EDEXCEL NATIONAL CERTIFICATE
UNIT 4 – MATHEMATICS FOR TECHNICIANS
OUTCOME 1
TUTORIAL 2 – STRAIGHT LINE GRAPHS
Determine the fundamental algebraic laws and apply algebraic manipulation techniques to the
solution of problems involving algebraic functions, formulae and graphs
CONTENTS OF OUTCOME 1
Algebraic laws and algebraic manipulation techniques
Algebraic laws and manipulation: introduction to algebraic expressions and equations, algebraic
operations, rules for manipulation and transposition of formulae; manipulation and algebraic
solution of algebraic expressions;
direct and inverse proportion and constants of proportionality;
linear, simultaneous and quadratic equations; graphical solution of simple equations; use of
standard formulae to solve surface areas and volumes of regular solids; use of calculator for
algebraic problems
Indices, logarithms and functions: the index as a power, numerical and literal numbers in index
form, laws of indices and logarithms, common and Naperian logarithms, functional notation and
manipulation, logarithmic and exponential functions, other simple algebraic functions.
This tutorial is not specifically stated in the contents but it is implied that the student should
understand graphs so it is used as a starting point for this module.
It is assumed that the student is familiar with arithmetical operations, the use of brackets and
fractions.
CONTENTS OF TUTORIAL 2
©D.J.Dunn
•
Straight line laws
•
Gradients
•
Intercepts
•
Practical applications
1
STRAIGHT LINE GRAPHS
1. DIRECTLY PROPORTIONAL RELATIONSHIPS
In Engineering and Science, the relationship
between two quantities is often DIRECTLY
PROPORTIONAL and when one is plotted
against the other, a straight line graph is produced.
For example a mechanical spring usually behaves
such that the change in length ‘x’ is directly
proportional to the applied force ‘F’. With no
force the spring has its normal length. Suppose
that a force of 1 N stretches the spring 2 mm.
Being directly proportional it follows that 2 N will
stretch it 4 mm and 3 N 6 mm and so on. The
graph shows this. Because the graph goes through
the zero origin we see that what ever the force the
ratio F/x is 0.5 N/mm and this is the gradient of
the graph. The mathematical law is simply:
F = 0.5 x
Figure 1
In general such laws are written as y = m x where y is the quantity plotted vertically and x the
quantity plotted horizontally and m is the gradient simply found as m = y/x.
Now consider the graph that relates the Fahrenheit and
Celsius temperature scales. This may be used to convert
from one to the other. 212oF corresponds to 100oC so
one point on the graph goes there. 0oC corresponds to
32oF so another point goes there. Joint the two points
with a straight line and we have the complete graph. If
we want to use a formula we find that all the points on
the graph are related by the law:
o
F = oC x 9/5 + 32
Figure 2
9/5 is the gradient and is simply found as the ratio: (212 + 32) ÷ 100 = 1.8 or 9/5
Note that the graph intercepts the vertical axis at 32 and this is called the intercept. In general the
law is written
y = mx + C where m is the gradient and C the intercept with the y axis.
The symbol ∝ means “directly proportional to”
For example if y is directly proportional to x and we write this as y ∝ x
To get the equation, we replace ∝ with = m hence y = m x or y/x = m
m is the constant of proportionality.
©D.J.Dunn
2
WORKED EXAMPLE No.1
Convert 60oC into Fahrenheit
o
F = oC x 9/5 + 32 = 60 x 9/5 + 32 = 108 + 32 = 140oF
WORKED EXAMPLE No.2
The graph shows the relationship between the pressure ‘p’ of a gas plotted vertically and the
temperature ‘θ’ of a gas plotted horizontally when the volume is kept constant. Deduce the law
relating them. Calculate the temperature at which the pressure is zero.
Figure 3
SOLUTION
The vertical axis is p and the horizontal is θ so the graph has a law p = mθ + C
The gradient m is found from the two marked points.
Vertical change = 210 – 100 = 110 kPa Horizontal change = 300 – 0 = 300oC
m = 110/300 = 0.367 kPa/oC
The intercept with the vertical axis is at 100 kPa so C = 100 kPa
The law is p = 0.367θ + 100 where p is in kPa and θ is in oC
Now put p = 0 and find θ
0 = 0.367θ + 100
0.367θ = -100
θ = -100/0.367 = -272.5oC
WORKED EXAMPLE No.3
The graph shows the relationship between
a variable y plotted vertically and x plotted
horizontally. Deduce the law relating them.
