function: arccosh x
Restrict domain of cosh x to be [0,∞) to give a 1-1 function.
Then
domain arccosh x = range cosh x = [1,∞).
range arccosh x = restricted domain cosh x = [0,∞).
cosh(arccosh x) = x, x ≥ 1.
arccosh(cosh x) = x, x ≥ 0.
x
y
1 2 3 4 5
1
2
3
4
5
y = arccosh(x)
y = cosh(x)
104 / 391
3. Inverse hyperbolic tangent function: arctanh x
Since tanh x is a 1-1 function
domain arctanh x = range tanh x = (−1, 1).
range arctanh x = domain tanh x = R.
tanh(arctanh x) = x, −1 < x < 1.
arctanh(tanh x) = x, x ∈ R.
x
y
−4 −3 −2 −1 1 2 3 4
−3
−2
−1
1
2
3
y = arctanh(x)
y = tanh(x)
105 / 391
The inverse hyperbolic functions can be expressed in terms of
natural logarithms.
arcsinh x = log
(
x +
√
x2 + 1
)
, x ∈ R
arccosh x = log
(
x +
√
x2 − 1
)
, x ≥ 1
arctanh x =
1
2
log
(1 + x
1 − x
)
, −1 < x < 1
We can also define inverse reciprocal hyperbolic functions:
• arcsech x (0 < x ≤ 1)
• arccosech x (x , 0)
• arccoth x (x < −1 or x > 1)
106 / 391
Example 2.5: Proof of arcsinh x relation.
Solution:
107 / 391 108 / 391
Example 2.6: Simplify cosh(arctanh x) for −1 < x < 1.
Solution:
109 / 391 110 / 391
111 / 391 112 / 391
Derivatives
d
dx
(arcsinh x) =
1√
x2 + 1
(x ∈ R)
d
dx
(arccosh x) =
1√
x2 − 1
(x > 1)
d
dx
(arctanh x) =
1
1 − x2 (−1 < x < 1)
Each formula is derived using implicit differentiation or by
differentiating the logarithm definition of each function.
113 / 391
Example 2.7: Prove that
d
dx
(arcsinh x) =
1√
x2 + 1
.
Solution:
114 / 391
115 / 391
Example 2.8: Find
d
dx
(arctanh(2x) cosh(3x)).
Solution:
116 / 391
Section 3: Complex Numbers
The Cartesian form of a complex number z ∈ C is
z = x + iy where x, y ∈ R
and
• x = Re(z) is the real part of z,
• y = Im(z) is the imaginary part of z,
• i2 = −1.
ReHzL
ImHzL
x
r
y
z=x+iy
Θ
117 / 391
The complex number can be written as
z = r(cosθ + i sinθ)
where
• r = |z| =
√
x2 + y2
• tanθ = y
x
Note:
The angle θ is not unique – only defined up to multiples of 2pi.
We choose θ such that −pi < θ ≤ pi and call this angle the
principal argument of z.
118 / 391
The Complex Exponential
We define the complex exponential using Euler’s formula
eiθ = cosθ + i sinθ
for θ ∈ R.
We can then write the polar form of a complex number as
z = reiθ
119 / 391
Example 3.1: Write z = −1 + i in polar form.
ReHzL
ImHzL
1
2
1
-1+i
Α
Θ
Solution:
120 / 391
Properties of the Complex Exponential
1. ei0 = 1
Proof:
ei0 = cos 0 + i sin 0 = 1.
2. eiθeiφ = ei(θ+φ)
Proof:
eiθeiφ = (cosθ + i sinθ)
(
cosφ + i sinφ
)
= cosθ cosφ + i cosθ sinφ + i sinθ cosφ − sinθ sinφ
=
(
cosθ cosφ − sinθ sinφ
)
+ i
(
cosθ sinφ + sinθ cosφ
)
= cos
(
θ + φ
)
+ i sin(θ + φ)
= ei(θ+φ).
121 / 391
Products and Division in Polar Form
If z = r1eiθ and w = r2eiφ then
zw = r1r2ei(θ+φ)
z
w
=
r1
r2
ei(θ−φ)
122 / 391
Example 3.2: Using the complex exponential, simplify
(
√
3 − i)(1 + √3i) and
√
3 − i
1 +
√
3i
.
ReHzL
ImHzL
3
1
2
1+ 3 i
3 -i
2
1
3Α1
Α2
123 / 391
Solution:
124 / 391
125 / 391 126 / 391
De Moivre’s Theorem:
If z = reiθ and n is a positive integer then
zn =
(
reiθ
)n
= rneinθ
127 / 391
Example 3.3: Evaluate
(
1 +
√
3i
)15
.
Solution:
128 / 391
Exponential Form of sinθ and cosθ
Now eiθ = cosθ + i sinθ (1)
⇒ e−iθ = cos(−θ) + i sin(−θ)
⇒ e−iθ = cosθ − i sinθ (2)
Equation (1) + (2) gives
eiθ + e−iθ = 2 cosθ
⇒ cosθ = 1
2
(
eiθ + e−iθ
)
129 / 391
Equation (1) − (2) gives
eiθ − e−iθ = 2i sinθ
⇒ sinθ = 1
2i
(
eiθ − e−iθ
)
Note:
These formulae give a connection between the hyperbolic and
trigonometric functions.
cosh(iθ) =
1
2
(
eiθ + e−iθ
)
= cosθ
sinh(iθ) =
1
2
(
eiθ − e−iθ
)
= i sinθ
130 / 391
Example 3.4: Express sin5 θ in terms of the functions
sin(nθ) for integers n.
Solution:
131 / 391
Differentiation via the Complex Exponential
If z = x + yi where x, y ∈ R then we define
ez = ex+iy = ex eiy = ex(cos y + i sin y).
Derivatives of functions from R to C are defined similarly as
those from R to R.
Differentiation to functions from R to C is also linear and follows
the product law.
Show that
d
dt
(
ekt
)
= kekt when k = a + bi ∈ C.
d
dt
[
e(a+bi)t
]
=
d
dt
[
eateibt
]
132 / 391
=
d
dt
[
eat (cos(bt) + i sin(bt))
]
= aeat [cos(bt) + i sin(bt)] + eat [−b sin(bt) + bi cos(bt)]
= aeat [cos(bt) + i sin(bt)] + eat
[
bi2 sin(bt) + bi cos(bt)
]
= aeat [cos(bt) + i sin(bt)] + bieat [cos(bt) + i sin(bt)]
= (a + bi)eat [cos(bt) + i sin(bt)]
= (a + bi)eateibt
= (a + bi)e(a+ib)t.
