1
Contents
1 Course Outline and Administration
1.1 Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Lecture outline – projected. . . . . . . . . . . . . . . . . .
2
4
5
2 Exercises
2.1 Difference Equations
2.2 Differentiation . . . .
2.3 Integration . . . . . . .
2.4 Differential Equations
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7
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. 28
3 Answers
3.1 Difference Equations
3.2 Differentiation . . . .
3.3 Integration . . . . . . .
3.4 Differential Equations
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35
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39
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2
1
Introductory Applied Mathematics
Course Outline and Administration
Welcome!
Introductory Applied Mathematics 620-163 is a mathematics subject tailored to students who have not done Specialist Mathematics at VCE (or an
equivalent) or those who feel they need to strengthen their mathematics in
order to master the material in Mathematics A 620-141 a core mathematics
subject. With this role in mind Mathematics A 620-141 is a corequisite for
Introductory Applied Mathematics 620-163 that is if 620-163 is undertaken,
620-141 must be done in the same semester.
In broad terms Introductory Applied Mathematics 620-163 bolsters the
calculus skills of students (both differential and integral) to better equip them
for both Mathematics A 620-141 and the succeeding core subject Applied
Mathematics 620-143.
On successful completion of Introductory Applied Mathematics 620-163,
students should:
• be able to solve linear first order difference equations and most second
order linear difference equations.
• be able to model a range of real life situations with difference equations.
• understand the structure of the solution space for homogenous linear
difference equations.
• understand equilibria points and their nature and appreciate how cycles
may behave in the same fashion as equilibria.
• understand and apply the fundamental tools of differential calculus –
the product and chain rule.
• be able to differentiate implicitly and using these techniques to derive
the derivatives of inverse trigonometric functions.
• understand and apply the fundamental tools of integral calculus – substitution and integration by parts and the ability to apply these to a
wide range of settings including the calculation of areas, volumes of
revolution and arclengths.
620 163
3
• have synthesized the previous calculus techniques to solve a variety of
first order differential equations as well as first and second order linear
differential equations.
• understand the structure of the solution space for homogenous linear
differential equations.
• be able to apply differential equations to some real world settings including population growth and springs.
Classes
Introductory Applied Mathematics has 3 lectures a week in a single lecture
stream, the first lecture is on Monday at 12:00pm - 1:00pm in theatre 3 the
Medical Centre, the second and third lectures are on Wednesday and Friday
at 12:00pm - 1:00pm in the Medley theatre in the Redmond Barry building.
The Lecturer is Dr Dave Coulson; whose room is G46, in the Richard
Berry Building (the Mathematics Department).
Additionally each student should be enrolled in and attend 1 tutorial a
week in which the past weeks topics will be discussed and practised. Tutorials
begin in the second lecturing week (note that Thursday 25 April – Anzac Day
is a university holiday).
To enrol in a tutorial you need to use ALLOC8 on the web (as you do
for all Science subjects). Once a tutorial has been alloc8ed you may find the
location by visitting the departmental website.
Textbooks and Lecture Notes
The texts for the subject are those that Mathematics A 620-141 requires.
As the text has been in use for a number of years there may well be second
hand copies available. The lecturer plans to make copies of overheads used
during lectures available prior to the lecture (via the subject web page) so
that students may invest their time in understanding (perhaps adding extra
comments) rather than transcribing the material. A good practice is to
read through the notes prior to the lecture (do not expect to understand
everything) to pave the way for the development of a conceptual framework.
Problem sheets are given and cover all the material of the course. They are
for students private work. Students who master the problems (even without
4
Introductory Applied Mathematics
those flagged as ‘hard’) should have no trouble passing the mid semester test
or final exam.
During each tutorial a tutorial exercise will be given out and worked on
individually or in groups, these exercises will be constructed to cover most
of the topical points from lectures. Additionally on the first 10 tutorials you
will be given a short assignment to hand back into the succeeding tutorial.
1.1
Assessment
Assessment will consist of an assignment component (15%), a mid semester
test component (up to 10%) and an end of semester examination (between 75
% and 85%). The final mark is max(0.15H + 0.1M + 0.75E, 0.15H + 0.85E)
where H is the homework percentage, M is the mod semester test percentage
and E is the exam percentage. Thus each assignment will be worth 1.5% and
the mid semester test will only count if the mark gained for it is greater than
the mark gained for the exam. In particular students who miss the mid
semester test will have their mark calculated by 0.15H + 0.85E – no extra
sittings will be given. It is expected however that most students will do
better on the mid semester test and that the feedback from this test will be
useful, thus sitting the test is doubly advantageous.
In the mathematics department we believe that some students will learn
more effectively in small groups and so we see no harm in students brainstorming maths problems together HOWEVER when it comes to assignments
(for assessment) all work should be written up (and understood) by students
seperately using their own expression. For each assignment you will be required to sign a declaration to this effect. The consequences for students
breaking this signed declaration can be very serious.
Late assignments will only be accepted by the lecturer and then only in
exceptional circumstances.
Calculators
A calculator will be
√of some use doing problems in the problem sheet. The
functions needed are x, log, sin, cos, ex . A calculator will not be allowed
into the examination though – answers may be left in the form log 3×sin(π/5)
rather than numerically evaluated to 0.645748101.
The use of spreadsheets will be useful in answering some of the recursive
620 163
5
difference equations given in the problem sheets. The lecturer used Excel
of the ubiqitous MicroSoft suite (this is on the macs in the departments
mac labs) but any spread sheet should suffice. Alternatively you may like
to program the answer in your favourite language. You may choose any
methodology you like. Please if you need to, come and see the lecturer if
you are not yet able to do any of these. The use of computers is a skill that
is very important and any mathematician no matter how theoretical will at
every opportunity try to verify their results computationally – computers are
an invaluable aid for this.
Getting help
In addition to lectures and tutorials there will be hours set aside when
students may get 1-1 help with problems. The lecturer will be making some
hours available as well as extra hours being done by some tutors. The specific
times will be made available later. There is no need to ‘book’ an appointment
during these hours nor do you need to see only your regular tutor – to the
contary an alternate explanation may broaden your understanding.
Still having trouble.
Do touch base with the lecturer. If the matter is serious enough, you
should also apply for ‘special consideration’ at the faculty office.
1.2
Lecture outline – projected.
Week 1 (March 4th):
Difference equations. Verification of solutions,
linear equations – superposition,
general solution of order 1 difference equations.
Week 2 (March 11th): Difference equations. Population modelling,
linear second order difference equations
(with distinct real roots), equilibrium solutions,
strange attractors/chaos.
6
Introductory Applied Mathematics
Week 3 (March 18th): Differentiation. limits, trigonometric derivatives via limits,
calculation of tangents, linearity of the differential operator,
product rule.
Week 4 (March 25th): Differentiation. Chain rule and applications,
quotient rule, implicit differentiation.
( Good Friday - No Class )
Easter Break Week
(April 1st – 5th)
Week 5 (April 8th):
Differentiation. Derivatives of inverse trigonometric
functions, stationary points, graph sketching.
Week 6 (April 15th):
Integration. Antidifferentiation of polynomial, trigonometric
and exponential functions. Integration by substitution,
areas by integration, integration by parts.
Week 7 (April 22nd):
Mid semester test Monday 22nd April Integration.
Volumes of revolution, arclengths.
UNIVERSITY HOLIDAY, Thursday April 25.
Week 8 (April 29th):
Integration. Centres of gravity, mean and variance,
partial fractions, inverse trigonometric substitutions.
Week 9 (May 6th):
Differential equations. Substitution of solutions,
superposition for linear DEs, Initial value problems,
1st order linear DEs and their applications.
Week 10 (May 13th):
Differential equations. Separable DEs, 2nd order linear DEs.
Week 11 (May 20th):
Differential equations. Applications – springs,
population modelling. Equilibria.
Week 12 (May 29th):
Revision.
620 163
2
2.1
7
Exercises
Difference Equations
1 Fibonacci sequences. Write down P0 , . . . , P10 for the Fibonacci difference
equation (Pn+1 = Pn + Pn−1 ) for initial conditions
a) P0 = 1, P1 = 2
b) P0 = 2, P1 = 1
(Notice how the set of subsequent values – the solution, is dependent on
initial conditions).
