Grade by Grade
B. Exponential Functions · Exponential Functions

Growth and Decay Models, and What Makes Sense

B3.2About 70 minutes
Learning goals
  • I can identify exponential growth and decay in tables, graphs and equations.
  • I can write growth and decay models using percent rates, doubling times and half-lives.
  • I can explain the restrictions a real situation places on the domain and range of a model.
  • I can interpret each number in a cooling model with a non-zero asymptote.

Words to know

exponential growth
An increase by the same factor (greater than 1) in each equal time period: y = a(1 + r)ᵗ, r > 0.
exponential decay
A decrease by the same factor (between 0 and 1) in each equal time period: y = a(1 − r)ᵗ.
growth or decay rate (r)
The percent change per period, as a decimal. 3% growth: b = 1.03. 12% decay: b = 0.88.
doubling time
The time it takes a growing quantity to double: y = a(2)^(t/d), where d is the doubling time.
half-life
The time it takes a decaying quantity to fall to half: y = a(1/2)^(t/h), where h is the half-life.
ambient temperature
The temperature of the surroundings (for example, the room). A cooling object approaches it.

Growth or decay?

Every exponential model in the real world is either growing (b > 1) or decaying (0 < b < 1). Bank balances, populations and the number of people who have heard a rumour grow. Radioactive material, the value of a car, the heat of a cup of tea and the air pressure in a slowly leaking tire decay. Today you will recognize growth and decay in tables, graphs and equations, and think carefully about which inputs and outputs actually make sense in each situation.

Remember from Grade 9

A percent increase of r means multiplying by (1 + r). A 5% raise multiplies by 1.05; a 20% discount multiplies by 0.80. In Grade 9 you also used A = P(1 + i)ⁿ for compound interest.

FormMeaningExample
y = a(1 + r)ᵗgrows by r (a percent) each periodP = 1200(1.03)ᵗ: grows 3% per period
y = a(1 − r)ᵗdecays by r each periodV = 500(0.88)ᵗ: loses 12% per period
y = a(2)^(t/d)doubles every d time unitsN = 40(2)^(t/5): doubles every 5 hours
y = a(1/2)^(t/h)halves every h time units (half-life h)m = 80(1/2)^(t/12): half-life 12 years
Common forms of exponential growth and decay models. In each, a is the initial amount.
ty0123402468growth: y = 1(2)ᵗdoubles each step (b > 1)ty0123402468decay: y = 8(1/2)ᵗhalves each step (0 < b < 1)
Growth multiplies by a number greater than 1 each step; decay multiplies by a number between 0 and 1. Both curves stay above the asymptote y = 0.
Worked example

Example 1: Identify growth or decay

For each model, say whether it is growth or decay and give the rate, doubling time or half-life. (a) P = 1200(1.03)ᵗ (b) V = 500(0.88)ᵗ (c) N = 40(2)^(t/5) (d) m = 80(1/2)^(t/12)

  1. (a) b = 1.03 > 1: growth of 3% per period. Initial value 1200.
  2. (b) b = 0.88 < 1: decay. 1 − 0.88 = 0.12, so it loses 12% per period.
  3. (c) Growth: the amount doubles each time t/5 goes up by 1, that is every 5 time units.
  4. (d) Decay: the amount halves every 12 time units (half-life 12).

(a) growth 3%; (b) decay 12%; (c) growth, doubling time 5; (d) decay, half-life 12.

Worked example

Example 2: From a table

A table shows the value of a snowmobile: year 0: $12 000; year 1: $10 200; year 2: $8670; year 3: $7369.50. Is this growth or decay? Write a model.

  1. Ratios: 10 200 ÷ 12 000 = 0.85, 8670 ÷ 10 200 = 0.85, 7369.50 ÷ 8670 = 0.85.
  2. Constant ratio 0.85 < 1: exponential decay, losing 15% of its value each year.
  3. Model: V = 12 000(0.85)ᵗ, with t in years.

