Mathematical Foundations, Supplementary notes

8/10/2019 Mathematical Foundations, Supplementary notes

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Mathematical FoundationsSupplementary notes

Juan Carlos Ponce Campuzano [email protected]

March 24, 2017

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Contents

1 Solving inequalities 7

1.1 Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Functions 14

2.1 Historical note . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 The denition of function . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3 How to nd the domain and range of functions . . . . . . . . . . . . . 16

2.4 Geogebra applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Logarithms 19

3.1 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3 Historical note: The origin of logarithms . . . . . . . . . . . . . . . . . 23

4 Applications of the derivative 27

4.1 Optimisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5 Integration 30

5.1 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . 30

5.2 Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6 Vectors: Some basic properties 35

7 Applications of vectors 36

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7.1 What is a Force? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

7.1.1 Weight, Normal reaction, and Friction . . . . . . . . . . . . . . . 36

7.1.2 Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

7.2 Motion in a straight line . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7.2.1 Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7.2.2 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

8 Sequences and series 47

9 Mathematical Induction 48

10 Miscellaneous problems 50

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Preface

Solving problems can be considered as a practical ability, like swimming, drivinga car or playing the guitar. It is something that you can learn by imitation andpractice. If you wish to learn swimming you have to go into the water, and if youwish to become a problem solver you have to solve problems.

In order to solve mathematical problems you may need more than just practice andimitation. Our knowledge about any subject consists of information and of know-how. If you have had experience of mathematical work at any level, elementaryor advanced, there will be no doubt that, in mathematics, know-how is much moreimportant than mere possession of information.

Of course, there is not a magic key that opens all the doors and solves all the prob-lems, but for years people have tried to establish some methods. For example, GeorgePlya (1887-1985) suggests the following steps when solving a mathematical problem[3]:

1. First, you have to understand the problem.

2. After understanding, then make a plan.

3. Carry out the plan.

4. Look back on your work. How could it be better?

If such technique fails, try to solve rst some related problem. Could you imagine amore accessible related problem? Or, if you can not solve a problem, then there is aneasier problem you can solve: nd it! A solution that you have obtained by your owneffort or one that you have read or heard, but have followed with real interest andinsight, may become a pattern for you, a model that you can imitate with advantagein solving similar problems. Just keep in mind that the ability to solve problems not merely routine problems but problems requiring some degree of independence, judgment, originality, creativity is not easy to achieve and takes time to improve;however, the more you practice, the better you become.

The present notes cannot offer you (and no book will ever be able to offer you) a uni-versal perfect method for solving problems, but it offers you examples for imitationand many opportunities for practice.

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Important:

This notes are a supplementary material for the course Mathematical Foundationsand are constantly updated. If you nd a typo or a factual error, by all means let meknow: [email protected]

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1 Solving inequalities

Solving inequalities algebraically is similar to solving equations, except for one small but important detail: you ip the inequality sign whenever you multiply or dividethe inequality by a negative . The easiest way to show this is with some examples.

Example 1.1. Consider the problem of solving the inequality 2x + 3 < 1. So wehave

2x + 3 < 12x + 3 3 < 13

2x < 22x2

> 22

x > 1.We can also write the solution as: x (1, ). Notice that when we divided bythe negative two, we had to ip the inequality sign. Graphically the solution isrepresented in Figure 1.

Figure 1 : Number line representation of x>

1.

Note that the solution to a "less than, but not equal to" inequality is graphed withan open dot at the endpoint, indicating that the endpoint is not included within thesolution.

Example 1.2. Consider now the problem of solving the inequality 3 12x 52x.So we have

3 12x 52x

3 5 2x + 12

x

2 32

x

2(2) 32

x(2)4 3x43 x.

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In this case, when we multiplied by the negative two, we had to ip the inequalitysign. Graphically the solution is represented in Figure 2.

Figure 2 : Number line representation of x(4/3,

).

Note that the solution to a "less than or equal to" inequality is graphed with a closeddot at the endpoint, indicating that the endpoint is included within the solution.