2 Verifying difference equation solutions. Show that the following difference equations (given on the left) have the following particular solutions
(given on the right) by using substitution.
difference equation
a) Pn+1 − Pn = 0
alleged solution
Pn = 1
b) Pn+1 − 43 Pn = −400000
Pn = 1.2 × 106
c) Pn+1 − Pn = 4n
Pn = 13 4n
d) Pn+1 − Pn = 2n
Pn = n2 − n
e) hard Pn+1 − Pn = 2 cos[(2n + 1)π/4]
firstly show that
2 cos[(2n + 1)π/4] =
Pn =
√
√
2 sin(nπ/2)
2(cos[nπ/2] − sin[nπ/2])
f) Pn+3 − 2Pn+2 − 5Pn+1 + 6Pn = 0
Pn = 1
g) Pn+3 − 2Pn+2 − 5Pn+1 + 6Pn = 0
Pn = (−2)n
h) Pn+3 − 2Pn+2 − 5Pn+1 + 6Pn = 0
Pn = 3n
8
Introductory Applied Mathematics
i) Pn+3 − 2Pn+2 − 5Pn+1 + 6Pn = −6 Pn = n
j) Pn+1 − Pn = 3 × 2n with P0 = 3
Pn = 3 × 2n
k) Pn+1 − Pn = an
Pn = an /(a − 1) + c
l) hard Show that the Fibonacci difference equation Pn+1 = Pn + Pn−1
with P0 = 1 and P1 = 1 has solution
√
√
1 1 + 5 n+1
1 − 5 n+1
−(
)
) ].
Pn = √ [(
2
2
5
√
√
Hint [(1 ± 5)/2]2 = [(3 ± 5)/2]
(We will later see how to derive this solution to the difference equation.)
3 Linear or not. Classify the following difference equations as linear or
non-linear.
a) Pn+1 − Pn = 3
b) Pn+1 − Pn = sin n
c) Pn+1 − Pn = (sin n)n
d) Pn+1 − Pn = 0
e) Pn+1 − (sin n)Pn = n2
f) Pn+1 − sin(Pn ) = 0
g) Pn+1 − 1/Pn = 7
2
h) Pn+1
− Pn = sin n
i) Pn+2 − 7n Pn+1 + Pn = cos n
j) Pn+1 + arctan(Pn ) = 0.
4 Superposition Using the superposition principle and answers to the previous question, find as general a solution as you can to
a)
Pn+1 − Pn
b) Pn+3 − 2Pn+2 − 5Pn+1 + 6Pn
c)
Pn+1 − Pn
d)
Pn+1 − Pn
e)
Pn+1 − Pn
f ) Pn+3 − 2Pn+2 − 5Pn+1 + 6Pn
=
=
=
=
=
=
0
0
3 × 4n
− cos[(2n − 1)π/4]
3n
−6A; A ∈ R
620 163
9
5 First order linear difference equations. Solve the following for general
solutions
a) Pn+1 − 3Pn
=0
b) Pn+1 + (1/2)Pn = 0
c) Pn+1 − Pn
=0
d) Pn+1 + 5Pn
=0
e)
f)
g)
h)
Pn+1 + 2Pn
Pn+1 − 4Pn
Pn+1 − 2Pn
Pn+1 + 3Pn
= (−1)n
= 2n
=n+2
= n + 2n
Solve the following for a particular solution satisfying the initial condition
i) Pn+1 + 2Pn
= (−1)n ,
P0 = 3
j) Pn+1 − 4Pn
= 2n ,
P0 = 0
k) Pn+1 − 2Pn
= n + 2,
P0 = 1
n
l) Pn+1 + 3Pn
=n+2 ,
P0 = 171/80.
6 Population modelling.
a) exponential growth
At the moment the human population is increasing by 1% per annum.
i) Write down a difference equation describing the behaviour of the human
population Py where y is the year.
ii) If the human population in 2001 is 3.5 × 109 what will it be in one
generation’s time (i.e. in 2026)?
iii) How many years does it take to double the human population?
iv) What happens (under this model) to the human population in the
long term?
v) List shortcomings of this model for the human population.
b) exponential growth with harvest
A lake has an initial population of one million fish. Due to natural growth,
this population would increase by one-third a year. If currently 400 000 fish
are caught per year,
i) write down a difference equation describing the behaviour of the fish
population Py where y is the year.
10
Introductory Applied Mathematics
ii) find the size of the fish population after n years (hint question 2 part
b may help).
iii) and calculate how many years it will take before the population is
depleted to less than half a million?
iv) How many years will it be until the lake is empty of fish?
c) exponential growth with variable harvest
A certain animal population has an initial size of 1000 and increases by
20% per year. Solve the following either by solving the difference equation
or enumerating Pn .
i) If each year 100 of the population are harvested, how many years is it
before the population exceeds 2000?
ii) Repeat if the harvest increases by 10 a year i.e. 110 in the second year
etc.
7 Linear second order, real distinct roots.
a) Solve the following homogenous linear difference equations with constant coefficients by finding two independent solutions and then using the
superposition principle.
i)
ii)
iii)
iv)
Pn+2 − 2Pn+1 − 3Pn
Pn+2 + 6Pn+1 + 8Pn
Pn+2 − 36Pn
Pn+2 + 2Pn+1 − 15Pn
=0
=0
=0
=0
b) Solve the following non-homogenous linear difference equations with
constant coefficients by finding a particular solution and using this with the
complementary function (your answers to the previous part).
i)
ii)
iii)
iv)
Pn+2 − 2Pn+1 − 3Pn
Pn+2 + 6Pn+1 + 8Pn
Pn+2 − 36Pn
Pn+2 + 2Pn+1 − 15Pn
= 5 × 4n
= 12 × 2n
= 70 + 32 × (−2)n
= 24n − 8 + 24 × (−3)n
620 163
11
c) Solve the following initial value problems (IVPs) by using the initial
conditions with the known general solutions.
i)
ii)
iii)
iv)
Pn+2 − 2Pn+1 − 3Pn
Pn+2 + 6Pn+1 + 8Pn
Pn+2 − 36Pn
Pn+2 + 2Pn+1 − 15Pn
= 5 × 4n
= 12 × 2n
= 70 + 38 × (−2)n
= 24n − 8 + 24 × (−3)n
P0
P0
P0
P0
= 1, P1 = 8
= 1, P1 = 3
= −1, P1 = −4
= 1, P1 = 21
8 Equilibrium solutions.
a) Find the equilibrium solutions of the following difference equations.
ii) Pn+1 = (−0.7)Pn iii) Pn+1 = (0.3)Pn
i) Pn+1 = (1.5)Pn
iv) Pn+1 = (−1.0)Pn v) Pn+1 = 12 (Pn2 − 3) vi) Pn+1 = 12 (Pn + Pn−1 )
b) Using cobweb diagrams investigate the stability of the previous equilibrium solutions.
c) For the above difference equations that are linear solve for Pn explicitly
and verify your assertions on stability by taking lim n → ∞.
9 Strange attractors. Consider the logistic difference equation
7
Pn+1 = Pn − Pn2 .
2
a) Investigate by evaluating Pn (P0 , P1 , . . . , P10 ) what happens when
i) P0 = 0;
ii) P0 = 5/2;
iii) P0 = 1;
In the following 4 parts using a spreadsheet or otherwise evaluate P500 , P501 , . . . , P510
iv) P0 = 2 (a spreadsheet is recommended);
v) P0 = 2.01 (a spreadsheet is recommended);
vi) P0 = 1.01 (a spreadsheet is recommended);
vii) P0 = .01 (a spreadsheet is recommended).
b) What are the equilibrium points?
12
Introductory Applied Mathematics
c) Are the equilibrium points stable/unstable?
d) harder Prove your assertion above by linearizing the solution. Use
the substitution Pn = e + Qn where Pn = e is an equilibrium solution.
e) For typical values of P0 what is the solution Pn attracted to?
10 Chaos. Consider the non-linear difference equation
Pn+1 =
31
Pn − Pn2 .
8
a) Investigate by evaluating Pn (use a spread sheet) what happens when
i) P0 = 0;
ii) P0 = 23/8;
iii) P0 = 1;
√
1
(39 + 273) then
iv) Without using a spreadsheet show that if P0 = 16
√
√
1
1
P1 = 16
(39 − 273) and P2 = 16
(39 + 273). Hence what happens long
term?
√
1
v) P0 = 3.470169478 ≈ 16
(39+ 273) (enumerate 500 terms with a spread
sheet)
vi) Why do the answers to the previous two parts differ so markedly after
400 terms?
vii) P0 = .01;
b) Are their any discernable patterns in the previous part?
