Exponential Growth & Decay Functions

Recall from unit 1 that the graph of f(x) = ax (a>1)

looks like

y = ax

As x then y but as x - then y 0

GROWTH FUNCTION

While the graph of f(x) = a-x ie 1 (a>1) ax

looks like

y = a-x

As x then y 0 but as x - then y

DECAY FUNCTION

From function transformations in Unit 1

If f(x) = ax then f(-x) = a-x and f(-x) is obtained by reflecting f(x) in the Y-axis.

Example1

(a) It is estimated that the value of a house increases by 3.5% per annum. If a house is bought for £70000 then how much will it be worth in 5 years time?

(b) At this rate of increase how long does it take for the value to double? (answer to nearest year!)

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(a) An increase of 3.5 % gives us 103.5% or 1.035 X original.

Let Vn be the value in year n then …

V0 = £70000

V1 = 1.035V0

V2 = 1.035V1 = (1.035)2V0

etc

So V5 = (1.035)5V0 = (1.035)5 X £70000 = £83100 to nearest £100

(b) If the value doubles then Vn = 2V0

And since Vn = (1.035)nV0

Then we need (1.035)nV0 = 2V0

Or simply (1.035)n = 2

Using trial & error

Take n =10 then (1.035)10

= 1.41.. too small

Take n =20 then (1.035)20 = 1.99.. too small

Take n =21 then (1.035)21 = 2.06.. too large

**

To the nearest year time required to double value = 20 years

Example 2

In chemistry if it takes 10 years for half of a sample of radioactive material to decay then we say that it has a half-life of 10 years.

If we started with 80g of the material then …

Time in Years Mass Remaining

0 80g

10 40g

20 20g

30 10g

40 5g

50 2.5g

etc

Example 3

A radioactive substance has a half-life of 7 years. Find how long it takes for any given sample to decay to only 10% of its initial mass.

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Suppose the An is the amount remaining after n half-lives.

If A0 is the original amount then we have

A1 = 0.5A0

A2 = 0.5A1 = (0.5)2A0

A3 = 0.5A2 = (0.5)3A0

etc

In general An = (0.5)nA0

We also need An = 10% of A0 or 0.1A0

This means that (0.5)nA0 = 0.1A0

Or (0.5)n = 0.1

Again by trial & error

n = 5 (0.5)5 = 0.03…. too small

n = 3 (0.5)3 = 0.125 too large

n = 4 (0.5)4 = 0.0625 too small

n = 3.2 (0.5)3.2 = 0.11… too large

n = 3.1 (0.5)3.1 = 0.12… too large

n = 3.3 (0.5)3.3 = 0.10… closest

So required time = 3.3 half-lives

= 3.3 X 7 years

= 23.1 years

The long period required for the radioactive substance to decay significantly should give us an indication of the dangers of radioactive materials and the potential problems they can create for the environment.