Paradoxes and Sophisms in Calculus

8/14/2019 Paradoxes and Sophisms in Calculus

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From Mahobe Resources (NZ) Ltd

We hope you enjoy these sample pages of the book Paradoxes and Sophisms in Calculus. We

have provided the first 42 pages of the book. The actual book can be purchased from our

website www.mahobe.co.nz

This book is a supplementary resource intended to enhance the teaching and learning of a

first-year university Calculus course. It can also be used in upper secondary school. It consists

of selected paradoxes and sophisms that can be used as a pedagogical strategy by creating

surprise and interest in the subject.

In this book the following major topics from a typical single-variable Calculus course are

explored: Functions, Limits, Derivatives and Integrals. As with the authors previous book

Counter-Examples in Calculus (Maths Press, Auckland, New Zealand, 2004, ISBN

0-476-01215-5, 116 p.) the intention of this book is to encourage teachers and students to use

it in the teaching and learning of Calculus, with these purposes: for deeper conceptualunderstanding to reduce or eliminate common misconceptions to advance ones

mathematical thinking, that is neither algorithmic nor procedural to enhance generic critical

thinking skills analysing, justifying, verifying, checking, proving which can benefit students

in other areas of life to expand the example set - a number of examples of interesting

functions for better communication of ideas in mathematics and in practical applications to

make learning more emotional, active and creative The book can be useful for: upper

secondary school teachers and university lecturers as a teaching resource upper secondary

school and first-year university students as a learning resource.

If you enjoy the book then why not purchase it! Go to the Mahobe Resources (NZ) Ltdwebsite www.mahobe.co.nz

Go ahead and enjoy the first pages of this very informative and thought provoking book.
http://www.mahobe.co.nz/http://www.mahobe.co.nz/http://www.mahobe.co.nz/http://www.mahobe.co.nz/


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Copyright 2007 Sergiy Klymchuk

This e-book is copyright. All rights reserved. Apart from any fair dealingfor the purpose of study, teaching, review, or otherwise permitted underthe Copyright Act, no part of this publication may be reproduced by anyprocess without written permission from the author.

ISBN 978-0-473-12798-5

The printed version first published

December 2005 ISBN 0-473-10550-0Maths PressPO Box 109-760

NewmarketAuckland 1031New Zealand

Klymchuk, Sergiy.Paradoxes and Sophisms in Calculus


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Contents

Preface.7

Paradoxes

1. Functions and Limits.11

2. Derivatives and Integrals........15

Sophisms



Solutions to Paradoxes



Solutions to Sophisms




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7

Preface

Dear God,

If I have just one hour remaining to live,

Please put me in a Calculus class

So that it will seem to last forever.

A bored students prayer

There are many interesting non-routine problems, puzzles, paradoxes

and sophisms. Some of them can change the life of students forever.

They can enthuse, enlighten and inspire. They can open the mind and

drive a person into the labyrinths of knowledge. They can encourage

passion for invention and discovery.

The teaching/learning process loses its effectiveness without emotional

involvement. What we tend to remember most are the events,

situations, news and facts that bring with them strong feelings and

emotions. The purpose of this book is to create and use such feelings

and emotions as a pedagogical strategy in a Calculus course.

The book is a supplementary resource intended to enhance the teaching

and learning of a first-year university Calculus course. It can also be

used in upper secondary school. The following major topics from a

typical single-variable Calculus course are explored in the book:

Functions, Limits, Derivatives and Integrals.

The book consists of two parts: paradoxes and sophisms. The word

paradox comes from the Greek word paradoxon which meansunexpected. There are several usages of this word, including those that

deal with contradiction. In this book, the word means a surprising,

unexpected, counter-intuitive statement that looksinvalid but in fact is

true. The word sophism comes from the Greek word sophos which

means wisdom. In modern usage it denotes intentionally invalid

reasoning that looks formally correct, but in fact contains a subtle

mistake or flaw. In other words, it is a false proof of an incorrect

statement. Each such proof contains some sort of error in reasoning.


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Most students are exposed to sophisms at school. Often finding and

analyzing the mistake in a sophism can give a student deeper

understanding than a recipe-based approach in solving standard

problems. The former works on a psychological level whereas the latter

on a procedural level. Some basic examples of the sophism include

division by zero, taking only a non-negative square root and so on.

These tricks are used to prove statements like 1 = 2. In this book the

tricks from Calculusare used to prove such statements.

Some of the paradoxes presented in the book (such as Cat on a Ladder

and Encircling the Earth) are at a lower level than Calculus, but they

can still be used in Calculus classes to demonstrate that sometimes our

intuition fails, even when we are dealing with very familiar shapes (likecircles for example). Much of the books content can be used as

edutainment both education and entertainment.