V = 12 000(0.85)ᵗ (decay, 15% per year).

Restrictions from the context

The mathematical function y = a(bˣ) + c has domain ℝ. A model usually does not. Ask:

  • Can the input be negative? Time before the start of an experiment often makes no sense, so the domain is usually t ≥ 0 (and it may have an upper limit, when the situation ends).
  • Must values be whole numbers? Bounces, rounds, years in which interest is paid, and numbers of animals are discrete. The model may give 14.7 animals, but the real answer is a whole number.
  • What limits the output? The range of the model is limited by the starting value and by the asymptote. A cooling drink can never get colder than the room it sits in (the ambient temperature). A leaking tire's gauge pressure cannot go below 0, because then the air inside is at the same pressure as the air outside.
t (min)T (°C)0510152025300102030405060708090(0, 80)(10, about 41)room temperature 20°C (asymptote)T = 20 + 60(0.9)ᵗ
A cup of hot chocolate at 80°C cools in a 20°C room: T = 20 + 60(0.9)ᵗ. The temperature drops quickly at first and then levels off toward the room temperature, which acts as the asymptote.
Worked example

Example 3: A cooling curve

Hot chocolate at 80°C is set on a table in a 20°C kitchen. Its temperature t minutes later is modelled by T = 20 + 60(0.9)ᵗ. (a) Explain each number. (b) Find T after 10 minutes. (c) State a reasonable domain and range.

  1. (a) 20 is the room temperature (the asymptote). 60 = 80 − 20 is the starting difference between the drink and the room. 0.9 means the difference shrinks by 10% each minute.
  2. (b) T = 20 + 60(0.9)¹⁰ ≈ 20 + 60(0.349) ≈ 40.9°C.
  3. (c) Domain: t ≥ 0 (time after it was set down; the model is for this cooling period). Range: 20 < T ≤ 80. It starts at 80°C and never quite reaches 20°C.
  4. The model is approximately how real cooling behaves (it is based on Newton's law of cooling); real measurements wobble a little, as in Lesson 12.

About 40.9°C after 10 minutes; domain t ≥ 0; range 20 < T ≤ 80.

Worked example

Example 4: Discrete growth

A lab starts with 500 bacteria, and the population doubles every 3 hours: N = 500(2)^(t/3). (a) Find N after 12 hours. (b) Discuss the domain and range.

  1. (a) N = 500(2)^(12/3) = 500(2⁴) = 8000 bacteria.
  2. (b) Domain: t ≥ 0, up to some time when food or space runs out and the model stops working (real populations cannot double forever).
  3. Range: N ≥ 500, and in reality N is a whole number, so model values like 707.1 (at t = 1.5) are rounded.

8000 bacteria; domain 0 ≤ t ≤ (end of the growth period); range N ≥ 500, whole numbers.

Worked example

Example 5: A leaking tire

A bicycle tire's gauge pressure is 220 kPa and it loses 4% of it each day: P = 220(0.96)ᵗ. (a) Growth or decay? (b) Find P after one week. (c) State the domain and range of the model and explain the restriction.

  1. (a) b = 0.96 < 1: decay, 4% per day.
  2. (b) P = 220(0.96)⁷ ≈ 165 kPa.
  3. (c) Domain t ≥ 0 (days since it was pumped up). Range 0 < P ≤ 220: the gauge pressure starts at 220 kPa and approaches 0, which is when the pressure inside matches the air outside.

Decay; about 165 kPa after a week; range 0 < P ≤ 220.

Just for fun

A funny-but-true observation: hot chocolate left on the table on a snowy January afternoon never gets colder than the kitchen.

It isn't being stubborn. It's just respecting its asymptote.

Hands-on activity

Cooling-water investigation

You need: a mug, hot water from the tap (not boiling; a grown-up should handle anything hotter than warm tap water), a kitchen or candy thermometer, a timer and paper.

