Interpreting the numbers in a model
A model says a rowing club has 40(1.15)^t members t years after it was founded. What do the 40 and the 1.15 tell you about the club?
Where you are 5 in Modeling with functions
When you reach for it
A model is a rule with numbers in it, and each number stands for something in the situation. A canoe hire shop charges C = 8h + 12 dollars for h hours. A rowing club has P = 40(1.15)^t members t years after it was founded. Before a model can be used, each number in it has to be read back as a full sentence with units: “each hour of hire adds 8 dollars”, “the club was founded with 40 members”.
For a line, Linear functions in context reads the slope and intercept of one model, and Interpreting the slope and intercept does the same for a line fitted to data. This lesson is mostly about exponential models, where the meaning is easier to miss: a factor of 0.97 hides a loss of 3\%, and a 4 inside an exponent can be a length of time. It also sets the two kinds side by side.
The procedure
Name the quantities first. Every reading needs the input and the output with their units: h in hours and C in dollars, t in years and P in members. Without them a sentence can only say “it goes up by 8”, which is not yet a meaning.
The starting value comes from an input of 0. In y = mx + b, putting x = 0 leaves y = b. In y = a \cdot b^t, putting t = 0 leaves a \cdot b^0 = a \cdot 1 = a. So the number standing alone is the output when the input is 0, in the output’s units. Then ask whether an input of 0 belongs to the situation at all. If it does not, the number still fixes the model, but it may describe nothing real (A reasonable domain for a real situation).
The change comes from raising the input by 1. For a line,
so m is the amount added for each one unit of input, measured in output units per input unit, such as dollars per hour. For an exponential,
so b is the factor the output is multiplied by for each one unit of input. A factor has no units: it is a ratio of two amounts measured in the same units. Its distance from 1 is the percent rate, as Growth and decay factors explains: b = 1.15 is growth of 15\% per period, and b = 0.97 is a loss of 3\% per period.
The two forms use the letter b for different things, a start in one and a factor in the other, so read each number by its position, not its letter.
| line y = mx + b | exponential y = a \cdot b^t | |
|---|---|---|
| value at input 0 | b | a |
| each one unit of input | adds m | multiplies by b |
| units of that number | output per input | none |
| percent rate per unit | changes as y changes | b - 1, always the same |
Rewrite a disguised form before reading it. A model is often printed in a form that hides its parts, and reading the parts of an expression is how they are found. Three rewrites cover most cases.
- P = 40(1.15)^t is P = 40 \cdot 1.15^t. The exponent applies only to the 1.15 beside it, since powers are worked out before multiplication. It is not 46^t.
- C = 8(h + 1.5) is 8h + 12 by the The distributive property. The factored form says something too: every hire is charged as if it lasted 1.5 hours longer.
- A = 2000(0.5)^{t/3} has t/3 as its exponent, and t/3 counts blocks of 3 units of t. So 0.5 is the factor for every 3 units: the amount halves every 3 units of t. The factor for one unit is 0.5^{1/3}, by Powers of powers, products and quotients.
Changing the unit of the input changes the rate, never the start. The input 0 is the same moment whether t counts years or months, so the start does not move. A rate of m per year is m/12 per month, because twelve equal monthly additions make one yearly one. A factor of b per year is b^{1/12} per month, because twelve equal monthly factors must multiply to b.
Figure 1 shows the two kinds side by side.
Worked examples
Canoe hire in two forms
A canoe hire shop charges C = 8(h + 1.5) dollars to hire a canoe for h hours. Say what each number means, and find the cost of a 3-hour hire.
SolutionBy the distributive property, 8(h + 1.5) = 8h + 12. The slope 8 is in dollars per hour: each extra hour adds 8 dollars. The intercept 12 is the cost when h = 0. Nobody hires a canoe for 0 hours, so 12 dollars is not the price of any real hire; it is a fixed fee that is part of every hire, whatever its length.