133 / 391
Example 3.5: Find
d56
dt56
(
e−t cos t
)
.
Solution:
134 / 391
135 / 391
Note:
Example 3.5 also gives the answer to
d56
dt56
(
e−t sin t
)
.
136 / 391
Integration via the Complex Exponential
Since
d
dx
(
ekx
)
= k ekx if k = a + bi (a, b ∈ R) , then
∫
k ekx dx = ekx + C
⇒
∫
ekx dx =
1
k
ekx + D
137 / 391
Example 3.6: Evaluate
∫
e3x sin(2x) dx.
Solution:
138 / 391
139 / 391
Note:
Example 3.6 also gives the answer to
∫
e3x cos(2x) dx.
140 / 391
Section 4: Integral Calculus
Derivative Substitutions
To evaluate ∫
f [g(x)]g′(x)dx
put u = g(x)⇒ du
dx
= g′(x).
Then ∫
f [g(x)]g′(x)dx =
∫
f (u)
du
dx
dx
=
∫
f (u) du
141 / 391
Example 4.1: Evaluate
∫
(6x2 + 10) sinh(x3 + 5x − 2)dx.
Solution:
142 / 391
Example 4.2: Evaluate
∫
sech2(3x)
10 + 2 tanh(3x)
dx.
Solution:
143 / 391
Trigonometric and Hyperbolic Substitutions
We can use trigonometric and hyperbolic substitutions to
integrate expressions containing
√
a2 − x2,
√
a2 + x2,
√
x2 − a2,
where a is a positive real number.
Method:
Put x = g(θ). Then∫
f (x) dx =
∫
f [g(θ)]g′(θ) dθ
144 / 391
Integrand Substitution
√
a2 − x2, 1√
a2 − x2
,
(
a2 − x2
) 3
2 etc.
x = a sinθ
or
x = a cosθ
√
a2 + x2,
1√
a2 + x2
,
(
a2 + x2
)− 32 etc. x = a sinhθ
√
x2 − a2, 1√
x2 − a2
,
(
x2 − a2
) 5
2 etc. x = a coshθ
1
a2 + x2
x = a tanθ
145 / 391
Example 4.3: Evaluate
∫
1√
x2 + 25
dx using a substitution.
Solution:
146 / 391
147 / 391
Example 4.4: Evaluate
∫
1
x2 + 2
dx using a substitution.
Solution:
148 / 391
149 / 391
Example 4.5: Evaluate
∫ √
9 − 4x2 dx if |x| ≤ 32 .
Solution:
150 / 391
151 / 391
Example 4.6: Evaluate
∫ (
x2 − 1
) 3
2 dx if x ≥ 1.
Solution:
152 / 391
Powers of Hyperbolic Functions
Consider the integral:∫
sinhm x coshn x dx
where m,n are integers (≥ 0).
• If m or n is odd, create a “derivative” substitution by rewriting
one of the odd power terms using identities.
• If m and n are even, use double angle formulae.
153 / 391
Example 4.7: Evaluate
∫
sinh4 θ dθ.
Solution:
154 / 391
155 / 391
Finish Example 4.6:
156 / 391
Example 4.8: Evaluate
∫
sinh5 x cosh6 x dx.
Solution:
157 / 391 158 / 391
Example 4.9: Evaluate I =
∫
sinh5 x cosh7 x dx.
Solution:
159 / 391 160 / 391
Partial Fractions
Let f (x) and g(x) be polynomials, then
f (x) −→ degree n
g(x) −→ degree d
can be written as the sum of partial fractions.
Case 1: n < d
1. Factorise g over the real numbers.
2. Write down partial fraction expansion.
3. Find unknown coefficients
A,A1,A2, . . . ,Ar,B,B1,B2, . . . ,Br.
161 / 391
Denominator Factor Partial Fraction Expansion
(x − a) A
x − a
(x − a)r A1
x − a +
A2
(x − a)2 + · · · +
Ar
(x − a)r
(x2 + bx + c)
Ax + B
x2 + bx + c
(x2 + bx + c)r A1x+B1x2+bx+c +
A2x+B2
(x2+bx+c)2 + · · · + Arx+Br(x2+bx+c)r
162 / 391
Example 4.10: Evaluate
∫
4
x2(x + 2)
dx (x , 0,−2).
Solution:
163 / 391 164 / 391
165 / 391
Example 4.11: Evaluate
∫
4x
(x2 + 4)(x − 2) dx (x , 2).
Solution:
166 / 391
167 / 391 168 / 391
Example 4.12: Evaluate
∫
2x4 + 3x2
(x2 + 1)2(x2 + 2)
dx.
Solution:
169 / 391 170 / 391
171 / 391
Note:
In general, for a positive integer n if we put x = tanθ then∫
1
(x2 + 1)n
dx =
∫
cos2n−2 θ dθ.
172 / 391
Case 2: n ≥ d
Use long division, then apply case 1.
Example 4.13: Find∫
5x4 + 13x3 + 6x2 + 4
x3 + 2x2
dx (x , 0,−2).
Solution:
173 / 391 174 / 391
Integration by Parts
The product rule for differentiation is
d
dx
(uv) =
du
dx
v + u
dv
dx
Integrate
∫
d
dx
(uv) dx =
∫ (
du
dx
v + u
dv
dx
)
dx
⇒ uv =
∫
du
dx
v dx +
∫
u
dv
dx
dx
⇒
∫
u
dv
dx
dx = uv −
∫
v
du
dx
dx
175 / 391
Example 4.14: Evaluate
∫
x2 log x dx (x > 0).
Solution:
176 / 391
Example 4.15: Evaluate
∫
xe5x dx.
Solution:
177 / 391
Example 4.16: Evaluate
∫
log x dx (x > 0).
Solution:
178 / 391
Note:
This technique can also be used to integrate inverse
trigonometric functions and inverse hyperbolic functions.
179 / 391
Example 4.17: Evaluate
∫
e3x sin(2x) dx.
Solution:
180 / 391
181 / 391 182 / 391
Section 5: First Order Differential Equations
Ordinary Differential Equations
(1) An equation of the form
f
(
x, y,
dy
dx
,
d2y
dx2
, . . . ,
dny
dxn
)
= 0
is an ordinary differential equation (o.d.e) of order n.
Example 5.1: What order is 3
d4y
dx4
=
(
dy
dx
)2
+ 2x2y?
