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 graph Here’s some data that is linear x y

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 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 theny = 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 = ax 2 + b x y

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 y=Y

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 = (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

Equations of the form y = kx n (Power Law growth / decay y=kx -n )

y = kx n 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 R

Here we try the relationship R = kd m where k is a constant logR = mlogd + logk Compare with y = mx + c log d (x) log R (y)

Reading from the graph we can see that c = = so k = = 0.02 gradient m = (change in y)/(change in x) = 4.0 The relationship is then R = 0.02d 4

Equations of the form y = ka x (Exponential relationships)

y = ka x 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 θ

If the relationship is of the form  = ka t where k and a are constants log  = (loga)t + logk (y = mx + c) t log θ

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

Summary Plot Y vs X where X=x 2 Plot Log(y) vs Log(x) Plot Log(y) vs x