620 163
2.2
13
Differentiation
11 Secant gradients. Calculate the gradients of the secants of the following
functions between the following points.
a) h(x) = (4 + x2 )−1
x=0
x=1
b) f (x) = x3 + 2x
x = −1
x=1
c) g(x) = (3/2)x
x=0
x=1
d) h(x) = 10x
x = −1/2 x = 0
12 Limits. Calculate the following limits.
a) lim 2x + 1
b) lim x4
c)
d)
x→1
x→−2
lim sin t
t→π/3
lim cos(3s)
s→π/6
e) lim h2 − 4h + 4
f) lim (a2 + 5a)/a
g) lim (y − 1)/(y 2 − 2)
h) lim {(2 + h)2 − 22 }/h
i) lim {(1 + h)3 − 13 }/h
j) lim {cos h − cos 0}/h
a→0
h→0
y→0
h→0
h→0
h→0
Use values h = ±0.1, ±0.01, . . . to guess a value for the last limit.
13 Trigonometric limits. Recall from lectures the limit
lim
h→0
sin h
= 1,
h
where h is measured in radians, the limit (from the previous question)
cos h − cos 0
=0
h→0
h
lim
14
Introductory Applied Mathematics
and from trigonometry
cos(t + h) − cos t = (cos t cos h − sin t sin h) − cos t
= cos t (cos h − 1) − sin t sin h
cos(t + h) − cos t
cos h − 1
sin h
so
= cos t (
) − sin t (
).
h
h
h
a) Using these results calculate the limit
cos(t + h) − cos t
h→0
h
lim
b) Using techniques similar to the previous question part calculate
lim
h→0
sin(t + h) − sin t
.
h
14 Trigonometric derivatives Using the limits calculated in the question
above find
a)
d
(cos t)
dt
b)
d
(sin t).
dt
15 Polynomial derivatives. Find the derivatives of the following polynomial like functions
a) f (x) = x7 − 7x4
b) g(x) =
c) f (t) = t1.5
d) g(s) =
e) f (t) = t−1
f) g(s) = s4 +
1 11
x
11
√
+ 16 x6 − 5x + 101
s = s1/2
√
√
s + 1/ s5
620 163
15
16 Applications of derivatives.
a) Find the equation of the tangent to the curve y = x3 − 4x at the point
x = 2.
b) Find the equation of the tangent to the curve y = 1/x2 at the point
x = −1.
c) Find the equation of the tangent to the curve y = sin x + cos x at the
point x = π/4.
d) Find the points on the curve y = 12x − x3 where the tangents are
parallel to the x−axis.
e) Find the points on the curve y = 13 (x + 1)3 where the tangents are
parallel to the line y = x.
f) Consider the curve y = 12 (1 − (x − 1)2 ).
i) Find the two x-intercepts of the curve.
ii) Show that the tangents to the curve at these x-intercepts meet each other
at right angles.
In lectures we have talked about there being a unique base a so that
evaluated at x = 0 is 1, that is the tangent to the curve f (x) = ax at
(x, y) = (0, 1) has slope 1.
This unique base is denoted e, (and you will meet e on many different
occasions in mathematics) and can be shown to be
d x
a
dx
1+
1
1
1
1
+ + + ... +
+ ...
1! 2! 3!
n!
an infinite but converging sum
= 2.718281828 to 10 sig. figs
an irrational number (despite the appearance of a pattern in the digits - this
pattern does not continue indefinitely).
Now since
eh − e0
= 1 (by the slope of the tangent), we have
h→0
h
ex+h − ex
ex eh − ex
= lim
lim
h→0
h→0
h
h
h
e
− e0
= ex lim
h→0
h
= ex
lim
16
Introductory Applied Mathematics
So the derivative of ex is ex , that is
d x
e = ex .
dx
17 Mixed derivatives. Evaluate the following
a)
d
(5ex )
dx
b)
d
(ex
dx
c)
d −3
(t
dt
d)
d
(2 sin t
dt
e)
d
(s−3/2
ds
+ es )
f)
d
(s
ds
g)
d
(uπ
du
+ πeu )
h)
d
[(u2
du
i)
d
(w2
dw
+ w + 1)2
j)
d w+1
e .
dw
+ cos t)
+ x10 )
− et )
+ 10 + sin s)
+ u)(u2 − u + 1)]
18 Product rule. Using the product rule for differentiation
d
d
d
[f (x)g(x)] = [ f (x)]g(x) + f (x)[ g(x)]
dx
dx
dx
evaluate the following
a)
d
[(u2
du
c)
d
[ses ]
ds
+ u)(u2 − u + 1)]
b)
d
[(w2
dw
d)
d
[r sin r]
dr
+ w + 1)(w2 + w + 1)]
The following 4 parts are a little contrived but demonstrate that the product
rule may be used to derive results more commonly derived by other means.
e)
d
[(cos t)(cos t)]
dt
f)
d 2v
e
dv
g)
d 3v
e
dv
h)
d
[w10 w−10 ]
dw
i)
(=
d
[e2v ev ])
dv
d
[(sin u)eu ]
du
j)
(=
d
[ev ev ])
dv
d
[(cos s)es ]
ds
620 163
17
trickier
k) dtd [t3 e3t ]
m)
l)
d
[(sin t)/t2 ]
dt
d r √
[e / r]
dr
n)
d
{[w(sin w)]ew }
dw
Recall from lectures the derivative of log x,
1
d
log x = .
dx
x
Recall also the chain rule
d
[f (u(x))] = f (u(x)).u (x)
dx
or using a different notation,
df du
d
[f (u(x))] =
.
dx
du dx
19 Chain rule. Using the chain rule for differentiation evaluate the following
a)
d 3u
e
du
b)
d w2
e
dw
c)
d
(s2
ds
d)
d
dr
log(r + 1)
e)
d
dt
f)
d
dv
sin 4v
g)
d
dv
cos(2v + 3)
h)
d
dw
cos3 w
i)
d
du
sin(eu )
j)
d sin s
e
ds
k)
d
(sin t)−1/2
dt
l)
d
(log r)3
dr
log(sin t)
n)
d
dw
o)
d (s2 −s)
e
ds
log(er )
q)
d
dt
r)
d
dv
m)
p)
d
dt
d
dr
log(t2 )
√
ew
sin(cos t)
+ s + 1)2
sin2 (cos v)
20 Applications of the chain rule. Using the chain rule solve the following. They will require a little thought.
a) A stone is dropped into a lake causing circular waves whose radii
increase at a constant rate of 0.5 cm/sec. At what rate is the circumference
of a wave changing when its radius is 4m?
18
Introductory Applied Mathematics
b) A rectangular swimming pool is 10m long, 6m wide and has uniform
depth 1.5m. Winter rain has caused the water level in the pool to be too
high. If water is pumped out at a rate of 0.5 m3 /min, at what rate is the
water level dropping?
c) A cubic crystal grows so that its side length is increasing at a constant
rate of 0.1 cm per day. At noon on April 17th the crystal has a side length
of 3.0 cm. Find
i) the rate at which the volume of the crystal is increasing at noon on
April 17th;
ii) the rate at which the volume of the crystal is increasing at noon on
April 18th;
iii) the exact increase in the volume of the crystal between noon on April
17th and noon on April 18th.
d) A child makes a spherical snowball with radius 12cm and whilst sitting,
waiting for a friend notices that as the snowball is melting the radius is
decreasing at a constant rate, changing from 12cm to 8cm in 30 minutes.
How quickly is the volume decreasing when the radius is 10cm?
e) A ladder of length 5m is resting against the wall when the bottom slips
out. When the top of the ladder is 3m above the ground it is falling with a
velocity of 8m/sec. At what rate is the foot of the ladder sliding out from
the wall at this time?
f) Water is filling an inverted (pointy end down) conical tank of height
10m and top radius 5m at a constant rate of 0.1 m3 /minute. (In a tank
pressure at a point is proportional to the height of water above that point.
Volume for a cone is V = 1/3πhr2 .) The pressure at the tap at the bottom
of the tank is 1kg/sec2 /m when the water has a height of 2m. Find the rate
of change of pressure when the water height is
i) 1m;
ii) 4m.
21 Quotient rule. Using either the quotient rule
f × g − f × g
d f
( )=
dx g
g2
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19
or the product and chain rule in unison
d
[f × g −1 ] = f × g −1 − f × g −2 × g dx
evaluate the following derivatives.
a)
d
[sin u/u2 ]
du
b)
d
[ew /(w2
dw
d)
d
dr
e)
g)
d
[ve−v ]
dv
h)
sec r
c)
d
ds
d
[1/(t log t)]
dt
f)
d √
[ v/ log v]
dv
d
[(3w2
dw
i)
d
[eu /(log3
du
+ 1)]
+ w)/(3w2 + w + 1)]
tan s
u)]
22 Implicit differentiation.
a) Find the gradient of the curve y 2 = x at the point (9,–3).
b) Find the gradient of the curve y + log y = x4 /16 at the point (2, 1).
c) Find the gradient of the curve x2 y + y 2 x + ex−1 = 7 at the point (1, 2).