As with my previous book, Counter-Examples in Calculus (Maths

Press, Auckland, New Zealand, 2004, ISBN 0-476-01215-5) the

intention of this book is to encourage teachers and students to use it in

the teaching/learning of Calculus with these purposes:

For deeper conceptual understanding

To reduce or eliminate common misconceptions

To advance ones mathematical thinking, that is neither

algorithmic nor procedural

To enhance generic critical thinking skills analysing, justifying,

verifying, checking, proving which can benefit students in other

areas of life

To expand the example set - a number of examples of interesting

functions for better communication of ideas in mathematics and

in practical applications

To make learning more emotional, active and creative

The book can be useful for:

Upper secondary school teachers and university lecturers as a

teaching resource

Upper secondary school and first-year university students as a

learning resource


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Acknowledgement

My thanks to Dr Farida Kachapova and Brody Radford from the

Auckland University of Technology for proofreading and formatting the

text and putting the material on AUT Online, the Universitys web-based

learning system.

Contact Details for Feedback

Please send your questions and comments about the book to the postal

address:

Dr Sergiy Klymchuk

Associate Professor

School of Mathematical Sciences

Faculty of Design and Creative Technologies

Auckland University of Technology

Private Bag 92006

Auckland 1020

New Zealand

or e-mail to:

[email protected]

Sergiy Klymchuk

June 2006


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11

Paradoxes

I see it but I don't believe it!

Georg Cantor (1845 1918), in a letter to R. Dedekind (1877)

1. Paradoxes: Functions and Limits

1. Laying bricks

Imagine you have an unlimited amount of the same ideal

homogeneous bricks. You are constructing an arc by putting the

bricks one on top of another without using any cementing solution

between them. Each successive brick is further to the right than

the previous (see the diagram below).

How far past the bottom brick can the top brick extend?

2. Spiral curves

Construct two similar-looking spiral curves that both rotate

infinitely many times around a point, with one curve being of a

finite length and the other of an infinite length.

3. A tricky curve

Construct a curve that is closed, not self-crossing, has an infinite

length and is located between two other closed, not self-crossing

curves of a finite length.


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4. A tricky area

A square with sides of 1 unit (and therefore area of 1 square unit)

is divided into 9 equal squares, each with sides 1/3 unit and areas

of 1/9 square unit, then the central square is removed. Each of the

remaining 8 squares is divided into 9 equal squares and the central

squares are then removed. The process is continued infinitely many

times. The diagram below shows the first 4 steps. At every step 1/9

of the current area is removed and 8/9 is left, that is at every step

the remaining area is 8 times bigger than the area removed. After

infinitely many steps what would the remaining area be?

5. A tricky next term

What is the next term in the sequence 2, 4, 8, 16?

6. A tricky shape

It looked like the cross-section of an object of circular shape. To

determine whether it was a circle, a student suggested measuring

its several diameters (the length of line segments passing through

the centre of symmetry of the figure and connecting its two

opposite boundary points). The student reasoned that if they all

appeared the same, the object had a circular shape. Was the

students reasoning correct?

7. Rolling a barrel

A person holds one end of a wooden board 3 m long and the otherend lies on a cylindrical barrel. The person walks towards the

barrel, which is rolled by the board sitting on it. The barrel rolls


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without sliding. This is shown in the following diagram:

What distance will the person cover before reaching the barrel?

8. A cat on a ladder

Imagine a cat sitting half way up a ladder that is placed almost

flush with a wall. If the base of the ladder is pushed fully up

against the wall, the ladder and cat are most likely going to fall

away from the wall (i.e. the top of the ladder falls away from the

wall).

Part 1: If the cat stays on the ladder (not likely perhaps?) what will

the trajectory of the cat be? A, B or C?

A B C

Part 2: Which of the above options represents the cats trajectory if

instead of the top of the ladder falling outwards, the base is pulled

away? A, B or C?

9. Sailing

A yacht returns from a trip around the world. Different parts of the

yacht have covered different distances. Which part of the yacht has

covered the longest distance?


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10. Encircling the Earth

Imagine a rope lying around the Earths equator without any bends

(ignore mountains and deep-sea trenches). The rope is lengthened

by 20 metres and the circle is formed again. Estimate how high

approximately the rope will be above the Earth:

A) 3 mm B) 3 cm C) 3 m?

11. A tricky equation

To check the number of solutions to the equationx

x

16

1log16

1 one can sketch the graphs of two inverse functions

xy16

1log and

x

y

16

1

.