In the previous examples, the inequalities are called linear inequalities because weare dealing with linear expressions like " x 2" ("x > 2" is just "x 2 > 0", beforeyou nished solving it). Moreover, in both examples, we have used two important

properties of the real numbers:

Let a, b and c be real numbers. We have that

1. If a < b and c < 0 , then ac > bc.

2. If a < b and c < 0 , then a/ c > b/ c.

When we have an inequality with " x2" as the highest-degree term, it is called aquadratic inequality . In this case, the method of solution is different and not unique.

Example 1.3. Consider the problem of solving the following inequality:

2x2 + 5x + 12 < 0.Factoring, we get y = (2x 3) (x 4). Thus we have

2x2 + 5x + 12 < 0(2x 3) (x 4) < 0

We need to analyse two cases due to the next property of the real numbers.

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If a, b are real numbers such that ab < 0 , then we have two possible cases:

1. a < 0 and b > 0; or 2. a > 0 and b < 0.

Case 1:

2x 3 < 0 and x 4 > 0

2x < 3 x > 4

2x

2 >

3

2x >

3

2Both inequalities are represented in Figure 3. In this case, we have that x > 4, orx(4,

).

Figure 3 : Number line representation of x(4, ).

Case 2:

2x 3 > 0 and x 4 < 0

2x > 3 x < 4

2x2

4. In other words,x ,

32

(4, ).

Figure 5 : Number line representation of x( , 32 )(4, ).

The following examples with rational inequalities are more complicated.

Example 1.4. Consider the inequality

3x + 4 < 2.

In order to determine the values of x, rst we need to change the format of theinequality as follows:

3x + 4

< 2 3x + 4

+ 2 < 0 3 + 2x + 8

x + 4 < 0

2x + 11x + 4

< 0

Now we can proceed to solve the inequality. As we did in Example 1.3, we need to

analyse two cases due to the next property of the real numbers.

If a, b are real numbers such that ab < 0 , then we have two possible cases:

1. a < 0 and b > 0; or 2. a > 0 and b < 0.

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Case 1:

2x + 11 < 0 and x + 4 > 0

2x < 11 x > 4

2x2

< 112

x < 112

In this case the values of x have to satisfy that x >

4 and x 4 and x < 112 .

Case 2:

2x + 11 > 0 and x + 4 < 0

2x > 11 x < 4

2x2

> 112

x > 112

In this case the values of x have to satisfy that x < 4 and x > 112 . Therefore

112 < x < 4. This is represented in Figure 7.11


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Figure 7 : Number line representation of x < 4 and x > 112 .

In other wordsx( , 4) 112 , = 112 ,4 .

Finally, considering just Case 2, we have that 3x+ 4 < 2 provided that 112 < x < 4.That is to said

x112

,4 .

Observation: To solve an inequality is also equivalent to determine whichvalues of the domain of a function satisfy a particular condition. [ 4, p. 47]

Consider the function f (x) = 3/ (x + 4) which domain is

D = ( , 4)(4, ) .To solve the inequality

3x + 4

< 2means that we have to determine which values of the domain D satisfy that

f (x) < 2.

In this case, the function f (x) = 3/ (x + 4) is less than 2 only in the interval112 ,4 . In Figure 8, we can appreciate this condition.

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Figure 8 : The green segment represents the interval 112 ,4 .

1.1 Practice

Exercise 1.1. Solve the following inequalities:

a) 3x > 5 2x Answer a.

b) 1x2 4

0 Answer b.

c) |x 3| < |x + 4| 3 Answer c.d) 15x

2x

> 1 Answer d.

e) x2

5xx2 9 0 Answer e.f)

y + 12

0 and b = 1. If x is a positive real number, we write

logb x

to designate the logarithm of x to the base b (or the logarithm to the base

b of x), which is the (unique) real number y that satisesx = b y.

The most common bases for logarithms are 10 and e (the irrational mathematicalconstant 2.71828). The notations for logarithms to these bases are:

log10

x = log x and loge x = ln x

These are called common logarithm (or just logarithm ) and natural logarithm , re-spectively. Logarithms with respect to any base b can be determined using either of these two logarithms by the formula:

logb x = log xlog b

= ln xln b

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Example 3.2. The following results are obtained using the preceding formula:

a) log2 9 = log9log2

= ln 9ln 2 3.1699 . . . c) log 27 =

ln27ln10 1.4313...

b) log7 12 = log 12

log7 = ln12

ln 7 1.2769...

