# “Teach A Level Maths” Vol. 2: A2 Core Modules

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“Teach A Level Maths” Vol. 2: A2 Core Modules
48: Growth and Decay © Christine Crisp

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

a is called the growth factor.
If 0 < a < 1, the function decays. e.g. We are going to use functions of the type to model some practical situations where we have growth or decay. a is called the growth factor.

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

Since these are practical problems, one of the variables is time
Since these are practical problems, one of the variables is time. We use the letter t and since it is time, We use any meaningful letter for the other variable.

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 Solution: At birth, t = 0, so

“ the baby has a mass of 6 kg at 10 weeks
“ the baby has a mass of 6 kg at 10 weeks. Find the value of the growth factor “ When t = 10, m = 6: So, Mass at birth is kg and the growth factor is kg per week.

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 Solution: We want to find t given T = 60 Divide by 80: A log is just an index, so but this is no help, since we don’t know the value of logs with base We “take” logs: Using the 3rd law of logs: It doesn’t matter whether we use base 10 or base e Time taken is approximately 4 minutes 39 seconds.

where t is the time in years and is the initial mass
3. Find the half-life of a radio-active element that decays according to the following model: where t is the time in years and is the initial mass The half-life is the time taken for the mass to halve. Solution: The initial mass is so we want t when “Take” logs: Notice that the half-life doesn’t depend on the initial mass. Using the 3rd law: Half-life is 69 years to nearest year.

k is the initial value ( the value of y when t = 0 ),
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, and t 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

Exercise 1. (i) Find the growth factor, a, in the following equation, given that y = 70 when t = 2 Give your answer to 2 s.f. (ii) Hence, find the value of t, to the nearest integer, when y reaches 100. 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.

Solutions: 1. (i) y = 70 when t = 2

Solutions: (ii) Hence, find the value of t, to the nearest integer, when y reaches 100. “Take” logs: ( nearest integer )

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. Solution: Either substitute : Or go directly to this stage: “Take” logs: Half-life is s. ( 1 d.p. )

If we want to find a rate of change, we need to differentiate.
Using implicit differentiation we saw that So, if we have the model This is of the form we get This is an awkward result and we can avoid it by using e in the model.

Using e and natural logs
Suppose we have (1) We want to replace by an expression using e. Let Changing to log form: Substituting in (1) :

b > 0 gives exponential growth b < 0 gives exponential decay
Using e and natural logs So, can be replaced by ( 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. N.B. If , b > 0 gives exponential growth b < 0 gives exponential decay

The number is increasing at 306 per hour.
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) (ii) We are being asked to find a rate of increase. The number is increasing at 306 per hour.

Exercise 1. The number of fish, n, in a pond is given approximately by the equation where t is the time in months. (i) How many fish were originally in the pond? (ii) How many fish were there after 1 year? (iii) At what rate are the numbers declining after 1 year? ( Give the ans. to the nearest integer ). Solution: (i) (ii) (ii) They are declining at a rate of 1 per month.

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.

k is the initial value ( the value of y when t = 0 ),
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, and t 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

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 Solution: At birth, t = 0, so

When t = 10, m = 6: So, “ the baby has a mass of 6 kg at 10 weeks. Find the value of the growth factor “ Mass at birth is kg and the growth factor is kg per week.

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 Solution: We want to find t given T = 60 Divide by 80: We “take” logs: Using the 3rd law: Time taken is approximately 4 minutes 39 seconds.

The half-life is the time taken for the mass to halve.
Solution: e.g. 3. Find the half-life of a radio-active element that decays according to the following model: 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 “Take” logs: Using the 3rd law: Half-life is 69 years to the nearest year.

Using e and natural logs
Let Changing to log form: We want to replace by an expression using e. Suppose we have (1) Substituting in (1) :

The number is increasing at 306 per hour.
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) (ii) We are being asked to find a rate of increase. The number is increasing at 306 per hour.