420-2

3

2

1

0

-1

-2

420-2

3

2

1

0

-1

-2

From the graphs we can see that there is one intersection point and

therefore one solution to the equation

x

x

16

1log16

1 . But it is easy

to check by substitution that both2

1x and

4

1x satisfy the

equation. So how many solutions does the equation have?


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2. Paradoxes: Derivatives and Integrals

1. An alternative product rule

The derivative of the product of two differentiable functions is the

product of their derivatives: vuuv )( . In which cases is this rule

true?

2. A strange integral

Evaluate the following integral .dxdx

3. Missing information?

At first glance it appears there is not enough information to solvethe following problem: A circular hole 16 cm long is drilled through

the centre of a metal sphere. Find the volume of the remaining part

of the sphere.

4. A paint shortage

To paint the area bounded by the curvex

y 1

, the

x-axis and the line x= 1 is impossible. There is not enough paint in

the world, because the area is infinite: .)1ln(lnlim1

1

bdxx bHowever, one can rotate the area around the x-axis and the

resulting solid of revolution would have a finite volume:

.)1

11(lim

1

1

2

bdx

x bThis solid of revolution contains the area

which is a cross-section of the solid. One can fill the solid with

cubic units of paint and thus cover the area with paint. Can you

explain this paradox?


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Sophisms

1 + 1 = 3 for large values of 1.

A student joke

1. Sophisms: Functions and Limits

1. 1 = 0

Let us find the limit

n

kn kn1

2

1lim using two different methods.

a)

00...001lim...2

1lim1

1lim1lim222

1 2

nnnnkn nnn

n

kn

b) The lower and upper boundaries for the sum

n

k kn1 21

are:

n

k nn1 2

1


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3. 1 = 0

Let us find the limit )(lim0

x

xx

using two different methods.

a) First find the limit of the base and then the other limit:

00lim)lim()(lim000

x

x

x

x

x

x

xx .

b) First find the limit of the power and then the other limit:

.1lim)(lim 0

0

)lim(

0

0

xxxx

xx

x

x

Comparing the results in a) and b) we conclude that 1 = 0.

4. 1 =

Let us find the limit nn

nlim using two different methods.

a) First find the limit of the expression under the radical and then

the other limit:

n

nn

n

n

nnn limlimlim .

b) First find the limit of the nthroot and then the other limit:

11limlimlimlim 0

1lim

1

nn

nn

n

n

nnnnn n .

Comparing the results in a) and b) we conclude that 1 = .

5. xkkx sinsin

It is known that 1

sin

lim0 x

x

x .

Using this fact we can find the following two limits:

a) ku

uk

kx

kxk

x

kx

uxx

sinlim

sinlim

sinlim

000.

b) kx

xk

x

xk

xx

sinlim

sinlim

00.

Sincex

xk

x

kx

xx

sinlim

sinlim

00 then

x

xk

x

kx sinsin and xkkx sinsin .

6. 1 = 0

We know that the limit of the sum of two sequences equals the sum

of their limits, provided both limits exist. We also know that this is

true for any number kof sequences in the sum.

Let us take n equal sequencesn

1 and find the limit of their sum


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when n:

00...001lim...

1lim1lim)

1...11

(lim nnnnnn nnnn

.

On the other hand, the sum

n

nnn

1...11

is equal to 11

n

n .

So we receive 1 = 0.

7. 1 = -1

Let us find two limits of the same function:

a) 1)1(lim

1

1

limlimlimlim

xyxyx

y

x

y

x

yx

yx.

b) 11lim

1

1

limlimlimlim

yxyxy

x

yx

y

yx

yx.

Sinceyx

yx

yx

yx

xyyx

limlimlimlim , the results from a) and b) must

be equal. Therefore we conclude that 1 = -1.

8. a= 1/a

Let abe any non-zero number.

Let us find two limits of the same function:

a)aa

ay

x

y

ax

ayx

yax

xyxyx

11lim

1

limlimlimlim

.

b) aa

x

ayx

y

a

ayx

yax

yxyxy

lim

1

limlimlimlim .

Sinceayx

yax

ayx

yax

xyyx

limlimlimlim , the results from a) and b) must

be equal. Therefore we conclude that a= 1/a.


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9. 1 = 2

a) Take two line segments of length 1 unit and 2 units and

establish a one-to-one correspondence between their points as

shown on the diagram below:

The number of points on the line segment of length 1 unit is the

same as the number of points on the line segment of length 2 units,meaning 1 = 2.

b) Take two circles of radius 1 unit and 2 units and establish a

one-to-one correspondence between their points as shown on the

diagram below:

The number of points on the circumference of the inner circle is the

same as the number of points on the circumference of the outer

circle, so we can conclude that 1 = 2.