Basic logarithmic identities

Formula Example

log a 1 = 0 because a0 = 1 , where a > 0 log9 1 = 0 (or 90 = 1)

log a a = 1 because a1 = a log7 7 = 1 (or 7

1 = 7)

alog a x = x 11log11 x = x

log a(ax) = x log3(3

x) = x

The next table contains some useful rules which allow us to manipulate expressionsinvolving logarithms:

Formula Example

Product loga(xy) = log a x + log a y log2 (3 5) = log2 3 + log2 5

Quotient logax y

= log a x log a y log357

= log3 5 log3 7

Power loga (x p) = p log a x log7 3

11 = 11 log7 3

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3.1 Solved problems

Problem 3.1. Find the value of x if 7x+ 3 = 75x3.

Solution: Take the logarithm to base 7 of both sides. Thus

log7 7x+ 3 = log7 75x3(x + 3) log7 7 = ( 5x 3) log7 7

x + 3 = 5x 3x = 6

4

x = 32

.

Problem 3.2. Solve for x:

log 10x3 2 = log 110 .

Solution: First, notice that 2 = log 100. Then

log10x

3 log 100 = log 110

log

10x3

100 = log 110

10x300

= 110

100x = 300x = 3.

Problem 3.3. Find the value of x, if 2x 32x = 102x1.Solution:

Here we can apply logarithm to base 2, 3 or 10. It does not matter which base you use, all the answers will be equivalent. Lets use common logarithm tosolve this equation.

log 2x 32x = log 102x1log (2x) + 32x = (2x 1) log 10

x log2 + 2x log3 = 2x 1.Now, solving for x we obtain

x = 1log2 + 2log3 2 = 1

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Check the details! Do you get the same answer?

In this case, we can nd an equivalent answer using the formula:

logb = ln xln b

Considering that 2 = log 100 = ln100ln10

and 1 = log 10 = ln10ln10

, we have

x = 12 log2 2log3

=

ln10ln10

ln 100ln10

ln 2ln10 2

ln 3ln10

=

1ln10

ln 10

1ln10

(ln100 ln 2 2ln3)= ln10

ln 100 ln 2 2ln3= ln10

ln 1002

2ln3

= ln10ln50 2ln3

3.2 Practice

Exercise 3.1. Find de value of x for the following:

a) 2x2x = 1. Answer a. b) 4x 52x+ 1 = 113x1. Answer b.c) log2(x + 1) + log2(x) = 1. Answer c.

d) log 3(6x2 44x + 14) log3(2x 14) = 3. Answer d.

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3.3 Historical note: The origin of logarithms

Let us say it is the year 1900 and we are asked to compute the expression

x = 3 493.8(23.67)25.104 .

Of course, this is not an easy task and, probably, would take a long time. However,we can compute this expression using logarithms, which will be easier and faster. Inparticular, for this task, we need a table of four-place common logarithms, as the oneshown in Figure 16 (which can still be found in books of algebra or calculus or in theinternet).

Figure 16 : Four-Place Logarithms.

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We also need to use the laws of logarithms:

log (ab) = log a + log b,log (a/ b) = log a log b,

log (an) = n log a.

where a and b denote any positive numbers and n any real number; here log standsfor the common logarithm (that is, logarithm base 10), although any other base forwhich tables are available could be used.

We begin by writing the above expression in a form more suitable for logarithmiccomputation by replacing the radical with a fractional exponent:

x = 493.8 (23.67)2

5.104

1/3

.

Taking the logarithm of both sides, we have

log x = 13

(log (493.8) + 2log (23.67) log (5.104)) .We now nd each logarithm, using the Proportional Parts section of the table (Fig-ure 16) to add the value given there to that given in the main table. Thus, to ndlog (493.8) we locate the row that starts with 49, move across to the column headed by 3 (where we nd 6928), and then look under the column 8 in the Proportional Parts

to nd the entry 7. We add this entry to 6928 and get 6935. Since 493.8 is between100 and 1,000, the characteristic is 2; we thus have log 493.8 =2.6935. We do the samefor the other numbers. The complete computation is shown in the following table:

N log N 23.67 1.3742 2

2.7484

493.8 + 2.69355.4419

5.104 0.70794.7340

3

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For the last step we used a table of antilogarithms-logarithms in reverse (Figure17). We look up the number .5780 (the mantissa) and nd the entry 3784; since thecharacteristic of 1.5780 is 1, we know that the number must be between 10 and 100.Thus x = 37.84, rounded to two places.

