Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 7: what is 35% of 40?
Answer
14. 10% is 4, so 30% is 12 and 5% is 2. - From Lesson 1: in y = 3 × 2x, what is y when x = 5?
Answer
96. 3 × 32. - From Year 8: what is 360 ÷ 6?
Answer
60.
Words used in this lesson
- half-life
- the time it takes for a quantity to fall to half. For example: 6 days for a fading dye.
- doubling time
- the time it takes for a quantity to double. For example: the days for a patch to cover twice the area.
- growth factor
- the number you multiply by for each period. For example: 1.15 for a rise of 15%.
- percentage rate
- the percentage change in one period. For example: a rise of 4% a month.
- period
- one step of time in the model. For example: one week.
- model
- an equation that describes a real situation. For example: y = 400 × 1.1x.
Coffee at midnight
Caffeine leaves the blood of a healthy adult at a steady proportion: about half of what is there goes every 5 hours. Someone in Addis Ababa drinks a strong coffee at 8 in the morning, and there is 100 mg in the blood a little later.
Predict first: at midnight, 16 hours later, will there be almost none, or a noticeable amount? Make a guess for the fraction of the 100 mg that is left, and say whether your guess depends on the number 5 or only on the word “half”. Use the first figure to see how the time and the half-life work together.
Half-life 4 hours, after 0 hours: 80 mg is left, which is 100% of the start.
Show me the steps
- Half-life 4 hours, 0 hours: 80 mg, the start. The model is y = 80 × 0.5t/4, where t/4 counts the half-lives so far.
- Slide to 8 hours: that is 2 half-lives, so 80 → 40 → 20 and the amount is 20 mg, which is 25% of the start.
- Slide to 10 hours: that is 2.5 half-lives, and 80 × 0.52.5 is about 14.1 mg. It is between 20 and 10, and not their midpoint 15, because the curve bends.
Three ways to find a and b.
- A percentage rate r% each period: b = 1 + r/100 for growth and b = 1 − r/100 for decay. A 3% rise is b = 1.03 and a 12% fall is b = 0.88.
- A doubling or half-life time T: y = a × 2t/T for doubling and y = a × 0.5t/T for halving. The exponent t/T counts how many doublings or halvings have happened.
- Two values one period apart: b = later value ÷ earlier value, and a is the earlier one at t = 0.
In every case, a is the amount at t = 0 and the exponent counts periods, not hours or years.
Start 200, then 240: b = 240 ÷ 200 = 1.2, a rise of 20% each period.
Show me the steps
- Start 200, then 240: the factor is 240 ÷ 200 = 1.2, which is a rise of 20% each period. The next values are 240 × 1.2 = 288 and then 345.6.
- Slide to 100: the factor is 100 ÷ 200 = 0.5, a fall of 50% each period. The values are 200, 100, 50, 25.
- Slide to 400: the factor is 400 ÷ 200 = 2, a rise of 100% each period. A rise of 100% doubles the value.
The rate is not the base. A 40% rise is b = 1.4, not 0.4, and a 35% fall is b = 0.65, not 0.35 and not 35. And half-lives are not equal amounts: a substance that halves every 2 hours loses less in each later 2 hours, because it halves a smaller amount. Check the base: a fall must have b below 1, a rise above 1.
Worked examples
Algae in a pond starts at 5 g and doubles every 3 days. Find the mass after 12 days.
- The model is y = 5 × 2t/3, with t in days.
- At t = 12 the exponent is 12 ÷ 3 = 4, which is 4 doublings.
- 24 = 16, so y = 5 × 16 = 80 g.
A sample of 500 g loses 20% of its mass each hour. Find its mass after 2 hours.
- A 20% fall is b = 1 − 0.2 = 0.8, so y = 500 × 0.8t.
- 0.82 = 0.64, and 500 × 0.64 = 320 g.
Your go. The first step is given. A sample of 360 mg halves every 6 hours. Find the mass after 18 hours.
- The model is y = 360 × 0.5t/6.
- Work out how many half-lives 18 hours is, then halve that many times.
Check the answer
18 ÷ 6 = 3 half-lives, so 360 → 180 → 90 → 45, which is 45 mg. A common slip is to halve once and then divide by 6, or to use 18 as the exponent.
With a calculator
When the time is not a whole number of half-lives, work out the exponent first: for 7 hours with a half-life of 4 hours it is 7 ÷ 4 = 1.75. Then use the power key: 0.5, power, 1.75, equals, and multiply by the start amount.
Use the answer memory or write the exponent on paper, so that a mistyped bracket cannot turn 7 ÷ 4 into 7 ÷ 4 × something else. Round only at the end.
Check before you go on
1. A quantity falls by 2.5% each hour. What base b goes in y = a × bx?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. A quantity starts at 6000 and is multiplied by 0.9 each week. What is it after 2 weeks?
3. A quantity of 7 units doubles every 4 hours. How many units are there after 12 hours?
Level 2 · Building
4. A value rises by 0.5% each month. What base b goes in y = a × bx? Give it as a decimal.
5. A count was 250 last week and 325 this week, and it changes by the same factor each week. What is the factor, to 2 decimal places?
6. A substance has a half-life of 5 hours. There is 120 mg at the start. How many mg are left after 15 hours?
Level 3 · Stretch
7. The model is y = 800 × bx with b above 0, and y = 288 when x = 2. Find b.
8. Two machines are each worth 5000 rand today. Machine A loses 10% of its value each year. Machine B loses 400 rand each year. After 3 years, how many rand more is B worth than A?
Something to think about
A quantity falls by 50% each period. Another falls by 100% in one period. What are their bases, and what is strange about the second?
One way to think about it
The first has base 0.5. The second has base 0, which means it is gone after one period: 1 − 100/100 = 0. A base of 0 is not allowed in an exponential function, because every value after the first would be 0 and there would be no curve at all.
Exit ticket
9. A quantity starts at 10 and triples every 2 days. Which model gives its value after t days?