EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS ...2007/outcome1t1.pdf · edexcel national certificate unit 28 – further mathematics for technicians outcome 1 advanced

© D.J.Dunn www.freestudy.co.uk 1

EDEXCEL NATIONAL CERTIFICATE

UNIT 28 – FURTHER MATHEMATICS FOR TECHNICIANS

OUTCOME 1

ADVANCED GRAPHICAL TECHNIQUES

CONTENTS

1 Be able to use advanced graphical techniques

Advanced graphical techniques: graphical solution e.g. of a pair of simultaneous equations with

two unknowns, to find the real roots of a quadratic equation, for the intersection of a linear and a

quadratic equation, non-linear laws such as

x

bayb,axy 2 by the use of logarithms to reduce laws of type y = ax

n to straight line

form, of a cubic equation such as 2x3

- 7x2

+ 3x + 8 = 0 , recording, evaluating and plotting e.g.

manual, computerised

It is assumed that the student has completed the module MATHEMATICS FOR

TECHNICIANS.


1. SIMULTANEOUS EQUATIONS - GRAPHICAL SOLUTION

Let's remind ourselves what a simultaneous equation is about. If we have a problem with two

unknowns (typically x and y), then we need two equations with the same variables in order to solve

them. In general simultaneous equations take the form:

ax + by = f1(x,y)

αx + βy = f2(x,y) Where a, b, c and d are known coefficients.

Remember that 'f ' means "a function of " and to solve x and y we need to have a value for the

function. It is easier to follow if we write the equations as:

ax + by = C

αx + βy = D

This tutorial only deals with two unknown variables. If we had three unknown variables we would

need three simultaneous equations and so on.

We can rearrange our equation to give :

α

y βD x and

a

y bCx

or as

β

xαDy and

b

axCy

We could plot y against x for both equations and determine the point where x and y are the same for

both as the point where they cross.

WORKED EXAMPLE No. 1

Solve x and y given the following simultaneous equation

x + 7y = 39 …………………..(1)

2x + 3y = 23 …………………..(2)

SOLUTION

Rearrange to make y the subject (you could make x the subject).

y = (39 – x)/7 …………….(3)

y = (23 – 2x)/3 ……………(4)

Plot x against y using both equations and we get:

There is only one value of x and y that are the same for both graphs so x = 4 and y = 5 satisfies

both equations.



Solve x and y given the following simultaneous equation

2x2 - y = 0…………………..(1)

4x - y = 0 …………………..(2)

SOLUTION

Rearrange to make y the subject

(you could make x the subject).

y = 2x2…………….(3)

y = 4x ……………(4)

Plot x against y using both equations and we

get the graphs shown.

We find that x = 2 and y = 8 satisfies both equations

but x = 0 and y = 0 is also a solution.


When two bodies travelling towards each other on the same line collide, the resulting velocities

v1 and v2 after the collision are represented by the simultaneous equations:

v1 - v2 = -6 and 5v1 + 3v2 = -2

Solve the velocities using the graphical method.

SOLUTION

Rearrange the equations to make v1 the subject.

v1 = -6 + v2 and 5

3v2v 2

1

Plot and we get the graph shown.

v1 = -2.5 m/s and v2 = + 3.5 m/s

SELF ASSESSMENT EXERCISE No. 1

Solve the variables in the following simultaneous equations by plotting suitable graphs.

1. x + y = 7 and 2x + 3y = 19 (Answers x = 2, y = 5)

2. x - y = -9 and 3x + y = 5 (Answers x = -1, y = 8)

3. 3x2 + 2x - y = -4 and 8x - y = -4 (Answers x = 2, y = 20 but 0,0 is also a solution)

You might observe from the previous examples that when one graph is a curve of x2, the possibility

exists that there are two solutions. In fact this is a form of quadratic equation.


2. QUADRATIC EQUATIONS - GRAPHICAL SOLUTION

Most relationships in Engineering and Science are anything but proportional and many are

quadratic.