10. R= r

Two wheels of differentradius are attached to each other and put

on the same axis. Both wheels are on a rail (see the diagrams on

the next page). After one rotation the large wheel with radius R

covers the distance AB which is equal to the length of its

circumference 2R. The small wheel with radius r covers the

distance CD which is equal to the length of its circumference 2r. It

is clear that AB = CD, therefore 2R= 2r and R= r.


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Wheel

Rail

Cross-section showing the wheel and the rail.

11. 2 > 3

We start from the true inequality:

8

1

4

1 or

32

2

1

2

1

Taking natural logs of both sides:32

2

1ln

2

1ln

Applying the power rule of logs:

2

1ln3

2

1ln2

Dividing both sides by

2

1ln :

2 > 3.

C D

A B


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12. 2 > 3

We start from the true inequality:

8

1

4

1

or

32

2

1

2

1

Taking logs with the base2

1of both sides:

3

2

1

2

2

12

1log

2

1log


2

1log3

2

1log2

2

1

2

1

Since 12

1log2

1

we obtain: 2 > 3.

13.2

1

4

1

We start from the true equality:

2

1

2

1

Taking natural logs of both sides:

2

1ln2

1ln

Doubling the left hand side we obtain the inequality:

2

1ln2

1ln2


2

1ln

2

1ln

2

Since y= ln x is an increasing function2

1

2

1 2

or2

1

4

1 .


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14. 2 = 1

Let us take an equilateral triangle with sides of 1 unit. Divide the

upper sides by 2 and transform them into a zig-zagging, segmented

line as shown on the diagram below:

a) The length of this segment line is 2 units because it is

constructed from two sides of 1 unit each. We continue halving the

triangle sides infinitely many times. At any step the length of the

segment line equals 2 units.

b) On the other hand from the diagram we can see that with more

steps, the segment line gets closer and closer to the base of the

triangle which has length 1 unit. That is 1lim nn S , where nS is the

length of the segment line at step n.

Comparing a) and b) we conclude that 2 = 1.

15. = 2

Let us take a semicircle with diameter d. We divide its diameter into

nequal parts and on each part construct semicircles of diametern

d

as shown on the following diagram:

1

11

1 1 1


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a) The arc length of each small semicircle isn

d

2

. The total length Ln

of n semicircles is22dn

ndLn

. Therefore the limit of Lnwhen

nis:22

limlim dd

Ln

nn

.

b) From the diagram we can see that when n increases, the curve

consisting of nsmall semicircles gets closer to the diameter, which

has length d. That is dLnn

lim .

Comparing a) and b) we see that dd 2

and conclude that = 2.

16. = 0

Let us find the lateral surface area of a cylinder with height 1 unit

and radius 1 unit using the following approach. Divide the cylinder

into n horizontal strips. Divide the circumference of each

cross-section by points into mequal parts. Rotate all odd-numbered

circumferences in such a way that the points on them are exactly

midway between the points on the even-numbered circumferences.

Form 2mn equal isosceles triangles by joining any two adjacent

points on each circumference with a point midway between them

on the circumferences above and below.


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Using simple geometry of a right-angled triangle it can be shown

that the area of the resulting polyhedral surface is:

mn

mmSmn

2sin41sin2

42 .

When both m and n tend to infinity this area tends to the lateral

surface area of the cylinder. The limit of Smnis found using the well

known formula 1sinlim0

x

x

x. Let us consider 3 cases.

a) n = m

.

2

2sin

41

sin

2lim2sin41sin2limlim

4

2446

m

mm

m

m

mmmmS mmm m

b) n = m2

4

4

44

2

2sin

41

sin

2lim2

sin41sin2limlim

m

m

m

m

mm

mmS

mmm

m

412

4 .


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c) n = m3

.

2

2sin

41

sin

2lim2

sin41sin2limlim

4

2446

m

mm

m

m

mm

mmS

mmmm

In all three cases a), b) and c) when m the polyhedral surface

tends to the lateral surface of the cylinder. So the limit mm

Slim in all

three cases a), b) and c) must be the same.

This is only possible if = 0.

17. Achilles and the Tortoise

This is one of the sophisms created by the Greek philosopher Zenoin the 5thcentury B.C. Sometimes they are called Zenos paradoxes,

but in the sense of paradox and sophism accepted in this book

they are considered sophisms.

In a race between Achilles, the fastest of Greek warriors, and a

tortoise that had a head start, Achilles will never pass the tortoise.