Solution:
183 / 391
(2) A solution of an o.d.e is a function y that satisfies the o.d.e
for all x in some interval.
Example 5.2: Verify that y(x) = x2 +
2
x
is a solution of
dy
dx
+
y
x
= 3x for all x ∈ R \ {0}.
Solution:
184 / 391
First Order O.D.E’s
The general form of a first order o.d.e is
dy
dx
= f (x, y).
Example 5.3: Solve
dy
dx
= x3.
Solution:
This is the general solution where c ∈ R is an arbitrary constant.
Each value of c corresponds to a different solution of the o.d.e.
185 / 391
-4 -2 2
x
-4
-2
2
4
6
y
c=0
c=-2
c=2
c=4
186 / 391
Initial value problem for a first order o.d.e
Solve
dy
dx
= f (x, y) subject to the condition y(x0) = y0.
Example 5.4: Solve
dy
dx
= x3 given y(0) = 2.
Solution:
187 / 391
Separable O.D.E’S
A separable first order o.d.e has the form:
dy
dx
=M(x)N(y), (M(x) , 0, N(y) , 0)
To solve use separation of variables
dy
dx
=M(x)N(y)
⇒ 1N(y)
dy
dx
=M(x)
⇒
∫
1
N(y)
dy
dx
dx =
∫
M(x)dx
⇒
∫
1
N(y)dy =
∫
M(x) dx
188 / 391
Example 5.5: Solve
dy
dx
=
y
1 + x
(x , −1).
Solution:
189 / 391 190 / 391
-6 -4 -2 2 4
x
-6
-4
-2
2
4
6
y
c=1
c=2
c=-1
c=-2
H-1,0L
191 / 391 192 / 391
Example 5.6: Solve
dy
dx
=
1
2y
√
1 − x2
(−1 < x < 1, y , 0)
if y(0) = 3.
Solution:
193 / 391 194 / 391
x
y
c=9
c=4
c=1
c=0
c=9
c=4
c=1
c=0
y= arcsin x + c
y=- arcsin x + c
x=-1 x=1
* Solution
195 / 391
Linear First Order O.D.E’s
Example 5.7: Solve x
dy
dx
+ y = ex.
Solution:
196 / 391
A linear first order o.d.e has the form:
dy
dx
+P(x)y = Q(x)
To solve:
Multiply o.d.e by I(x)
I(x)dy
dx
+P(x)I(x)y = Q(x)I(x)
If the left side can be written as the derivative of y(x)I(x), then
d
dx
[
y(x)I(x)] = Q(x)I(x)
which can be solved by integrating with respect to x.
197 / 391
Aim:
Find an integrating factor I so the left side will be the derivative
of yI. Then
d
dx
(
yI) ≡ Idy
dx
+PIy
⇒ dy
dx
I + ydI
dx
= Idy
dx
+PIy
⇒ ydI
dx
= PIy
To solve for all y
⇒ dI
dx
= PI (separable)
198 / 391
⇒ 1I
dI
dx
= P
⇒
∫
1
IdI =
∫
Pdx
⇒ log |I| =
∫
P dx + c
⇒ |I| = e
∫ Pdx+c
= e
∫ Pdx · ec
⇒ I = ±ec︸︷︷︸
constant
·e
∫ Pdx
199 / 391
So one integrating factor is
I(x) = e
∫ Pdx
Note:
Since we only need one integrating factor I, we can neglect
the ‘+c’ and modulus signs when calculating I.
200 / 391
Example 5.8: Find the general solution of
dy
dx
+
y
x
= sin x (x , 0).
Solution:
201 / 391 202 / 391
203 / 391
x
y
c=0 c=1
c=-2
c=-2
c=1
-2 2
-2
2
y(x) = − cos x + 1
x
sin x +
c
x
204 / 391
Example 5.9: Solve
1
2
dy
dx
− xy = x if y(0) = −3.
Solution:
205 / 391 206 / 391
Note:
207 / 391
x
y
c=1
H0,1L
c=2
c=-2
c=-1
H0,-2L
y = −1 + cex2
208 / 391
Other First Order O.D.E’s
Sometimes it is possible to make a substitution to reduce a
general first order o.d.e to a separable or linear o.d.e.
• A homogeneous type o.d.e has the form
dy
dx
= f
(y
x
)
Substituting u = yx reduces the o.d.e to a separable o.d.e.
• Bernoulli’s equation has the form
dy
dx
+ P(x)y = Q(x)yn
Substituting u = y1−n reduces the o.d.e to a linear o.d.e.
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Example 5.10: Solve the homogeneous type differential
equation
dy
dx
=
y
x
+ cos2
(y
x
) (
−pi2 < yx < pi2
)
by substituting u =
y
x
.
Solution:
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211 / 391 212 / 391
Example 5.11: Solve the Bernoulli equation
dy
dx
+ y = e3xy4 (y , 0)
by substituting u = y−3.
Solution:
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215 / 391
Population Models
Malthus (Doomsday) model
Rate of growth is proportional to the population p at time t.
dp
dt
∝ p
⇒ dp
dt
= kp (separable/linear)
where k is a constant of proportionality representing net births
per unit population per unit time.
If the initial population is p(0) = p0, then the solution is
p(t) = p0ekt
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Note:
The Doomsday model is unrealistic since if
• k > 0 − unbounded exponential growth
• k < 0 − population dies out
• k = 0 − population stays constant
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Equilibrium Solutions
1. An equilibrium solution is a solution that does not change
with time.
i.e.
dx
dt
= 0 ⇒ x(t) = x0
2. Stable equilibrium – solutions that start nearby move closer
as t increases.
t
x
x0 Stable
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3. Unstable equilibrium – solutions that start nearby move
further away as t increases.
t
x
x0 Unstable
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4. Semistable equilibrium – on one side of x0 solutions that
start nearby move closer as t increases whereas on the other
side of x0 solutions move further away as t increases.
t
x
x0 Semistable
t
x
x0 Semistable
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5. Phase plots:
If
dx
dt
= f (x), a plot of
dx
dt
as a function of x will give the
equilibrium solutions and the behaviour of solutions close by.
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Doomsday model with harvesting.
Remove some of the population at a constant rate.
dp
dt
= kp − h, h > 0.
Example 5.12:
dp
dt
= 3p − 2 (k = 3, h = 2)
Solution:
• Equilibrium solutions
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• Phase plot
p
dp
dt
23
-2
dp
dt
>0
dp
dt
<0
dies out!
