THE UNIVERSITY OF MELBOURNE
SCHOOL OF MATHEMATICS AND STATISTICS
MAST10006 Calculus 2
Lecture Notes
© School of Mathematics and Statistics, University of Melbourne, 2022.
STUDENT NAME:
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These notes are for the use of students of the University of Melbourne enrolled in the subject MAST10006 Calculus 2.
Edition 22, July 2022.
Table of Contents
Section 0 - Notation used in MAST10006 Calculus 2
Section 0 - Notation used in MAST10006 Calculus 2
2
Section 1 - Limits, Continuity, Sequences and Series
7
Section 2 - Hyperbolic Functions
Standard Abbreviations
1. such that or given that: |
2. for all: ∀
88
Section 3 - Complex Numbers
117
3. there exists: ∃
Section 4 - Integral Calculus
142
4. equivalent to: ≡
Section 5 - First Order Ordinary Differential Equations
181
Section 6 - Second Order Ordinary Differential Equations
249
Section 7 - Functions of Two Variables
311
5. that is: i.e
6. approximate: ≈
7. much smaller than: ≪
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Standard Notation for Sets of Numbers
Standard Notation for Intervals
1. element of: ∈
so a ∈ X means “a is an element of the set X ”
1. natural numbers: N = {1, 2, 3, . . . }
2. integers: Z = {0, ±1, ±2, . . . }
2. open interval: (a, b)
so x ∈ (0, 1) means “0 < x < 1”
3. rational numbers: Q = { mn | m, n ∈ Z, n , 0}
3. closed interval: [a, b]
so x ∈ [0, 1] means “0 ≤ x ≤ 1”
4. real numbers: R (rational numbers plus irrational numbers)
5. complex numbers: C = {x + iy | x, y ∈ R, i2 = −1}
4. partial open and closed interval: (a, b] or [a, b)
so x ∈ [0, 1) means “0 ≤ x < 1”
6. R2 = {(x, y) | x, y ∈ R} (xy plane)
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”.
7. R3 = {(x, y, z) | x, y, z ∈ R} (3 dimensional space)
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More Standard Notation
Greek Alphabet
1. natural logarithm (base e): log x
base 10 logarithm: log10 x
α
β
γ
δ
ϵ or ε
ζ
η
θ
ι
κ
λ
µ
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”
alpha
beta
gamma
delta
epsilon
zeta
eta
theta
iota
kappa
lambda
mu
ν
ξ
o
π
ρ
σ
τ
υ
ϕ
χ
ψ
ω
nu
xi
omicron
pi
rho
sigma
tau
upsilon
phi
chi
psi
omega
5. approaches: →
so f (x) → 1 as x → 0 means “f (x) approaches 1 as x
approaches 0”
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Section 1: Limits, Continuity, Sequences, Series
Definition of limit of a function
Let f be a real-valued function.
Why do we study limits and continuity in calculus?
The limit of f (x) as x approaches a is L, written
lim f (x) = L
x→a
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.
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(
Example 1.1: If f (x) =
2x x , 1
, evaluate lim f (x).
4 x=1
x→1
Example 1.2: If f (x) =
1
, evaluate lim f (x).
x→0
x2
f HxL
f HxL
4
3
2
1
-3
Solution:
-2
-1
1
2
3
x
x
-1
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.
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(
Example 1.3: If f (x) =
We can describe this behaviour in terms of one-sided limits.
We write
1 x<0
, evaluate lim f (x).
2 x≥0
x→0
f HxL
2
1
Theorem:
x
lim f (x) = L if and only if lim− f (x) = L and lim+ f (x) = L.
Solution:
x→a
x→a
x→a
Thus the limit exists if and only if the left and right hand limits
exist and are equal.
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Limit Laws
x3 + 2x2 − 1
Example 1.4: Use the limit laws to evaluate lim
.
x→2
5 − 3x
Let f and g be real-valued functions and let c ∈ R be a constant.
If lim f (x) and lim g(x) exist, then
x→a
x→a
Solution:
1. lim [f (x) + g(x)] = lim f (x) + lim g(x).
x→a
x→a
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x→a
2. lim [cf (x)] = c lim f (x).
x→a
x→a
3. lim [f (x)g(x)] = lim f (x) · lim g(x).
x→a
x→a
# lim f (x)
f (x)
x→a
=
4. lim
x→a g(x)
lim g(x)
x→a
"
x→a
provided lim g(x) , 0.
x→a
5. lim c = c.
x→a
6. lim x = a.
x→a
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Limits as x Approaches Infinity
Example 1.5: Evaluate lim e−x .
x→∞
The limit of f (x) as x approaches positive infinity is L,
y = e-x
lim f (x) = L
x→∞
H0,1L
if f (x) gets arbitrarily close to L whenever x is sufficiently large
and positive.
The limit of f (x) as x approaches negative infinity is M,
lim f (x) = M
x
x→−∞
Solution:
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 to limits as x → ±∞.
3. lim f (x) = lim f (−x).
x→−∞
x→∞
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Some Standard Limits
Terminology
The following are all equivalent ways of expressing lim f (x) = L:
x→a
▶ The limit of f (x) as x approaches a is L
▶ f (x) converges to L as x approaches a
We can use the following standard limits without further proof:
▶ f (x) → L as x → a
1
(1) lim p = 0 (p > 0)
x→∞ x
These all mean the same thing.
If the limit lim f (x) exists, we say that
(2) lim rx = 0
x→∞
x→a
(0 ≤ r < 1)
f (x) converges as x approaches a.
If the limit lim f (x) does not exist, we say that
x→a
f (x) diverges as x approaches a.
We will see more standard limits later.
Similarly for limits as x → ±∞.
Note:
Diverges simply means does not converge.
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More about divergence
Example 1.6: Explain why the following limits diverge.
1. lim sin x
How do we show that a limit lim f (x) diverges?
x→a
x→∞
2. lim
x→0
1
x2
Solution:
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f (x)
∞
has indeterminate form
as x → a if
g(x)
∞
f (x) → ∞ and g(x) → ∞.
Notation and ∞
We say a function
∞ is not a number, and should not appear as a number in your
writing.
In other textbooks, you will often see the notation lim
x→0
mean that
1
x2
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3x2 − 2x + 3
.
x→∞ x2 + 4x + 4
Example 1.7: Evaluate lim
1
= ∞ to
x2
Solution:
diverges to infinity.
You should not use this notation in Calculus 2 and will lose
notation marks for this!
Why?
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Sandwich Theorem (version for x → a)
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→∞
If g(x) ≤ f (x) ≤ h(x) when x is near a but x , a, and
lim g(x) = lim h(x) = L
x2 + 1 − x .
x→a
then
x→a
lim f (x) = L.
Solution:
x→a
y
hHxL
f HxL
L
a
gHxL
x
Note: “x is near a but x , a” means that x ∈ (b, a) ∪ (a, c) for
some b < a < c.
We will finish this example later.
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Sandwich Theorem (version for x → ∞)
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1
Example 1.9: Evaluate lim x2 sin
.
x→0
x
A version of the Sandwich Theorem also works for limits as
x → ∞.
Solution:
If g(x) ≤ f (x) ≤ h(x) when x is sufficiently large, and
lim g(x) = lim h(x) = L
then
x→∞
x→∞
lim f (x) = L.
x→∞
Similarly, the theorem holds for x → −∞.
Note: “x is sufficiently large” means x ∈ (c, ∞) for some real
number c.
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1
Example 1.10: Evaluate lim x sin
.
x→0
x
Example 1.8 (continued)
lim
Solution:
x→∞
√
x2 + 1 − x =
lim √
x→∞
1
x2 + 1 + x
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Continuity
f HxL
Example 1.11: Let
(
2x x , 1
f (x) =
4 x = 1.
4
3
2
Is f continuous at x = 1?
Definition of continuity
1
-3
Let f be a real-valued function.
-2
-1
1
2
3
x
-1
The function f is continuous at x = a if
Solution:
lim f (x) = f (a).
x→a
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 2
x −4




 x−2 x,2
Example 1.12: Let f (x) = 




4 x = 2.
Theorem:
The following function types are continuous at every point in
their domains:
▶ polynomials
Is f continuous at x = 2?
▶ trigonometric functions: sin x, cos x, tan x, sec x, cosec x,
cot x, arcsin x, arccos x, arctan x
▶ exponential functions: ax for a > 0
Solution:
▶ logarithm functions: loga x for a > 0
√
▶ nth root functions: n x for n ∈ {2, 3, 4, ...}
▶ hyperbolic functions: sinh x, cosh x, tanh x, sech x, cosech x,
coth x, arcsinh x, arccosh x, arctanh x
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Example 1.13: Evaluate lim sin (e−x ) .
x→∞
Let f and g be real-valued functions and b ∈ R.
Solution:
Theorem:
If lim g(x) = b and f is continuous at b then
x→a
lim f g(x) = f lim g(x) = f (b).
x→a
x→a
Note:
This theorem also holds for limits as x → ∞.