Quadratic equations have the general form:- y = f(x) = ax2 + bx

1 + cx

0

This is normally written as :- y = f(x) = ax2 + bx + c

a, b, c and d are coefficients and note that x1 = x and x

o =1.

There are two values of x that satisfy the equations. Depending on the data, these may be solvable

as real numbers or not. We will only deal with equations that you can solve at this level.

When y = 0, the values of x that satisfy the equation are called the ROOTS of the equation.

If y is a non zero number, we only need to subtract y from both sides of the equation to produce a

new equation that equates to zero.

The roots may be solved by plotting suitable graphs.


Solve the values of x that satisfy the equation :

y = 0 = 2x2 - 4x – 6

SOLUTION

Plot x against y over a suitable range we get the

graph shown.

From the graph determine the value of x where y = 0.

We find that x = -1 and x = 3.

It is always a good idea to check that putting these values into the equation produces y = 0.

x = -1 y = 2(-1)2- 4(-1) – 6 = 2 + 4 – 6 = 0

x = 3 y = 2(3)2 - 4(3) – 6 = 18 – 12 – 6 = 0

An alternative method is to rearrange the equation

and make x2 the subject.

In this case we get 2x2 = (4x + 6)/2 = 2x + 3

If we let f1(x) = x2 and f2(x) = 2x + 3 and plot both functions against x, the point where the

graphs cross is the point where the functions are equal and give the answer for x.



A closed metal cylindrical canister has a mean radius R and length L. The surface area of the

metal used is given by the formula A = 2π(R2 + RL)

Given that the area is 1000 mm2 and the length is 20 mm what is the radius?

SOLUTION

Put the values into the formula. 1000 = 2π(R2 + 20R)

Rearrange into standard form. F(R) =0 = 2π(R2 + 20R) - 1000

0 = 2πR2 + 40πR - 1000

Plot the graph.

We see that the value of R required is 6.1 mm.

The negative solution is not a possible solution

for a real problem.


1. Solve the following equations by plotting a suitable graph.

2x2 – 3x -2 = 0 (Answers x = 2 or -0.5)

-x2 + x + 2 = 0 (Answers x = 2 or -1)

x2 – 5.2x + 6.4 = 0 (Answers x = 3.2 or 2)

3 x2 -20 log(x) -7.077 = 0 (Answer x = 2.15)

2. A closed metal cylindrical canister has a mean radius R and length L. The surface area of the

metal used is given by the formula A = 2π(R2 + RL)

Given that the area is 2000 mm2 and the length is 30 mm what is the radius?

(Answers 8.3 mm, the negative answer is ignored as it is not physically possible)


3. GRAPHICAL SOLUTION OF CUBIC EQUATIONS

Cubic equations have the general form :- y = f(x) = 0 = ax3 + bx

2 + cx + d

If a solution exists then it may be found graphically in the same way as for a quadratic equation. A

cubic equation has three roots. Many cubic plots do not cross the y axis and there may be three, two,

one or no real answers that satisfy the equation.


Find the values of x that satisfy the equation 2x3 + 10x

2 - 20x – 50 = 0

SOLUTION

The solution is the roots of f(x) = 2x3 + 10x

2 - 20x – 50

To solve x we must plot f(x) = y = 2x3 + 10x

2 - 20x – 50 and see if any value of x produces a

result y = 0

The plot is shown. The graph crosses the y axis three times so there are three values of x that

give a solution and these are x = -6, x = -1.61 and 2.59

Check them out to see if they are correct.

x = -6 2x3 + 10x

2 - 20x – 50 = -2 so the answer is not exact but close.

x = -1.61 2x3 + 10x

2 - 20x – 50 = -0.23 so the answer is not exact but close.

x = 2.59 2x3 + 10x

2 - 20x – 50 = 0.03 so the answer is not exact but close.

The graphical method will not usually give exact answers depending on the scale of the graph.

We can rearrange the equation to make x3 the subject and plot two graphs like before.


WORKED EXAMPLE No.7

Find the values of x that satisfy the same equation 2x3 + 10x

2 - 20x – 50 = 0 (example 6) by

plotting two graphs.