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t
p
Unstablep(0)=2/3
Note:
Solving
dp
dt
= 3p − 2 with p(0) = p0 gives p(t) = 23 +
(
p0 − 23
)
e3t
which agrees with predicted behaviour.
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Logistic model.
Include “competition” term in Malthus’ model since
overcrowding, disease, lack of food and natural resources will
cause more deaths.
dp
dt
= kp − k
a
p2 = kp
(
1 − p
a
)
↗︷ ︸︸ ︷
net birth rate
↖︷ ︸︸ ︷
competition term
where a > 0 is the carrying capacity.
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Example 5.13:
dp
dt
= p
(
1 − p
4
)
(k = 1, a = 4)
Solution:
• Equilibrium solutions
• Phase plot
p
dp
dt
1
0 2 4
dp
dt
>0
dp
dt
<0
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t
p
pH0L=4
pH0L=0
227 / 391
• Exact solution
dp
dt
=
p
4
(
4 − p) (separable)
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Suppose p(0) = 1
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t
p
1
4
limiting growth
initial exponential growth
Note:
Logistic model accurately describes
• population in a limited space (e.g. bacteria culture).
• population of USA from 1790-1950.
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Logistic model with harvesting.
Remove some of the population at constant rate:
dp
dt
= kp
(
1 − p
a
)
− h, h > 0, a > 0
Example 5.14:
dp
dt
= p
(
1 − p
4
)
− 3
4
(
a = 4, k = 1, h =
3
4
)
Solution:
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• Phase plot
p
dp
dt
1
4
-
3
4
1 3
dp
dt
>0
dp
dt
<0 dp
dt
<0
dies out!
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tp
pH0L=3
pH0L=1
dies out!
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Find the time taken until the population dies out if p(0) =
1
2
.
dp
dt
= −1
4
(p − 3)(p − 1) (separable)
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Mixing Problems
Example 5.15: Effluent (pollutant concentration 2g/m3)
flows into a pond (volume 1000m3, initially 100g pollutant)
at a rate of 10m3/min. The pollutant mixes quickly and
uniformly with pond water and flows out of pond at a rate
of 10m3/min.
Find the concentration of pollutant in the pond at any time.
Solution:
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Let x be the amount (grams) of pollutant in pond at time t
minutes. Then C =
x
V
is the concentration of pollutant in pond
(grams/m3), where V is the volume of the pond (m3) at time t.
dx
dt
= rate pollutant flows in - rate pollutant flows out
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x
dx
dt
20
2000
dx
dt
>0
dx
dt
<0
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Expect solution (for all initial conditions) to look like
t (min)
x (grams)
xH0L=2000 Stable
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• Exact solution
dx
dt
+
x
100
= 20 (Linear/Separable)
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tHminL
conc. Hgm3L
2
0.1
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Definitions
1. Transient terms: terms decaying to 0 as t→∞.
2. Steady state terms: terms NOT decaying to 0 as t→∞.
The solution for the concentration can be classified as follows.
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Example 5.16: Find the concentration of pollutant in pond
if input flow rate is decreased to 5m3/min.
Solution:
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Let V be volume in pond (m3) at time t minutes.
245 / 391
Note:
There are no equilibrium solutions for x.
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247 / 391
tHminL
conc. Hgm3L
2
0.1
200
248 / 391
Section 6: Second Order Differential Equations
A second order o.d.e has the form
F
(
x, y,
dy
dx
,
d2y
dx2
)
= 0
The general form of a linear second order o.d.e is
d2y
dx2
+P(x)dy
dx
+ Q(x)y = R(x)
• If R(x) = 0, the o.d.e is homogeneous (H).
• If R(x) , 0, the o.d.e is inhomogeneous (IH).
Note:
A homogeneous linear o.d.e is different to a homogeneous type
first order o.d.e.
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Initial value problem for a second order o.d.e
Solve
d2y
dx2
+P(x)dy
dx
+ Q(x)y = R(x)
subject to the conditions y(x0) = y0 and y′(x0) = y1.
Boundary value problem for a second order o.d.e
Solve
d2y
dx2
+P(x)dy
dx
+ Q(x)y = R(x)
subject to the conditions y(a) = y0 and y(b) = y1.
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Homogeneous 2nd Order Linear O.D.E’s
Theorem:
The general solution of
y” +P(x)y′ + Q(x)y = 0
is the function y given by
y(x) = c1y1(x) + c2y2(x)
where
• y1, y2 are two linearly independent solutions of the
homogeneous o.d.e,
• c1, c2 ∈ R are arbitrary constants.
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Definition:
Two functions y1 and y2 are linearly independent if
c1y1(x) + c2y2(x) = 0 ⇒ c1 = c2 = 0.
Example 6.1: Are y1(x) = x2, y2(x) = 2x2 linearly
independent?
Solution:
Example 6.2: Are y1(x) = e2x, y2(x) = xe2x linearly
independent?
Solution:
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Homogeneous 2nd Order Linear O.D.E’s with Constant
Coefficients
General form:
ay” + by′ + cy = 0
where a, b, c are constants.
To solve for y(x):
Try y(x) = eλx
⇒ y′(x) = λeλx, y”(x) = λ2eλx
so
(
aλ2 + bλ + c
)
eλx︸︷︷︸
,0
= 0
⇒ aλ2 + bλ + c = 0 Characteristic Equation
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⇒ λ = −b ±
√
b2 − 4ac
2a
Case 1: b2 − 4ac > 0
• 2 distinct real values λ1, λ2
• 2 linearly independent solutions
eλ1x, eλ2x
• General Solution:
y(x) = Aeλ1x + Beλ2x
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Example 6.3: Solve y” + 7y′ + 12y = 0 for y(x).
Solution:
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Case 2: b2 − 4ac = 0
• 1 real value λ = −b
2a
• 1 solution is eλx
• 2nd linearly independent solution is xeλx (found using variation
of parameters — not in syllabus).
• General Solution:
y(x) = Aeλx + Bxeλx
We now verify that xeλx is a solution:
If y(x) = xeλx, then
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y′(x) = (λx + 1) eλx,
y”(x) =
(
λ2x + 2λ
)
eλx.
So ay” + by′ + cy
= a
(
λ2x + 2λ
)
eλx + b (λx + 1) eλx + cxeλx
= xeλx
(
aλ2 + bλ + c
)
︸ ︷︷ ︸
=0
+ (2λa + b)︸ ︷︷ ︸
=0
eλx
= 0
So y(x) = xeλx is a solution.