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Let f and g be real-valued functions and c ∈ R be a constant.
Theorem:
If the functions f and g are continuous at x = a, then the
following functions are continuous at x = a:
Recall that ( g ◦ f )(x) = g( f (x)).
1. f + g,
Theorem:
2. cf ,
If f is continuous at x = a and g is continuous at x = f (a), then
g ◦ f is continuous at x = a.
3. fg,
4.
f
if g(a) , 0.
g
Note:
The theorem follows from limit laws.
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Example 1.14: Let f (x) =
sin x2 + 1
log x
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.
For which values of x is f continuous?
Solution:
Note:
This set can also be written as
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 3

x − cx + 8, x ≤ 1



Example 1.15: f (x) = 


 x2 + 2cx + 2, x > 1.
For which values of c is f continuous at all x ∈ R?
Justify your answer.
Solution:
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Where does this definition come from?
Differentiability
Before seeing what is filled in from the lectures, try to come up
with a picture that explains where the definition comes from.
Definition of derivative
Let f be a real-valued function whose domain contains an open
interval around x = a. 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.
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Geometrically, f is differentiable at x = a if the graph y = f (x)
has a tangent line at x = a.
f HxL
Differentiability and Continuity
f HxL
Theorem:
If f is differentiable at x = a, then f is continuous at x = a.
x
a
Differentiable at x=a
x
What about the converse?
Not differentiable
at x=0
f HxL
If f is differentiable at x = a, then
f HxL
y = f (a) + f ′ (a)(x − a)
is the tangent line to the curve y = f (x) at the point x = a, and
f (x) ≃ f (a) + f ′ (a)(x − a)
x
a
Differentiable at x=a
when x is close to a.
x
Not differentiable
at x=0
That is, the tangent line is a linear approximation for the
function close to the point x = a.
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L’Hôpital’s Rule
Let f and g be differentiable functions near x = a, and
at all points x near a with x , a. If
g′ (x)
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Example 1.16: Evaluate lim
,0
x→0
sin x
.
x
Solution:
f (x)
lim
x→a g(x)
has the indeterminate form
lim
x→a
0
∞
or
then
0
∞
f (x)
f ′ (x)
= lim ′
x→a g (x)
g(x)
if the limit involving the derivatives exists.
Note:
L’Hôpital’s Rule also holds for limits as x → ∞, and for
one-sided limits x → a+ and x → a− for a ∈ R.
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3x2 − 2x + 3
Example 1.17: Evaluate lim 2
.
x→∞ x + 4x + 4
1
Example 1.18: Evaluate lim x− 3 log x .
Solution:
Solution:
x→∞
(0 · ∞)
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Aside - What is a limit really?*
Aside - What is a limit really?*
Recall our definition of limit:
Example 1.19: Using the definition, prove that
lim f (x) = L if f (x) gets arbitrarily close to L whenever x is close
lim 2x = 2
x→a
enough to a but x , a.
x→1
Solution:
What do ‘arbitrarily close’ and ‘close enough’ mean?
For an arbitrary positive real number ε,
More formally,
1
|f (x) − 2| = 2|x − 1| < ε if and only if |x − 1| < ε = δ.
2
This shows that |f (x) − 2| can be arbitrarily small whenever
|x − 1| is small enough but not equal to 0.
for any arbitrary positive real number ε, there is a positive real
number δ such that
|f (x) − L| < ϵ whenever 0 < |x − a| < δ.
In other words, f (x) can be arbitrarily close to 2 whenever x is
close enough to 1 but x , 1.
This formal definition of limit is covered in MAST20026 Real
Analysis, but as a taster we do two examples.
Therefore,
lim f (x) = 2.
x→1
* Slides 49-51 are not examinable in MAST10006.
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Aside - What is a limit really?*
Sequences
Definition of sequence
Example 1.20: Sketch a proof of the limit law
A sequence is a function f : N → R.
It can be thought of as an ordered list of real numbers
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
a1 , a2 , a3 , a4 , . . . , an . . .
Solution:
Suppose lim f (x) = L and lim g(x) = M.
x→a
Thus, f (n) = an .
The sequence is denoted by (an ), where an is the nth term.
x→a
For an arbitrary positive real number ε, to make
|f (x) + g(x) − (L + M)| < ε
ε
ε
we only need to make |f (x) − L| < and |g(x) − M| < .
2
2
These will be satisfied whenever x is close enough to a but
x , a since lim f (x) = L and lim g(x) = M.
x→a
Sometimes the notation {an } is also used.
Example
1 1 1 1
1, , , , , . . .
2 3 4 5
x→a
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
x→a
1
n
Example
1, −1, 1, −1, 1, −1, . . .
lim [f (x) + g(x)] = L + M = lim f (x) + lim g(x).
x→a
an =
x→a
an = (−1)n−1
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Limits of Sequences
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 =
Definition of limit of sequence
Example: an = (−1)n−1
1
n
A sequence (an ) has the limit L if an can be made arbitrarily
close to L by making n sufficiently large. We write
an
an
1
1
lim an = L
n→∞
1
−1
2
3
4
5
n
1
2
3
4
5
or an → L as n → ∞.
n
If the limit exists we say that the sequence converges.
Otherwise, we say that the sequence diverges.
−1
Note:
If it exists, L must be a unique finite real number.
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Theorem (Limit Laws):
Example 1.21: Determine whether
the following sequences
1
converge or diverge: (a)
(b) (−1)n−1 (c) (n)
n
Let (an ) and (bn ) be sequences of real numbers and c ∈ R a
constant.
If lim an and lim bn exist, then
n→∞
Solution:
n→∞
1. lim [an + bn ] = lim an + lim bn .
n→∞
n→∞
n→∞
2. lim [can ] = c lim an .
n→∞
n→∞
3. lim [an bn ] = lim an · lim bn .
n→∞
n→∞
4. lim
n→∞
n→∞
lim an
an
n→∞
=
bn
lim bn
provided lim bn , 0.
n→∞
n→∞
5. lim c = c.
n→∞
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The Factorial Function
Standard Limits
The factorial function n! (n = 0, 1, 2, ...) is defined by
n! = n(n − 1)! ,
or
(1) lim
n→∞
1
=0
np
1
0! = 1
(3) lim a n = 1
n→∞
n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1
(p > 0)
(2) lim rn = 0
(a > 0)
(4) lim n n = 1
an
= 0 (a ∈ R)
n→∞ n!
a n
(7) lim 1 +
= ea (a ∈ R)
n→∞
n
(5) lim
Therefore
1! = 1
2! = 2 × 1 = 2
3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
(9) lim arctan(cn) =
n→∞
Example
(2n + 2)! = (2n + 2) × (2n + 1) × (2n) × (2n − 1) × ... × 3 × 2 × 1
or
n→∞
(|r| < 1)
1
n→∞
(6) lim
log n
=0
np
(8) lim
np
=0
an
n→∞
n→∞
(p > 0)
(p ∈ R, a > 1)
π
(c > 0)
2
Note:
(2n + 2)! = (2n + 2) × (2n + 1) × (2n)!
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Standard limits (1), (3), (4), (6), (7), (8), (9) also hold for limits of
real-valued functions as x → ∞ by replacing n with x.
Standard limit (2) also holds for x → ∞ when 0 ≤ r < 1.
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"
Example 1.22: Evaluate lim
n→∞
n−2
n
n
#
4n2
+ n .
3
Example 1.23: Find the limit of the sequence
an =
Solution:
3n + 2
, n ≥ 1.
4n + 2n
Solution:
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Sandwich Theorem for sequences
Let (an ), (bn ) and (cn ) be sequences of real numbers.
Note:
If an ≤ cn ≤ bn for all n > N for some N, and
The order hierarchy can be used to help identify the largest
term in an expression:
lim an = lim bn = L
log n ≪ np ≪ an ≪ n!
then
n→∞
n→∞
lim cn = L.
n→∞
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1 + sin2
Example 1.24: Evaluate lim
√
n→∞
n
nπ
3
Continuity theorem for sequences
.
Solution:
Let f (x) be a real-valued function, (an ) a sequence of real
numbers and b ∈ R.
Theorem:
If lim an = b and f is continuous at x = b then
n→∞
lim f (an ) = f lim an = f (b).
n→∞
n→∞
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The only difference between lim an = L and lim f (x) = L is
h
i
Example 1.25: Evaluate lim log(3n2 + 2) − log n2 .
n→∞
n→∞
x→∞
that n is a natural number whereas x is a real number.
Solution:
Theorem:
Let f (x)) be a real-valued function and (an ) be a sequence of
real numbers such that an = f (n).
If
lim f (x) = L then
x→∞
lim an = L.
n→∞
This means that we can use the techniques for evaluating limits
of functions to evaluate limits of sequences.
Note:
lim an = L
n→∞
=⇒
̸
lim f (x) = L.
x→∞
eg. an = sin(2πn), f (x) = sin(2πx).