SOLUTION

Rearrange to make x3 the subject.

x3 = (- 10x

2 + 20x + 50)/2 = (- 5x

2 + 10x + 25)

If we let f1(x) = x3 and f2(x) = (- 5x

2 + 10x + 25) and plot both functions against x, the point

where the graphs cross is the point where the functions are equal and give the answer for x.

The answers are the same as in example 6.


Solve the following equation graphically. (4x3 – 3x

2 - 36x + 27) = 4.67(x+1)

SOLUTION

Rearrange to make two functions that can be plotted.

Let f1(x) = (4x3 – 3x

2 - 36x + 27) and f2(x) = 4.67(x + 1)

Plot both over a suitable range and pick off the answers from the points where the graphs cross

(the common abscissa) x = -3.1 ,x = 0.56 and x = 3.3



1. Find any real values of x that satisfy the equation y = 0 = x3 - 1.87x

2 - 4.54x + 6.41

(Answers x = -2 or 1.2 or 2.67)

2. Find any real values of x that satisfy the equation y = 0 = 5x3 +10x

2 -2x – 6

(Answer -1.875, -0.865 and 0.735)

3. A cylindrical vessel of diameter D and length L has hemispherical ends. The volume of the

vessel is hence given by V = π(DL + D3/6). If the volume is 200 x 10

3 mm

3 and the length L is

50 mm, what is the diameter?

(Answer 71.18 mm)


4. LINEARISING EQUATIONS WITH LOGARITHMS

Many relationships between two variable quantities are non-linear. This means that the graph of one

plotted against the other is not a straight line. For example consider the simple equation:

f(x) = y = x2

If we plot them we get the curved graph shown.

It is possible to represent a formula like this in a form that produces

a straight line graph and the simplest way is use logarithms. If we

take logs we get:

log(y) = 2 log x

Plotting this we get a straight line graph as shown. Note that the gradient is 2.

The general equation y = C xn may be reduced to a straight

line in the same way. Taking logs we have:

log(y) = log(C) + nlog(x)

This gives a straight line with a gradient n and an intercept C


Deduce the relationship between y and x from the graph shown.

SOLUTIONS

The intercept is 0.7 so log(C) = 0.7. Find the antilog and C = 5

The gradient is (3.7 – 0.7)/1 = 3 so n = 3

The equation is y = 5x3



The graph shows the results of an experiment in which a variables x and y are recorded and

plotted. When log(x) and log(y) are plotted the straight line graph shown is produced.

Determine the function f(x).

SOLUTION

From the straight line graph we have an intercept of 0.7 and a gradient of (3.7 – 0.7)/1 = 3

φ(x) = 0.7 + 3 log(x) Take antilogs f(x) = 5x3


The graph shows the results of an experiment in which a variables x and y are recorded and

plotted as logs. Determine the function f(x).

SOLUTION

From the straight line graph we have an intercept of 0.845.

The gradient is (2.345– 0.845)/(-15) = - 0.1

(x) = 0.845 – 0.1 log(x) Take antilogs f(x) = 7x-0.1



1. Determine the function f(x) for each of the graphs below,

Answers f(x) = 1.5x2 and f(x) = 0.5x

-0.5

2. When a pulley belt slips on a wheel it is believed the largest force F1 on one side and the smaller

force F2 on the other side is in the form:

F1 = F2 eμθ

μ is the coefficient of friction which is constant and unknown. θ is the angle of contact.

If an experiment is conducted with various forces and angles to verify this relationship and the

only unknown is μ, how should the data be plotted so that a straight results if the equation is

true? How may the value of μ be found from the straight line graph?

3. The pressure loss in a given pipe is believed to be related to the flow rate by the formula:

p = kQn

An experiment is conducted and the pressure is measured over a range of flow rates. The results

and plot are as shown.

Process the data and plot it so that a straight line is produced and deduce the values of k and n.

(Answers 0.032 and 1.8 note that you will have to project the graph to find the intercept)

Documents

EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS ...2007/outcome1t1.pdf · edexcel national certificate unit 28 – further mathematics for technicians outcome 1 advanced