√
d) Find the gradient of the curve sin y = x at the point ( 3/2, π/3).
e) Find the√equation of the tangent to the curve x2/3 + y 2/3 = 1 at the
point (−1/8, 3 3/8).
f) Consider the curve x2 + y 3 − 3y = 162.
i) Find the tangent to the curve at the point (12,3).
√
ii) What can you say about the slope of the tangent at
√ the point ( 160, −1)?
iii) What kind of line is the tangent at the point ( 160, −1)?
g) Find the two points on the curve 2x2 − 4xy + 5y 2 = 48 where the
tangent is parallel to the x-axis.
23 Derivatives of inverse functions. Using implicit differentiation calculate the following derivatives.
a)
d
dx
arccos x
d) Using
d
dy
b)
log y =
d
dx
1
y
arctan x
c)
d
arccosec
dx
x
and implicit differentiation show that
d x
e
dx
= ex .
20
Introductory Applied Mathematics
24 Stationary points.
a) Find the stationary points of the following functions.
i) 13 u3 − u2 − 3u + 4
ii) −t3 + 6t2 − 9t + 100
iii) r4 − 8r2 + 2
iv) s +
v) sin w cos w on [−π, π]
vi)
vii)
sin t
(2+cos t)
on [−π, π]
ix) 12 w + cos w on [−π, π]
9
(s+1)
x2
(9−x2 )
viii) v log v − 2v
x)
√
u/(4 + u)
b) Using the second derivative classify the stationary points of the previous
functions as either local maxima or local minima where ever possible.
25 Graph Sketching. Sketch the graphs of the following functions.
a) f (x) = 2x3 + 6x2 − 18x + 10.
√
√
b) f (w) = w(w + 3)(w − 3).
c) g(t) = t4 − 26t2 + 25.
d) h(v) = sin v + cos v on [−π, π].
2
e) f (x) = e−x .
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2.3
21
Integration
26 Antidifferentiation – polynomial like. Write down antiderivatives
for the following functions.
a) x3 + x
b) 5x9 − x2
c) − x12
d)
e) 7x6 −
2
x3
1
x7
f) xπ − xe
27 Antidifferentiation – exponentials. Write down antiderivatives for
the following functions.
a) ex
b) e5x
c) e−x
d) e2x + e−x
e) x9 + e10x
f) e7x −
1
x4
28 Antidifferentiation – trigonometric functions. Write down antiderivatives for the following functions.
a) cos x
b) sin x
c) cos 3x
d) sin(−2x)
e) 3 cos 2x − 4 sin 2x
f) (a tricky but standard result)
1
cos2 x
22
Introductory Applied Mathematics
29 Antidifferentiation – log functions. Find antiderivatives for the following functions.
a)
1
x
c)
1
x
e)
10
x
+x
+ sin x
b)
1
2x
d)
1
3x
+ ex
f)
1
3x
+ x2 + sin x
30 Antidifferentiation by substitution. Find antiderivatives of the following functions
a)
1
2x+3
√
b) x x2 + 1
c)
d)
t3
(t4 +1)2
e) s(1 + s)5
f) xex
x
x2 +1
2 +3
g) cos t sin3 t
h)
es +e−s
es −e−s
i) sin(2w + 3)
j) x2 (1 − x)4
k)
es
es +2
l)
m) ev sin(ev )
n)
cos s
sin s+2
o)
1
w log w
r)
1 − 1t
e
t2
p)
√x−2
x+1
√
q) 3x2 x3 + 4
sin t
cos t
31 Areas using integrals.
a) Find the area between the curve y = 2 + x − x2 and the x-axis.
b) Find the area between the curves y = −x2 + 5x − 4 and y = −x − 4.
c) Find the area between the curves y = x4 + 1 and y = 2x2 .
d) Find the area between the curves y = x2 and y = x3 .
e) Find the area between the curves x = y 2 − 2 and y = x.
f) Find the area (with ‘vertices’ (1,1), (2,4) and (3,5)) between the three
curves y = x2 , y = 2x − 1 and y = x + 2.
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23
32 Volumes of revolution.
a) Find the volume of the conical region generated by rotating the area
bounded by y = x, x = 3 and the x-axis about the x-axis.
b) Find the volume of√the solid wineglass region generated by rotating
the area bounded by y = x, x = 4 and the x-axis about the x-axis.
c) Find the volume of the spherical region generated by rotating the area
inside x2 + y 2 = 1 about the x-axis. (Using calculus).
d) A hemispherical water container of radius 13cm is filled with water to
a height of 6cm. What volume of water is held by the container?
e) Find the volume of the spindle shaped region generated by rotating the
area bounded by y = cos x and the x-axis between x = −π/2 and x = π/2
about the x-axis.
f) Find the volume of the anvil shaped region generated by rotating the
area bounded by y = cos x, y = 2, x = −π/2 and x = π/2 about the line
y = 2.
g) Find
√ the volume2 of the region generated by rotating the area bounded
by y = x and y = x about the x-axis.
h) Find the volume of the region remaining after drilling a cylindrical hole
of radius 8cm through the centre of a sphere of radius 10cm.
i) (harder) Find the volume of the region remaining after drilling a
cylindrical hole of length L through the centre of a sphere of any radius.
(The volume is [surprisingly] independent of the radius of the sphere)
j) (harder) A spherical container with radius 13cm which contains a
spherical shotput of radius 10cm and has just enough water to lap at the top
of the shotput. Find the volume of of the water contained within. (You may
be able to check your answer using part d).
24
Introductory Applied Mathematics
33 Integration by parts. Using integration by parts find antiderivatives
for the following functions.
a) xex
b) x cos x
c) x log x
d) x2 sin x
e) log x
f) (log x)2
2
g) x3 ex
h) arctan x
harder
i) e2x sin x
34 Centroids/Centres of gravity.
a) Find the centroid of the semicircular region bounded by x2 + y 2 = 1
and the x-axis.
b) Find the centroid of the parabolic region bounded by y = x2 , y = 4
and the y-axis.
c) Find the centre of gravity of a quadrant shaped laminate (of uniform
density) bounded by x2 + y 2 = 1 the y-axis and the x-axis.
d) Find the centre of gravity of the leaf like laminate (of uniform density)
bounded by y = x3 and x = y 3 .
e) (harder) Find the centroid of the region bounded by y = cos x (between x = −π/2 and x = π/2) and the x-axis. Hint – use symmetry for the
x coordinate, and a substitution y = cos θ then integration by parts for the
y-coordinate.
35 Continuous probability distributions.
a) Find the mean and variance of the uniform distribution p(x) = 1
0 ≤ x ≤ 1.
b) Find the mean and variance of the exponential distribution p(x) = e−x
0 ≤ x. Note that lim x → ∞ xn e−x = 0 for all n.
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25
c) Find the mean and variance of the normal distribution
e−(x /2)
p(x) = √
2π
2
for x real.
Note that
∞
−∞
e−(x
2 /2)
dx =
√
2π and lim xn e−(x
2 /2)
x→±∞
= 0 for all n.
(This first result requires vector calculus techniques to establish.)
36 Arclengths.
Find the lengths of the following curves. (In the case where two endpoints
are given the distance between these two points will be a lower bound for the
arclength as well as a rough estimate).
a) y = 3x + 5 between the points (1,8) and (5,20).
b) y = x3/2 between the points (0,0) and (1,1).
c) y = 16 (x3 + 3/x) between the points (1,2/3) and (3,14/3).
d) y = x4 /4 + x−2 /8 between the points (1,3/8) and (2,129/32).
e) 3y = 2(x + 1)3/2 between the points (0,2/3) and (3,16/3).
harder – Use implicit differentiation.
√
f) (y + 1)2 = 4x3 between the points (0,–1) and (8/9,32 2/27 − 1).
√
√
g) (3y−6)2 = (x2 +2)3 between the points (0,2+2 2/3) and (3,2+11 11/3).
h) The entire curve of x2/3 + y 2/3 = 1. A curve whose graph is a diamond
with inward bulging sides.
26
Introductory Applied Mathematics
37 Partial fractions.
a) Decompose the following into partial fractions.
i)
1
x(x+1)
ii)
2
x3 −x
iii)
4x2 +x+1
(x+2)(x2 +1)
v)
3t2 +4t+3
t4 +2t2 +1
vii)
1
t3 +t
iv)
2x2 +8x−2
(x+2)(x2 +1)
vi)
2x2 +3x+2
x3 +2x2 +x
viii)
2x4 −2x3 +7x2 −11x+2
x3 −x2 −4
b) Write down the template form you would use to try to decompose the
following into partial fractions. DO NOT ATTEMPT TO SOLVE.