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Example 1.26: Prove Standard Limit 6:
lim
n→∞
log n
=0
np
(p > 0)
Solution:
Note:
We must change from a limit of a sequence (in terms of n) to a
limit of a real-valued function (in terms of x) before applying
L’Hôpital’s rule.
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n
1
Example 1.27: Find the sum S of an =
, n ≥ 1.
2
Solution:
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 sn = lim (a1 + a2 + . . . + an )
n→∞
n→∞
the sum of the an ’s.
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Series
A series with terms an is denoted by the sum
∞
X
an .
n=1
If the associated sequence of partial sums (sn ) converges then
∞
X
we say that the series
an converges. Otherwise we say that
n=1
the series diverges.
Example
The constant sequence with an = 2 is the sequence 2, 2, 2, 2, . . .
∞
∞
X
X
an =
2 represents the sum 2 + 2 + 2 + · · ·
The series
n=1
n=1
The sequence of terms converges, but the series diverges to
infinity.
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Application: Decimals
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Properties of Series
The decimal representation of a number is actually a series.
Let
Example
The sequence
1
10n
If
= 0.1, 0.01, 0.001, . . ..
∞
X
1
The series
= 0.1 + 0.01 + 0.001 + . . . = 0.11111111 . . .
10n
an and
n=1
∞
X
n=1
n=1
an and
n=1
2.
If
For a number x ∈ (0, 1) with decimal digits d1 , d2 , d3 , d4 , . . .
∞
X
dn
x = 0.d1 d2 d3 d4 · · · =
10n
bn be series, and c ∈ R\{0} a constant.
bn converge then
∞
X
∞
X
n=1
n=1
(an + bn ) =
an +
∞
X
bn converges.
n=1
∞
∞
X
X
(can ) = c
an converges.
n=1
In General
∞
X
n=1
∞
X
1.
1
The sequence converges to 0 while the series converges to .
9
∞
X
∞
X
n=1
an diverges then
n=1
∞
X
(can ) diverges.
n=1
Note:
n=1
These follow from the properties of the limits of sequences.
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Geometric Series
Example 1.28: What does the series
A geometric series has the form
∞
X
arn =
n=0
∞
X
∞ n
X
1
n=0
arn−1 = a + ar + ar2 + ar3 + . . .
2
=1+
2 3
1
1
1
+
+ ...
+
2
2
2
converge to?
n=1
where a ∈ R\{0} and r ∈ R.
Solution:
The series converges if |r| < 1 and diverges if |r| ≥ 1.
If |r| < 1, we have
∞
X
arn =
n=0
a
.
1−r
Note:
This follows from the fact that
n
X
ark =
k=0
a(1 − rn+1 )
for r , 1.
1−r
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Harmonic p Series
Divergence Test
If lim an , 0 then
A harmonic p series has the form
n→∞
∞
X
1
.
np
∞
X
an diverges.
n=1
Note:
n=1
lim an , 0 includes the case that the limit lim an does not exist.
n→∞
The series converges if p > 1 and diverges if p ≤ 1.
n→∞
If lim an = 0 then
n→∞
∞
X
Example
1.
an may converge or diverge.
n=1
∞
∞
X
X
1
1
converges BUT
diverges.
2
n
n
n=1
n=1
2. The Divergence Test is not applicable, so we need to use
∞
X
another test to determine if
an converges or diverges.
n=1
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Example 1.29: Does the series
∞
X
n+1
n=1
n
converge?
Solution:
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Comparison Test
Let
∞
X
Example 1.30: Does
n=1
an be a series with non-negative terms (i.e. an ≥ 0).
1. If
bn is another series such that
n=1
0 ≤ bn ≤ an for all n, then
∞
X
∞
X
3+
5
n
2n2 + n + 2
converge or diverge?
Solution:
n=1
∞
X
∞
X
bn diverges and
n=1
an also diverges.
n=1
2. If
∞
X
cn is another series such that
n=1
cn ≥ an for all n, then
∞
X
∞
X
cn converges and
n=1
an also converges.
n=1
To apply the comparison test we compare a given series to a
harmonic p series or geometric series.
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Example 1.31: Does
∞
X
n2 + 4
n=1
n3 + 5
converge or diverge?
Solution:
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Ratio Test
Let
∞
X
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Example 1.32: Does
∞
X
10n
n=1
converge or diverge?
Solution:
an be a positive term series (i.e. an > 0 for all n). Let
n=1
n!
an+1
.
n→∞ an
L = lim
1. If L < 1,
∞
X
an converges.
n=1
2. If L > 1,
∞
X
an diverges.
n=1
3. If L = 1, the ratio test is inconclusive.
The ratio test is useful if an contains an exponential or factorial
function of n.
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Example 1.33: Does
∞
X
(2n)!
n=1
n! n!
converge or diverge?
Solution:
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Section 2: Hyperbolic Functions
y = cosh(x)
y
4
Even Functions
y
y = x2
4
A function f is an even function if
We define the hyperbolic cosine
function:
3
3
2
f (−x) = f (x)
2
1 x
e + e−x , x ∈ R
cosh x =
2
1
Example
1
x
f (x) = cos x and f (x) = x2
−2
−1
1
y
x
2
−3
Odd Functions
1
2
3
Properties
4
A function f is an odd function if
−1
−1
y = x3
8
−2
x
f (−x) = −f (x)
−2
−1
1
2
−4
Example
f (x) = sin x and f (x) = x3
−8
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y
y
We define the hyperbolic
tangent function:
y = sinh(x)
4
3
We define the hyperbolic sine
function:
sinh x =
1 x
e − e−x ,
2
x∈R
tanh x =
2
=
1
x
−3
−2
−1
1
2
3
=
−1
2
sinh x
cosh x
1 x
−x
2 (e − e )
1 x
−x
2 (e + e )
ex − e−x
ex + e−x
,
y = tanh(x)
1
x
−3
−2
−1
1
2
3
−1
x ∈ R.
−2
−2
Properties
Properties
−3
−4
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Why call them hyperbolic functions?
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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.
Let x = cosh t and y = sinh t then
y
x2-y2=1
t>0
4
t
x=cosh t
y=sinh t
2
4
2
-4
-2
-2
-4
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x
t=0
t<0
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Application: Catenary
Example 2.1: If cosh x =
13
12
and x < 0 find sinh x and tanh x.
Solution:
A flexible, heavy cable of uniform mass per length ρ and
tension T at its lowest point has shape
ρgx T
y=
cosh
ρg
T
where g is the acceleration due to gravity.
y
Support
Support
x
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Addition Formulae
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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
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Double Angle Formulae
Example 2.2: Prove the sinh(x + y) addition formula.
Solution:
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.
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Reciprocal Hyperbolic Functions
Reciprocal Hyperbolic Functions
We define the three reciprocal hyperbolic functions:
coth x =
sech x =
1
, x∈R
cosh x
cosech x =
1
, x ∈ R \ {0}
sinh x
cosh x
1
=
,
tanh x
sinh x
x ∈ R \ {0}
y
y
y
3
1.0
2
0.5
1
2
y = sech(x)
x
−3
−2
−1
1
−0.5
y = coth(x)
1
2
3
−3
−2
−1
1
2
y = cosech(x)
x
3
x
−3
−2
−1
1
2
3
−1
−1
−2
−2
−3
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Basic Identities
Derivatives of Hyperbolic Functions
cosh2 x − sinh2 x
=
1
coth2 x − 1
=
cosech2 x
1 − tanh2 x
=
sech2 x
d
(cosh x)
dx
=
sinh x,
x∈R
d
(sinh x)
dx
=
cosh x,
x∈R
d
(tanh x)
dx
=
sech2 x,
x∈R
d
(sech x)
dx
=
− sech x tanh x,
d
(cosech x)
dx
=
− cosech x coth x,
d
(coth x)
dx
=
− cosech2 x,
x∈R
x ∈ R \ {0}
x ∈ R \ {0}
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Example 2.3: Prove that
d (cosh x)
= sinh x.
dx
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Example 2.4: Let y =
Solution:
p
sinh(6x), x > 0. Find
dy
.
dx
Solution:
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Inverses of Hyperbolic Functions
2. Inverse hyperbolic cosine function: arccosh x
We define three inverse hyperbolic functions.
Restrict domain of cosh x to be [0, ∞) to give a 1-1 function.
Then
1. Inverse hyperbolic sine function: arcsinh x
Since sinh x is a 1-1 function
domain arcsinh x =
domain arccosh x =
range sinh x = R.
range arcsinh x =
range arccosh x =
domain sinh x = R.
arcsinh(sinh x) = x,
x ∈ R.
sinh(arcsinh x) = x,
x ∈ R.
x ≥ 1.
arccosh(cosh x) = x,
x ≥ 0.
y
4
3
y = arcsinh(x)
2
−2
y = arccosh(x)
2
1
−3
y = cosh(x)
5
3
−4
restricted domain cosh x = [0, ∞).
cosh(arccosh x) = x,
y = sinh(x)
y
range cosh x = [1, ∞).
