© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

48: Growth and 48: Growth and DecayDecay

Growth and Decay

growth

Functions of this type, with a > 1, are called functions.

xay We’ve already met the functione.g.

xy 2

xey xy 4

)1,0(

Growth and Decay

If 0 < a < 1, the function decays.e.g.

xy 50

xy 10xy 30

)1,0(

We are going to use functions of the typeto model some practical situations where we have growth or decay.

xkay

a is called the growth factor.

Growth and Decay

e.g. 1) could give the area, A, of a stain at time t seconds after it had an area of . The value of a, the growth factor, is greater than 1 so the stain is growing.

tA )41(3

23cm

e.g. 2) could give the mass of a radio-active element t years after it had a mass of 500 grams. The value of a is less than 1 so the element is decaying.

tm )80(500

e.g. 3) could give the value of £400 which has been earning compound interest of 5% for t years.

tS )051(400

Growth and Decay

We use any meaningful letter for the other variable.

Since these are practical problems, one of the variables is time. We use the letter t and since it is time, .0t

Growth and Decay

where m is mass in kg and t is time in weeks and if the baby has a mass of 6 kg at 10 weeks.

e.g. 1. In the first few weeks of life an exponential model is a reasonably good fit for the mass of a baby. Find the mass of the baby at birth and the value of the growth factor if tam 43

Solution:

At birth, t = 0, so

tam 43 043 am 43 m

Growth and Decay

When t = 10, m = 6: 10436 a tam 43

10

436 a

So,101

436

a 061

“ the baby has a mass of 6 kg at 10 weeks. Find the value of the growth factor “

tam 43

Mass at birth is kg and the growth factor is kg per week.

43061

Growth and Decay

e.g. 2. Find the time taken for a cup of tea to cool to Celsius if its temperature, Celsius, t minutes after it was poured is given by

60 T

tT )940(80 Solutio

n:

We want to find t given T = 60 tT )940(80 t)940(8060

Divide by 80:t)940(750

t

)940log(

)750log(

A log is just an index, sobut this is no help, since we don’t know the value of logs with base .

)750(log 940 t

940

We “take” logs:

t)940log()750log( )940log()750log( tUsing the 3rd law of

logs:

654 tTime taken is approximately 4 minutes 39

seconds.

xkx ak

a loglog It doesn’t matter whether we use base 10 or base e

Growth and Decay

Solution:

The half-life is the time taken for the mass to halve.

The initial mass is so we want t when

0m2

0mm

tmm )90(0tmm )990(50 00

“Take” logs:

Using the 3rd law:

t)990log()50log(

t

)990log(

)50log()990log()50log( t

Half-life is 69 years to nearest year.

Notice that the half-life doesn’t depend on the initial mass.

3. Find the half-life of a radio-active element that decays according to the following model:

tmm )990(0 where t is the time in years and is the initial mass

0m

Growth and Decay

tkay

SUMMARY• An exponential model is of the form

where, k is the initial value ( the value of y when t

= 0 ),a is the growth factor, andt is the time

• If a > 1, the model is for exponential growth.• If 0 < a < 1, the model is for exponential

decay.• The half-life of a radio-active element is the time taken for the mass to halve. It is independent of the initial mass. We always get ta50

Growth and DecayExercise1. (i) Find the growth factor, a, in the

following equation, given that y = 70 when t = 2 tay 50Give your answer to 2 s.f.

2. Find the half-life of a radio-active element whose mass m is given by

where t is measured in seconds. Give the answer correct to 1 d.p.

tkm )70(

(ii) Hence, find the value of t, to the nearest integer, when y reaches 100.

Growth and Decay

1. (i) y = 70 when t = 2;50 tay

Solutions:

25070 a

2

50

70a

a41

)2(21 s.f. a

Growth and DecaySolutions

:

t)21(50100 :100yt)21(2

“Take” logs:

t)21log()2log(

ty )21(50

)21log()2log( t

t )21log(

)2log(4 t ( nearest

integer )

(ii) Hence, find the value of t, to the nearest integer, when y reaches 100.

Growth and Decay

Solution:tkk )70(50

Or go directly to this stage:

t)70(50 “Take” logs: t)70log()50log(

)70log()50log( t

t

)70log(

)50log(

2. Find the half-life of a radio-active element whose mass m is given by tkm )70( where t is measured in seconds. Give the answer correct to 1 d.p.

Half-life is s. 91

( 1 d.p. )

Either substitute :km 50

Growth and Decay

If we want to find a rate of change, we need to differentiate.

