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The equation of a straight line

For any straight line:

y = mx + c

where: m

= gradient= (yPyR) / (xRxQ)

and c

= y-intercept

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The power law relationshipThis has the general form:

y = k x nwhere kand nare constants.

An example is the distance, stravelled aftertime, twhen an object is undergoing

acceleration, a.s = at 2

s = y ; t = x ; 2 = n ; a = k

To prove this relationship:

Draw a graph of yagainst x n The graph should be a straight line

through the origin and have a gradient

equal to k

y

xn

gradient = k

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Common examplespower, n = 1:

direct proportion relationship: y = k xprove by plotting yagainst xpower, n = 2:

square relationship: y = k x2plot yagainst x2

power, n = 3:

cube relationship: y = k x3plot yagainst x3

power, n = :square root relationship: y = k x = k xplot yagainst x

power, n = - 1:

inverse proportion relationship: y = k x -1= k / xplot yagainst 1 / x

power, n = - 2:

inverse square relationship: y = k x -2= k / x2plot yagainst 1 / x2

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

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QuestionQuantity Pis thought to be related to quantities Q, Rand T

by the following equation: P = 2Q R2

T 3

What graphs should be plotted to confirm the relationships

between Pand the other quantities?

State in each case the value of the gradient.

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When n is unknown

EITHER - Trial and errorFind out what graph yieldsa straight line.

This could take a longtime!

OR - Plot a log (y) against

log (x) graph.

Gradient = n

y-intercept = log (k)

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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 bethe LOGARITHM of the left hand number to theBASE OF 10

For example: log10(100) = 2 log10(50) = 1.699 etc..

(on a calculator use the lg button)

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Natural Logarithms

Logarithms can have any base number but in

practice the only other number used is

2.718281,

Napiers constant e.

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

(on a calculator use the ln button)

These are called natural logarithms

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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 1.301x 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

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Division with logarithms

log (A B) = log (A) - log (B)

Consider: 100 20 = 5

this can be written in terms of powers of 10:10 210 1.301 = 10 0.699

Note how the powers relate to each other:

2 - 1.301 = 0.699

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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 (23) = 3 x log10 (2)

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Another logarithm relationship

log B(Bn) = n

Example: log10 (103) = log10 (1000) = 3

The most impo rtant example of th is is :

ln (en) = n

[ loge (en) = n ]

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How log-log graphs work

The power relationship has the general form:

y = k x n

where kand nare 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)

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log (y) = n log (x) + log (k)

This has the form of theequation 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)

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QuestionDependent variable Pwas measured for various values of independentvariable Q. They are suspected to be related through a power law

equation: P = k Qn

where kand nare constants. Use themeasurements below to plot a log-log graph and from this graph find thevalues of kand 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)

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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.

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Exponential decay over time has the general form:

x = xo

e - t

where:

t is the time from some initial starting point

xis the value of the decaying variable at time t

xo is the initial value of xwhen t= 0

eis Napiers constant 2.718

is called the decay constant.

It is equal to the fraction of xthat decays in a unit time.

The higher this constant the faster the decay proceeds.

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In the radioisotope example:

t = the time in minutes.

x= the mass in grams of theisotope remaining at this time

xo = 100 grams (the startingmass)

e= Napiers constant 2.718

= the decay constant is equal tothe fraction of the isotope thatdecays over each unit time period(1 minute in this case).

About 0.11 min-1in this example.

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Proving exponential decay graphically

x = xo e- t

To prove this plot a graph ofln (x)against t .

If true the graph will be a straight

line and have a negativegradient.Gradient = -

y-intercept = ln (xo)

NOTE: ONLY LOGARITMSTO THE BASE e CAN BE USED.

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How ln-t graphs work

Exponential decay has the general form:

x = xo e- t

Taking logs TO THE BASE eon both sides:

ln (x ) = ln (xo e- t)

ln (x ) = ln (xo) + ln (e- t)

ln (x ) = ln (xo) - t

which is the same as:ln (x ) = - t + ln (xo)

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ln (x) = - t + ln (xo)

This has the form of the equation of astraight line:

y = m x + cwith:

y= ln (x )

x = t

m, the gradient

= the negativeof the decay constant

= -

c, the y-intercept = ln (xo)

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QuestionThe marks Mof a student are suspected to decay exponentially with time t.

They are suspected to be related through the equation: M = Moek t.

Use the data below to plot a graph of ln(M)against tand so verify the abovestatement. Also determine the students initial mark Mo(t = 0 weeks) and thedecay constant k, of the marks.

t / weeks 1 2 3 4 5 6

M 72 59 48 40 32 27

ln (M)