−1
2
1
3
4
x
1
−1
2
1
−2
3
4
5
x
−3
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3. Inverse hyperbolic tangent function: arctanh x
The inverse hyperbolic functions can be expressed in terms of
natural logarithms.
Since tanh x is a 1-1 function
domain arctanh x =
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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.
y
3
arcsinh x
=
√
log x + x2 + 1 ,
x∈R
arccosh x
=
√
log x + x2 − 1 ,
x≥1
arctanh x
=
1
1+x
log
,
2
1−x
−1 < x < 1
y = arctanh(x)
2
y = tanh(x)
1
−4
−3
−2
We can also define inverse reciprocal hyperbolic functions:
• arcsech x
(0 < x ≤ 1)
−1
• arccosech x
(x , 0)
−2
• arccoth x
(x < −1 or x > 1)
−1
1
2
3
4
x
−3
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Example 2.5: Proof of arcsinh x relation.
Solution:
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Derivatives
Example 2.6: Simplify cosh(arcsinh x) for x ∈ R.
Solution:
d
(arcsinh x)
dx
=
1
√
2
x +1
d
(arccosh x)
dx
=
√
d
(arctanh x)
dx
=
1
x2
1
1 − x2
−1
(x ∈ R)
(x > 1)
(−1 < x < 1)
Each formula is derived using implicit differentiation or by
differentiating the logarithm definition of each function.
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Example 2.7: Prove that
d
1
(arcsinh x) = √
.
dx
x2 + 1
Solution:
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Example 2.8: Find
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Section 3: Complex Numbers
d
(arctanh(2x) cosh(3x)).
dx
The Cartesian form of a complex number z ∈ C is
Solution:
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.
ImHzL
z=x+iy
r
y
Θ
x
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ReHzL
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The Complex Exponential
The complex number can be written as
z = r(cos θ + i sin θ)
where
q
• r = |z| = x2 + y2
y
• tan θ =
x
Note:
The angle θ is not unique – only defined up to multiples of 2π.
We choose θ such that −π < θ ≤ π and call this angle the
principal argument of z.
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The Complex Exponential
Example 3.2: Write z = −1 + i in polar form.
We define the complex exponential using Euler’s formula
ImHzL
eiθ = cos θ + i sin θ
-1+i
for θ ∈ R.
2
1
Α
We can then write the polar form of a complex number as
1
z = reiθ
Θ
ReHzL
Solution:
Example 3.1: Simplify eiπ + 1.
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Properties of the Complex Exponential
Products and Division in Polar Form
1. ei0 = 1
Proof:
ei0 = cos 0 + i sin 0 = 1.
If z = r1 eiθ and w = r2 eiϕ then
2. eiθ eiϕ = ei(θ+ϕ)
Proof:
eiθ eiϕ = (cos θ + i sin θ) cos ϕ + i sin ϕ
= cos θ cos ϕ + i cos θ sin ϕ + i sin θ cos ϕ − sin θ sin ϕ
= cos θ cos ϕ − sin θ sin ϕ + i cos θ sin ϕ + sin θ cos ϕ
= cos θ + ϕ + i sin(θ + ϕ)
zw
=
r1 r2 ei(θ+ϕ)
z
w
=
r1 i(θ−ϕ)
e
r2
= ei(θ+ϕ) .
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Solution:
Example 3.3: Using the complex
exponential, simplify
√
√
√
3−i
( 3 − i)(1 + 3i) and
√ .
1 + 3i
ImHzL
1+ 3 i
2
3
Α2
1
Α1
2
ReHzL
3
1
3 -i
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De Moivre’s Theorem:
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√ 15
Example 3.4: Evaluate 1 + 3i .
Solution:
If z = reiθ and n is a positive integer then
n
zn = reiθ = rn einθ
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Exponential Form of sin θ and cos θ
Equation (1) − (2) gives
Now eiθ = cos θ + i sin θ
(1)
eiθ − e−iθ = 2i sin θ
⇒ e−iθ = cos(−θ) + i sin(−θ)
⇒ e−iθ = cos θ − i sin θ
⇒ sin θ =
1 iθ
e − e−iθ
2i
(2)
Note:
Equation (1) + (2) gives
These formulae give a connection between the hyperbolic and
trigonometric functions.
1 iθ
e + e−iθ = cos θ
cosh(iθ) =
2
eiθ + e−iθ = 2 cos θ
⇒ cos θ =
1 iθ
e + e−iθ
2
sinh(iθ) =
1 iθ
e − e−iθ = i sin θ
2
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Differentiation via the Complex Exponential
Example 3.5: Express sin5 θ in terms of the functions
sin(nθ) for integers n.
If z = x + yi where x, y ∈ R then we define
ez = ex+iy = ex eiy = ex (cos y + i sin y).
Solution:
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 kt e = kekt when k = a + bi ∈ C.
dt
d h (a+bi)t i
d h at ibt i
e
=
e e
dt
dt
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=
i
d h at
e (cos(bt) + i sin(bt))
dt
= aeat [cos(bt) + i sin(bt)] + eat [−b sin(bt) + bi cos(bt)]
h
i
= 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)eat eibt
= (a + bi)e(a+ib)t .
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d56 −t
Example 3.6: Find 56 e cos t .
dt
Solution:
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Note:
Example 3.6 also gives the answer to
d56 −t
e
sin
t
.
dt56
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Integration via the Complex Exponential
Z
Example 3.7: Evaluate
e3x sin(2x) dx.
Solution:
d kx Since
e = k ekx if k = a + bi (a, b ∈ R) , then
dx
Z
k ekx dx = ekx + C
Z
⇒
ekx dx =
1 kx
e +D
k
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Note:
Z
e3x cos(2x) dx.
Example 3.7 also gives the answer to
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Section 4: Integral Calculus
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Derivative Substitutions
Review of integration
To evaluate
Z
f [g(x)]g′ (x)dx
put u = g(x) ⇒
du
= g′ (x).
dx
Then
Z
Z
f [g(x)]g (x)dx =
′
f (u)
du
dx
dx
Z
=
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f (u) du
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Z
Example 4.1: Evaluate
Z
(6x + 10) sinh(x + 5x − 2)dx.
2
3
Example 4.2: Evaluate
Solution:
sech2 (3x)
dx.
10 + 2 tanh(3x)
Solution:
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Integration by Parts
Z
Example 4.3: Evaluate
xe5x dx.
The product rule for differentiation is
Solution:
d
du
dv
(uv) =
v+u
dx
dx
dx
Integrate
Z
d
(uv) dx =
dx
Z
⇒ uv =
Z
⇒
Z
!
dv
du
v+u
dx
dx
dx
du
v dx +
dx
Z
dv
u dx = uv −
dx
u
dv
dx
dx
Z
v
du
dx
dx
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Z
Example 4.4: Evaluate
Z
2
x log x dx
(x > 0).
log x dx
Example 4.5: Evaluate
Solution:
(x > 0).
Solution:
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Z
Example 4.6: Evaluate
e3x sin(2x) dx.
Solution:
Note:
This technique can also be used to integrate inverse
trigonometric functions and inverse hyperbolic functions.
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Trigonometric and Hyperbolic Substitutions
Integrand
√
a2 − x2 ,
√
a2 + x2 ,
√
We can use trigonometric and hyperbolic substitutions to
integrate expressions containing
√
a2 − x2 ,
√
√
a2 + x2 ,
√
x2 − a2 ,
1
a2 + x2
32
,
,
a2 + x2
− 32
,
52
a2 − x2
1
Substitution
a2 − x2
etc.
x = a sin θ
or
x = a cos θ
etc.
x = a sinh θ
etc.
x = a cosh θ
where a is a positive real number.
√
Method:
Z
Put x = g(θ).
Then
Z
f (x) dx =
f [g(θ)]g′ (θ) dθ
x2 − a2 ,
1
√
x2 − a2
1
,
a2 + x2
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x2 − a2
1
(a2
2
+ x2 )
etc.
x = a tan θ
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Z
Example 4.7: Evaluate
1
√
x2 − 25
dx using a substitution.
Solution:
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Z √
Example 4.8: Evaluate
9 − 4x2 dx if |x| ≤ 32 .
Solution:
Note:
Note the identity
√
x2 = |x|
(x ∈ R)
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Z 32
Example 4.9: Evaluate
x2 + 1 dx .
Solution:
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Powers of Hyperbolic Functions
Consider the integral:
Z
sinhm x coshn x dx
where m, n are integers (≥ 0).
• If m or n is odd, use a “derivative substitution” after rewriting
one of the odd power terms using identities if necessary.
• If m and n are even, use double angle formulae.
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Z
Example 4.10: Evaluate
cosh4 θ dθ.
Solution:
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Z
Finish Example 4.9:
Example 4.11: Evaluate
sinh5 x cosh6 x dx.
Solution:
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Z
Example 4.12: Evaluate I =
sinh5 x cosh7 x dx.
Solution:
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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 .