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Example 6.4: Solve y” + 2y′ + y = 0 for y(x).
Solution:
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Case 3: b2 − 4ac < 0
• 2 complex conjugate values
λ1 = α + iβ, λ2 = α − iβ
• 2 complex linearly independent solutions
e(α+iβ)x, e(α−iβ)x
• General Solution:
y(x) = C1e(α+iβ)x + C2e(α−iβ)x where C1,C2 ∈ C
= C1eαx
(
cos(βx) + i sin(βx)
)
+ C2eαx
(
cos(βx) − i sin(βx))
= (C1 + C2)︸ ︷︷ ︸
A
eαx cos(βx) + (C1i − C2i)︸ ︷︷ ︸
B
eαx sin(βx)
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Put A = C1 + C2 and B = (C1 − C2)i. If C1 = C2, then A,B ∈ R.
Note:
Imposing real conditions on the o.d.e will always lead to real
coefficients A and B.
• 2 real linearly independent solutions
eαx cos(βx), eαx sin(βx)
Real General Solution:
y(x) = Aeαx cos(βx) + Beαx sin(βx)
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Example 6.5: Solve y”− 4y′ + 13y = 0 for y(x) if y(0) = −1
and y′(0) = 2.
Solution:
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This solution has the form y(x) = e2x (a cosθ + b sinθ).
In general,
a cosθ + b sinθ =
√
a2 + b2
(
a√
a2 + b2
cosθ +
b√
a2 + b2
sinθ
)
=
√
a2 + b2
(
cosφ cosθ + sinφ sinθ
)
=
√
a2 + b2 cos(θ − φ).
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Hence we can rewrite the solution from Example 6.5 as
This form is sometimes preferable for graphing or further
manipulation.
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Inhomogeneous 2nd Order Linear O.D.E’s
Theorem:
The general solution of
y” +P(x)y′ + Q(x)y = R(x)
is the function y given by
y(x) = yH (x) + yP(x)
where
• yH (x) = c1y1(x) + c2y2(x) is the general solution of the
homogeneous o.d.e (called the homogeneous solution, GS(H)),
• yP(x) is a solution of the inhomogeneous o.d.e (called a
particular solution, PS(IH)),
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Inhomogeneous 2nd Order Linear O.D.E’s with
Constant Coefficients
General form:
ay” + by′ + cy = R(x)
where a, b, c are constants.
Example 6.6: Solve y” + 2y′ − 8y = R(x) where
(a) R(x) = 1 − 8x2
(b) R(x) = e3x
(c) R(x) = 85 cos x
(d) R(x) = 3 − 24x2 + 7e3x.
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Solution:
Step 1: Find the general solution of y” + 2y′ − 8y = 0.
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Step 2: Find a particular solution of y” + 2y′ − 8y = R(x).
(a) R(x) = 1 − 8x2 : y” + 2y′ − 8y = 1 − 8x2
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271 / 391
(b) R(x) = e3x : y” + 2y′ − 8y = e3x(
e3x is NOT part of GS(H)
)
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273 / 391
(c) R(x) = 85 cos x : y” + 2y′ − 8y = 85 cos x
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275 / 391
Superposition of Particular Solutions
Theorem:
A particular solution of
ay” + by′ + cy = c1R1(x) + c2R2(x)
is yP(x) = c1y1(x) + c2y2(x) where
• y1(x) is a particular solution of ay” + by′ + cy = R1(x),
• y2(x) is a particular solution of ay” + by′ + cy = R2(x),
• a, b, c, c1, c2 are constants.
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Example 6.6 (d): R(x) = 3 − 24x2 + 7e3x.
Solution:
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Example 6.7: Solve y” − y = ex.
Solution:
GS(H) : yH (x) = Aex + Be−x
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Example 6.8: Solve y” + 2y′ + y = e−x.
Solution:
GS(H) : yH (x) = (A + Bx) e−x
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Example 6.9: Solve y” + 49y = 28 sin(7t).
Solution:
GS(H) : yH (t) = A cos(7t) + B sin(7t)
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Springs - Free Vibrations
An object (mass m kg) stretches a spring (natural length L m)
hanging from a fixed support by s m.
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The forces are:
• gravitational force = mg (g = 9.8 m/s2)
• restoring force in spring (from Hooke’s Law)
T = k · extension (k > 0)0
spring constant
At equilibrium, forces balance so:
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Suppose the mass is set in motion. Let y be the displacement
of the object from the equilibrium position (y = 0) at any time t.
Assume
• downward direction is positive
• spring is stretched below equilibrium
• mass is moving down (so damping is upwards)
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Extra force:
• damping force is proportional to velocity
R = βy˙ (β ≥ 0)0
damping constant
Using Newton’s Law (F = ma)
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To solve, try y(t) = eλt
⇒ mλ2 + βλ + k = 0
⇒ λ = −β ±
√
β2 − 4mk
2m
• If β = 0 : λ = ±ib simple harmonic motion
• If 0 < β < 2√mk : λ = a ± ib underdamped, weak damping
• If β = 2√mk : λ = a, a critical damping
• If β > 2√mk : λ = a, b overdamped, strong damping
287 / 391 288 / 391
Example 6.10: A 4049 kg mass stretches a spring hanging
from a fixed support by 0.2m. The mass is released from
the equilibrium position with a downward velocity of 3m/s.
Find the position of the mass y below equilibrium at any
time t, if the damping constant β is:
(a) 0 (b) 16049 (c)
80
7 (d)
2000
49
Solution:
289 / 391 290 / 391
(a) β = 0 : y¨ + 49y = 0
291 / 391
t
y
Simple harmonic motion
3
7
-
3
7
Π
7
2 Π
7
3 Π
7
Equil.
292 / 391
(b) β = 16049 : y¨ + 4y˙ + 49y = 0
293 / 391
t
y
Weak damping
1
5
-
1
5
21
Equil.
294 / 391
(c) β = 807 : y¨ + 14y˙ + 49y = 0
295 / 391
t
y
critical damping
0.15
0.1
0.05
0.5 1
Equil.
296 / 391
(d) β = 200049 : y¨ + 50y˙ + 49y = 0
297 / 391
t
y
strong damping
0.06
0.03
2 4
Equil
298 / 391
Springs - Forced Vibrations
If an external downwards force f is applied to the spring-mass
system at time t, the forces acting on the mass are:
299 / 391
Example 6.11: Apply an external downwards force
f (t) = 1607 sin(7t) in Example 6.10.