The factored form reads the fee another way: 12 = 8 \times 1.5, so the fee is the price of one and a half hours. A 3-hour hire costs 8(3 + 1.5) = 8 \times 4.5 = 36 dollars, the same as 8 \times 3 + 12 = 24 + 12.
The rowing club
The membership of a rowing club t years after it was founded is modeled by P = 40(1.15)^t. Say what 40 and 1.15 mean, and estimate the membership after 5 years.
SolutionThe exponent belongs to the 1.15 alone, so the model is 40 \cdot 1.15^t. At t = 0 it gives 40 \cdot 1 = 40: the club was founded with 40 members. The base 1.15 is the factor per year, and 1.15 = 1 + 0.15, so membership grows 15\% a year.
The 15\% is of the current membership, so the yearly gain grows. P(1) = 46, a gain of 6 members, and P(2) = 40 \times 1.3225 = 52.9, a gain of 6.9. After 5 years, 1.15^5 \approx 2.01136, so P(5) \approx 80.45: about 80 members, roughly twice the start.
Reading the model as (40 \times 1.15)^t = 46^t would give 46^0 = 1 member at founding, which shows it is a different function.
A shrinking glacier
In a made-up example, a glacier’s area is modeled by A = 120(0.97)^t square kilometers, t years after 2000. Say what each number means. Then find the factor for ten years and the percent of the area lost in a decade.
SolutionAt t = 0, A = 120: in 2000 the glacier covered 120 square kilometers. The factor 0.97 is less than 1, so the area shrinks, and 1 - 0.97 = 0.03: the glacier loses 3\% of its area each year. In the first year that is 0.03 \times 120 = 3.6 square kilometers; later years take 3\% of a smaller area, so they lose less.
Ten years multiply by 0.97 ten times: 0.97^{10} \approx 0.7374. About 73.7\% of the area remains, so about 26.3\% is lost in a decade, not 10 \times 3\% = 30\%. In 2010 the model gives 120 \times 0.97^{10} \approx 88.49 square kilometers.
Halving every three weeks
After a field is sprayed, the number of beetles in it is modeled by B = 2000(0.5)^{t/3}, where t is in weeks. Say what 2000, 0.5 and 3 mean, find B(12), and find the factor per week.
SolutionAt t = 0 the exponent is 0, so B = 2000: there were 2{,}000 beetles when the field was sprayed. The exponent t/3 counts three-week blocks: after 3 weeks it is 1, after 6 weeks it is 2. Each block multiplies by 0.5, so the number halves once every three weeks. The 3 is not a rate: it is a time, the number of weeks the beetles take to halve.
Twelve weeks are 12/3 = 4 blocks, so B(12) = 2000 \times 0.5^4 = 2000 \times 0.0625 = 125 beetles.
To read the weekly change, write (0.5)^{t/3} = \left(0.5^{1/3}\right)^t by the power of a power. A calculator gives 0.5^{1/3} \approx 0.7937, so the model is about B = 2000(0.7937)^t: the number falls by about 20.6\% each week. As a check, 0.7937^3 \approx 0.5000, one halving.
Per year or per month
(a) Rewrite the rowing club model P = 40(1.15)^t with n counting months, and say what its numbers mean. (b) Rewrite the canoe hire model C = 8h + 12 with n counting minutes.
Solution(a) n months are t = n/12 years, so
A calculator gives 1.15^{1/12} \approx 1.0117, so P \approx 40(1.0117)^n. The start is still 40 members, because n = 0 is the same founding day. The club grows about 1.17\% a month. That is not 15\% \div 12 = 1.25\%: twelve monthly factors multiply, and 1.0125^{12} \approx 1.161, which overshoots 1.15.
(b) n minutes are h = n/60 hours, so C = 8 \cdot \tfrac{n}{60} + 12 = \tfrac{2}{15}n + 12: about 0.133 dollars per minute, with the same 12-dollar fee. For a line the rate is divided by 60; for an exponential the factor takes a root. In both, the start stays.