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Denominator Factor
Partial Fraction Expansion
(x − a)
A
x−a
(x − a)r
A1
A2
Ar
+
+ ··· +
x − a (x − a)2
(x − a)r
(x2 + bx + c)
Ax + B
x2 + bx + c
(x2 + bx + c)r
A1 x+B1
x2 +bx+c
A2 x+B2
(x2 +bx+c)2
+
+ ··· +
Ar x+Br
(x2 +bx+c)r
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Z
Example 4.13: Evaluate
4
dx
+ 2)
x2 (x
(x , 0, −2).
Solution:
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Z
Example 4.14: Evaluate
(x2
4x
dx (x , 2).
+ 4)(x − 2)
Solution:
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Case 2: n ≥ d
Use long division, then apply case 1.
Example 4.15: Find
Z
5x4 + 13x3 + 6x2 + 4
dx
x3 + 2x2
Solution:
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(x , 0, −2).
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Section 5: First Order Differential Equations
A solution of an ODE is a function y(x) that satisfies the ODE
for all x in some interval.
Ordinary Differential Equations
Example 5.2: Verify by substitution that y(x) = x2 +
An ordinary differential equation (ODE) is an equation of the
form
!
dy d2 y
dn y
f x, y, , 2 , . . . , n = 0
dx dx
dx
solution of
dy y
+ = 3x for all x ∈ R \ {0}.
dx x
2
is a
x
Solution:
The order of an ODE is the order of the highest derivative
occurring in the ODE.
dy
d4 y
Example 5.1: What order is 3 4 =
dx
dx
!2
+ 2x2 y?
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First Order ODEs
Sketch of the family of solutions of
dy
The general form of a first order ODE is
= f (x, y).
dx
Example 5.3: Solve
dy
= x3
dx
y
dy
= x3 .
dx
6
Solution:
4
2
c=4
c=2
c=0
-4
A solution like this which contains an arbitrary constant c ∈ R is
called a general solution.
-2
2
-2
x
c=-2
-4
The general solution represents a family of solutions, where
each value of c corresponds to a different solution of the ODE.
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Initial value problem for a first order ODE
Solve
Separable ODEs
A separable first order ODE has the form:
dy
= f (x, y) subject to the condition y(x0 ) = y0 .
dx
dy
= M(x)N(y),
dx
dy
Example 5.4: Solve
= x3 given y(0) = 2.
dx
To solve, use separation of variables
dy
dx
Solution:
1 dy
N(y) dx
⇒
Z
Z
⇒
dy
y
=
dx 1 + x
= M(x)N(y)
= M(x)
1 dy
dx =
N(y) dx
⇒
Example 5.5: Solve
(M(x) , 0, N(y) , 0)
1
dy
N(y)
(N(y) , 0)
Z
M(x)dx
Z
=
M(x) dx
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(x , −1).
Solution:
Family of solutions
y
c=2
6
c=1
4
H-1,0L
-6
-4
2
-2
2
x
4
-2
-4
c=-1
-6
Example 5.6: Solve
dy −x
=
dx
y
c=-2
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(y , 0)
if y(0) = 3.
Solution:
Family of solutions
Linear First Order ODEs
y
5
Example 5.7: Solve x
4
Solution:
d = 16
3
2
1
dy
+ y = ex .
dx
d=9
d=4
d=1
x
-4
-3
-2
-1
0
1
2
3
4
5
-1 d = 1
-2
-3
d= 4
d=9
d = 16
-4
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A linear first order ODE has the form:
Aim:
dy
+ P(x)y = Q(x)
dx
Find an integrating factor I so the left side will be the derivative
of yI. Then
To solve:
dy
d
yI ≡ I + PIy
dx
dx
dy
dy
dI
⇒
I+y
= I + PIy
dx
dx
dx
Multiply ODE by I(x)
I(x)
dy
+ P(x)I(x)y = Q(x)I(x)
dx
If the left side can be written as the derivative of y(x)I(x), then
⇒ y
dI
= PIy
dx
⇒
dI
= PI
dx
To solve for all y
d y(x)I(x) = Q(x)I(x)
dx
which can be solved by integrating with respect to x.
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(separable)
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⇒
Z
⇒
1 dI
I dx
= P
1
dI =
I
Z
So one integrating factor is
Pdx
I(x) = e
Z
⇒ log |I| =
P dx + c
⇒ |I| = e
R
= e
R
⇒I =
R
Pdx
Note:
Pdx+c
Pdx
Since we only need one integrating factor I, we can neglect
the ‘+c’ and modulus signs when calculating I.
· ec
±ec ·e
|{z}
R
Pdx
constant
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Example 5.8: Find the general solution of
dy y
+ = sin x
dx x
(x , 0).
Solution:
Example 5.9: Solve
1 dy
− xy = x
2 dx
if y(0) = −3.
Solution:
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Note:
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Other First Order ODEs
Example 5.10: Solve the homogeneous type differential
equation
y
dy y
y
= + cos2
− π2 < x < π2
dx x
x
Sometimes it is possible to make a substitution to reduce a
general first order ODE to a separable or linear ODE.
y
by substituting u = .
x
• A homogeneous type ODE has the form
y
dy
=f
dx
x
Solution:
y
Substituting u = reduces the ODE to a separable ODE.
x
• Bernoulli’s equation has the form
Substituting u = y1−n
dy
+ P(x)y = Q(x)yn
dx
reduces the ODE to a linear ODE.
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Example 5.11: Solve the Bernoulli equation
dy
+ y = e3x y4
dx
(y , 0)
by substituting u = y−3 .
Solution:
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Population Models
Malthus (Doomsday) model
Note:
The Doomsday model predicts:
Rate of growth is proportional to the population p at time t.
dp
dt
dp
⇒
dt
• k > 0 : unbounded exponential growth
∝ p
= kp
• k < 0 : population dies out
(separable/linear)
where k is a constant of proportionality representing net births
per unit population per unit time.
• k = 0 : population stays constant
If the initial population is p(0) = p0 , then the solution is
Unbounded exponential growth is unrealistic in the long term.
p(t) = p0 ekt
You should check that you can derive this on your own!
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Equilibrium Solutions
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Example 5.12: Find all equilibrium solutions of the ODE
Definition
An equilibrium is a constant solution of an ODE.
Solution:
dx
= 3x − 2.
dt
Note:
For the ODE
dx
= f (x, t), this means
dt
▶ x(t) = C where C is a constant
▶
dx
=0
dt
Terminology
The plural form of equilibrium is equilibria.
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Phase plots
A phase plot is a plot of
Example 5.13: For the ODE
dx
= 3x − 2,
dt
dx
as a function of x.
dt
A phase plot will give
▶ the equilibria
(a) Draw a phase plot
(b) Sketch the family of solutions of the ODE, including
any equilibria.
▶ the behaviour of solutions close to the equilibria
(c) Describe the long-term behaviour of solutions with
initial conditions
1
i. x(0) =
2
Note:
Phase plots are only useful for ODEs of the form
dx
= f (x)
dt
ii. x(0) = 1
i.e., when the right-hand side has no explicit dependence on t.
ODEs of this form are called autonomous.
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(b).
Solution:
(a). The phase plot is
dx
dt
Family of solutions:
x
dx >0
dt
23
-2
1-
x
2/3-
dx <0
dt
equilibrium
x(t) = 2/3
1/2-
t
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(c) From the phase plot and family of solutions, we can see that
Stability of equilibria
An equilibrium is stable if solutions that start nearby move
closer to the equilibrium as t increases.
x
Note:
x0
Stable
We do not need to solve the ODE analytically to determine the
qualitative behaviour of the solutions.
t
When drawing a family of solutions, we include at least one
example of a solution with each possible qualitative behaviour
(unless stated otherwise in the question).
On a phase plot:
Two solutions have the same qualitative behaviour if their
shape and long-term behaviour are the same. Otherwise they
are qualitatively different.
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Stability of equilibria
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Stability of equilibria
An equilibrium is semistable if on one side of the equilibrium,
solutions that start nearby move closer as t increases, whereas
on the other side, solutions move further away as t increases.
An equilibrium is unstable if solutions that start nearby move
further away as t increases.
x
x
x
x0
Unstable
x0
Semistable
x0
t
t
Semistable
t
On a phase plot:
On a phase plot:
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Example 5.13 (continued):
(d) Determine the stability of the equilibrium.
Solution:
Doomsday model with harvesting.
Remove some of the population at a constant rate.
dp
= kp − h, h > 0.
dt
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Solution:
From Example 5.13 we know
Example 5.14: A pharmaceutical company grows
engineered yeast to produce a drug. The yeast is
continuously harvested to collect the drug.
The population p (in millions of yeast cells) at time t days
is described by
dp
= 3p − 2
dt
•
the equilibrium is
•
the phase plot is
dp
dt
dp
>0
dt
(p ≥ 0, t ≥ 0)
23
For what initial population sizes p(0) will the yeast
population eventually die out?
-2
p
dp
<0
dt
dies out!
