MATHEMATICS FOR ENGINEERING
BASIC ALGEBRA
TUTORIAL 2 – STRAIGHT LINE GRAPHS
This tutorial is useful to anyone studying engineering. It uses the principle of learning
by example.
•
Straight line laws
•
Gradients
•
Intercepts
•
Practical applications
1. DIRECTLY PROPORTIONAL RELATIONSHIPS
In Engineering and Science, the relationship
between two quantities is DIRECTLY
PROPORTIONAL and when one is plotted
against the other, a straight line 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.
©D.J.Dunn
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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
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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
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.
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
©D.J.Dunn
p = 333.3 (2) = 666.7 N/m2 and the graph bears this out.
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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)
3. Sketch the graph of the following relationships.
(a) y = 0.4x – 2 (b) y = -3x -2 (c) s = -2u + 4
Solutions on next page
©D.J.Dunn
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Figure 9(a)
Figure 9 (b)
Figure 9 (c)
©D.J.Dunn
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