1. Reducing Equations to a Linear Form (Linear Law)

2. x y Here’s some data that is linear 3.1 5.6 7.0 10.4 12.5 2.3 9.7 14.2
24.2
30.4

3. If these values are plotted on a graph they lie on a straight line and so obey the law
y = mx + c

5. We can calculate m and c from the graph as follows:
The gradient, m
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 = (to 2 s.f)
The y intercept, x
The graph cuts the y axis at (0, -7) ,
therefore y intercept b = -7
Equation is then y = 2.9x – 7.0

6. Equations of the form y = ax²+ b

7. 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

8. 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.

9. We need to make a table with values of x² and plot these on a graph, (a BIG graph if 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

11. 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 or
y = -0.2x

12. 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

14. Equations of the form y = kxn

15. y = kxn
Plot log y against log x
Intercept is log k
Gradient is n

16. A water pipe is being laid between two points
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
R = kdm d 1 2 3 5 10 R
0.02
0.32
1.62
12.53
199.8

17. Here we try the relationship R = kdm where k is a constant
logR = mlogd + logk
Compare with y = mx + c
log d
0.3
0.48
0.70
1.00 (x)
log R
-1.70
-0.49
0.21
1.10
2.30 (y)

19. Reading from the graph we can see that
so k = = 0.02
gradient m = (change in y)/(change in x)
= 4.0
The relationship is then
R = 0.02d4

20. (Exponential relationships)
Equations of the form y = kax
(Exponential relationships)

21. y = kax
Plot log y against x
intercept is log k
gradient is log a

22. The temperature θ in ºC of a cup of coffee after t minutes is recorded below2 4 6 8 10 12 θ 81 70 61 52 45 38

23. 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

25.  Reading from the graph gives logk = 1.98 k = 101.98 so k = = 95.5
gradient = y / x = -0.03
loga = so a = = 0.93
Relationship is
= 95.5 x 09.3t 

26. Summary
Plot Y vs X where X=x2
Plot Log(y) vs Log(x)
Plot Log(y) vs x