Figure 17 : Four-Place Antilogarithms.

Sounds complicated? Yes, if you have been spoiled by the calculator or the personalcomputer. But with some experience, the above calculation can be completed in a

couple of minutes; on a calculator it should take no more than a few seconds. Try itwith WolframAlpha.

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Figure 18 : John Napier (1550-1617).

Before the advent of computers and hand-held calculators, logarithms (or their mechani-cal equivalent, the slide rule) were practicallythe only way to perform calculations, as the oneshown above. Logarithms were devised as amethod for rapid and accurate computations re-lated to different problems in astronomy, naviga-tion, engineering, and other areas. No wonderthe scientic community embraced them withsuch enthusiasm. The Scottish mathematician John Nappier (1550-1617), who worked for about20 years on the theory, is generally acknowl-edge as the founder of logarithms. Although herst used the word articial number, he nallyadopted the term logarithm which in Greek lit-erally means ratio number.

In 1614, Nappier published his book entitled Mirici Logarithmorum Canonis Descriptio(Description of the Wonderful Rule of Logarithms), which contained fty seven pagesof explanatory matter and ninety pages of tables related to logarithms using the baseb = 0.9999999 = 1 107.The English mathematician Henry Briggs (1561-1631) visited Napier in 1615, andproposed a re-scaling of Napiers logarithms to form what is now known as thecommon or base-10 logarithms. Napier readily agreed to Briggs suggestions, but bythen he was already advanced in years and lacked the energy to compute a new setof tables. Briggs undertook this task publishing his results in 1624 under the title Arithmetica Logarithmica, a work containing the logarithms of thirty thousand naturalnumbers to fourteen decimal places (1-20,000 and 90,001 to 100,000) [ 2, pp. 256-258].

From 1960s onwards, electronic calculators and computer rendered logarithms obso-lete for purposes of calculation. However, the concept remained vital to mathematics, because logarithms had found fundamental roles in many parts of mathematics, in-cluding calculus and complex analysis. Also many physical and biological processesinvolve logarithmic behaviour. For example, logarithms are frequently used to poseand solve problems related to interest rate, population growth, radioactive decay,mortgage loan, and the list goes on.

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4 Applications of the derivative

4.1 Optimisation

Problem 4.1. A company produces juice containers which have the shape of a box with squares

on the base and the top, and a capacity of 2 litres.

Figure 19 : Box with a constant capacity of 2 litres.

a) Find a formula for the surface area, A, of the container in terms of the apothem a of thebase. The apothem is the segment from the centre of the square to the midpoint of one

of its sides.b) Draw a rough sketch of A versus a.

c) Find the apothem of the container which will minimise the cost of the material required tomake the box.

d) Find the corresponding height and side of the square.

Solution: If a represents the apothem and h the height of the box, then we have thefollowing:

Surface area: A = 4 (2a) h + 2(2a)2Volume of box: 2 = ( 2a)2 hFrom the Volume of box , we obtain: h =

24a2

= 12a2

.

Thus, substituting the value of h in the equation of the Surface area , we obtain:

A = 4(2a) 12a2

+ 2(2a)2 = 4a + 8a2.

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This is the Surface area in terms of the apothem a. Notice that the function A(a)is dened for all a = 0, which means that a > 0 or a < 0. However, we are justinterested in the positive values of a, see Figure 20. To play with the dynamic repre-sentation of the box click Here .

Figure 20 : Graph of A(a) = 4a + 8a2 on the interval (0, 2).

Now let us nd the minimum of the function A(a). In this case we have that:

A (a) = 16a 4a2

If A (a) = 0, then

16a 4a2

= 0

16a = 4a2

16a3 = 4a3 = 4

16 = 1

4 = 1

22

a = 3 122 = 13 22 0.6299 . . .Remark: Notice that we are using the laws of exponents:

n am = am/ n, an am = an+ m an / am = anm and (an)m = anm.28
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The second derivative of A(a) is:

A (a) = 16 + 8a3

Substituting the value a = 13 22 in A (a), we obtain

A 13 22 = 16 +

8 13 22

3 = 16 + 8122

= 16 + 32 = 48

which is a positive value. Hence, the function of the surface area has a minimum

when a = 13 22 .