Using implicit differentiation we saw that

xay aadx

dy x ln

So, if we have the model

tA )41(3

This is an awkward result and we can avoid it by using e in the model.

dt

dAwe get 41ln)41(3 t

This is of the form

tkaA

Growth and DecayUsing e and natural

logs

Let

be41Changing to log form:

be41 41ln b

We want to replace by an expression using e.

41

Suppose we have tA )41(3 - - - - - - (1)

3360

Substituting in (1) :

336041 e

teA )(3 3360teA 33603

Growth and DecayUsing e and natural

logs tA )41(3 So, can be replaced byteA 33603

( The result isn’t exact but could be made as accurate as we like by taking more decimal places for the value of b ).

In the same way we can replace any value of a, the growth factor, by a power of e.You probably won’t be asked to change from a to e, but the above explains why e appears in most equations for growth and decay.

b > 0 gives exponential growth

If ,btaey

b < 0 gives exponential decay

N.B.

Growth and Decay

tex 530200

e.g. The equation below gives the number, x, of bacteria in a solution t hours after counting started:

(i) How many bacteria are there after 2 hours?

(ii) How fast is the number increasing after 2 hours?

Solution:

(i) 2t 061200 ex 577(ii) We are being asked to find a rate of

increase.te

dt

dx 530530200

2t 306dt

dx The number is increasing at 306 per

hour.

Growth and Decay

(i) How many fish were originally in the pond?

1. The number of fish, n, in a pond is given approximately by the equation

ten 05050 where t is the time in

months.(ii) How many fish were there after 1

year?

Solution: (i) 500 nt(ii)

2712 nt

(iii) At what rate are the numbers declining after 1 year? ( Give the ans. to the nearest integer ).

(ii)

tedt

dn 050)050(50 41dt

dn

They are declining at a rate of 1 per month.

Exercise

Growth and Decay

Growth and Decay

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Growth and Decay

tkay

SUMMARY• An exponential model is of the form

where, k is the initial value ( the value of y when t

= 0 ),a is the growth factor, andt is the time

• If a > 1, the model is for exponential growth.• If 0 < a < 1, the model is for exponential

decay.• The half-life of a radio-active element is the time taken for the mass to halve. It is independent of the initial mass. We always get ta50

Growth and Decay

where m is mass in kg and t is time in weeks and if the baby has a mass of 6 kg at 10 weeks.

e.g. 1. In the first few weeks of life an exponential model is a reasonably good fit for the mass of a baby. Find the mass of the baby at birth and the value of the growth factor if tam 43

Solution:

At birth, t = 0, so

tam 43 043 am 43 m

Growth and Decay

When t = 10, m = 6: 10436 a tam 43

10

436 a

So,101

436

a 061

“ the baby has a mass of 6 kg at 10 weeks. Find the value of the growth factor “

tam 43

Mass at birth is kg and the growth factor is kg per week.

43061

Growth and Decaye.g. 2. Find the time taken for a cup of tea to

cool to Celsius if its temperature, Celsius, t minutes after it was poured is given by

60 T

tT )940(80 Solutio

n:

We want to find t given T = 60 tT )940(80 t)940(8060

Divide by 80:t)940(750

t

)940log(

)750log(

We “take” logs:

t)940log()750log( )940log()750log( tUsing the 3rd

law:

654 tTime taken is approximately 4 minutes 39

seconds.

xkx ak

a loglog

Growth and Decay

Solution:

e.g. 3. Find the half-life of a radio-active element that decays according to the following model: tmm )990(0 where t is the time in years and the initial mass.

The half-life is the time taken for the mass to halve.

The initial mass is so we want t when

0m2

0mm

tmm )90(0tmm )990(50 00

“Take” logs:

Using the 3rd law:

t)990log()50log(

t

)990log(

)50log()990log()50log( t

Half-life is 69 years to the nearest year.

0m

Growth and DecayUsing e and natural

logs

Let

be41Changing to log form:

be41 41ln b

We want to replace by an expression using e.

41

Suppose we have tA )41(3 - - - - - - (1)

3360

Substituting in (1) :

336041 e

teA )(3 3360teA 33603

Growth and Decay

tex 530200

e.g. The equation below gives the number, x, of bacteria in a solution t hours after counting started:

(i) How many bacteria are there after 2 hours?

(ii) How fast is the number increasing after 2 hours?

Solution:

(i) 2t 061200 ex 577(ii) We are being asked to find a rate of

increase.te

dt

dx 530530200

2t 306dt

dx The number is increasing at 306 per

hour.