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•
Logistic model.
the family of solutions is
p
Include “competition” term in Malthus’ model since
overcrowding, disease, lack of food and natural resources will
2/3-
Unstable
cause more deaths.
dp
k
= kp − p2 =
dt
a
t
dies out!
↗
The population will die out if
p
kp 1 −
a
↖
z }| {
z }| {
net birth rate
competition term
where a > 0 is the carrying capacity.
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• Phase plot
Example 5.15: Find the equilibrium solutions and their
stability for the ODE
dp
p
=p 1−
dt
4
dp
dt
(k = 1, a = 4)
1
Solution:
dp
>0
dt
• Equilibrium solutions
0
2
4
p
dp
<0
dt
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• Family of solutions:
p
Note:
The logistic model accurately describes
pH0L=4
• population in a limited space (e.g. bacteria culture).
• population of USA from 1790-1950.
pH0L=0
t
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Solution:
Logistic model with harvesting.
Remove some of the population at constant rate:
p
dp
= kp 1 −
− h, h > 0, a > 0
dt
a
Example 5.16: For the logistic model with harvesting
dp
p 3
3
= p 1−
−
a = 4, k = 1, h =
dt
4
4
4
determine the long-term consequences for the population
predicted by the model.
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•
•
Stability and family of solutions:
Phase plot:
1dp
dt
dp
>0
dt
1
4
1
3
4
dp
<0
dt
3
p
p
dp
<0
dt
pH0L=3
pH0L=1
t
dies out!
dies out!
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•
Interpretation:
Example 5.16 (continued): Find the time taken until the
1
population dies out if p(0) = .
2
Solve analytically for p(t) and set p(t) = 0.
dp
1
(separable - use
= − (p − 3)(p − 1)
separation
of variables)
dt
4
Note:
Case (2) is unrealistic in practice, as even a small deviation
from p(0) = 1 will result in the solution following case (1) or (3).
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Mixing Problems
Example 5.17: 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,
and interpret the long-term behaviour of the system.
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Derive the ODE:
Let x be the amount (grams) of pollutant in pond at time t
x
minutes. Then C =
is the concentration of pollutant in pond
V
(grams/m3 ), where V is the volume of the pond (m3 ) at time t.
dx
= rate pollutant flows in - rate pollutant flows out
dt
Solve analytically:
Interpret:
x
dx
+
= 20
dt 100
(Linear/Separable)
General solution is
conc. Hgm3L
−t
x(t) = 2000 + ce 100
(working omitted).
Initial condition is x(0) = 100
⇒
2
c = −1900
−t
⇒ x(t) = 2000 − 1900e 100
Find concentration:
0.1
tHminL
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Example 5.18: Derive an ODE describing the amount x
(g) of pollutant in the lake at time t (minutes), if the input
flow rate is decreased to 5m3 /min.
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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Section 6: Second Order Differential Equations
Let V be the volume in pond (m3 ) at time t minutes.
A second order ODE has the form
!
dy d2 y
F x, y, , 2 = 0
dx dx
The general form of a linear second order ODE is
dy
d2 y
+ P(x) + Q(x)y = R(x)
2
dx
dx
• If R(x) = 0, the ODE is homogeneous (H).
• If R(x) , 0, the ODE is inhomogeneous (IH).
Note:
A homogeneous linear ODE is different to a homogeneous type
first order ODE.
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Homogeneous 2nd Order Linear ODEs
The general solution of a second order ODE typically has two
arbitrary constants.
Theorem:
Initial value problem for a second order ODE
The general solution of
Solve
d2 y
dy
+ P(x) + Q(x)y = R(x)
2
dx
dx
y” + P(x)y′ + Q(x)y = 0
is the function y given by
subject to the conditions y(x0 ) = y0 and y′ (x0 ) = y1 .
y(x) = c1 y1 (x) + c2 y2 (x)
where
Boundary value problem for a second order ODE
Solve
• y1 , y2 are two linearly independent solutions of the
homogeneous ODE,
d2 y
dy
+ P(x) + Q(x)y = R(x)
2
dx
dx
• c1 , c2 ∈ R are arbitrary constants.
subject to the conditions y(a) = y0 and y(b) = y1 .
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Aside - A connection with Linear Algebra*
Definition:
Why are these ODEs called linear?
Two functions y1 and y2 are linearly independent if
Because they can be written in terms of a linear transformation.
The ODE
c1 y1 (x) + c2 y2 (x) = 0 ⇒ c1 = c2 = 0
d2 y
dy
+ P(x) + Q(x)y = R(x)
2
dx
dx
or equivalently, if neither function is a non-zero constant
multiple of the other function.
Example 6.1:
(a) Are y1 (x) = x2 , y2 (x) = 2x2 linearly independent?
can be written as
T(y) = R(x)
where
T(y) =
(b) Are y1 (x) = e2x , y2 (x) = xe2x linearly independent?
d2 y
dy
+ P(x) + Q(x)y
2
dx
dx
is a linear transformation on a vector space of functions.
These concepts are covered in MAST10007 Linear Algebra.
* This slide is not examinable in MAST10006.
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Homogeneous 2nd Order Linear ODEs with Constant Coefficients
General form:
ay” + by′ + cy = 0
where a, b, c are constants.
To solve for y(x):
Try y(x) = eλx .
Then y′ (x) = λeλx and y′′ (x) = λ2 eλx .
Substitute into the ODE:
ay′′ + by′ + cy = aλ2 eλx + bλeλx + ceλx = 0
aλ2 + bλ + c eλx = 0
|{z}
,0
⇒ aλ2 + bλ + c = 0
Characteristic Equation
√
−b ± b2 − 4ac
⇒λ=
2a
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Example 6.2: Solve y” + 7y′ + 12y = 0 for y(x).
Solution:
Case 1: b2 − 4ac > 0
• 2 distinct real values λ1 , λ2
• 2 linearly independent solutions
eλ1 x ,
eλ2 x
• General Solution:
y(x) = Aeλ1 x + Beλ2 x
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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
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We now verify that xeλx is a solution:
Example 6.3: Solve y” + 2y′ + y = 0 for y(x).
If y(x) = xeλx , then
Solution:
λx
y (x) = (λx + 1) e ,
′
y”(x) =
λ2 x + 2λ eλx .
So ay” + by′ + cy
= a λ2 x + 2λ eλx + b (λx + 1) eλx + cxeλx
= xeλx aλ2 + bλ + c + (2λa + b) eλx
|
{z
} | {z }
= 0
=0
=0
So y(x) = xeλx is a 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) = C1 e(α+iβ)x + C2 e(α−iβ)x
where C1 , C2 ∈ C
= C1 eαx cos(βx) + i sin(βx) + C2 eαx cos(βx) − i sin(βx)
= (C1 + C2 ) eαx cos(βx) + (C1 i − C2 i) eαx sin(βx)
| {z }
| {z }
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A
B
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Example 6.4: Solve y” − 4y′ + 13y = 0 for y(x) if y(0) = 1
and y′ (0) = 6.
Put A = C1 + C2 and B = (C1 − C2 )i. If C1 = C2 , then A, B ∈ R.
Solution:
Note:
Imposing real conditions on the ODE 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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This solution has the form y(x) = e2x (a cos θ + b sin θ).
Hence we can rewrite the solution from Example 6.4 as
In general, for a, b > 0,
!
√
a
b
2
2
a cos θ + b sin θ = a + b √
cos θ + √
sin θ
a2 + b2
a2 + b2
=
√
a2 + b2 cos ϕ cos θ + sin ϕ sin θ
=
√
a2 + b2 cos(θ − ϕ).
This form is sometimes preferable for graphing or further
manipulation.
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Inhomogeneous 2nd Order Linear ODEs with Constant
Coefficients
Inhomogeneous 2nd Order Linear ODEs
Theorem:
General form:
The general solution of
ay” + by′ + cy = R(x)
where a, b, c are constants.
y” + P(x)y′ + Q(x)y = R(x)
is the function y given by
Example 6.5: Solve y” + 2y′ − 8y = R(x) where
y(x) = yH (x) + yP (x)
where
(a) R(x) = 1 − 8x2
• yH (x) = c1 y1 (x) + c2 y2 (x) is the general solution of the
homogeneous ODE (called the homogeneous solution, GS(H)),
(b) R(x) = e3x
• yP (x) is a solution of the inhomogeneous ODE (called a
particular solution, PS(IH)),
(c) R(x) = 85 cos x
(d) R(x) = 3 − 24x2 + 7e3x .
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Solution:
Step 2: Find a particular solution of y” + 2y′ − 8y = R(x).
Step 1: Find the general solution of y” + 2y′ − 8y = 0.
(a) R(x) = 1 − 8x2 :
y” + 2y′ − 8y = 1 − 8x2
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(b) R(x) = e3x :
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y” + 2y′ − 8y = e3x
e3x is NOT part of GS(H)
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(c) R(x) = 85 cos x :
y” + 2y′ − 8y = 85 cos x
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Superposition of Particular Solutions
Theorem:
A particular solution of
ay” + by′ + cy = α R1 (x) + β R2 (x)
is
yP (x) = αy1 (x) + βy2 (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, α, β are constants.