If S denotes the side of the square, then

S = 2a = 2 13 22 = 3 2 1.2599 . . .

Finally, the height of the box is:

h = 12a2

= 1

2 13 2

2

2 = 3 2 1.2599...

To check this values, let us calculate the volume of the box which is

Volume of box = S2 hwe get

Volume of box = 3 2 2 3 2 = 22/3 21/3 = 22/3 + 1/3 = 2

Exercise 4.1. Find two nonnegative numbers whose sum is 9 and so that the productof one number and the square of the other number is a maximum.

Answer: Here

Exercise 4.2. A container in the shape of a right cylinder with no top has surface area3 m2. What height h and base radius r will maximise the volume of the cylinder?

To play with a dynamic representation of the cylinder click: Here

Answer: Here

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Part b): The function f (x) = 3/ (x 2) is not dened for x = 2. Notice also that2 [2, 3]. This means that the function f (x) is not continuous on the interval[2, 3]. Therefore, we can not apply the FTC.This function behaves similar to the function 1/ x. Hence, we can not calculate thedenite integral:

32 3x 2dx

Figure 22 : Graph of f (x) = 3x 2

.

Remark: It is possible to use the FTC to calculate the denite integral overan interval where f (x) is continuous and well-dened. For example, usingthe antiderivative

F(x) = 3 ln |x 2| ,we have that

75 3x 2dx = 3 ln |(7) 2|3 ln |(5) 2|= 3 ln |5|3 ln |3| = 3 ln 5 3ln3= 3 ln

53 1.532

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Part c): The function f (x) =

x1x2 1

is not dened for x = 1 or x = 1. Notice also that 1, 1 [2, 1]. Because f (x) isnot continuous on the interval [2, 1], we can not apply the FTC.As in the previous problem, this function behaves similar to the function 1/ x, because

x 1x2 1

= x 1(x 1)( x + 1)

= 1x + 1

Hence, we can not calculate the denite integral: 12 x 1x2 1dx

Figure 23 : Graph of f (x) = x1x2 1

.

Part d): We need to use the method of substitution. If u = e2x

e3x

+ 7, thendudx

= 2e2x 3e3x = du = ( 2e2x 3e3x)dx.Multiplying by 3 we have

3du = ( 6e2x 9e3x)dx.Thus,

6e2x

9e3x

e2x e3x + 7dx = 3du

u = 3 1u du = 3 ln |u|+ C = 3 ln e

2x

e3x

+ 7 + C

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Problem 5.2. Calculate the following indenite integral

2x3 2x x4 2x23

dx

Solution: Let u = x4 2x2. Thendudx

= 4x3 4x = du2 = ( 2x3 2x)dxThus

2x3 2x x4 2x23

dx = ( x4 2x2 This is u)3

This is du/2

2x3 2x dx=

u3

du

2= 1

2 u3du= 1

2 u4

4 + C = 1

8u4 + C

= 18

(x4 2x2)4 + C.Problem 5.3. Calculate the following indenite integral

e2x

e2x 4dx

Solution: Let u = e2x 4. Thendudx

= 2e2x = du

2 = e2xdx

Thus

e2x

e2x 4dx = e2x

dxe2x 4=

du2u

= du2u= 1

2 1u du= 1

2 ln |u|+ C

= 12 ln |e

2x

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5.2 Practice

Exercise 5.1. Determine the following integrals, if possible. If not, state why.

a)

1

5

4

x3 + 2x

4

2x2

1 dx Ans. a) Here

b) /2 /2 sin (2x)cos (2x) dx Ans. b) Herec) 2xex

2

ex2 dx Ans. c) Here

d) 7x + 8x2 + x 2 dx Ans. d) Heree) sin5 x dx Ans. e) Here

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6 Vectors: Some basic properties

Matrix addition of vectors:

If v = v1

v2 and w = w1

w2,

thenv + w = v1 + w1v2 + w2

Scalar multiplication:

If v = v1v2 and t

R ,

thentv = t v1t v2

Norm of a vector:

If v = v1v2 then Norm of v = ||v|| = v21 + v22.