For example
10x2 + 3x + 5
b
c
=a+
+
.
(x + 2)(x − 1)
x+2 x−1
i)
2x4 +1
(x2 +x+3)3
2x6 +1
(x2 +x+3)3
ii)
17
(x+1)3 (x2 −x+4)2
iii)
c) Using your answers to part a) find the indefinite integrals (antiderivatives) of the following.
i)
iii)
v)
1
dx
x(x+1)
vii)
2
dx
x3 −x
4x2 +x+1
dx
(x+2)(x2 +1)
3t2 +4t+3
dt
t4 +2t2 +1
ii)
iv)
vi)
1
dt
t3 +t
2x2 +8x−2
(x+2)(x2 +1)
dx
2x2 +3x+2
dx
viii)
x3 +2x2 +x
2x4 −2x3 +7x2 −11x+2
dx
x3 −x2 −4
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27
38 Inverse trigonometric substitutions.
Using the derivatives of inverse trigonometric functions calculated in the
previous section calculate the following;
a)
1
√
dx
1 − x2
b)
c)
1
dx
1 + x2
1
dx.
x x2 − 1
√
d) By substituting x = 5w evaluate
√
1
dx.
25 − x2
e) By substituting x + 2 = w evaluate
1
dx.
5 + 4x + x2
f) harder (no hints) Evaluate
√
1
dx.
5 − 12x − 9x2
28
Introductory Applied Mathematics
2.4
Differential Equations
39 Substitution and verification. By substituting the alleged solution
into the DE (differential equation) verify that the alleged solution is indeed
a solution for the following
a) Linear with constant coefficients
.
x −3x = 0
i)
x = e3t
.
x +2x = 0
x = e−2t
ii)
.
x −Rx = 0
iii)
x = eRt , R a constant
..
.
iv)
v)
x −5 x +4x = 0
..
.
x −5 x +4x = 0
x = et
x = e4t
vi)
vii)
x − x +20x = 0
..
.
x − x +20x = 0
..
x = e5t
x = e−4t
...
x = e2t
x = e−2t
x = e3t
viii)
ix)
x)
.
..
.
x −3 x −4 x +12x = 0
..
.
x −3 x −4 x +12x = 0
...
..
.
x −3 x −4 x +12x = 0
...
An important pair of examples (the characteristic equation has
complex conjugate roots a ± ib)
..
.
x −2a x +(a2 + b2 )x = 0
xi)
x = eat cos[bt]
..
.
xii) x −2a x +(a2 + b2 )x = 0
x = eat sin[bt]
b) Non-linear
.
2
x +2tx = 0
x = te−t
i)
t −3)
.
ii) x= −x2 + 7x − 12
x = (4e
t −1) with t > 0
(e
.
N
K
) K real
N = (βe−at
iii) N = aN (1 − K
+1)
c) First order system of DEs (show x and y solve the DEs simultaneously)
.
i) x= −x + 4y
x = 2et + 2e−3t
.
y= x − y
y = et − e−3t
620 163
29
40 Superposition to General. Using the superposition principle (applicable to linear, homogenous DEs) write down the most general solution you
can to the following DEs (which you have seen in the previous question)
..
.
a) x −5 x +4x = 0
..
.
b) x − x +20x = 0
...
..
.
c) x −3 x −4 x +12x = 0
..
.
d) x −2a x +(a2 + b2 )x = 0
41 IVPs. Using the general solutions you found to the previous question find
the particular solution satisfying the following IVPs (initial value problems)
DE
..
.
a) x −5 x +4x = 0
..
.
b) x − x +20x = 0
...
..
.
c) x −3 x −4 x +12x = 0
..
.
d) x −2a x +(a2 + b2 )x = 0
Initial conditions
.
x(0) = −1 x (0) = 2
.
x(0) = 2 x (0) = 1
.
..
x(0) = 0 x (0) = 0 x (0) = 0
.
x(0) = 0 x (0) = −b
42 Classification. Apply any of the following descriptors where applicable
to the following DEs (linear, homogenous, linear with constant coefficients,
separable, order n) DO NOT SOLVE
a)
c)
e)
g)
i)
.
x +tx = 0
.
x +tx2 = t
..
.
x + x +x = cos t
...
x −x = 3
.
x −x2 = 0
b)
d)
f)
h)
j)
.
x +tx2 = 0
..
.
x + x +x = 0
.
x + cosx t = 0
.
x + tan x = t2
.
(x)2 = e2t
43 First order DEs with constant coefficients.
a) Solve the following DEs
.
x −2x = 0
i)
.
iii) x +4x = 0
b) Solve the following IVPs
.
i) x= (1.5)x
.
ii) x= (−0.8)x
.
ii) x= 3x
.
iv) x= (1.05)x
x(0) = 10
x(0) = 100
c) A university lecturer has $10 000 in a bank account that earns an
interest rate of 4% per annum (calculated continuously). She is not going to
touch this account until she retires in 25 years time.
30
Introductory Applied Mathematics
i) Describe this with a DE with an appropriate initial condition.
ii) By solving this DE find out how much money will be in this account
when the lecturer retires in 25 years time.
d) A new car has plastics which give off chemicals (possibly carcenogenic)
causing the signature ‘new car’ odour. The amount of chemical leeched into
the air is proportional to the amount of chemical remaining in the plastic.
i) Show that dP
= −kP. (A is proportional to B written A ∝ B exactly
dt
when A = kB where k is a real constant).
ii) The half life of the chemical emmision is 2 months. Use this to solve
for the constant of proportionality in your DE.
iii) If the concentration of chemical when the car is made is 40 parts per
million and the recognized upper bound for safety is 5 parts per million. (By
solving the DE with an initial condition) how long must the company keep
the cars before they are safe to release (with respect to the chemicals)? Does
this concur with common sense?
44 Separable DEs.
a) Solve the following separable DEs
.
.
x= et
i)
ii) x= xt
.
.
iii) x= −2tx2
iv) x= e−x sin t
.
.
v) x= et−x
vi) x= √2tx
1+t2
b) Solve the following separable IVPs
.
x= xt
x(1) = 5
i)
.
ii) x= xt + t
x(0) = 2
.
iii) xt x= log t
x(1) = 1
.
2
x=
x(0) = 1
iv) x
t(1 + x )
c) Logistic revisited A simple model for the spread of disease states
that the spread of disease is proportional to the product of the proportion of
infected people with the proportion of uninfected people.
i) Show that dP
= kP (1 − P ).
dt
ii) Solve this DE to show that P/(1 − P ) = Aekt where k is the constant
of proportionality and A depends on initial conditions.
iii) Find A and k if at the time measurements are started 1/3 of the
population is infected and after 1 week e/(2+e) of the population are infected
(measure time in weeks).
620 163
31
iv) Find P as an explicit function of time and find out how long it is until
90% of the population is infected.
v) Draw a rough sketch of P versus t.
45 second order homogenous linear DEs with constant coefficients.
a) Assuming solutions of the form x(t) = eλt find the values of λ giving
solutions to the following DEs.
..
.
..
.
x −5 x +6x = 0 ii)
x −2 x −15x = 0
i)
..
.
..
x −25x = 0
iii) x + x −6x = 0 iv)
..
.
..
.
x −6 x +9x = 0 vi)
x −4 x +3x = 0
v)
..
.
..
.
vii) x −4 x +4x = 0 viii) x −4 x +5x = 0
b) Using the heuristic that second order DEs will have 2 degrees of freedom (arbitrary constants) and the superposition principle write down the
general solutions (also known as the complementary function) to the following equations
..
.
x −5 x +6x = 0
i)
..
.
ii) x −2 x −15x = 0
..
.
iii) x + x −6x = 0
..
iv) x −25x = 0
..
.
v) x −4 x +3x = 0
c) Solve the following second order IVPs.
i)
ii)
iii)
iv)
..
.
x +5 x +6x = 0
..
.
x − x −6x = 0
..
x −25x = 0
..
.
x +2 x −15x = 0
x(0) = 2
x(0) = 1
x(0) = 2
x(0) = 8
.
x (0) = −5
.
x (0) = 8
.
x (0) = 0
.
x (0) = 0
46 Springs.
a) non-damped motion.
Hooke’s Law says that the force exerted by a spring is proportional to the
compression/extension of the spring beyond its natural length and (the force)
acts to restore the spring to its equilibrium position.
Hence F = −kx where F is the force, x is the displacement from the
equilibrium position and k is the spring constant (the constant of proportionality).