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Example 6.6: Solve y” − y = ex .
Example 6.5 (d): R(x) = 3 − 24x2 + 7e3x .
Solution:
Solution:
GS(H) : yH (x) = Aex + Be−x
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Example 6.7: Solve y” + 2y′ + y = e−x .
Solution:
GS(H) : yH (x) = (A + Bx) e−x
Example 6.8: Solve y” + 49y = 28 sin(7t).
Solution:
GS(H) : yH (t) = A cos(7t) + B sin(7t)
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The forces are:
Springs - Free Vibrations
Consider an object (of mass m kg) attached to a spring hanging
from a fixed support.
• gravitational force, given by the gravitational law:
W = mg
Suppose that the natural length of the spring when unloaded is
L m, and that when the object is attached, it causes the spring
to stretch by s m downwards at equilibrium.
(g = 9.8m/s2 on Earth)
• restoring force in spring, given by Hooke’s Law:
T = −k · extension
(k > 0)
0
spring constant
Note:
The negative sign in Hooke’s law is because the direction of the
force T is opposite to the direction of extension:
▶ if the spring is stretched downwards, then it pulls up
▶ if the spring is compressed upwards, then it pushes down.
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Suppose the object is set in motion.
Let y(t) be the displacement of the object below the equilibrium
position (y = 0) at time t seconds.
At equilibrium, the net force is zero, so
Note:
Since displacement y is measured below the equilibrium
position, downwards is the positive direction.
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Derive the equation of motion
Newton’s Law states that
When the object is moving, there is an extra force acting:
net force = mass · acceleration
• damping force is proportional to velocity
R = −βẏ
Fnet =
m
·
ÿ
(β ≥ 0)
Apply Newton’s Law to the object:
0
damping constant
Note:
The negative sign is because the direction of the damping force
is opposite the direction of motion.
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To solve, try y(t) = eλt
⇒ mλ2 + βλ + k = 0
⇒ λ=
• If β = 0 :
λ = ±ib
√
• If 0 < β < 2 mk :
−β ±
β2 − 4mk
2m
p
simple harmonic motion
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
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Example 6.9: A 40
kg mass stretches a spring hanging
49
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)
160
49
(c)
80
7
(d)
2000
49
Solution:
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(a) β = 0 :
ÿ + 49y = 0
y
3
7
Equil.
t
Π
7
-
2Π
7
3Π
7
3
7
Simple harmonic motion
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(b) β =
160
49
:
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ÿ + 4ẏ + 49y = 0
y
1
5
Equil.
t
1
-
1
2
Weak damping
5
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(c) β =
80
7
ÿ + 14ẏ + 49y = 0
:
y
0.15
0.1
critical damping
0.05
Equil.
t
0.5
1
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(d) β =
2000
49
:
ÿ + 50ẏ + 49y = 0
y
0.06
0.03
strong damping
Equil.
t
2
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4
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Springs - Forced Vibrations
Example 6.10: Apply an external downwards force
f (t) =
If an external downwards force f is applied to the spring-mass
system at time t, the forces acting on the mass are:
160
7
sin(7t) in Example 6.9.
Solution:
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(a) β =
GS(IH) :
80
7
:
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ÿ + 14ẏ + 49y = 28 sin(7t)
y(t) = (A + Bt) e−7t − 72 cos(7t)
y
0.4
GSHHL contributes
Equil.
1
-0.4
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2
3
t
4
oscillatory motion due to PSHIHL
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(b)
β=0:
GS(IH) :
ÿ + 49y = 28 sin(7t)
y
10
y(t) = A cos(7t) + B sin(7t) − 2t cos(7t)
5
Equil.
2
4
6
t
-5
Resonance
-10
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
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RLC series electric circuit
The circuit components are
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
Let q(t) be the charge on the capacitor (measured in Coulomb)
at time t seconds.
The charge satisfies the second-order ODE
L
d2 q
dq q
+R + =V
2
dt C
dt
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.
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Section 7: Functions of Two Variables
We can represent the function f by its graph in R3 . The graph
of f is:
n
o
(x, y, z) ∈ R3 : (x, y) ∈ D and z = f (x, y) .
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).
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.
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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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: Sketch the plane 4x + 3y + z = 2.
Solution:
Level Curves
Sketching Functions of Two Variables
The key steps in drawing a graph of a function of two variables
z = f (x, y) are:
A curve on the surface z = f (x, y) for which z is a constant is a
contour.
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.
The same curve drawn in the xy plane is a level curve.
So a level curve of f has the form
2. Draw the y − z cross section.
{(x, y) : f (x, y) = c}
3. Draw some level curves and their contours.
4. Draw the x − z cross section.
where c ∈ R is a constant.
5. Label any x, y, z intercepts and key points.
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Example 7.2: Find the level curves,
and cross sections in
p
2
the x-z and y-z planes of z = 1 − x − y2 . Hence sketch
the surface and identify it.
Solution:
y
H0,1L
H-1,0L
c=1
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c=0
Level curves
H1,0L
x
H0,-1L
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Consider cross sections (slices) to help sketch graph.
z
z
H0,1L
H-1,0L
Surface is a hemisphere radius 1, centre at (0, 0, 0) for z ≥ 0.
H0,1L
H1,0L
H-1,0L
x
H1,0L
y
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Example 7.3: Sketch the graph of z = 4x2 + y2 .
Consider cross sections (slices) to help sketch graph.
Solution:
y
H0,2L
c=4
z
4
Level curves
H-1,0L
H1,0L
z
4
z=4x2
x
z=y2
c=0
-1
1
x
-1
1
y
H0,-2L
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Example 7.4: Sketch the graph of z =
The surface is an elliptic paraboloid (parabolic bowl).
p
4x2 + y2 .
Solution:
y
H0,2L
c=2
Level curves
H-1,0L
H1,0L
x
c=0
H0,-2L
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Cross sections
The surface is an elliptic cone.
z
z
2
-1
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2
x
1
z=2ÈxÈ
-1
y
1
z=ÈyÈ
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Limits
Continuity
Let f : D → R be a real-valued function, where D ⊆ R2 .
We say f has the limit L as (x, y) approaches (x0 , y0 )
lim
(x,y)→(x0 ,y0 )
Let f : R2 → R be a real-valued function.
f (x, y) = L
f is continuous at (x, y) = (x0 , y0 ) if
if when (x, y) approaches (x0 , y0 ) along ANY path in the domain,
f (x, y) gets arbitrarily close to L.
lim
(x,y)→(x0 ,y0 )
f (x, y) = f (x0 , y0 )
Note:
Note:
The continuity theorems for functions of one variable can be
generalised to functions of two variables.
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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Example 7.5: Let f (x, y) = x2 + y2 . For which values of x
and y is f continuous?
Example 7.6: Evaluate
lim
(x,y)→(2,1)
log(1 + 2x2 + 3y2 ).
Solution:
Solution:
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First Order Partial Derivatives
Geometric Interpretation of
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
f (x + h, y) − f (x, y)
= lim
h
∂x
h→0
fy =
∂f
f (x, y + h) − f (x, y)
= lim
h
∂y
h→0
∂f
∂f
and
∂x
∂y
Note:
•
•
∂f
measures the rate of change of f with respect to x
∂x
when y is held constant.
∂f
measures the rate of change of f with respect to y
∂y
when x is held constant.
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Example 7.7: Let f (x, y) = xy2 . Find
∂f
from first
∂y
principles.
Let C1 be the curve where the vertical plane y = y0 intersects
∂f
the surface. Then
gives the slope of the tangent to C1
∂x (x0 ,y0 )
at (x0 , y0 , z0 ).
Solution:
Let C2 be the curve where the vertical plane x = x0 intersects
∂f
the surface. The
gives the slope of the tangent to C2 at
∂y (x0 ,y0 )
(x0 , y0 , z0 ).
•
T1 and T2 are the tangent lines to C1 and C2 .
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Example 7.8: Let f (x, y) = 3x3 y2 + 3xy4 . Find
Example 7.9: Let f (x, y) = y log x + x tanh(3y). Find fx , fy at
(1, 0).
∂f
∂f
and
.
∂x
∂y
Solution:
Solution:
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The tangent line T1 has equation (y = y0 fixed):
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.
z − z0 =
∂f
∂x
(x0 ,y0 )
(x − x0 )
The tangent line T2 has equation (x = x0 fixed):
z − z0 =
This holds if fx and fy exist and are continuous near (x0 , y0 ).
∂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 )
∂f
∂x
(x0 ,y0 )
(x − x0 ) +
∂f
∂y
(x0 ,y0 )
(x − x0 ) +
∂f
∂y
(x0 ,y0 )
(y − y0 )
or equivalently,
z = z0 +
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(y − y0 ).
