EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS ?· EDEXCEL NATIONAL CERTIFICATE UNIT 4 – MATHEMATICS…

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  • D.J.Dunn www.freestudy.co.uk 1

    EDEXCEL NATIONAL CERTIFICATE

    UNIT 4 MATHEMATICS FOR TECHNICIANS

    OUTCOME 1

    TUTORIAL 1 - INDICES, LOGARITHMS AND FUNCTION

    Determine the fundamental algebraic laws and apply algebraic manipulation techniques to the

    solution of problems involving algebraic functions, formulae and graphs

    Learning outcomes

    On completion of this unit a learner should: 1 Know how to use algebraic methods

    2 Be able to use trigonometric methods and standard formula to determine areas and volumes

    3 Be able to use statistical methods to display data

    4 Know how to use elementary calculus techniques.

    OUTCOME 1- Know how to use algebraic methods

    Indices and logarithms: laws of indices (am

    x an = a, = am+n, nmn

    m

    aa

    a , (am

    )n = a

    mn)

    laws of logarithms (log A + log B = log AB, log An = n log A,

    B

    AlogB logA log )

    e.g. common logarithms (base 10), natural logarithms (base e), exponential growth and decay

    Linear equations and straight line graphs: linear equations e.g. y = mx + c;

    straight line graph (coordinates on a pair of labelled Cartesian axes, positive or negative

    gradient, intercept, plot of a straight line);

    experimental data e.g. Ohms law, pair of simultaneous linear equations in two unknowns

    Factorisation and quadratics: multiply expressions in brackets by a number, symbol or by another expression in a bracket; by

    extraction of a common factor

    e.g. ax + ay, a(x + 2) + b(x + 2);

    by grouping e.g. ax - ay + bx - by,

    quadratic expressions e.g. a2 + 2ab + b

    2;

    roots of an equation e.g. quadratic equations with real roots by factorisation, and by the use of

    formula

  • D.J.Dunn www.freestudy.co.uk 2

    1. INTRODUCTION

    Before we had electronic calculators, accurate calculations involving multiplication and division

    were done with the aid of logarithms, pen and paper. To do this, the logarithms of numbers had to

    be looked up in tables and added or subtracted. This is easier than multiplying and dividing. The

    greater the accuracy needed, the larger the tables. To do less accurate calculations we used slide

    rules and these were devices based on logarithms. Although we do not need these today, we do use

    logarithms widely in mathematics as part of the wider understanding of the relationship between

    variables so this is an important area work. In order to understand logarithms, it is necessary to

    understand indices and we should start with this.

    2. INDICES

    In algebra, a way of writing a number or symbol such as 'a' that is multiplied by itself 'n' times is an

    n is called the index. For example a3 is the shorthand for a x a x a or a.a.a to avoid use of the

    multiplication sign. an is called the n

    th power of a. There are four laws that help us use this to solve

    problems.

    Law of Multiplication notable results

    a x . a

    y = a

    x + y

    Law of Division

    yx

    y

    x

    aa

    a

    0xx

    x

    x

    aaa

    a1

    xx

    aa

    1

    Law of Powers

    ax . a

    y . a

    z = a

    x+y+z

    a . a . a ....n times = an

    y = a n then a =

    ny

    ax . a

    x . a

    x ....n times = a

    nx (a

    x)n = a

    nx

    Law of Roots

    y = a1/n

    + a1/n

    + a1/n

    ....n times a1/n

    = ny

    y = (a1/n

    ) n

    a1/n

    = na

    WORKED EXAMPLE No.1

    Simplify the following

    5

    332

    a

    aas

    SOLUTION

    s = a 2 . a

    3 . 3 a

    - 5

    s = a 2 + 9 - 5

    = a 6

  • D.J.Dunn www.freestudy.co.uk 3

    SELF ASSESSMENT EXERCISE No.1

    Simplify the following.

    C = x 4 . x

    3 (C = x

    7 )

    4

    5

    d

    dF (F = d)

    6

    35

    b

    .bbA (A = b

    2 )

    D = a . a 3 (D = a

    3.5 )

    a

    aS

    3

    (S = a2.5

    )

    2

    52

    xy

    yxS (S = x. y

    3)

    3. DEFINITION of a LOGARITHM

    The logarithm of a number is the power to which a base number must be raised in order to produce

    it. The most commonly used base numbers are 10 and the natural number e which has a rounded

    off value of 2.7183 (also known as Naperian Logarithms). If you have not come across this number

    yet, dont worry about where it comes from but you need to know that it possesses special

    properties.