Solution:
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(a) β = 807 : y¨ + 14y˙ + 49y = 28 sin(7t)
GS(IH) : y(t) = (A + Bt) e−7t − 27 cos(7t)
301 / 391
t
y
GSHHL contributes
oscillatory motion due to PSHIHL
0.4
-0.4
1 2 3
Equil.
302 / 391
(b) β = 0 : y¨ + 49y = 28 sin(7t)
GS(IH) : y(t) = A cos(7t) + B sin(7t) − 2t cos(7t)
303 / 391
t
y
Resonance
10
-10
5
-5
2 4 6
Equil
Definition
Resonance: Resonance occurs when the external force f has
the same form as one of the terms in the GS(H).
If β = 0, then the PS(IH) will grow without bound as t→∞.
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RLC series electric circuit
An RLC series electric circuit is an electric circuit with 4
components connected sequentially in a loop:
Circuits such as this are common in radio communications.
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RLC series electric circuit
The circuit components are
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RLC series electric circuit
Let q(t) be the charge on the capacitor (measured in Coulomb)
at time t seconds.
The charge satisfies the second-order ODE
L
d2q
dt2
+ R
dq
dt
+
q
C
= V
This equation has the same form as the equation of motion for
a spring with external driving force, and can exhibit all the same
solution types as the spring system.
307 / 391 308 / 391
Section 7: Functions of Two Variables
Example
The temperature T at a point on the Earth’s surface at a given
time depends on the latitude x and the longitude y. We think of
T being a function of the variables x, y and write T = f (x, y).
In general
A function of two variables is a mapping f that assigns a unique
real number z = f (x, y) to each pair of real numbers (x, y) in
some subset D of the xy plane R2. We also write
f : D→ R
where D is called the domain of f .
Example
If f (x, y) = x2 + y3 then f (2, 1) = 4 + 1 = 5.
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We can represent the function f by its graph in R3. The graph
of f is: {
(x, y, z) : (x, y) ∈ D and z = f (x, y)
}
.
This is a surface lying directly above the domain D. The x and y
axes lie in the horizontal plane and the z axis is vertical.
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Equations of a Plane
The Cartesian equation of a plane has the form
ax + by + cz = d
where a, b, c, d are real constants.
n = ai + bj + ck is a normal vector to the plane.
In fact, the plane passing through a point (x0, y0, z0) with a
normal vector (a, b, c) consists of the points (x, y, z) such that
(a, b, c) is perpendicular to (x − x0, y − y0, z − z0) and thus has
equation
a(x − x0) + b(y − y0) + c(z − z0) = 0,
that is,
ax + by + cz = ax0 + by0 + cz0.
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Example 7.1: The plane 4x + 3y + z = 2 can be written as
z = 2 − 4x − 3y, so is the graph of the function f : R2 → R
given by f (x, y) = 2 − 4x − 3y. Sketch the plane.
Solution:
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Level Curves
A curve on the surface z = f (x, y) for which z is a constant is a
contour.
The same curve drawn in the xy plane is a level curve.
So a level curve of f has the form
{(x, y) : f (x, y) = c}
where c ∈ R is a constant.
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Sketching Functions of Two Variables
The key steps in drawing a graph of a function of two variables
z = f (x, y) are:
1. Draw the x, y, z axes.
For right handed axes: the positive x axis is towards you,
the positive y axis points to the right, and the positive z axis
points upward.
2. Draw the y − z cross section.
3. Draw some level curves and their contours.
4. Draw the x − z cross section.
5. Label any x, y, z intercepts and key points.
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Example 7.2: Find the level curves of z =
√
1 − x2 − y2.
Hence identify the surface and sketch it.
Solution:
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x
y
c=1
c=0
H0,1L
H0,-1L
H-1,0L H1,0L
Level curves
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Consider cross sections (slices) to help sketch graph.
x
z
H-1,0L H1,0L
H0,1L
y
z
H-1,0L H1,0L
H0,1L
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Surface is a hemisphere radius 1, centre at (0, 0, 0) for z ≥ 0.
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Example 7.3: Sketch the graph of z = 4x2 + y2.
Solution:
x
y
c=0
c=4
H0,2L
H0,-2L
H-1,0L H1,0L
Level curves
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Consider cross sections (slices) to help sketch graph.
x
z
-1 1
4
z=4x2
y
z
-1 1
4
z=y2
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The surface is an elliptic paraboloid (parabolic bowl).
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Example 7.4: Sketch the graph of z =
√
4x2 + y2.
Solution:
x
y
c=0
c=2
H0,2L
H0,-2L
H-1,0L H1,0L
Level curves
324 / 391
Cross sections
x
z
-1 1
2
z=2ÈxÈ
y
z
-1 1
2
z=ÈyÈ
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The surface is an elliptic cone.
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Limits
Let f : R2 → R be a real-valued function.
We say f has the limit L as (x, y) approaches (x0, y0)
lim
(x,y)→(x0,y0)
f (x, y) = L
if when (x, y) approaches (x0, y0) along ANY path in the domain,
f (x, y) gets arbitrarily close to L.
Note:
1 L must be finite.
2 The limit can exist if f is undefined at (x0, y0).
3 The usual limit laws apply.
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Continuity
Let f : R2 → R be a real-valued function.
f is continuous at (x, y) = (x0, y0) if
lim
(x,y)→(x0,y0)
f (x, y) = f (x0, y0)
Note:
The continuity theorems for functions of one variable can be
generalised to functions of two variables.
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Example 7.5: Let f (x, y) = x2 + y2. For which values of x
and y is f continuous?
Solution:
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Example 7.6: Evaluate lim
(x,y)→(2,1)
log(1 + 2x2 + 3y2).
Solution:
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First Order Partial Derivatives
Let f : R2 → R be a real-valued function. The first order partial
derivatives of f with respect to the variables x and y are defined
by the limits:
fx =
∂f
∂x
= lim
h→0
f (x + h, y) − f (x, y)
h
fy =
∂f
∂y
= lim
h→0
f (x, y + h) − f (x, y)
h
Note:
• ∂f
∂x
measures the rate of change of f with respect to x
when y is held constant.
• ∂f
∂y
measures the rate of change of f with respect to y
when x is held constant.
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Geometric Interpretation of
∂f
∂x
and
∂f
∂y
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Let C1 be the curve where the vertical plane y = y0 intersects
the surface. Then
∂f
∂x
∣∣∣∣
(x0,y0)
gives the slope of the tangent to C1
at (x0, y0, z0).