32
Introductory Applied Mathematics
Using Force=mass × acceleration we see that
..
m x= −kx
where m is the mass of an object attached to the end of the spring.
i) A mass of 2kg is attached to a spring with spring constant 8 kg sec−2 .
What DE governs the motion of the spring?
ii) The mass is initially stationary in the equilibrium position and is set
in motion with a hammer strike which imparts a velocity of 10 m sec−1 (in
the extension direction). What initial conditions are satisfied by the system?
iii) By solving the IVP (using the result outlined in question ) find the
subsequent behaviour of the mass.
iv) What is the amplitude and frequency of the oscillation?
v) Does the amplitude of the oscillation change with time?
vi) Does this seem reasonable?
vii) harder What do you think happens to the frequency and amplitude
of oscillation if the spring gets stiffer (ie the spring constant increases)? Check
your common sense/intuition by multiplying the spring constant k above by
4 and resolving the problem.
47 Population modelling.
a) The exponential model.
A simple model for population is based on the premise that the population
has a constant propensity to breed so overall the growth rate of the population
is proportional to the population size, that is
dP
= (α − β)P
dt
where P is the population size, α is the average birth rate per head of population per year and β is the average death rate per head of population per
year.
At the moment for the human population α − β is about 0.02.
i) Find the general solution to the DE dP
= 0.02P.
dt
9
ii) If P (0) = 3.5 × 10 (the current human population for the world
approximately), solve for a particular solution of P (t) and find the world’s
population in 1 generation’s time (25 years).
620 163
33
iii) Under this model how long will it take the population to reach 1 000
000 000 000 000=1015 ?
iv) Suggest a shortcoming with this population model.
b) The logistic model (simple).
i) Solve the DE dP
= P (1 − P ).
dt
ii) Draw a rough sketch of 3 solutions to the above DE satisfying P1 (0) =
0.1, P2 (0) = 3, P3 (0) = −0.2 respectively.
c) The logistic model (real life).
An equivalent approach to the logistic model is with the DE
dP
P
= R0 P (1 − )
dt
K
(where R and K are constants) giving solution
P (t) =
K
1 + αe−R0 t
which has long term solution K.
In 1920 the logistic model was fitted to the U.S. population over the years
1790-1910, showing good agreement with the parameters R0 = 0.03134 and
K = 197273000 where time is measured in years.
We are now investigating how well this model can be extrapolated (extended) to the years 1920 to 1980 with the actual values being given in the
table below.
i) Before quantitatively solving as requested below can you tell how well
the model will fit the data?
U.S. Population Data (in 1000’s) 1920-1980.
Year Population
1920 105 711
1930 122 775
1940 131 670
1950 150 679
1960 179 323
1970 203 235
1980 226 546
34
Introductory Applied Mathematics
ii) Compare the predictions of the logistic model with the actual values
below. Use as an initial condition the 1910 population being 91 972 000.
d) The logistic model with harvesting.
Left to nature the worlds population of orange roughy (a deep sea fish)
would grow according to
dP
1
= P (1 − P/4)
dt
10
where time t is measured in years and population P is measured in 107 fish.
This was the case until two decades ago when the fish was discovered. Now
the orange roughy (a.k.a. deep sea perch) is being fished (a.k.a. harvested)
for arguments sake at a constant rate of h per year.
i) If h = 0.075 (ie 750 thousand orange roughy are caught a year) show
that
dP
40
= −P 2 + 4P − 3
dt
describes the growth of population.
ii) By drawing a graph of dP
versus P find and classify the equilibrium
dt
points of P.
iii) Solve the two IVPs with P (0) = 2 and P (0) = 1/2 and sketch the two
solutions on the same P versus t graph.
iv) Using a graph of dP
versus P or otherwise find the maximum value of
dt
the harvest h that will support any viable population of the orange roughy.
620 163
3
35
Answers
3.1
Difference Equations
1 Fibonacci sequences.
a) 1,2,3,5,8,13,21,34,55,89,144
b) 2,1,3,4,7,11,18,29,47,76,123
2 No answers required only verification by substitution.
3 Linear or not.
a) linear
c) linear
e) linear
g) non-linear
i) linear
b) linear
d) linear
f) non-linear
h) non-linear
j) non-linear.
4 Superposition
a) α × 1
b) α + β(−2)n + γ × 3n
c) c + 4n
d) c − √12 sin(nπ/2)
e) c + 12 × 3n
f) α + β(−2)n − γ × 3n + A × n
5 First order linear difference equations.
a) A × 3n
b) A × ( −1
)n
2
c) c
d) A × (−5)n
e)
g)
A × (−2)n + (−1)n
A × 4n − n − 3
i) 2 × (−2)n + (−1)n
k) 4 × 2n − n − 3
f) A × 4n − 12 × (2n )
h) A × (−3)n + 15 × 2n + 14 n −
j)
l)
− 12 2n
2 × (−3)n + 15 .2n + 14 n −
1 n
4
2
1
.
16
1
16
36
Introductory Applied Mathematics
6 Population modelling.
a) i)
Py+1 − Py =
1
Py .
100
)n where A = 3.5 × 109 , giving P25 = 4.49 × 109 .
ii) Py + A( 101
100
iii) The human population will be slightly more than double after 70
years.
iv) The human population grows without bound under this model.
v) This unbounded nature of growth is unrealistic (for example given
enough time the population would be greater than the mass of the earth in
kilograms a contradiction given that people weigh more than 1 kg even at
birth).
b) i) Py+1 − Py = 13 Py − 400000.
ii) Py = A( 43 )y + 1200000 where A = −200000.
iii) After 5 years the population is 357 202.
iv) After 7 years.
c) i) Py = 500×(6/5)y +500 so by enumeration after 7 years the population
is 2291.
ii) Py = 250 × (6/5)y + 50y + 750 so by enumeration after 8 years the
population is 2224.
7 Linear second order, real distinct roots.
a) i) A × 3n + B × (−1)n
ii) A × (−4)n + B × (−2)n
iii) A × (−6)n + B × 6n
iv) A × (−5)n + B × 3n
b) i) A × 3n + B × (−1)n + 4n
ii) A × (−4)n + B × (−2)n + 12 × 2n
iii) A × (−6)n + B × 6n − 2 − (−2)n
iv) A × (−5)n + B × 3n − 2 × (−3)n − 2n
c) i) 3n − (−1)n + 4n
ii) −( 32 ) × (−4)n + 2 × (−2)n + ( 12 ) × 2n
iii) −2 × (−6)n + 6n − 2 − (−2)n
iv) 3 × (−5)n − 5 × 3n − 2 × (−3)n − 2n
620 163
37
8 Equilibrium solutions.
a), b) & c)
i) 0 unstable, Pn = A × 1.5n Pn → ∞ as n → ∞
ii) 0 stable, Pn = A × (−0.7)n Pn → 0 as n → ∞
iii) 0 stable, Pn = A × 0.3n Pn → 0 as n → ∞
iv) 0 unstable, Pn = A × (−1)n Pn has no limit
v) -1 stable, 3 unstable, a non-linear equation
vi) -1 stable, 1 stable, a non-linear equation
9 Strange attractors.
a) i) Pn is fixed at 0.
ii) Pn is fixed at 5/2.
iii) P0 = 1 and thereafter (n > 0) Pn is fixed at 5/2.
iv) Pn alternates between 3 and 1.5 a 2-cycle.
v) Pn converges to a 4-cycle with values (to 2 significant figues) 1.34, 2.89,
1.75 and 3.06. (This implies the 2-cycle is unstable)
vi) Pn once again converges to the 4-cycle with values (to 2 significant
figues) 1.34, 2.89, 1.75 and 3.06.
vii) Pn once again converges to the 4-cycle with values (to 2 significant
figues) 1.34, 2.89, 1.75 and 3.06. (Suggesting that this 4-cycle is stable)
b) & c) 0 unstable and 5/2 unstable.
d) harder Linearizing at the equilibrium point 0 gives Qn+1 = 7/2Qn ,
Linearizing at the equilibrium point 5/2 gives Qn+1 = −3/2Qn hence both
equilibria are unstable.
e) Generally Pn converges to the 4-cycle with values (to 2 significant
figures) 1.34, 2.89, 1.75 and 3.06.
10 Chaos.
a) i) Pn is fixed at 0.
ii) Pn is fixed at 23/8.
iii) P0 = 1 and thereafter (n > 0) Pn is fixed at 23/8.
√
273) and 1/16(39+
iv)
We
have
a
2-cycle
alternating
between
1/16(39−
√
273).
v) Initially we have a 2-cycle alternating between 1.40 and 3.47 but by
the 400th term this has gone haywire to give a sequence with no discernable
pattern.