Let v, w be vectors and tR . Verify the following results:

0 ||v||, ||v + w | | | |v||+ ||w ||, ||t v|| = t ||v||

Component form of vectors:

If v = ab then v = ai + bj,

wherei = 10 and j =

01

Converting vectors from geometric to component form:

v = ||v||cos i + ||v||sin j

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7 Applications of vectors

7.1 What is a Force ?

A force is any interaction which tends to change the motion of an object.Thus, a force can cause an object with mass to change its velocity, that is,to accelerate. A force has both magnitude and direction, making it a vectorquantity. As a reminder:

Force = mass acceleration .The stander unit of measurement for forces is the newton, denoted by N.

7.1.1 Weight, Normal reaction, and Friction

Forces are given many names, such as push, pull, thrust, lift, weight, friction, andtension. Traditionally, forces have been grouped into several categories and givennames relating to their source, how they are transmitted, or their effects.

Weight: is a force due to gravity, and it is denoted by W. The magnitud of the vector weight, associated with an object of mass m kilograms (kg) is

||W|| = m gwhere g = 9.8 m/s 2 is the acceleration due to gravity.

Normal reaction: is a force that pushes at right-angle to a surface and itis denoted by N . The word normal means perpendicular to a surface. Thenormal reaction force can be less than the objects weight if the object is onan inclined plane.

Friction: is a force in the direction parallel to a surface. It is the force thatcounteracts an object sliding along a surface. This force is denoted by F.

Consider an object resting on an inclined plane that makes an angle with the hori-

zontal as shown in Figure 24. This means that the forces acting on that object are in balance. The force of gravity acting on the object is divided into two components: a

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force acting perpendicular to the plane, denoted by W y, and a force acting parallelto the plane, denoted by W x.

Figure 24 : An object rests on an inclined plane that makes an angle with the horizontal.

The perpendicular force of weight, W y, is typically equal in magnitude and oppositein direction to the normal reaction, N . The force acting parallel to the plane, Wx,causes the object to accelerate down the plane. The force of friction, F, opposes themotion of the object, so it acts upward along the plane (see Figure 24).

If the magnitude of the weight vector W is w, then the magnitudes of the weightcomponents are

Wx = w sin and W y = w cos

Instead of memorising the above equations, it is helpful to be able to determinethem from reason. To do this, consider the right triangle formed by the three weightvectors (see Figure 24). Notice that the angle of the inclined plane is the same asthe angle formed between the vectors W and W y.

Then you can use trigonometry to determine the magnitude of the weight compo-nents:

sin = Wxw and cos = W y

w

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Problem 7.1. A 3 kg brick is sitting on a inclined plane at /6 radians to the horizontal. Theonly forces acting on it are weight, normal reaction from the plane and friction. Determine themagnitude of each force.

Solution:

First, we have thatW = w = m g

where m = 3 kg and g = 9.8 m/s 2. Thus

w = 3 9.8 N = 29.4 N

ThusF = w

sin

6 = ( 29.4 N)

1

2 = 14.7 N

and

N = w cos 6

= ( 29.4 N) 32

= 14.7 3 N

Exercise 7.1. A 12 kg brick is sitting on a inclined plane at /3 radians to the horizontal. Theonly forces acting on it are weight, normal reaction from the plane and friction. Determine themagnitude of each force. Click here to check your answer

7.1.2 Tension

Now let us consider a system where a mass hangs from a wooden beam via a singlerope.

Tension: is a force along the length of a medium, especially a force carried

by a exible medium, such as a rope, string or cable. This force is denoted by T . The word tension comes from a Latin word meaning to stretch.

If neither the mass nor the rope are moving, the entire system is at rest. Because of this, we know that, for the mass to be held in equilibrium, the tension force mustequal the force of gravity on the mass, tension = force of gravity (see Figure 25).Thus

T = W

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Figure 25 : Representation of the vectors tension and weight: T = W .

Example 7.1. Assuming that the mass is 10 kg, then the magnitude of the tension force is:

T = 10 kg 9.8 m/s 2 = 98 N.

Gravity isnt the only force that can affect the tension in a rope. If, for instance, asuspended object is being pulled upwards by a force on the rope or cable, this force(mass acceleration) is added to the tension caused by the weight of the object. Inother words

T = W + F (7.1)where F is the force applied to pull the mass upwards.