Let C2 be the curve where the vertical plane x = x0 intersects
the surface. The
∂f
∂y
∣∣∣∣
(x0,y0)
gives the slope of the tangent to C2 at
(x0, y0, z0).
• T1 and T2 are the tangent lines to C1 and C2.
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Example 7.7: Let f (x, y) = xy2. Find
∂f
∂y
from first
principles.
Solution:
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Example 7.8: Let f (x, y) = 3x3y2 + 3xy4. Find
∂f
∂x
and
∂f
∂y
.
Solution:
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Example 7.9: Let f (x, y) = y log x + x tanh(3y). Find fx, fy at
(1, 0).
Solution:
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Tangent Planes and Differentiability
Let f : R2 → R be a real-valued function. We say that f is
differentiable at (x0, y0) if the tangent lines to all curves on the
surface z = f (x, y) passing through (x0, y0, z0) form a plane,
called the tangent plane.
This holds if fx and fy exist and are continuous near (x0, y0).
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The tangent line T1 has equation (y = y0 fixed):
z − z0 = ∂f∂x
∣∣∣∣
(x0,y0)
(x − x0)
The tangent line T2 has equation (x = x0 fixed):
z − z0 = ∂f∂y
∣∣∣∣
(x0,y0)
(y − y0)
Since a plane passing through (x0, y0, z0) has the form
z − z0 = α(x − x0) + β(y − y0)
the tangent plane has equation
z − z0 = ∂f∂x
∣∣∣∣
(x0,y0)
(x − x0) + ∂f∂y
∣∣∣∣
(x0,y0)
(y − y0).
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Example 7.10: Find the equation of the tangent plane to
the surface z = f (x, y) = 2x2 + y2 at (1, 1, 3).
Solution:
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Linear Approximations
If f is differentiable at (x0, y0), we can approximate z = f (x, y) by
its tangent plane at (x0, y0, z0).
This linear approximation of f near (x0, y0) is:
f (x, y) ≈ f (x0, y0) + ∂f∂x
∣∣∣∣
(x0,y0)
(x − x0) + ∂f∂y
∣∣∣∣
(x0,y0)
(y − y0)
Let ∆x = x − x0, ∆y = y − y0, ∆f = z − z0 = f (x, y) − f (x0, y0).
Then the approximate change in f near (x0, y0), for given small
changes in x and y, is:
∆f ≈ ∂f∂x
∣∣∣∣
(x0,y0)
∆x + ∂f∂y
∣∣∣∣
(x0,y0)
∆y
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Example 7.11: Let z = f (x, y) = x2 + 3xy − y2. If x changes
from 2 to 2.05 and y changes from 3 to 2.96, estimate the
change in z.
Solution:
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Note:
The actual change in f is
∆f = f (2.05, 2.96) − f (2, 3)
= 13.6449 − 13
= 0.6449
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Example 7.12: Find the linear approximation of
f (x, y) = xexy at (1, 0). Hence, approximate f (1.1,−0.1).
Solution:
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Note:
The actual value is
(1.1)e−0.11 ≈ 0.98542
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Second Order Partial Derivatives
Let f : R2 → R be a real-valued function. The second order
partial derivatives of f with respect to x and y are defined by:
• fxx =
(
fx
)
x
=
∂
∂x
(
∂f
∂x
)
=
∂2f
∂x2
• fyy =
(
fy
)
y
=
∂
∂y
(
∂f
∂y
)
=
∂2f
∂y2
• fxy =
(
fx
)
y
=
∂
∂y
(
∂f
∂x
)
=
∂2f
∂y∂x
• fyx =
(
fy
)
x
=
∂
∂x
(
∂f
∂y
)
=
∂2f
∂x∂y
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Theorem:
If the second order partial derivatives of f exist and are
continuous then fxy = fyx.
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Example 7.13: Find the second order partial derivatives of
f : R2 → R given by f (x, y) = x sin(x + 2y).
Solution:
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Note:
fxy = fyx as expected since trigonometric functions and
polynomials are continuous for all (x, y) ∈ R2.
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Chain Rule
1. If z = f (x, y) and x = g(t), y = h(t) are differentiable
functions, then z = f (g(t), h(t)) is a function of t, and
dz
dt
=
∂z
∂x
dx
dt
+
∂z
∂y
dy
dt
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Example 7.14: If z = x2 − y2, x = sin t, y = cos t. Find dz
dt
at
t =
pi
6
.
Solution:
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351 / 391
2. If z = f (x, y) and x = g(s, t), y = h(s, t) are differentiable
functions, then z is a function of s and t with
∂z
∂s
=
∂z
∂x
∂x
∂s
+
∂z
∂y
∂y
∂s
∂z
∂t
=
∂z
∂x
∂x
∂t
+
∂z
∂y
∂y
∂t
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Example 7.15: If z = ex sinh y, x = st2, y = s2t.
Find
∂z
∂s
and
∂z
∂t
.
Solution:
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Directional Derivatives
Let uˆ = (u1,u2) be a unit vector in the xy-plane (so u21 + u
2
2 = 1).
The rate of change of f at P0 = (x0, y0) in the direction uˆ is the
directional derivative Duˆf
∣∣∣
P0
.
Geometrically this represents the slope of the surface z = f (x, y)
above the point P0 in the direction uˆ.
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The straight line starting at P0 = (x0, y0) with velocity uˆ = (u1,u2)
has parametric equations:
x = x0 + tu1, y = y0 + tu2.
Hence,
Duˆf
∣∣∣
P0
= rate of change of f along the straight line at t = 0
= value of
d
dt
f (x0 + tu1, y0 + tu2) at t = 0
= fx(x0, y0)x′(0) + fy(x0, y0)y′(0) by the chain rule
= fx(x0, y0)u1 + fy(x0, y0)u2.
We can also write this as a dot product
Duˆf
∣∣∣
P0
=
(
∂f
∂x
∣∣∣∣∣
P0
,
∂f
∂y
∣∣∣∣∣
P0
)
· (u1,u2).
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Gradient Vectors
If f : R2 → R is a differentiable function, we can define the
gradient of f to be the vector
grad f = ∇f = ∂f
∂x
i +
∂f
∂y
j =
(
∂f
∂x
,
∂f
∂y
)
Then the directional derivative of f at the point P0 in the
direction uˆ is the dot product
Duˆf
∣∣∣
P0
= ∇f
∣∣∣
P0
· uˆ
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Example 7.16: Find the directional derivative of f (x, y) = xey
at (2, 0) in the direction from (2, 0) towards
(1
2
, 2
)
.