    BASE 10

    This is usually shown as log on calculators but more correctly it should be written as log10. Since it

    is the most widely used, it is always assumed that log means with a base of 10.

    Suppose we want the log of 1000. We should know that 1000 = 103 so the log of 1000 is 3.

    Similarly : The log of 100 is 2 since 102 is 1000.

    The log of 10 is 1 since 101 is 10.

    The log of 1 is 0 since 100 is 1.

    The log of 0.1 is -1 since 10-1

    is 0.1 and so on.

    This is all well and good if we are finding the log of multiples of 10 but what about more difficult

    numbers. In general if y = 10n then n is the log of y and without calculators we would have to look

    them up in tables. You can use your calculator.

    SELF ASSESSMENT EXERCISE No.2

    Use your calculator to find the log of the following numbers. Just enter the number and press the

    log button.

    260 (2.415) 70 (1.845) 6 (0.778) 0.5 (-0.301)

  • D.J.Dunn www.freestudy.co.uk 4

    BASE e

    Consider the expression

    n

    n

    11x

    and try evaluating it for bigger and bigger values of n. For

    example:

    n = 1, x = 2

    n = 10, x = 2.594

    n = 100, x =2.705

    n = 1000, x = 2.717.

    n = , x2.7183 (rounded off)

    The value when n = is called the natural number or exponential and denoted with 'e'. For practical

    purposes we take e = 2.7183. Further explanation is not given here but this number has very special

    properties and a set of logarithms with the base of e was originally devised by Robert Napier and

    are used for good reasons. These are called NATURAL LOGARITHMS or sometimes Naperian

    Logarithms and they are denoted loge or ln (as it appears on most calculators).

    If y = en then n is the natural logarithm. You get it from your calculator by simply entering the

    number and pressing the button marked ln.

    SELF ASSESSMENT EXERCISE No.3

    Use your calculator to find the ln of the following numbers. (answers in red)

    260 (5.561) 70 (4.248) 6 (1.792) 0.5 (-0.693)

    OTHER BASES

    Your calculator may allow you to find the logarithms to other bases by programming in the base

    number but this wont be covered here. Here are some simple examples.

    WORKED EXAMPLE No.2

    Find the log2 of the number 8. Since 23 = 8 then log2(8) = 3.

    Find log3(81). Since 34 = 81 then log3(81) = 4

    SELF ASSESSMENT EXERCISE No.4

    Find the following (answers in red)

    log2(16) (4) log3(27) (3) log5(625) (4)

  • D.J.Dunn www.freestudy.co.uk 5

    4. ANTILOGS

    An Antilog is the number that gives us the logarithm or put another way, the number resulting from

    raising the base number to the power of the logarithm.

    For example if the base is 10, Antilog(2) = 102 = 100 Antilog(3) = 10

    3 = 1000

    On a calculator this is usually shown as 10x and is often the second function of the same key as

    log10.

    If the base is e then for example anti-ln(5.561) = e5.561

    = 260 and so on. On a calculator this is the

    button marked ex and is often the second function of the same key as ln.

    5. POWER LAWS

    If A = (xn) and B = (x

    m) then AB = x

    n+m

    If x is the base of our logarithms then n = logx(A) and m = log x(B) and log x(AB)= log xA+ log xB

    This is useful because if we can look up the logs of numbers we can solve multiplication problems

    by adding the logs.

    NEGATIVE POWERS (INDICES)

    If mnm

    n

    xx

    x

    B

    Ay then mnlog(B)log(A)

    B

    Aloglog(y)

    WORKED EXAMPLE No.3

    Solve y = (36.5)(17.72)

    Taking logs we have log y = log(36.5) + log(17.72) = 1.562 + 1.248 = 2.811

    y = antilog(2.811) = 102.811

    = 646.78

    WORKED EXAMPLE No.4

    Solve y = (36.5) (17.72)

    Taking logs we have log y = log(36.5) - log(17.72) = 1.562 - 1.248 = 0.314

    y = antilog(0.314) = 100.314

    = 2.060

    Of course we can get the same answers on our calculators without this process but it is very useful to change multiplication into adding and division into subtraction.