Two models of the glacier
Inigo models the glacier’s area as L = 120 - 3.6t and Noemi as G = 120(0.97)^t, both in square kilometers t years after 2000. (a) Say what the numbers in each model mean, and compare the two in 2010. (b) Mordecai rewrites Inigo’s model with Y the calendar year. Find his model and say what its numbers mean.
Solution(a) Both start at 120 square kilometers in 2000. Inigo’s slope -3.6 means the glacier loses 3.6 square kilometers every year, the same amount each year. Noemi’s factor 0.97 means it loses 3\% of its area each year, which is 3.6 square kilometers in the first year and less after that. So the two agree over the first year and then part: in 2010, L = 120 - 36 = 84 and G \approx 88.49 square kilometers. Inigo’s model reaches 0 when 3.6t = 120, at t \approx 33.3, and gives negative areas after that; Noemi’s never reaches 0.
(b) The years since 2000 are t = Y - 2000, so
The slope is unchanged: 3.6 square kilometers lost per year. The intercept 7{,}320 is what the line gives when Y = 0, the year 0, two thousand years before the model starts. No glacier was measured then. The number only fixes where the line crosses the vertical axis, and it has no meaning in the situation. That is why Inigo counted years from 2000: then the intercept, 120, is a real area.
Boundaries
A start needs input 0 in the domain. When 0 is not a sensible input, the constant still fixes the model but is not the value of anything real, as with the 7{,}320 above.
A factor belongs to its period. b is the factor for one unit of the input as written. With t/3 in the exponent, the period is three units. Changing the unit raises the factor to a power; it never divides the percent.
The form must be exactly a \cdot b^t before a and b are read off. (40 \times 1.15)^t is a different function, and in y = 5 \cdot 2^t + 3 the value at t = 0 is 8, not 5.
An exponential model needs b > 0 and b \ne 1. With b = 1 the model is a constant.
A reading is about the model. Real clubs and glaciers do not change by exactly the same amount or percent every year, and a model’s numbers are often rounded. A sentence about a parameter says what the model claims, which is only as good as its fit (Judging and comparing models).
Where this goes next
Where a model holds: domain, interpolation and extrapolation asks how far beyond its data a model’s sentences can be trusted. Building a model from a description runs this lesson backwards, turning sentences into the numbers of a rule. Graphs of exponential functions shows a as the vertical intercept and b as how steeply the curve climbs or falls, and Linear versus exponential growth follows a line and an exponential far enough to see the exponential overtake. The time a factor takes to double or halve an amount is Doubling time and half life.
Exercises
Linear models
- A canoe hire shop charges C = 6h + 10 dollars for h hours. Say what 6 and 10 mean.
- Write C = 5(h + 2) in the form mh + b, and say what the 2 means for a canoe hire charged this way.
- Odhran bails water out of a canoe. The water left is W = 30 - 2.5m liters after m minutes. Say what 30 and -2.5 mean, and find when the canoe is empty.
Exponential models
- A weekly park run has Q = 500(1.08)^t registered runners t years after 2020. Say what 500 and 1.08 mean, and estimate the number of runners in 2023.
- A made-up glacier has area A = 65(0.985)^t square kilometers t years after 2010. Say what 65 and 0.985 mean, and find the area in 2020.
- Give the percent change per period hidden in each factor: 0.92, 1.2, 0.08.
- Perdita reads P = 40(1.15)^t as “the club gains 15 members a year.” Find the gains in the first two years and correct her sentence.
Disguised forms
- Find 30(1.2)^t and 36^t at t = 0 and at t = 2. Then say what 30 and 1.2 mean in 30(1.2)^t.
- A rumor has been heard by N = 50(2)^{t/6} people t hours after it starts. Say what 50, 2 and 6 mean, find N(18), and find the factor per hour to four decimal places.
- A new shop has K = 200(1.69)^{t/2} customers t years after it opens. Find the factor and the percent growth per year.