Suppose the initial distance between them is 1 unit and Achilles is

moving 100 times faster than the tortoise. When Achilles covers the

distance of 1 unit the tortoise will have moved 100

1th

of a unit

further from its starting point. When Achilles has covered the

distance of100

1 th of a unit the tortoise will move2100

1 of a unit

further, and so on. The tortoise is always ahead of Achilles byn100

1

of a unit no matter how long the race is. This means that Achilles

will never reach the tortoise.

18. A snail

Imagine a snail moving at a speed of 1 cm/min along a rubber rope

1 m long. The snail starts its journey from one end of the rope. After

each minute the rope is uniformly expanded by 1 m. Below is a

proof that at some stage the snail will eventually reach the other

end of the rope.

In the first minute the snail will cover the first100

1thof the rope.

In the second minute the snail will cover2001 thof the rope.


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In the third minute the snail will cover300

1thof the rope, and so on.

The distance covered by the snail after nminutes will be

n

1...

3

1

2

11

100

1. The sum in the brackets represents the first n

terms of the harmonic series, which is divergent. The sum in the

brackets (and therefore the distance) can be made bigger than any

number. So no matter how big the length of the rope is, the

distance covered by the snail at some stage will be bigger than the

length of the rope. This means that the snail will reach the other

end of the rope.

19. 1,000,000

2,000,000If we add 1 to a big number the result would be approximately

equal to the original number. Let us take 1,000,000 and add 1 to it.

That is 1,000,000 1,000,001.

Similarly 1,000,001 1,000,002.

And 1,000,002 1,000,003.

And so on

1,999,999 2,000,000.

Multiplying the left-hand sides and the right-hand sides of the

above equalities we receive:

.000,000,2...002,000,1001,000,1999,999,1...001,000,1000,000,1

Dividing both sides by 999,999,1...001,000,1 we conclude that.000,000,2000,000,1

20. 1 = -1

Since baba , it follows that

.111)1()1(11 2 iii

21. 2 = -2Two students were discussing square-roots with their teacher.

The first student said: A square root of 4 is -2.


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The second student was sceptical, and wrote down 24 .

Their teacher commented: You are both right.

The teacher was correct, so 2 = -2.

22. 2 = 1

Let us find the equation of a slant (or oblique) asymptote of the

function1

42

x

xxy using two different methods.

a) By performing long division:1

62

1

42

xx

x

xx. The last term,

1

6

x tends to zero as x . Therefore as the function

approaches the straight line 2 xy , which is its slant asymptote.

b) Dividing both numerator and denominator by x we receive:

x

xx

x

xx

11

41

1

42

. Bothx

4and

x

1tend to zero as x . Therefore

as x the function approaches the straight line 1 xy which is

its slant asymptote.

The function1

42

x

xxy has only one slant asymptote. Therefore

from a) and b) it follows that 12 xx .

Cancelling xwe receive 2 = 1.


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2. Sophisms: Derivatives and Integrals

1. 1 = C, where C is anyreal number

Let us apply the substitution method to find the indefinite integral

xdxxcossin using two different methods:

a) 1

2

1

2

2

sin

2cos

sincossin C

xC

uudu

xdxdu

xuxdxx

b)

2

2

2

2

2

cos

2sin

coscossin C

xC

uudu

xdxdu

xuxdxx ,

where C1and C2are arbitrary constants. Equating the right hand

sides in a) and b) we obtain

1

2

2sin Cx = 2

2

2cos Cx .

Multiplying the above equation by 2 and simplifying we receive

12

2222cossin CCxx or Cxx 22 cossin since the difference of

two arbitraryconstants is an arbitrary constant. On the other hand

we know the trigonometric identity 1cossin 22 xx . Therefore 1 = C.

2. 1 = 0

Let us find the indefinite integral dxx1

using the formula for

integration by parts :vduuvudv

dx

xdx

xxx

xxvdxdv

xdu

xu

dxx

11

1111

12

2 .

That is, dxx

dxx

11

1.

Subtracting the same expression dxx1

from both sides we receive

0 = 1.

3. Division by zero

Let us find the indefinite integral 12x

dx by the formula

Cxfxf

dxxf)(ln

)(

)(using two different methods:

a)

121ln

21

2

12112

Cx

x

dxxdx .


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b)

212ln

2

1

12

2

2

1

12Cx

x

dx

x

dx.

Equating the right hand sides in a) and b) we obtain

21

12ln2

1

2

1ln2

1CxCx

Since C1and C2are arbitrary constants that can take any values,

let C1= C2= 0. Then 12ln2

1

2

1ln2

1 xx .

Solving for xwe receive 122

1 xx ,

2

1x .

Substituting this value of x into the original integral gives zero in

the denominator, so division by zero is possible!