    SIMPLIFYING NUMBERS WITH POWERS (INDEXES)

    ROOTS

    You know that: n1

    n AA hence if log(A)n

    1log(y) then AAy n

    1

    n

  • D.J.Dunn www.freestudy.co.uk 6

    WORKED EXAMPLE No.5

    Find the fifth root of 600 using logarithms.

    594.310 566)antilog(0. y

    0.556 2.7785

    1 log(600)

    5

    1 y log 600600y

    0.556

    5

    1

    5

    This can be done directly on a calculator to check the answer but the basic transformation is

    very useful in derivations and manipulation of formulae.

    6. DECIBELS

    The ratio of two numbers can be expressed in decibels. The definitions is G(db) = 10 log(G) where

    G(db) is the ratio in decibels and R is the actual ratio. This is commonly applied to equipment in

    which there is a change in POWER such that G = Power Out/Power in and the G is the Gain. The

    reason for doing this is that if you put two such items in series the overall gain is:

    G(over all) = G1 G2

    Taking logarithms G(db) = G1(db) + G2(db)

    The gains in db are the sum of the individual

    gains.

    7. PRACTICAL EXAMPLES OF LOGARITHMS

    WORKED EXAMPLE No.6

    An electronic amplifier increases the power of the signal by a factor of 20. What is the gain in

    decibels?

    SOLUTION

    G(db) = 10 log 20 = 13 db

    The amplifier is fed into another amplifier with a gain of 5. What is the overall gain in decibels?

    The gain of the second amplifier is G = 10 log 5 = 7 db

    The total gain is 13 + 7 = 20 db

    Check this way:- Overall gain = 20 x 5 = 100 In decibels G = 10 log 100 = 20 db

    WORKED EXAMPLE No. 7

    The ratio of the tensions in a pulley belt is given by R = e2.5

    Find the value of when R is 5.

    SOLUTION

    5 = e2.5

    So take natural logs and ln(5) = 2.5 ln(e) and by definition ln(e) is 1

    ln(5) = 2.5 = 1.609 = 1.609/2.5 = 0.644

  • D.J.Dunn www.freestudy.co.uk 7

    WORKED EXAMPLE No. 8

    A well known formula used in the analysis of damped vibrations is -1

    2

    x

    xln

    22

    1

    Where x1 and x2 are the amplitude of two successive vibrations and is the damping ratio.

    Calculate when x1= 3 mm and x2 = 0.5 mm respectively. Calculate the amplitude reduction

    factor and the damping ratio.

    SOLUTION

    0.2740.075 and 0.07513.298

    1 and 1 13.298

    12.298-1 so -1

    39.4783.21

    sidesboth square -1

    2 1.792

    1.792 ln60.5

    3ln

    x

    xln

    22

    22

    2

    2

    2

    2

    1

    WORKED EXAMPLE No. 9

    When a gas is compressed from pressure volume V1 and temperature T1 to a final volume V2

    and temperature T2 the relationship is :

    C

    2

    1

    1

    2

    V

    V

    T

    T

    where C is a constant. Given 5.2

    T

    T

    1

    2 and

    8V

    V

    2

    1 determine C

    SOLUTION

    C85.2 Take logs and log(2.5) = C log(8) C = log(2.5)/log() = 0.398/0.903 = 0.441

    SELF ASSESSMENT EXERCISE No.5

    1. An amplifier has a gain of 32. What is this in decibels? (Answer 15 db)

    2. Given y = xn determine the value of n when y = 6 and x = 20 (Answer 0.598)

    3. Given 12 = e find . (Answer 2.485)

  • D.J.Dunn www.freestudy.co.uk 8

    8. EXPONENTIAL GROWTH AND DECAY

    It was shown earlier that the exponential 'e' is the value of

    11 2.7183

    Exponential growths refer to the way things change with time under certain circumstances. Here is

    an example:

    Suppose that you have 100 in a bank account. Every year you add 12% interest to the sum in your

    account.