Changing the unit of time
- The glacier in Example 3 shrinks by the factor 0.97 a year. Find the factor per month to four decimal places and the percent lost each month.
- A shop charges C = 9h + 12 dollars for h hours. Write the cost for n minutes.
- A quantity is M = 300(1.02)^t with t in months. Find the factor per year to four decimal places and the yearly percent growth.
Numbers with and without a meaning
- A glacier’s area is modeled by A = 5100 - 2.5Y square kilometers, with Y the calendar year, for 1990 \le Y \le 2030. Say what -2.5 and 5{,}100 mean, find the area in 2000, and rewrite the model with t years after 2000.
- The water in a canal lock is H = 0.4t - 1 meters deep t minutes after the gates close, for 5 \le t \le 20. Say what 0.4 and -1 mean.
Reasoning
- Explain why, in any model y = a \cdot b^t, the number a is the value at t = 0, and the output is multiplied by b each time t goes up by 1.
- Sigurd says a model with factor 1.5 grows 50\% each period, so it grows 100\% over two periods. Is he right? Explain.
Answers
- Each hour adds 6 dollars to the cost, and 10 dollars is a fixed fee charged on every hire. The 10 is the value at h = 0, which is not a real hire.
- C = 5h + 10. Each hour costs 5 dollars, and the 10-dollar fee is the price of 2 hours: a hire is charged as if it lasted 2 hours longer.
- The canoe holds 30 liters at the start, and -2.5 means 2.5 liters are removed each minute. It is empty when 30 - 2.5m = 0, at m = 12 minutes.
- The run had 500 registered runners in 2020, and the number grows 8\% a year. In 2023, t = 3: 500(1.08)^3 = 500 \times 1.259712 \approx 629.9, about 630 runners.
- The area was 65 square kilometers in 2010, and it loses 1.5\% a year, since 1 - 0.985 = 0.015. In 2020, 65(0.985)^{10} \approx 55.88 square kilometers.
- 0.92 is a loss of 8\%; 1.2 is a growth of 20\%; 0.08 is a loss of 92\%.
- P(1) = 46 and P(2) = 52.9, gains of 6 and 6.9 members. The club grows by 15\% of its current membership each year, so the yearly gain grows too.
- At t = 0: 30 and 1. At t = 2: 30 \times 1.44 = 43.2 and 36^2 = 1296. In 30(1.2)^t, 30 is the start and 1.2 the factor per period, a growth of 20\%.
- 50 people have heard the rumor when it starts, and the number doubles every 6 hours. N(18) = 50 \times 2^3 = 400. The factor per hour is 2^{1/6} \approx 1.1225, about 12.2\% growth an hour.
- 1.69^{1/2} = 1.3, so K = 200(1.3)^t: a factor of 1.3, growth of 30\% a year, or 69\% every two years.
- 0.97^{1/12} \approx 0.9975, so about 0.25\% is lost each month.
- C = 0.15n + 12, since 9 \div 60 = 0.15 dollars per minute. The fee stays 12 dollars.
- 1.02^{12} \approx 1.2682, growth of about 26.8\% a year.
- -2.5: the area falls 2.5 square kilometers a year. 5{,}100: the area in the year 0, far outside 1990 \le Y \le 2030, so it has no meaning. In 2000, 5100 - 2.5 \times 2000 = 100 square kilometers, and A = 100 - 2.5t.
- 0.4: the water rises 0.4 meters each minute. -1: the depth at t = 0, which is outside the domain and would be a negative depth, so it has no meaning.
- Putting t = 0 gives a \cdot b^0 = a \cdot 1 = a. Going from t to t + 1, a \cdot b^{t+1} = (a \cdot b^t) \cdot b by the Multiplying and dividing powers with the same base, so the new output is the old one times b, whatever t is.
- No. Two periods multiply by 1.5^2 = 2.25, a growth of 125\%. The second 50\% is taken of an amount that has already grown.