Solution:
• direction uˆ
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Example 7.17: Find the directional derivative of
f (x, y) = arcsin
(
x
y
)
at (1, 2) in the direction
pi
4
anticlockwise
from the positive x axis.
Solution:
• direction uˆ
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Properties of ∇f and Duˆf
The directional derivative of f is
Duˆf = ∇f · uˆ
= |∇f | |uˆ| cosθ
= |∇f | cosθ
where θ is the angle between ∇f and uˆ, and |v| denotes the
length of a vector v.
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So for fixed ∇f :
• Duˆf is maximum when cosθ = 1 so θ = 0
⇒ f increases most rapidly along ∇f .
• Duˆf is minimum when cosθ = −1 so θ = pi
⇒ f decreases most rapidly along −∇f .
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• Duˆf = 0 when cosθ = 0 so θ = pi2 and ∇f ⊥ uˆ.
But Duˆf = 0, whenever uˆ is tangent to a level curve of f (where
f = constant).
⇒ ∇f ⊥ level curves of f
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Example 7.18: Let f (x, y) = 4x2 + y2.
(a) Find ∇f at (1, 0) and (0, 2).
(b) Show that ∇f is perpendicular to the level curves, by
sketching ∇f at these points and the level curves of f .
Solution:
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(b)
0,2
1,0
4j
8i
Level curves of f x,y
y
x
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Example 7.19: In what direction does f (x, y) = xey
(a) increase (b) decrease
most rapidly at (2, 0)? Express direction as a unit vector.
Solution:
From Example 7.16
∇f (2, 0) = i + 2j
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Stationary Points
A stationary point of f is a point (x0, y0) at which
∇f = 0
So
∂f
∂x
= 0 and
∂f
∂y
= 0 simultaneously at (x0, y0).
Geometrically, this means that the tangent plane to the graph
z = f (x, y) at (x0, y0) is horizontal, i.e. parallel to the xy-plane.
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Three important types of stationary points are
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A function f has a
1. local maximum at (x0, y0) if f (x, y) ≤ f (x0, y0) for all (x, y) in
an open disk centred at (x0, y0),
2. local minimum at (x0, y0) if f (x, y) ≥ f (x0, y0) for all (x, y) in
an open disk centred at (x0, y0),
3. saddle point at (x0, y0) if (x0, y0) is a stationary point, and
there are points near (x0, y0) with f (x, y) > f (x0, y0) and
other points near (x0, y0) with f (x, y) < f (x0, y0).
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Any local maximum or minimum of f will occur at a critical point
(x0, y0) such that
1. ∇f (x0, y0) = 0 or
2.
∂f
∂x
and/or
∂f
∂y
do not exist at (x0, y0).
z =
√
x2 + y2. Minimum at (0, 0) BUT ∇f does not exist at (0, 0).
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Second Derivative Test
If ∇f (x0, y0) = 0 and the second partial derivatives of f are
continuous on an open disk centred at (x0, y0), consider the
Hessian function
H(x, y) = fxxfyy − (fxy)2
evaluated at (x0, y0).
Then (x0, y0) is a
1. local minimum if H(x0, y0) > 0 and fxx(x0, y0) > 0.
2. local maximum if H(x0, y0) > 0 and fxx(x0, y0) < 0.
3. saddle point if H(x0, y0) < 0.
Note: Test is inconclusive if H(x0, y0) = 0.
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Example 7.20: Find and classify the stationary points of
f : R2 → R given by f (x, y) = x3 + y3 + 3x2 − 3y2 − 8.
Solution:
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375 / 391 376 / 391
Example 7.21: Find and classify the stationary points of
f : R2 → R given by f (x, y) = y sin x.
Solution:
377 / 391 378 / 391
Partial Integration
Let f : R2 → R be a continuous function over a domain D in R2.
The partial indefinite integrals of f with respect to the first and
second variables (say x and y) are denoted by:∫
f (x, y) dx and
∫
f (x, y) dy.
•
∫
f (x, y) dx is evaluated by holding y fixed and integrating
with respect to x.
•
∫
f (x, y) dy is evaluated by holding x fixed and integrating
with respect to y.
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Example 7.22: Evaluate
∫
(3x2y + 12y2x3) dx.
Solution:
Note:
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Example 7.23: Evaluate
∫ 1
0
(3x2y + 12y2x3) dy.
Solution:
381 / 391
Double Integrals
Let f : R2 → R be a continuous function over a domain D in R2.
We can evaluate the double integral:"
D
f (x, y) dA =
"
D
f (x, y) dx dy
"
D
f (x, y) dA is the volume under the surface z = f (x, y) that
lies above the domain D in the xy plane, if f (x, y) ≥ 0 in D.
382 / 391
Δx
Δy
Volume of thin rod = (Area base)︸ ︷︷ ︸
‖
· (height)︸ ︷︷ ︸
‖
∆x∆y f (x, y)
383 / 391
The double integral is defined as the limit of sums of the
volumes of the rods:"
D
f (x, y) dA =
"
D
f (x, y) dx dy
= lim
∆x→0 lim∆y→0
n∑
i=1
[
f (x, y)∆x∆y
]
i
Note:
If f (x, y) = 1 then "
D
dA =
"
D
dx dy
gives the area of the domain D.
384 / 391
Double Integrals Over Rectangular Domains
Definitions
1. R = [a, b] × [c, d] is a rectangular domain defined by
a ≤ x ≤ b, c ≤ y ≤ d.
2.
∫ d
c
∫ b
a f (x, y) dx dy =
∫ d
c
[∫ b
a f (x, y) dx
]
dy means integrate
with respect to x first and then integrate with respect to y.
385 / 391
Fubini’s Theorem:
Let f : R2 → R be a continuous function over the domain
R = [a, b] × [c, d]. Then"
R
f (x, y)dA =
∫ d
c
∫ b
a
f (x, y) dx dy
=
∫ b
a
∫ d
c
f (x, y) dy dx
So order of integration is not important.
386 / 391
Example 7.24: Evaluate
"
R
(x2 + y2) dx dy if
R = [−1, 1] × [0, 1].
Solution:
387 / 391 388 / 391
Note:
As expected, the order of integration is not important since
polynomials are continuous for all (x, y) ∈ R2.
389 / 391
Example 7.25: Using double integrals, find the volume of
the wedge shown below.
Solution:
390 / 391
391 / 391

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