    After year 1 you would have 100 + 12% x 100 = 112

    Suppose instead you added 6% every 6 months.

    After 6 months you would have 100 + 100 x 6% = 106

    After a year you would have 106 + 106 x 6% = 112.36 which is more than previously.

    Suppose that you added 1% every month. This would take 12 calculations but after a year you

    would have 112.68.

    If you took it down to 1 second intervals the sum would be 112.75 after 1 year.

    If we took smaller and smaller intervals of time we would get no closer to the sum when rounded

    off to the nearest penny. This is known as continuous compound interest. It can be shown that the

    way to calculate the sum for continuous compound interest is

    Sum = (Original sum) x enp

    N is the number of years and p is annual rate of interest.

    Hence after 1 year compound interest the sum is

    100 x e0.12

    = 100 x 1.1275 = 112.75

    After 40 years it is 100 x e0.12x40

    = 12 151

    This is called an exponential growth. It starts off growing

    slowly but as time goes on it grows quicker and quicker.

    If you invested 100 at 12% compound interest it would

    grow over the years like this.

    In nature and in science many things grow

    and decay in this manner. A prime example

    is when something fills or empties through a

    resistance. For example if you allowed water

    to run out of a tank through a hole in the

    bottom, the level would fall with time like

    this. This is an exponential decay where the

    level falls quickly at the start but then falls

    slower and slower as time goes on.

    An exponential growth follows the mathematical law Tt

    Aex and an exponential decay follows the

    law Tt

    Aex

    where A is the starting value, T is a constant and t is time. Further studies will show

    that these curves and equations have special properties.

    There is one other graph that you should know about. This is when something grows quickly at the

    beginning and then slows down exponentially. This is given by the equation

    T

    t

    e-1Ax

    This is illustrated in the following example.

  • D.J.Dunn www.freestudy.co.uk 9

    When an electrical capacitor is charged through a resistance from a battery of voltage Vs, the

    voltage 'Vc' across the capacitor grows with time 't' in such a way that

    T

    t

    s e-1Vv . T is constant

    called the time constant (T = RC) and the larger the value of T the slower the change. The natural

    number 'e' is called the exponent in this case hence we have an exponential growth. A plot of the

    above equation shows the

    If we plotted the voltage 'VR' across the resistor, we would find that since Vs = VR + VC then VR

    falls with time so that Tt

    sR eVV

    and we have an exponential decay.

    THE MEANING OF THE TIME CONSTANT

    Starting with the equation

    T

    t

    sc e-1VV

    If we put t = T we have

    S1STt

    SC V 0.633e-1 Ve-1 VV

    So the time constant is the time taken for the

    voltage to change to 63.3% of the final value.

    This is useful when finding T from a graph.

    Note in theory that the final voltage is never reached but if we calculate the voltage at t = 4T we

    find V = 98.2%VS. We can say effectively that it takes 4T for a capacitor to almost completely

    charge or discharge.

  • D.J.Dunn www.freestudy.co.uk 10

    WORKED EXAMPLE No. 10

    A capacitor of value 50 F is charged from zero to 100 V through a 5 M resistor. Calculate the

    time constant and the time taken for the voltage to rise to 50 V.

    SOLUTION

    T = RC = 50 x 10-6

    x 5 x 106 = 250

    5.05.01e e-1 0.5 e-1 10050 e-1 VV 250-t

    250

    -t

    250

    -t

    T

    t

    SC

    0.6931ln(0.5)250

    t

    t = 250 x 0.6931 = 173.2 seconds

    SELF ASSESSMENT EXERCISE No. 6

    1. The graph shows a charging curve for a capacitor and resistance. Work out the time constant T

    and determine the capacitance if the resistor value is 6 k.

    (Answers 0.12 s and 2 F)

    2. A tank of compressed air contains a pressure of 20 bar. When a small leak is made, the pressure

    falls with time such that the Tt

    e02p

    and the resulting plot is shown below. Determine the time

    constant T in minutes from the graph and calculate how long it takes for the pressure to fall to 10

    bar. Confirm this from the graph.

    (Answer 2.2 minutes and 1.52 minutes)

    3. If 500 is invested at 5% annual compound interest, what is the total sum after 10 years?

    (Answer 824.36)

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