4. 1sin2 x for anyvalue of x

Let us differentiate the function y= tan xtwice:

x

xy

xy

32cos

sin2,

cos

1 .

The second derivative can be rewritten as

)(2cos

1tan2

coscos

sin2

cos

sin2 2223

yyyx

xxx

x

x

xy .

Integrating both sides of the equation )( 2 yy we receive

2yy or x

x

2

2 tan

cos

1 , or

x

x

x 2

2

2cos

sin

cos

1 .

From here 1sin 2 x .

5.

,2

0

Let us estimate the integral

0

2cos1 x

dx.

a) Since 1cos1

1

2

12

x

on ],0[ then

002

0 cos12

1dx

x

dxdx or

0

2cos12 x

dx.


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b) On the other hand

0

02

202

2

02

020

cos

00tan

1tan1

cos

cos1 t

dt

txx

dxdt

txxt

x

x

dx

x

dx

.

Comparing a) and b) we conclude that

,2

0 .

6. 2ln is not defined

Let us find the area enclosed by the graph of the functionx

y 1

,

the x-axis and the straight lines x = -2 and x = -1 using two

different methods (see the diagram on the next page).

a) On one hand, the derivative of the function y = ln xisx

y 1 and

therefore an antiderivative ofx

xf 1)( is xxF ln)( . We can apply

the Newton-Leibnitz formula to the integral

1

2

1dx

x (the limits are

finite and the function is continuous on [-2,-1]) to find the required

area: ))2ln()1(ln(11

2

dx

x

A . The area is undefined since the

logarithm of a negative number does not exist.

b) On the other hand, this area is the same as the area enclosed by

the graph of the functionx

y 1 , the x-axis, and the straight lines

x= 1 and x= 2 due to the symmetry of the graph about the origin.

Therefore the area equals: .2ln1ln2ln12

1

dxx

A

Comparing a) and b) we conclude that 2ln is not defined.


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20-2-4

1.5

1

0.5

0

-0.5

-1

-1.5

-2

20-2-4

1.5

1

0.5

0

-0.5

-1

-1.5

-2

20-2-4

1.5

1

0.5

0

-0.5

-1

-1.5

-2

20-2-4

1.5

1

0.5

0

-0.5

-1

-1.5

-2

7. is not defined

Let us find the limitxx

xx

x sin

sinlim

using two different methods.

a)

x

xx

x

xx

xx

xx sin1

sin

limsin

sinlim .

b) Since both numerator and denominator are differentiable we can

use the well known rule)(

)(lim

)(

)(lim

xg

xf

xg

xf

xx

which gives us:

x

x

xx

xx

xx cos1

coslim

sin

sinlim

, which is undefined.

Comparing the results in a) and b) we conclude that is not

defined.

8. 0 = C, where C is anyreal number

We know the property of an indefinite integral:

dxxfkdxxkf )()( where kis a constant.

Let us apply this property for k= 0.

a) The left hand side of the above equality is Cdxdxxf 0)(0 ,

where C is an arbitrary constant.

a) The right hand side is 0)(0 dxxf .

Comparing a) and b) we conclude that 0 = C, where C is anyreal

number.


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32

9. 1 = 2

Let us find the volume of the solid of revolution produced by

rotating the hyperbola 122 xy about the x-axis on the interval

[-2, 2] using two different methods.

a) 3

4)

3()1(

2

2

32

2

22

2

2

xx

dxxdxyV (cubic units).

b) Since the hyperbola is symmetrical about the y-axis we can find

the volume of a half of the solid of revolution, say on the right from

the y-axis and then multiply it by 2. Obviously the point (1,0) is a

vertex to the right of the origin and the right branch of the

hyperbola is to the right of the vertex (1,0). Therefore the volume of

the right half is 3

4)

3()1(

2

2

32

1

2

2

1

2

1

x

xdxxdxyV (cubic

units) and the total volume 3

82 1 VV (cubic units).

Comparing a) and b) we obtain 3

8

3

4 or 1 = 2.

10. An infinitely fast fall

Imagine a cat sitting on the top of a ladder leaning against a wall.

The bottom of the ladder is pulled away from the wall horizontallyat a uniform rate. The cat speeds up, until its falling infinitely fast.

The proof is below.

x

y l


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By the Pythagoras Theorem22

xly , where )(txx , )(tyy

are the horizontal and vertical distances from the ends of the

ladder to the corner at time t. Differentiation of both sides with

respect to t gives us 22 xl

xx

y

. Since the ladder is pulled

uniformly x is a constant. Let us find the limit of y when x

approaches l:

22limlim

xl

xxy

lxlx. When the bottom of the

ladder is pulled away by the distance lfrom the wall, the cat falls

infinitely fast.