Lets say that, in our example of the 10 kg mass suspended by a rope, instead of beingxed to a wooden beam, the rope is actually being used to pull the mass upwards atan acceleration of 1.5 m/s 2. In this case, the magnitude of the force applied to pullthe mass upwards is

F = ( 10 kg)( 1.5m/s2) = 15 N

Therefore, considering equation ( 7.1), we have

T = 98 N + 15 N = 113 N

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7.2 Motion in a straight line

7.2.1 Displacement

The displacement of an object moving in a straight line is dened as thevector from some initial point to a nal point.

If D represents the displacement of an object, then the distance covered bythe object is the magnitude of D . That is, distance covered = ||D ||.If D1, D 2, . . . , D n are displacements, the resulting displacement D r , is thesum

D 1 + D 2 + + D n

Figure 26 : Representation of displacement vectors.

Consider the following example: A person, from a point A, starts to walk 3.5 metersEast, then 2.3 meters South. After that, 3.5 meters West. And nally 2.3 meters North. Thisdisplacement is shown in Figure 26. Calculate his resulting displacement.

Even though the person has walked a total distance of 11.6 metres, that is, the sum

D 1 + D 2 + D 3 + D 4 ,

his resulting displacement is 0 (as a vector). In other words, there is no displacementfor his motion. Thus we have that

D R = D 1 + D 2 + D 3 + D 4 = 0i + 0j = 0

where D R is the resulting displacement and

D 1 = 3.5 i + 0 j, D2 = 0 i 2.3 j, D3 = 3.5 i + 0 j and D4 = 0 i + 2.3 j40


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Displacement, being a vector quantity, must give attention to direction. In the abovesituation the 3.5 meters East cancels with the 3.5 meters West; and the 2.3 metersSouth cancels with the 2.3 meters North.

Problem 7.2. A surveyor walks 174 metres due North. Then he turns clockwise through an

angle of 5 /6 radians and walks 234 metres. Finally he turns and walks 210 metres due West.Find his resulting displacement, relative to his starting point and the total distance covered

Solution: Figure 27 shows a geometrical representation of the displacement vectorsand its resulting displacement. Notice that 5 /6 is equal to 150 o.

Figure 27 : Geometrical representation of the displacement vectors (red) and its resulting displacement (blue).First, the total distance covered by the surveyor is 618 metres because

total distance = D 1 + D 2 + D 3 = 174 + 234 + 210 = 618

Now we need to nd D R, so we have to calculate

D 1 + D 2 + D 3.

For doing this, we need to nd the component form of each vector D i.

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7.2.2 Momentum

If an object is moving, then it has momentum or we can say that its mass is inmotion . The momentum of an object depends on two variables:

how much stuff is moving (its mass) and how fast the stuff is moving (its velocity).

Momentum is a vector quantity, possessing a direction as well as a magni-tude. It is dened as the product of the mass and velocity of an object. Inphysics, momentum is usually denoted by p . Thus

p = m vwhere m is the mass and v is the velocity of an object. The standard unitfor the magnitude of momentum is newton seconds, denoted Ns = kgm/s.The direction of the momentum vector p is the same as the direction of thevelocity vector v.

Principle of the conservation of momentum:

If objects collide, the total momentum before collision is equal to the to-tal momentum after collision (provided that no external forces act on thesystem). [ 4, p. 166]

Consider two objects with different masses m1 and m2. If the velocities of the twoobjects are u 1 and u 2 before the collision, and afterwards they are v1 and v2, then theabove principle establishes that

m1 u 1 + m2 u 2 Total momentum-Before

= m1 v1 + m2 v2 Total momentum-After

Problem 7.3. A car with a mass of 911 kg is moving East with speed of 60 km/h. A truck ismoving North. The car and truck collide, and after the collision the combined wreck heads in

a North-Easterly direction with speed 55 km/h. Using conservation of momentum, determinethe mass and the initial speed of the truck.

