The equation of a straight line

For any straight line:y = mx + c

where: m = gradient = (yP – yR) / (xR – xQ)

and c = y-intercept

The power law relationshipThis 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 examplespower, n = 1: direct proportion relationship: y = k x – prove by plotting y against xpower, n = 2: square relationship: y = k x2 – plot y against x2

power, n = 3: cube relationship: y = k x3 – plot y against x3

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 / x2 – plot y against 1 / x2

In all these cases the graphs should be straight lines through the origin having gradients equal to k.

QuestionQuantity 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 unknownEITHER - 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 = ny-intercept = log (k)

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 1.301200 = 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: log10(100) = 2 log10(50) = 1.699 etc..(on a calculator use the ‘lglg’ button)

Natural Logarithms

Logarithms can have any base number but in practice the only other number used is 2.718281…,

Napier’s constant ‘e’.

Examples: loge(100) = 4.605 loge(50) = 3.912 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:

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

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

Powers with logarithms

log (An) = n log (A)

Consider: 2 3 = 2 x 2 x 2

this can be written in terms of logs to base 10:

log10 (2 3) = log10 (2) + log10 (2) + log10 (2)

log10 (2 3) = 3 x log10 (2)

Another logarithm relationship

log B(Bn) = n

Example: log10 (10 3) = log10 (1000) = 3

The most important example of this is:

ln (en) = n

[ loge (en) = 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)

log (y) = n log (x) + log (k) This has the form of the equation of a straight line: y = m x + cwhere:y = log (y) x = log (x) m = the gradient

= the power nc = the y-intercept

= log (k)

QuestionDependent 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 1.0 2.0 3.0 4.0 5.0 6.0

P 2.00 16.0 54.0 128 250 432

log 10 (Q)

log 10 (P)

Exponential decayThis 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 = xo e - λ t

where:

t is the time from some initial starting point

x is the value of the decaying variable at time t

xo 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

xo = 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 = xo 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 (xo)

NOTE: ONLY LOGARITMS TO THE BASE e CAN BE USED.

How ln-t graphs workExponential decay has the general form:

x = xo e - λ t

Taking logs TO THE BASE e on both sides:

ln (x) = ln (xo e - λ t)

ln (x) = ln (xo ) + ln (e - λ t)

ln (x) = ln (xo ) - λ twhich is the same as:

ln (x) = - λ t + ln (xo )

ln (x) = - λ t + ln (xo ) This has the form of the equation of a straight line:y = m x + cwith:y = ln (x) x = t

m, the gradient = the negative of the decay constant = - λ

c, the y-intercept = ln (xo )

QuestionThe marks M of a student are suspected to decay exponentially with time t.They are suspected to be related through the equation: M = Mo 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 Mo (t = 0 weeks) and the decay constant k, of the marks.

t / weeks 1 2 3 4 5 6

M 72 59 48 40 32 27

ln (M)