Reducing Equations to a Linear Form

1. Reducing Equations to a Linear Form Andrew Robertson

2. 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...............................

4. Results that give a straight line graph Here’s some data that is linear

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

7. We can calculate a and b from the graphas 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

8. Equations of the form y = ax²+ b

9. 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.

10. 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

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

13. 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

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

16. y = kxn • Plot logy against logx • Intercept is logk • Gradient is n

17. 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

18. Here we try the relationship R = kdm where k is a constant logR = mlogd + logk Compare with y = mx + c

21. 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

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

23. y = kax Plot logy against x intercept is logk gradient is loga

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

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

27. 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 

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