Safety: hot water can scald. Keep the mug on a stable table, and do not hold the thermometer in the water with your fingers in it.

  1. Measure the room temperature first, and write it down.
  2. Fill the mug with hot tap water and measure the starting temperature.
  3. Measure the temperature every 2 minutes for 20 minutes.
  4. For each reading, subtract the room temperature to get the difference D. Find the ratio of each D to the one before it.
  5. Write a model T = (room temperature) + D₀(b)ⁿ, where D₀ is the starting difference, b is the mean ratio and n counts 2-minute steps.
  6. State the domain and range of your model, and explain why the range has a lower limit.

What to look for: the differences shrink by roughly the same factor every 2 minutes (often somewhere around 0.85 to 0.95, depending on the mug), so the cooling is approximately exponential. The water never cools below room temperature, so the range is (room temperature) < T ≤ (starting temperature).

Your turn

1

Growth or decay? Give the rate: A = 750(1.065)ᵗ.

2

Growth or decay? Give the rate: y = 90(0.7)ᵗ.

3

A substance has a half-life of 6 hours. Starting with 400 mg, write a model.

4

A population of 2500 grows by 4% per year. Write a model and find it after 5 years.

5

Why is the domain of a cooling model usually t ≥ 0?

6

A cup of tea cools in a 22°C room. Why can't the range include 15°C?

7

A table: 64, 48, 36, 27 at t = 0, 1, 2, 3. Growth or decay? Rate?

More practice: warm-up

1

Which base shows growth: 0.95 or 1.05?

2

What is b for a 9% decrease per year?

3

What is b for a 2.5% increase per month?

4

In y = a(1/2)^(t/h), what is h?

5

Give one reason a model's domain might be restricted.

More practice: on your own

1

A car bought for $32 000 loses 18% of its value per year. Write a model and find its value after 4 years.

2

Explain the meaning of each number in T = 21 + 74(0.92)ᵗ (t in minutes).

3

For N = 300(2)^(t/4) (bacteria, t in hours), find N at t = 10 and state a reasonable domain and range.

4

A ball's rebound height is h = 2(0.6)ⁿ metres after the nth bounce. Why is the domain {0, 1, 2, 3, …} and not all real numbers?

5

Write a model for a quantity that triples every 8 years starting at 50, and identify it as growth or decay.

Challenge

1

A model T = 20 + 70(0.85)ᵗ describes coffee cooling. A student says the range is {T ∈ ℝ | T > 20}. What is missing? Give the full range.

2

Show that a 50% decrease each period followed by a 50% increase does not get you back to the start, using exponential factors.

3

A town's model is P = 15 000(1.02)ᵗ (t in years since 2020). Why might the model fail by 2080, even if the math is correct?

Check your understanding

Quiz

1.

Which model shows exponential decay?

2.

A quantity grows 7% per year. What is b?

3.

In m = 120(1/2)^(t/30), the half-life is:

4.

Soup at 90°C cools in a 20°C room. Which range is reasonable?

5.

Why is the domain of a bouncing-ball model {0, 1, 2, …}?

6.

V = 8000(0.9)ᵗ. By what percent does V decrease each period?

For parents and teachers

Big idea: Expectation B3.2: identify exponential growth and decay (radioactive decay, population growth, cooling, a leaking tire) in tables, graphs and equations, and explain restrictions the context puts on the domain and range (for example, room temperature limits a cooling curve).

How to help: Point out growth and decay in daily life: a car's resale value, a savings account, a cooling cup of tea. Ask "what number is it multiplied by each year (or minute)?" and "what values make sense here?"

Common difficulties:

  • Writing a 12% decrease as b = 0.12 instead of 0.88.
  • Stating the domain of a real model as "all real numbers". Think about what the variable means.
  • Forgetting the upper limit of a range (a cooling drink never gets hotter than it started).
  • Safety: the cooling activity uses hot tap water only. Please supervise and keep boiling water out of it.