11. A positive number equals a negative number

a) The functionx

xf2cos1

sin)(

is continuous and non-negative on

the interval

4

3,0

. Therefore by the definition of the definite

integral the area enclosed by the functionf(x) and the x-axis on the

interval

4

3,0

is a positive number.

b) On the other hand, since the function )(sectan)( 1 xxF is an

antiderivative of the function f(x) (this is easy to check by

differentiation) calculating the area as the integral off(x) on

4

3,0

we receive a negative number: .4

2tancos1

sin 14

3

02

dxx

x

Hence a positive number equals a negative number.


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Solutions to Paradoxes

1. Solutions to Paradoxes: Functions andLimits

1. Laying bricks

The top brick can be infinitely far from the bottom brick! The

x-coordinate of the position of the centre of mass of a system of n

objects with masses m1, m2,, mnis defined by the formula:

n

nn

mmmxmxmxmx

......

21

22110 .

Let us consider two bricks. For the upper brick not to fall from the

lower brick the perpendicular distance from the centre of mass of

the upper brick should not be beyond the right edge of the lower

brick. That is, the maximum value of the xcoordinate of the centre

of mass of the upper brick is l: lx 0 . So the maximum shift is:

21

lx .

lx

yy


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Let us consider three bricks.

For the top brick we have the maximum possible shift:2

1

lx . Let

us find the maximum possible shift for the middle brick. Again, the

perpendicular distance from the centre of mass of the system of the

middle and top bricks should not extend past the right edge of the

lower brick. In other words, the maximum value of the xcoordinate

of the centre of mass of the system of the middle and the top bricksis l: lx 0 . Expressing x0 for the system of the middle and the top

bricks from (1) we obtain:

lm

llxm

lxm

2

)22

()2

( 22.

From here4

2

lx .

In a similar way we can obtain:nlxlxlx n2

,...,8

,6 43 .

Adding all shifts we receive: )1...2

11(2

...21n

lxxx n .

When n the sum in the brackets tends to infinity. This means

that the maximum possible shift of the top brick with respect to

the bottom brick can be made as large as we want.

Comment: In practice it is of course impossible. Starting from a

certain value of nwe will not be able to make shifts of the length

n

l

2as they will be too small to perform.

lx

y


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2. Spiral curves

a) Let us construct a spiral curve of a finite length.

Draw a line segment of length d. Draw a semicircle with diameter d

on one side of the line segment. Then on the other side of the line

segment draw a semicircle of diameter d/2. Then on the other side

draw a semicircle of diameter d/4, and so on.

The length of the curve is:

ddddd

...)8

1

4

1

2

1(...

842.

b) Let us construct a spiral curve of infinite length. Draw a line

segment AB of length d with midpoint P. Draw a circle with the

centre C on the line segment at distance afrom P. On one side of

the line segment draw a semicircle of the diameter d.On the other

side of the line segment draw a semicircle of the diameter AE,

where point E is the midpoint of PB. Then on the other side of the

line segment draw a semicircle of the diameter EF, where point F isthe midpoint of AD and so on (see the following diagram).


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The curve has infinitely many rotations around point C and each

rotation has a length bigger than the circumference 2a, so the

length of the curve is infinite.

3. A tricky curve

One example of such a curve is the famous Koch snowflake. We

start with an equilateral triangle and build the line segments on

each side according to a simple rule. At every step each line

segment is divided into 3 equal parts, then the process is repeated

infinitely many times. The resulting curve is called the Koch curve

or Koch snowflake and is an example of a fractal. The first four

iterations are shown below:

The initial triangle and all consecutive stars and snowflakes are

located between the circumferences inscribed into the triangle andcircumscribed around it. Both circumferences have finite lengths.

If the perimeter of the initial triangle is 1 unit, then the perimeter


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of the star in the first iteration is3

4

9

112 units. The perimeter of

the snowflake in the second iteration is

2

3

4

9

16

27

148

units.

The perimeter of the snowflake in the nthiteration is

n

3

4units.

As nthe perimeter of the snowflake tends to infinity. The Koch

curve has an infinite length but bounds a finite area, which is

between the area of the circle inscribed into the initial triangle and

the area of the circle circumscribed around it.

4. A tricky area

Although at every step the remaining area is 8 times bigger than

the area removed, after infinitely many steps the remaining area

will be zero and the area removed will be 1 square unit. Let us

show this. After the first step the remaining area equals .9

8

9

11

After the second step the remaining area is

2

9

8

81

64

81

189

8

.

Similarly, after the nth step the remaining area is

n

9

8 and if n

tends to infinity this area tends to zero. This figure is called the

Sierpinski carpet and is another example of a fractal.