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Linear Approximations
Example 7.10: Find the equation of the tangent plane to
the surface z = f (x, y) = 2x2 + y2 at (1, 1, 3).
Solution:
If f is differentiable at (x0 , y0 ), we can approximate z = f (x, y) by
its tangent plane at (x0 , y0 , z0 ), when (x, y) is close to (x0 , y0 ).
That is:
f (x, y) ≈ z0 +
∂f
∂x
(x0 ,y0 )
|
(x − x0 ) +
∂f
∂y
(x0 ,y0 )
{z
(y − y0 )
}
equation of tangent plane
when (x, y) is close to (x0 , y0 ).
This is called the linear approximation to f near (x0 , y0 ).
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Example 7.11: Find the linear approximation of
f (x, y) = xexy at (1, 0). Hence, approximate f (1.1, −0.1).
Solution:
Note:
The actual value is
(1.1)e−0.11 ≈ 0.98542
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Approximate Change
Example 7.12: 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.
Rearranging the linear approximation equation, we get
f (x, y) − f (x0 , y0 ) ≃
∂f
∂x
(x0 ,y0 )
(x − x0 ) +
∂f
∂y
Solution:
(x0 ,y0 )
(y − y0 ).
Let ∆x = x − x0 , ∆y = y − y0 , ∆f = z − z0 = f (x, y) − f (x0 , y0 ).
The approximate change in f near (x0 , y0 ), for small changes ∆x
and ∆y in x and y, is:
∆f ≈
∂f
∆x
∂x (x ,y )
0 0
+
∂f
∆y
∂y (x ,y )
0 0
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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
Note:
• fyy
The actual change in f is
∆f
• fxy
= f (2.05, 2.96) − f (2, 3)
= 13.6449 − 13
= 0.6449
• fyx
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!
∂2 f
∂ ∂f
= fx =
= 2
x
∂x ∂x
∂x
!
∂2 f
∂ ∂f
= 2
= fy =
y
∂y ∂y
∂y
!
∂2 f
∂ ∂f
= fx =
=
y
∂y ∂x
∂y∂x
!
∂2 f
∂ ∂f
= fy =
=
x
∂x ∂y
∂x∂y
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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:
Theorem:
If the second order partial derivatives of f exist and are
continuous then fxy = fyx .
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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 ∂z dx ∂z dy
=
+
dt ∂x dt ∂y dt
Note:
fxy = fyx as expected since trigonometric functions and
polynomials are continuous for all (x, y) ∈ R2 .
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dz
Example 7.14: If z = x2 − y2 , x = sin t, y = cos t. Find
at
dt
π
t= .
6
Solution:
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Example 7.15: If z = ex sinh y, x = st2 , y = s2 t.
∂z
Find .
∂s
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 ∂z ∂x ∂z ∂y
=
+
∂s ∂x ∂s ∂y ∂s
Solution:
∂z ∂z ∂x ∂z ∂y
=
+
∂t ∂x ∂t ∂y ∂t
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Directional Derivatives
The straight line starting at P0 = (x0 , y0 ) with velocity û = (u1 , u2 )
has parametric equations:
Let û = (u1 , u2 ) be a unit vector in the xy-plane (so + = 1).
The rate of change of f at P0 = (x0 , y0 ) in the direction û is the
directional derivative Dû f P .
u21
u22
x = x0 + tu1 , y = y0 + tu2 .
Hence,
0
Geometrically this represents the slope of the surface z = f (x, y)
above the point P0 in the direction û.
Dû f
= rate of change of f along the straight line at t = 0
P0
= value of
d
f (x0 + tu1 , y0 + tu2 ) at t = 0
dt
= 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
!
∂f
∂f
Dû f P =
,
· (u1 , u2 ).
0
∂x P0 ∂y P0
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Gradient Vectors
Example 7.16: Find the directional derivative of f (x, y) = xey
1
at (2, 0) in the direction from (2, 0) towards , 2 .
2
If f : R2 → R is a differentiable function, we can define the
gradient of f to be the vector
∂f
∂f ∂f
∂f
i+ j=
,
grad f = ∇f =
∂x
∂y
∂x ∂y
Solution:
•
!
direction û
Then the directional derivative of f at the point P0 in the
direction û is the dot product
Dû f
P0
= ∇f
P0
· û
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Example 7.17: Find
! the directional derivative of
x
π
f (x, y) = arcsin
at (1, 2) in the direction anticlockwise
y
4
from the positive x axis.
Solution:
•
direction û
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Properties of ∇f and Dû f
So for fixed ∇f :
•
Dû f is maximum when cos θ = 1 so θ = 0
The directional derivative of f is
Dû f
= ∇f · û
= |∇f | |û| cos θ
⇒ f increases most rapidly along ∇f .
= |∇f | cos θ
•
Dû f is minimum when cos θ = −1 so θ = π
where θ is the angle between ∇f and û, and |v| denotes the
length of a vector v.
⇒ f decreases most rapidly along −∇f .
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•
Dû f = 0 when cos θ = 0 so θ =
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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 .
π
and ∇f ⊥ û.
2
Solution:
But Dû f = 0, whenever û is tangent to a level curve of f (where
f = constant).
⇒ ∇f ⊥ level curves of f
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Example 7.19: In what direction does f (x, y) = xey
(b)
(a) increase
y
(b) decrease
most rapidly at (2, 0)? Express direction as a unit vector.
4j
Solution:
0,2
From Example 7.16
Level curves of f x,y
1,0
8i
∇f (2, 0) = i + 2j
x
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Stationary Points
A stationary point of f is a point (x0 , y0 ) at which
∇f = 0
So
∂f
∂f
= 0 and
= 0 simultaneously at (x0 , y0 ).
∂x
∂y
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
A function f has a
1. local maximum at (x0 , y0 ) if f (x, y) ≤ f (x0 , y0 ) for all (x, y) in
some disk centred at (x0 , y0 ),
2. local minimum at (x0 , y0 ) if f (x, y) ≥ f (x0 , y0 ) for all (x, y) in
some 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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Second Derivative Test
Any local maximum or minimum of f will occur at a critical point
(x0 , y0 ) such that
1. ∇f (x0 , y0 ) = 0
2.
If ∇f (x0 , y0 ) = 0 and the second partial derivatives of f are
or
continuous on an open disk centred at (x0 , y0 ), consider the
Hessian function
∂f
∂f
and/or
do not exist at (x0 , y0 ).
∂x
∂y
evaluated at (x0 , y0 ).
H(x, y) = fxx fyy − (fxy )2
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.
q
z = x2 + y2 . Minimum at (0, 0) BUT ∇f does not exist at (0, 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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Example 7.21: Find and classify the stationary points of
f : R2 → R given by f (x, y) = y sin x.
Solution:
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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:
Z
Z
f (x, y) dx and
f (x, y) dy.
Z
f (x, y) dx is evaluated by holding y fixed and integrating
•
with respect to x.
Z
•
f (x, y) dy is evaluated by holding x fixed and integrating
with respect to y.
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Z
Example 7.22: Evaluate
Z
(3x2 y + 12y2 x3 ) dx.
1
(3x2 y + 12y2 x3 ) dy.
Example 7.23: Evaluate
0
Solution:
Solution:
Note:
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Double Integrals
Let f : R2 → R be a continuous function over a domain D in R2 .
We can evaluate the double integral:
"
"
f (x, y) dx dy
f (x, y) dA =
Δx
D
D
"
Δy
f (x, y) dA is the volume under the surface z = f (x, y) that
D
lies above the domain D in the xy plane, if f (x, y) ≥ 0 in D.
Volume of thin rod = (Area base) · (height)
| {z } | {z }
∥
∆x∆y
∥
f (x, y)
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Double Integrals Over Rectangular Domains
The double integral is defined as the limit of sums of the
volumes of the rods:
"
Definitions
1. R = [a, b] × [c, d] is a rectangular domain defined by
a ≤ x ≤ b, c ≤ y ≤ d.
"
f (x, y) dA =
D
f (x, y) dx dy
D
=
lim lim
∆x→0 ∆y→0
n
X
f (x, y)∆x∆y i
i=1
Note:
If f (x, y) = 1 then
"
"
dA =
D
dx dy
D
2.
gives the area of the domain D.
R dR b
c
a
f (x, y) dx dy =
R d R b
c
a
f (x, y) dx dy means integrate
with respect to x first and then integrate with respect to y.
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"
Fubini’s Theorem
(x2 + y2 ) dx dy if
Example 7.24: Evaluate
R = [−1, 1] × [0, 1].
Solution:
Let f : R2 → R be a continuous function over the domain
R = [a, b] × [c, d]. Then
"
Z dZ b
f (x, y)dA =
f (x, y) dx dy
R
c
a
d
bZ
Z
R
=
f (x, y) dy dx
a
c
So order of integration is not important.
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Note:
As expected, the order of integration is not important since
polynomials are continuous for all (x, y) ∈ R2 .
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Example 7.25: Using double integrals, find the volume of
the wedge shown below.
Solution:
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