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Now, the principle of conservation of momentum establishes that

911 u 1 + m2 u 2 Total momentum-Before

= ( 911 + m2) v

Total momentum-AfterSubstituting and simplifying we get

911(16.67 i + 0 j) + m2(0 i + s2 j) = ( 911 + m2)15.28 2 i +

15.28 2 j

15 186.37 i + ( m2 s2) j = 15.28(911 + m2) 2 i +

15.28(911 + m2) 2 jThus we obtain the equations

15 186.37 = 15.28(911 + m2)

2 (7.2)m2 s2 =

15.28(911 + m2) 2 (7.3)From the rst equation ( 7.2) we obtain m2 = 494.548 and using this value for solvingequation ( 7.3), we obtain that s2 = 30.7076.

Therefore, the mass of the truck is 494.548 kg and its initial velocity is 30.7076 metresper second, or 110.55 km/h.

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9 Mathematical Induction

Mathematical induction is of no use for deriving formulas. But it is a good way toprove the validity of a formula that you might think is true.

Problem 9.1. Use mathematical induction to show

n

i= 1

(3i 2) = n(3n 1)

2 for n 1.

Solution:

Step one: We need to check that the identity is true for the smallest value of n, whichin this case is 1. So we have that

1

i= 1

(3i 2) = (1)( 3(1) 1)

2

3(1) 2 = 22

1 = 1

Step two: We assume that the identity is true for n = k . So we change n for k , and wehave

k

i= 1

(3i 2) = k (3k 1)

2

This step commonly known as the Hypothesis of Induction (H. I.).

Step three: We need to prove that the identity is true for n = k + 1. That is, we needto prove that

k + 1

i= 1

(3i 2) = (k + 1)( 3(k + 1) 1)

2

is true.

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In this case we start with the left hand side, so we have the following

k + 1

i= 1

(3i 2) =k

i= 1

(3i 2) + 3(k + 1) 2= k (3k 1)

2 + 3(k + 1)

2 Using the H. I.

= 3k 2 k 2

+ 3k + 3 2= 3k

2 k 2

+ 3k + 1

= 3k 2 k 2

+ 6k + 22

= 3k 2 + 5k + 2

2

= (k + 1)( 3k + 2)2

= (k + 1)( 3(k + 1) 1)2

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10 Miscellaneous problems

Exercise 10.1. Find the domain and range of

f (x) = 2

x 4Exercise 10.2. Find the range of f (x) = 7x 4

Exercise 10.3. Find the value of , if cos = 32 and 0 2 .

Exercise 10.4. Solve the inequality4x + 22x 3

32

Exercise 10.5. Solve the equation

1 + 2log2(x) = log2(7x + 9)

Exercise 10.6. If f (x) = x + 3

x + 2 and g(x) = x

+ 1x 1

then

a) Find f ( g(x)) .

b) What is the domain of f ( g(x)) ?Exercise 10.7. Suppose a gas is pumped into a spherical balloon at a constant rate of 50 cubiccentimetres per second. Assume that the gas pressure remains constant and that the balloonalways has a spherical shape. How fast is the radius of the balloon increasing when the radiusis 5 centimetres?

Exercise 10.8. A river ows due East at a speed of 1.3 metres per second. A girl in a rowingboat, who can row at 0.4 metres per second in still water, starts from a point on the South

bank and steers due North. The boat is also blown by a wind with speed 0.6 metres per secondfrom a direction of N20o E (see Figure 33).

a) Find the resultant velocity of the boat and its magnitude.

b) If the river has a constant width of 10 metres, how long does it take the girl to cross theriver, and how far upstream or downstream has she then travelled?

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Figure 33 : The red arrows represent the velocities of the boat (b ), wind (w ) and ow (r).

Exercise 10.9. Read the following answered problems. Can you determine whether they arecorrect? What do you think?

Problem 1.

8116

34= 81

16

34

= 813/4

163/4

=92 3/4

(42)3/4

= 278

Problem 2.

x1

x1

y2

y1

y2

y3

= x1

x1

y2

y1

y1

= x1 y2

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Problem 3. Solve for x: log1010x

3 2 = log10 110log10

10x3 log10 100 = log10

110

log10 10x

3100

= log10 110

log1010003x

= log10 110

10003x

= 110

3x = 10000x = 10000

3

Exercise 10.10. Show that the following identity is true:1

12 1/3 2

= 22/3

1/3

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Mathematical Foundations, Supplementary notes

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