Reducing Equations to a Linear Form

Andrew Robertson

Today

Reducing data relationships to straight lines

WHY ?

One good reason is to be able to make forecasts from data

It is much easier to work and predict future results when the data lies on a straight line

Another reason is that reducing to straight line form may help us understand the relationship between x and y variables better

Consider the following data...............................

Results that give a straight line graphHere’s some data that is linear

x 3.1 5.6 7.0 10.4 12.5

y 2.3 9.7 14.2 24.2 30.4

If these values are plotted on a graph

they lie on a straight line and so obey the law

y = mx + c

Experimental data 1

-15

-10

-5

0

5

10

15

20

25

30

35

-2 0 2 4 6 8 10 12 14

x

• We can calculate a and b from the graph as follows:

The gradient, a Choosing two points a long way apart, the

line passes through (2.3, 0) and (12.0, 28.5),

Therefore gradient a = 28.5/9.7 = 2.94 = 2.9 (to 2 s.f)

The y intercept, b The graph cuts the y axis at (0, -7) , therefore y intercept b = -7

Equation is then y = 2.9x – 7.0

Equations of the form y = ax²+ b

If in another experiment the data appears to satisfy a quadratic relationship; If we let

Y =y, X = x²

Then we can get a straight line by plotting

Y = aX + b

We can plot Y (=y) against X (= x²) we should get a straight line and we can find a and b from our graph.

Example A hosepipe squirts water and the height,

y metres of the water above a fixed level at a distance x m from the hose is measured as

This is thought to obey the law y = ax2 + b

x 2 4 5 6 7 8

y 6.1 3.6 2.2 -0.1 -2.9 -5.5

We need to make a table with values of x² and plot these on a graph, (a BIG graph if by plotted by hand!)

x²=X 4 16 25 36 49 64

y=Y 6.1 3.6 2.2 -0.1 -2.9 -5.5

Hosepipe: height (Y) against distance squared (X)

-8

-6

-4

-2

0

2

4

6

8

0 10 20 30 40 50 60 70

X (distance squared)

Y (h

eigh

t)

Plotting Y against X, gives a straight line Y = aX + b

From the graph, choosing 2 points e.g. (0, 6.9) and (35, 0) gives a = -0.20 (to 2 s.f.) (The gradient is -0.2) The line cuts the Y axis where Y = 6.9

and so b = 6.9

Therefore y = -0.2X + 6.9 or

y = -0.2x2 + 6.9

Hosepipe: height (Y) against distance squared (X)

-8

-6

-4

-2

0

2

4

6

8

0 10 20 30 40 50 60 70

X (distance squared)

Y (h

eigh

t)

Equations of the form y = kxn

(Power Law growth / decay y=kx-n )

y = kxn

• Plot logy against logx

• Intercept is logk

• Gradient is n

A water pipe is being laid between two points. The following data are being used to show how, for a given pressure difference, the rate of flow R litres per second, varies with the pipe diameter d cm

d 1 2 3 5 10

R 0.02 0.32 1.62 12.53 199.8

Here we try the relationship R = kdm where k is a constant

logR = mlogd + logk

Compare with y = mx + c

log d 0 0.3 0.48 0.70 1.00 (x)

log R -1.70 -0.49 0.21 1.10 2.30 (y)

Log flow rate against log pipe diameter

y = 4.0x - 1.7

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.2 0.4 0.6 0.8 1 1.2

Log d

Log

R

Log flow rate against log pipe diameter

y = 4.0x - 1.7

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.2 0.4 0.6 0.8 1 1.2

Log d

Log

R

Reading from the graph we can see that c = = - 1.70

so k = = 0.02

gradient m = (change in y)/(change in x)

= 4.0

The relationship is then

R = 0.02d4

Equations of the form y = kax

(Exponential relationships)

y = kax

Plot logy against x

intercept is logk

gradient is loga

The temperature θ in ºC of a cup of coffee after t minutes is recorded below

t 2 4 6 8 10 12

θ 81 70 61 52 45 38

• If the relationship is of the form = kat

where k and a are constants

• log = (loga)t + logk

• (y = mx + c)

t 2 4 6 8 10 12

log θ 1.91 1.85 1.79 1.72 1.65 1.58

Log theta against time

y = -0.03x + 1.98

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10 12 14

t (minutes)

Log

thet

a

Reading from the graph gives logk = 1.98 k = 101.98 so k =

= 95.5

gradient = y / x = -0.03 loga = - 0.03 so a = 10-0.03 = 0.93

Relationship is = 95.5 x 09.3t

Summary

x

n

kby

kxy

xy

2Plot Y vs X where X=x2

Plot Log(y) vs Log(x)

Plot Log(y) vs x