5. A tricky next term

a) The expected answer is 32, but there are infinitely many other

correctanswers. Actually the next term in the sequence 2, 4, 8, 16

can be anynumber.

Let the nthterm be xnnnna n

n )4)(3)(2)(1(2 .The first 4 terms are 2, 4, 8, 16. One can make 5a equal to any

number by determining xfrom the formula for 5a . This can be done

to obtain the formula for the nthterm. For example, let the 5thterm

be 4. From the equation x 123424 5 we obtain x = -1.5

and the formula for the nthterm is

)5.1)(4)(3)(2)(1(2 nnnna nn .

b) Another interesting example has a geometrical flavour. Draw acircle, and put two dots on the circumference and connect them

with a line segment. The circle is divided into 2 regions. Put a third


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dot and connect all dots. The circle is now divided into 4 regions.

Put a fourth dot and connect all dots. The circle is now divided into

8 regions. Put a fifth dot and connect all dots. The circle is now

divided into 16 regions. It looks like we have a clear pattern. But

when you put the sixth dot and connect all dots the circle is divided

into 30 regions!

6. A tricky shape

No, the student was not right. A figure can be of constant diameter

yet not be a circle. As an example, consider the following curve. In

an equilateral triangle draw circular arcs with the radius equal to

the side of the triangle from each vertex. The resulting figure is a

curved triangle, which is called the Reuleaux triangle (see thediagram below). One of its properties is that it has a constant

diameter. When it rolls on a horizontal surface its center moves

along a sine curve with ups and downs (unlike a circle whose

center does not move up and down only along a straight

horizontal line). For this reason it is not practical to use it as a

wheel, but it does have practical applications. It is used in the

Wankel rotary engine, and in some countries manhole covers are

shaped like the Reuleaux triangle.

7. Rolling a barrel

The barrel rolls as long as the person continues walking. The

velocity of the point on the top of the barrel equals the velocity of

the walking person and is twice the velocity of the axis of the barrel,

so the person will cover 6 m by the time he reaches the barrel.

8. A cat on a ladder

Part 1. Most people are confident that C is the answer to Part 1.

Without much difficulty, one can imagine the ladder rotating about

a central point, i.e. where the base of the ladder touches the wall.


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An arc is the result and in this case represents a quarter of the

circumference of a circle.

Part 2. However, Part 2 isnt so easy! Many people conclude that A

is the correct answer. It sounds reasonable that as the ladder

slides outwards away from the wall that it would appear to drop

quickly, then level out as it approaches the horizontal.

Surprisingly, the answer to this problem is also C. Try it out bymaking a model (see the sequential sketches below). With a paper

ladder, with a point marking half way, slowly slide the ladder down

and away from the wall. After each small amount of movement, put

a dot on the page at the place where the centre of the ladder lies.

Note that as the ladder approaches the horizontal, further lateral

movement is minimal.

Here is a simple proof:


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Let AB be the ladder. Point C (where the cat sits) is always at the

same distance (half the length of the ladder) from the point O

regardless of the position of the ladder. This comes from the fact

that diagonals in a rectangle are the same and are divided in half

by the point of their intersection.

You may well be very surprised to see that the trajectory is the

same in both cases. Do not be alarmed however in this case theintuition of many people fails. My colleague tried this test out with

a class of 100 4thyear engineering students in Australia, Germany,

New Zealand and Norway. These young men and women, aged

about 21, are expected to be able to quickly conceptualise shapes,

dimensions, movements and forces. The students were given 40

seconds to find the answer. They were told that it was a mental

exercise, with no calculations or drawings permitted. The results

were startling, for although 74% of the students gave C, the correct

answer to Part 1, 86% were wrong in Part 2. 14% and 34% gave B

as the answer in Parts 1 and 2 respectively.

9. Sailing

The top of the yacht has covered the longest distance. The shape of

the Earth is approximately spherical, so the top of the yacht has

the longest radius compared to lower parts and therefore has the

longest circumference.

O A

BC

D


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10. Encircling the Earth

Approximately 3 m high. This is a surprising answer for many

people. Let rbe the radius of the Earth and R be the radius of the

circle after adding 20 metres to the rope. The difference between

the two circumferences is 20 m: 2022 rR or .20)(2 rR

From here the difference between the two radii is .3mrR The

answer does not depend on the original length of the rope.

11. A tricky equation

The rough sketch of the graphs is too rough. Both functions are

decreasing for all xin their domains but they are very close to both

axes, and in fact have 3 intersection points. It can be shown that

the equation has 3 solutions.


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Documents

Paradoxes and Sophisms in Calculus