SOLUTION
The law is y = mx + C
The gradient m is found by choosing any two points
such as (17.5, 30) and (0, -15).
The vertical change in y is 30 – (-15) = 45 and the
horizontal change in x is 17.5 – 0 = 17.5
m = 45/17.5 = 90/35 = 2.571 and the intercept with the y axis is C = -15.
The law is y = (90/35)x – 15 or 2.571x - 15
Note 90/35 is an exact number and 2.571 is rounded off.
Figure 4
Check if it works by choosing any value of x say 10.
y = (2.571)(10) – 15 = 10.71 and this is correct when checked on the graph.
©D.J.Dunn
3
WORKED EXAMPLE No.4
The graph shows the relationship between a variable y plotted vertically and x plotted
horizontally. Deduce the law relating them.
Figure 5
SOLUTION
The law is y = mx + C The gradient m is found by choosing any two points such as (-2, 30)
and (7, -10).
The vertical change in y is -10 – (30) = -40 and the horizontal change in x is 7 - (-2) = 9
m = -40/9 = -4.444 and the intercept with the y axis is C = 21.
The law is y = -(40/9)x + 21 or -4.444x + 21
Note 90/35 is an exact number and 2.571 is rounded off.
Check if it works by choosing any value of x say 0.
y = (4.444) (0) + 21 = 21 and this is correct when checked on the graph.
Note that when the line slopes down to the right y is decreasing as x increases and the gradient
is always negative.
2. INVERSELY PROPORTIONAL
The 1/x button on your calculator is called the inverse button. If you enter 2 and press the 1/x button
you get the answer 0.5 because it evaluates ½. Inverse means “1 divided by”
For example inverse y means 1/y Inverse A means 1/A and so on.
Consider the equation x y = C
Rearrange and we get y = C/x or y = C (1/x)
1/x is the inverse of y. If we have an equation x ∝ 1/y it means that x is inversely proportional to y.
An example of this is Boyle’s law used in physics pV = C. In this law p is pressure and V is volume
and the relationship is true when the temperature is constant. If we rearrange the law into p = C/V
we may say that pressure is inversely proportional to volume or p = C (1/V). If we plot p against
(1/V) we will get a straight line passing through the origin and C (the constant) is the gradient of the
graph.
©D.J.Dunn
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WORKED EXAMPLE No.5
Determine the law relating p and V from the graph.
Determine the pressure when V = 0.5 m3.
Figure 6
SOLUTION
The law is p = C(1/V) and C is the gradient 1000/3
Hence p = (1000/3)(1/V) and pV = 1000/3 = 333.3 Nm
When V = 0.5 1/V = 2
p = 333.3 (2) = 666.7 N/m2 and the graph bears this out.
SELF ASSESSMENT EXERCISE No.1
1. Determine the law that relates the variables for the following graphs.
(a)
(b)
Figure 7 a and b
(Answers y = 2.286x + 2 and y = -20.833x + 250)
2. Find the relationship for y and x from the graph.
Figure 8
(Answer y = 57.1(1/x) + 10)
©D.J.Dunn
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3. Sketch the graph of the following relationships.
(a) y = 0.4x – 2
(b) y = -3x -2
(c) s = -2u + 4
SOLUTIONS
Figure 9(a)
Figure 9 (b)
Figure 9 (c)
©D.J.Dunn
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SELF ASSESSMENT EXERCISE No.2
1. In the equation y = 6 x what is the constant of proportionality?
2. If a ball is dropped and allowed to fall, it is observed that the velocity ‘v’ is directly proportional
to the time ‘t’ measured from the moment it was dropped.
Write out the equation linking v and t
It is observed that the velocity after 1 second is 9.81 m/s. What is the value and units of the
constant?
3. It is observed that the volume of a gas ‘V’ inside a balloon is inversely proportional to the
pressure ‘p’.
Write down the equation linking p and V
It is observed that when the volume is 2 m3 the pressure is 5 kN/m2.
What is constant of proportionality and what are its units?
Plot p against V over the range V = 0.1 to 2 cm3 and show that this is a curve.
What would you have to plot in order to get a straight line graph?
Plot this straight line graph.
SOLUTIONS
1.
2.
3.
6
v ∝ t hence v = Ct (Normally we use a ‘g’ for the gravitational constant)
C = 9.81 m/s2
p ∝ 1/V = C/V
C = pV = 5 x 2 = 10 kN/m
The graph is a curve known as a HYPERBOLA. To get a straight line we plot p vertically and
1/V horizontally. The gradient is 10.
Figure 10
©D.J.Dunn
7