38
Introductory Applied Mathematics
vi) The difference is√an example of the butterfly effect. The 2-cycle is
unstable and 1/16(39 + 273) is not exactly equal to 3.470169478 so initially
we have something close to the 2-cycle but this diverges away exponentially
to something qualitatively very different to the 2-cycle (for example we don’t
even have a 2-cycle with any numbers). This is a trademark of chaos.
vii) No pattern is discernable – the behaviour is chaotic. Note also this
suggests that the equilibrium point 0 is unstable.
b) As previously remarked the behaviour is chaotic. Contemplate how a
humble quadratic has given rise to chaos.
620 163
3.2
39
Differentiation
11 Secant gradients.
a)
c)
1
20
1
2
12 Limits.
a) 3√
c) 23
e) 4
g) 12
i) 3
b) 74
d) 2(1 −
√1 )
10
≈ 1.39
b) 16
d) 0
f) 5
h) 4
j) 0
13 Trigonometric limits.
a) − sin t
b) cos t
14 Trigonometric derivatives
a) − sin t
b) cos t
40
Introductory Applied Mathematics
15 Polynomial derivatives.
a) 7x6 − 28x3 b) x10 − x5 − 5
c) 1.5 × t0.5
d) 2√1 s = 12 × s−1/2
e) −t−2
f) 4s3 + 12 × s−1/2 − 52 s−7/2
16 Applications of derivatives.
a) y = 8x − 16
b) y = √
2x + 3
c) y = 2
d) (−2, −16), (2, 16)
e) (0, 13 )
f) i) x = 0 and x = 2
ii) The tangents have gradients 1 and -1 respectively, with -1×1=-1 meaning the two gradients are perpendicular.
17 Mixed derivatives.
a) 5ex
c) −3t−4 − sin t
e) − 32 s−5/2 + es
g) πuπ−1 + πeu
i) 4w3 + 6w2 + 6w + 2
b) ex + 10x9
d) 2 cos t − et
f) 1 + cos s
h) 4u3 + 1
j) ew+1 .
18 Product rule.
a) 4u3 + 1
c) (s + 1)es
e) −2 sin t cos t
g) 3e3v 3
i) (cos u + sin u)eu
k) 3(t3 + t2 )e3t
m) (t cos t − 2 sin t)/t3
b) 2(2w3 + 3w2 + 3w + 1)
d) sin r + r cos r
f) 2e2v
h) 0
j) (cos s − sin s)es
l) (r−1/2 − 1/2r−3/2 )er
n) (sin w + w cos w + w sin w)ew
620 163
41
19 Chain rule.
a) 3e3u
d) 1/(r + 1)
g) −2 sin(2v + 3)
j) cos sesin s
m) cos t/ sin t
p) 1
2
b) 2wew
e) 2/t
h) −3 sin w cos2 w
k) (− 12 ) cos t(sin t)−3/2
n) 12 ew/2
q) sin t × cos(cos t)
c) 2(2s + 1)(s2 + s + 1)
f) 4 cos 4r
i) eu cos(eu )
l) 3(log r)2 /r
2
o) (2s − 1)e(s −s)
r) −2 sin(cos v) × sin v
20 Applications of the chain rule.
a) π cm/sec
b)
1
120
cm/min
c) i) 2.7 cm3 /day
d)
160π
3
ii) 2.883 cm3 /day iii) 2.79 cm3
cm3 /min
e) 6 m/sec
f) i)
1
5π
kg/sec2 /m
ii)
1
80π
kg/sec2 /m
21 Quotient rule.
a)
2u sin u−u2 cos u
u4
b)
ew (w2 −2w+1)
(w2 +1)2
c)
1
cos2 s
d)
sin r
cos2 r
e)
−(1+log t)
(t log t)2
f)
−1+log v
√
v(log v)2
g)
1−v
ev
h)
6w+1
(3w2 +w+1)2
i)
u log u−3)eu
u log4 u
22 Implicit differentiation.
a) -1/6
b) 1
c) -9/5
d) 2 √
√
e) y = 3x + 3/2
f) dy/dx = −2x/[3(y 2 − 1)]
i) y = −x + 9
42
Introductory Applied Mathematics
ii) The slope is undefined or more loosely infinite.
iii) The tangent is vertical.
g) (4,4) and (-4,-4).
23 Derivatives of inverse functions.
√
a) −1/ 1 − x2
b) 1/(1 + x2 )
√
c) −1/[x x2 − 1]
d) The answer is given in the question.
24 Stationary points.
a) & b)
i) -1 max, 3 min
ii) 1 min, 3 max
iii) -2 min, 0 max, 2 min
iv) -4 max, 2 min
v) −3π/4, π/4 both max, −π/4, 3π/4 both min, vi) 0 min
vii) −2π/3 min, 2π/3 max
viii) e min
ix) π/6 max, 5π/6 min
x) 4 max
25 Graph Sketching.
a) f (x) = 2x3 + 6x2 − 18x + 10 has a maxima at (-3,64) a minima at (1,0),
zeros at (-5,0) and (1,0) (a double zero) and y intercept at (0,10).
√
√
b) f (w) √
= w(w + 3)(w − 3)
√ has a maxima at (-1,2) a minima at (1,-2),
zeros at (− 3, 0), (0, 0) and (− 3, 0) and y intercept at (0,0).
620 163
43
√
√
c) g(t) = t4 −26t2 +25 has a maxima at (0,20) and minimas at (− 13, −144), ( 13, −144)
zeros at (-5,0), (-1,0), (1,0) and (5,0) and y intercept at (0,20).
√
d) h(v) = √
sin v+cos v on [−π, π] has a maxima at (π/4, 2) and a minima
at (−3π/4, − 2) zeros at (−π/4, 0), (3π/4, 0) and y intercept at (0,1).
e) f (x) = e−x has a global maxima at (0,1) and f (x) → 0+ as x → ±∞,
no zeros and the y intercept is (0,1).
2
44
3.3
Introductory Applied Mathematics
Integration
26 Antidifferentiation – polynomial like.
b) 12 x10 − 13 x3 + c
a) 14 x4 + 12 x2 + c
−1
c) x1 + c
d) 6x
6 + c
1
1
7
2
e) x + 1/x + c
f) π+1 xπ+1 − e+1
xe+1 + c
27 Antidifferentiation – exponentials.
b) 15 e5x + c
a) ex + c
c) −e−x + c
d) 12 e2x − e−x + c
1
10
10x
e) 10 (x + e ) + c
f) 17 e7x + 1/(3x3 ) + c
28 Antidifferentiation – trigonometric functions.
a) sin x + c
b) cos x + c
c) 1/3 cos 3x + c
d) cos(−2x) + c
e) 3/2 sin 2x + 2 cos 2x + c f) sin x/ cos x + c = tan x + c
29 Antidifferentiation – log functions.
a) log x + c
b) 12 log x + c
c) log x + 12 x2 + c
d) 13 log x + ex + c
e) 10 log x − cos x + c
f) 13 (log x + x3 ) − cos x + c
30 Antidifferentiation by substitution.
b) 13 (x2 + 1)3/2 + c
a) 12 log(2x + 3) + c
1
2
c) 2 log(x + 1) + c
d) 4(t−1
4 +1) + c
2
1
1
1
e) 6 (1 + s)6 − 7 (1 + s)7 + c f) 2 ex +3 + c
g) 14 sin4 t + c
h) log(es − e−s ) + c
i) −1
cos(2w + 3) + c
j) −1
(1 − x)5 + 13 (1 − x)6 − 17 (1 − x)7 + c
2
5
s
k) log(e + 2) + c
l) − log(cos t) + c
v
m) − cos(e ) + c
n) log(sin s + 2) + c
√
√
o) log(log w) + c
p) 23 x3/2 − x − 2 x + 2 log( x + 1) + c
1
q) 23 (x3 + 4)3/2 + c
r) e− t + c
620 163
45
31 Areas using integrals.
a) 4 12 square units
b) 32 square units
c)
44
15
square units
d)
1
12
square units
e) 4 12 square units
f)
5
6
square units
46
Introductory Applied Mathematics
32 Volumes of revolution.
Answers and a rough picture of the region that is revolved.
a) 9π units cubed
b) 8π units cubed
c) 4π/3 units cubed
d) 396π cm3
e) π 2 /2 units cubed
f) 9π 2 /2 units cubed
620 163
47
g) 3π/10 units cubed
h) 288π cm3
i) L3 π/6 units cubed – the picture is similar to above except with general
dimensions.If the radius of the sphere is R, R > L/2 then the radius of the
cylinder is R2 − (L/2)2 by Pythagoras.
j) 1200π cm3 . Note that the volume above the shotput 396π (from part
d) + the volume calculated here 1200π is 1596π which in turn is 43 × 133 π −
4
× 103 π the difference in volume of the two spheres, ie the volume inside the
3
large sphere that is outside the shotput.
