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

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

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

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.

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.

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)

Logarithms Consider: 10 = = = = = = = = = 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) = etc.. lg (on a calculator use the ‘lg’ button)

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

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: x = 10 3 Note how the powers (the logs to the base 10) relate to each other: = 3.000

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 ÷ = Note how the powers relate to each other: = 0.699

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)

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 ]

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)

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)

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. Q P log 10 (Q) log 10 (P)

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.

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.

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.

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.

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 )

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 )

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 / weeks M ln (M)