Reducing Equations to a Linear Form (Linear Law)

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

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

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 = 2.9 (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

Equations of the form y = ax²+ b

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

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.

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

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

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...............................

Equations of the form y = kxn

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

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

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)

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

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

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

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

 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 Plot Y vs X where X=x2 Plot Log(y) vs Log(x) Plot Log(y) vs x