33 Integration by parts.
a) xex − ex + c
2
2
c) x2 log x − x4 + c
e) x log x − x + c
2
2
2
g) x2 ex − 12 ex + c
2x
i) e5 (2 sin x − cos x) + c
b) x sin x + cos x + c
d) −x2 cos x + 2x sin x + 2 cos x + c
f) x(log x)2 − 2x log x − 2x + c
h) x arctan x − 12 log(1 + x2 ) + c
34 Centroids/Centres of gravity.
4
)
a) (0, 3π
4
4
c) ( 3π
, 3π
)
π
e) (0, 8 )
b) ( 12
, 3)
5 4
d) ( 16
, 16 )
35 35
35 Continuous probability distributions.
a) mean 1/2, variance 1/12.
b) mean 1, variance 1.
48
Introductory Applied Mathematics
c) mean 0, variance 1.
36 Arclengths.
√
a) 4 10
c) 14
3 √
√
e) 2(5 5 − 2 2)/3
g) 12
√
b) (13 13 − 8)/27
d) 3 27
32
f) 52/27
1
0.5
-1
-0.5
0.5
1
-0.5
h) 6 (see graph)
37 Partial fractions.
a)
1
i) x1 − x+1
1
1
iii) x−1
− x2 + x+1
3
v) x+2(
+ xx−1
2 +1
4t
vii) (t2 +1)2 + t23+1
-1
t
ii) 1t − t2 +1
−2
iv) x+2
+ x24x+1
1
vi) x2 − (x+1)
2
3
viii) 2x + x−2 +
2(2x+1)
x2 +x+2
b)
i) x2ax+b
+ (x2cx+d
+ (x2ex+f
+x+3
+x+3)2
+x+3)3
ax+b
cx+d
ii) A + x2 +x+3 + (x2 +x+3)2 + (x2ex+f
+x+3)3
x+g
a
b
c
dx+e
iii) x+1 + (x+1)2 + (x+1)3 + x2 −x+4 + (x2f−x+4)
2
c)
x
|+c
i) log |x| − log |x + 1| + c = log | x+1
1
ii) log |t| − 2 log(t2 + 1) + c = log | √t2t+1 | + c
iii) log |x − 1| − 2 log |x| + log |x + 1| + c = log |1 − x12 | + c
2 +1
iv) −2 log |x + 2| + 2 log(x2 + 1) + c = 2 log | xx+2
|+c
1
2
v) 3 log |x + 2| + 2 log(x + 1) − arctan x + c
1
vi) 2 log |x| + x+1
+c
−2
vii) t2 +1 + 3 arctan t + c
viii) x2 + 3 log |x − 2| + 2 log |x2 + x + 2| + c
38 Inverse trigonometric substitutions.
620 163
a) − arccos x + c = arcsin x + c
b) arctan x + c
c) −arccosec x + c
d) − arccos(x/5) + c = arcsin(x/5) + c
e) arctan(x + 2) + c
f) substituting 3x + 2 = 3w we get 13 arcsin(x + 23 ) + c
49
50
3.4
Introductory Applied Mathematics
Differential Equations
39 Substitution and verification.
Answers given in the text of the questions.
40 Superposition to General.
a) x(t) = αet + βe4t
b) x(t) = αe5t + βe−4t
c) x(t) = αe2t + βe−2t + γe3t
d) x(t) = αeat cos[bt] + βeat sin[bt]
41 IVPs.
a) x(t) = −2et + e4t
b) x(t) = e5t + e−4t
c) x(t) = 0
d) x(t) = −eat sin[bt]
42 Classification.
a) homogenous, linear, separable, order 1
b) homogenous, separable, order 1
c) separable, order 1
d) homogenous, linear, constant coefficients, order 2
e) linear, constant coefficients, order 2
f) homogenous, separable, order 1
g) linear, constant coefficients, order 3
h) order 1
i) homogenous, separable, order 1
j) order 1
620 163
51
43 First order DEs with constant coefficients.
a)
i)
x(t) = αe2t
iii) x(t) = αe−4t
ii) x(t) = αe3t
iv) x(t) = αe1.05t
b)
i) x(t) = 10e1.5t
ii) x(t) = 100e−0.8t
c)
.
i) x= 0.04x, x(0) = 10000 ii) $27 182.81=e × 104 .
d)
.
i) P = −kP ii) k = 12 log 2 iii) 6 months, (after 2 months the decay is 20
pp million,. . . after 6 months the decay is 5 pp million).
44 Separable DEs.
a)
i)
x = αet
ii) x = at
2
iii) x = 1/(c + t ) iv) x = log(c
− cos t)
√
t
2 1+t2
v) x = log(c + e ) vi) x = Ae
b)
2
i)
x=
5t
ii) x = −1 + 3et /2
√
iii) x = (log t)2 + 1 iv) x = −1 + 2et2
c) Logistic revisited
i) dP
= kP (1 − P )
dt
ii) Answer given in the Question.
iii) A = 12 , k = 1
iv)
P (t) =
et
2 + et
90% or more of the population are infected when t ≥ log 18 ≈ 2.89 weeks.
52
Introductory Applied Mathematics
v)
45 second order homogenous linear DEs with constant coefficients.
a)
i)
2, 3
iii) -3, 2
v)
3, 3
vii) 2, 2
ii)
5, -3
iv)
5, -5
vi)
3, 1
viii) 2 ± i
b)
i)
ii)
iii)
iv)
v)
x(t) = αe2t + βe3t
x(t) = αe5t + βe−3t
x(t) = αe2t + βe−3t
x(t) = αe5t + βe−5t
x(t) = αe3t + βet
c)
i)
ii)
iii)
iv)
x(t) = e−2t + e−3t
x(t) = 2e3t − e−2t
x(t) = e5t + e−5t
x(t) = 3e−5t + 5e3t
46 Springs.
a) non-damped motion.
Answer given in the question.
..
i) x= −4x
.
ii) x(0) = 0, x (0) = 10
iii) x(t) = 5 sin 2t
iv) amplitude=5 and frequency=2 (cycles per unit time)
v) No – it is fixed at 5.
vi) No this does not seem realistic as it is an example of perpetual motion.
..
vii) x= −16x, x(t) = 2.5 sin 4t thus frequency increases and amplitude
decreases.
620 163
53
47 Population modelling.
a) The exponential model.
i) P = Ae0.02t .
ii) P = 3.5×109 e0.02t so in 25 years time the population will be 5.77×109 .
iii) About 628 years.
iv) The population grows without bound.
b) The logistic model (simple).
i) P = 1/(1 + αe−t ) = Aet /(1 + Aet ) with α = 1/A.
ii)
3
7.5
2
5
1
2.5
0.25
0.5
0.75
1
1.25
1.5
1
-1
2
3
4
5
-2.5
-2
-5
-3
iii)
-7.5
c) The logistic model (real life).
i) The constant K is an upper bound for the population (if 0 < P0 < K
which is the case here). This is clearly violated in 1970 and 1980.
ii)
U.S. Population Data (in 1000’s) 1920-1980.
Year Actual Population Modelled Population
1920
105 711
107 424
1930
122 775
122 454
1940
131 670
136 394
1950
150 679
148 766
1960
179 323
159 323
1970
203 235
168 034
1980
226 546
175 025
d) The logistic model with harvesting.
i) Answer is given in the question.
54
Introductory Applied Mathematics
ii) The equilibrium .points are P = 1 unstable
and P = 3 unstable. (See
.
graph for 1 < P < 3, P > 0 and for P > 3, P < 0)
0.02
1
2
3
4
-0.02
-0.04
-0.06
iii) P1 (0) = 2 gives (3 + e−t/20 )/(1 + e−t/20 ) and P2 (0) =
5e
)/(1 − 5e−t/20 ).
−t/20
1
2
gives (3 −
2
1.5
1
0.5
2
4
6
8
10
iv) Refering to the graph below harvesting 1 million fish a year (shifting
the parabola down until it touches the P axis) is the greatest sustainable
1
harvest giving DE dP
= − 10
(1 − P/2)2 though this leaves P = 2 as an
dt
unstable equilibrium point and any small change in the population giving
P < 2 will mean P → 0, that is the fish becomes extinct.
1.5
-0.005
-0.01
-0.015
-0.02
-0.025
2
2.5
3