1. 6 Coursework Power & Exponential Relationships Breithaupt pages 247 to 252 September 11 th, 2010 CLASS NOTES HANDOUT VERSION

2. The equation of a straight line For any straight line: y = mx + c where: m = gradient = (y P – y R ) / (x R – x Q ) and c = y-intercept

3. The power law relationship This has the general form: y = k x n where k and n are constants. An example is the distance, s travelled after time, t when an object is undergoing acceleration, a. s = ½ at 2 s = y; t = x; 2 = n; ½ a = k To prove this relationship: –Draw a graph of y against x n –The graph should be a straight line through the origin and have a gradient equal to k y x n gradient = k

4. Common examples power, n = 1: direct proportion relationship: y = k x – prove by plotting y against x power, n = 2: square relationship: y = k x 2 – plot y against x 2 power, n = 3: cube relationship: y = k x 3 – plot y against x 3 power, n = ½: square root relationship: y = k x ½ = k √x – plot y against x ½ power, n = - 1: inverse proportion relationship: y = k x -1 = k / x – plot y against 1 / x power, n = - 2: inverse square relationship: y = k x -2 = k / x 2 – plot y against 1 / x 2 In all these cases the graphs should be straight lines through the origin having gradients equal to k.

5. Question Quantity P is thought to be related to quantities Q, R and T by the following equation:P = 2π Q R 2 T 3 What graphs should be plotted to confirm the relationships between P and the other quantities? State in each case the value of the gradient.

6. When n is unknown EITHER - Trial and error Find out what graph yields a straight line. This could take a long time! OR - Plot a log (y) against log (x) graph. Gradient = n y-intercept = log (k)

7. Logarithms Consider: 10 = 10 1 100 = 10 2 1000 = 10 3 5 = 10 0.699 50 = 10 1.699 500 = 10 2.699 2 = 10 0.301 20 = 10 1.301 200 = 10 2.301 In all cases above the power of 10 is said to be the LOGARITHM of the left hand number to the BASE OF 10 For example: log 10 (100) = 2 log 10 (50) = 1.699 etc.. lg (on a calculator use the ‘lg’ button)

8. Natural Logarithms Logarithms can have any base number but in practice the only other number used is 2.718281…, Napier’s constant ‘e’. Examples: log e (100) = 4.605 log e (50) = 3.912 etc.. (on a calculator use the ‘ln’ button) These are called ‘natural logarithms’

9. Multiplication with logarithms log (A x B) = log (A) + log (B) Example consider: 20 x 50 = 1000 this can be written in terms of powers of 10: 10 1.301 x 10 1.699 = 10 3 Note how the powers (the logs to the base 10) relate to each other: 1.301 + 1.699 = 3.000

10. Division with logarithms log (A ÷ B) = log (A) - log (B) Consider: 100 ÷ 20 = 5 this can be written in terms of powers of 10: 10 2 ÷ 10 1.301 = 10 0.699 Note how the powers relate to each other: 2 - 1.301 = 0.699

11. Powers with logarithms log (A n ) = n log (A) Consider: 2 3 = 2 x 2 x 2 this can be written in terms of logs to base 10: log 10 (2 3 ) = log 10 (2) + log 10 (2) + log 10 (2) log 10 (2 3 ) = 3 x log 10 (2)

12. Another logarithm relationship log B (B n ) = n Example: log 10 (10 3 ) = log 10 (1000) = 3 The most important example of this is: ln (e n ) = n [ log e (e n ) = n ]

13. How log-log graphs work The power relationship has the general form: y = k x n where k and n are constants. Taking logs on both sides: log (y) = log (k x n ) log (y) = log (k) + log (x n ) log (y) = log (k) + n log (x) which is the same as: log (y) = n log (x) + log (k)

14. This has the form of the equation of a straight line: y = m x + c where: y = log (y) x = log (x) m = the gradient = the power n c = the y-intercept = log (k)

15. Question Dependent variable P was measured for various values of independent variable Q. They are suspected to be related through a power law equation: P = k Q n where k and n are constants. Use the measurements below to plot a log-log graph and from this graph find the values of k and n. Q1.02.03.04.05.06.0 P2.0016.054.0128250432 log 10 (Q) log 10 (P)

16. Exponential decay This is how decay occurs in nature. Examples include radioactive decay and the loss of electric charge on a capacitor. The graph opposite shows how the mass of a radioactive isotope falls over time.

17. Exponential decay over time has the general form: x = x o e - λ t where: t is the time from some initial starting point x is the value of the decaying variable at time t x o is the initial value of x when t = 0 e is Napier’s constant 2.718… λ is called the decay constant. –It is equal to the fraction of x that decays in a unit time. –The higher this constant the faster the decay proceeds.

18. In the radioisotope example: t = the time in minutes. x = the mass in grams of the isotope remaining at this time x o = 100 grams (the starting mass) e = Napier’s constant 2.718… λ = the decay constant is equal to the fraction of the isotope that decays over each unit time period (1 minute in this case). About 0.11 min -1 in this example.

19. Proving exponential decay graphically x = x o e - λ t To prove this plot a graph of ln (x) against t. If true the graph will be a straight line and have a negative gradient. Gradient = - λ y-intercept = ln (x o ) NOTE: ONLY LOGARITMS TO THE BASE e CAN BE USED.

20. How ln-t graphs work Exponential decay has the general form: x = x o e - λ t Taking logs TO THE BASE e on both sides: ln (x) = ln (x o e - λ t ) ln (x) = ln (x o ) + ln (e - λ t ) ln (x) = ln (x o ) - λ t which is the same as: ln (x) = - λ t + ln (x o )

21. This has the form of the equation of a straight line: y = m x + c with: y = ln (x) x = t m, the gradient = the negative of the decay constant = - λ c, the y-intercept = ln (x o )

22. Question The marks M of a student are suspected to decay exponentially with time t. They are suspected to be related through the equation: M = M o e – k t. Use the data below to plot a graph of ln(M) against t and so verify the above statement. Also determine the student’s initial mark M o (t = 0 weeks) and the decay constant k, of the marks. t / weeks123456 M725948403227 ln (M)