Algebra 1 · 8.F.B.4 · N.Q.A.2

Building a model from a description

A paragraph describes something changing. How do you turn its words into a rule you can calculate with?

You need

Choosing a function family for dataAlgebra 1 Independent and dependent variablesMath 6 Relations and functionsMath 8

Where you are 5 in Modeling with functions

Functions13

Modeling with functions5

Choosing a function family for data

Building a model from a description

Interpreting the numbers in a model Where a model holds: domain, interpolation and extrapolation Judging and comparing models

When you reach for it

Often all you have is a description: “the lift goes down at a steady 5 meters per second”, “the colony grows by 6\% a year”, “the dose of a made-up medicine halves every 8 hours”. No table, no graph. To predict anything, the words have to become a rule: one quantity written as a function of another. This is also how you read a function out of a real situation at all. You decide which quantity depends on which (Independent and dependent variables) and check that each input gives exactly one output, as a function must.

The words settle almost everything. They say what is measured and in what units, where the counting starts, and whether the change is a fixed amount or a fixed percentage, which decides between a line and an exponential. Day 1 turns a single description into a rule and tests it. Day 2 handles descriptions that need two facts combined, a period longer than one unit, numbers that do not belong, and a domain.

The procedure

Day 1: from words to a rule

Name every quantity in full, with its units. Write “t = seconds since the lift left the platform”, not “t = time”, and “d = depth of the lift below the surface, in meters”, not “d = distance”. The full name does two jobs. It fixes where the input starts counting, which decides the starting value, since the start of a rule is its output when the input is 0. And it fixes the units, so the rate has units too: meters per second, dollars per month.

Pick the input and the output. The output is the quantity the question asks about; the input is the one it depends on, usually a time or a count. “How deep is the lift after t seconds” makes depth the output. Descriptions often say it outright: “the total depends on the number of months”.

Let the words decide the family. Two kinds of phrase cover most descriptions.

The words say	Each period	Family	Rule
“goes up by 12 each”, “a fee of 40 plus 25 per month”, “a steady 5 meters per second”	the same amount is added	linear	y = b + mx
“grows by 6\% each year”, “doubles every year”, “halves every 8 hours”	the amount is multiplied by the same number	exponential	y = a \cdot b^x

The reason is the one behind Choosing a function family for data. Adding the same amount every period gives equal differences, which is a line. “Grows by 6\%” means 6\% of what is there now, so the amount added keeps changing while the multiplier 1.06 does not: equal ratios, an exponential. Growth and decay factors turns a percentage into that factor, 1 + r for growth and 1 - r for decay. One phrase needs care: “5\% of the original silt each pass” is a fixed amount, so it is linear (Exercise 20).

Write the rule. For a line, the start goes in as b and the rate as m, negative if the quantity falls; Modeling with linear relationships builds these in detail. For an exponential, the start is a and the factor per period is b, raised to the number of periods that have passed.

Test it on one value the description gives. If the description says “after 2 passes through the filter, 225 grams of silt remain”, put 2 into the rule and see whether 225 comes out. A rule built from the words can still be wrong: a factor written as 0.25 instead of 0.75, a fee and a rate swapped. A value the rule was not built from catches these.

Day 2: fuller descriptions

Two facts to combine. A description may not hand over the start and the rate directly. “A joining fee of 40 dollars, and 190 dollars paid in total after 6 months” gives the start and one later value; the rate is what the months added, \displaystyle \frac{190 - 40}{6}. “Growing by 10\% a year, with 1{,}331 pairs after 3 years” gives the factor and a later value, so the start a is the unknown in a(1.1)^3 = 1{,}331, and dividing by 1.1^3 works back to it.

A period that is not 1. “Halves every 8 hours” makes the factor \displaystyle \frac{1}{2} per 8 hours, not per hour. After t hours, \displaystyle \frac{t}{8} periods of 8 hours have passed, so

A = a\left(\tfrac{1}{2}\right)^{t/8}

Testing t = 8 gives \displaystyle a \cdot \frac{1}{2}, exactly one halving, as it should. “Doubles every 5 years” is a \cdot 2^{y/5} in the same way.

Numbers that do not belong. Every number in a rule is the start, the rate or factor, or the length of a period. A description may also give the height of a wall, the number of volunteers, the time on the clock. Ask of each number what it measures, with its units; if it is none of those three, it stays out of the rule. Some of those numbers still matter, because they set where the rule stops.

A domain with the rule. The rule accepts any input; the situation does not. State the inputs that make sense alongside the rule: whole numbers when the input counts months, passes or yearly counts, an end when the lift reaches the bottom or the next dose is due. That is the reasonable domain, and a rule without one is only half a model. If the description changes how the quantity changes partway, such as a price that drops after the first year, one rule is not enough: that is a piecewise model.

Worked examples

Example 1

Day 1: naming the quantities for a mine lift

A lift in a mine shaft leaves a loading platform 30 meters below the surface and descends at a steady 5 meters per second. It reaches the 180-meter level 30 seconds after leaving. Write a rule for the lift’s position.

Solution

Let t be the number of seconds since the lift left the platform, and d the depth of the lift below the surface, in meters. “Time” alone would not say when the clock starts, and “distance” would not say from where: distance from the platform and depth below the surface differ by 30.

The depth depends on the time, so d is the output. “A steady 5 meters per second” adds the same amount every second, so the rule is linear. At t = 0 the lift is at the platform, so the start is 30, and the depth grows by 5 each second:

d = 30 + 5t

Test with the value the description gives: 30 + 5(30) = 180. It matches.

Measuring height above the surface is just as correct if it is named: with h the height above the surface in meters, h = -30 - 5t. The two rules differ only because the names differ, which is why the names come first.

Example 2

Day 1: one colony, two descriptions

Two made-up reports describe a colony of gannets (a seabird) on a sea stack that had 800 nesting pairs when counts began. Report A says the colony gains 48 pairs each year. Report B says it grows by 6\% each year. Write a rule for each, and compare them after 1, 2 and 10 years.

Solution

Let y be the number of years since counts began and P the number of nesting pairs, so P = 800 when y = 0.

Report A adds the same 48 pairs each year: linear, P = 800 + 48y.

Report B adds 6\% of whatever is there, so each year multiplies the count by 1 + 0.06 = 1.06: exponential, P = 800(1.06)^y.

y	A: 800 + 48y	B: 800(1.06)^y
1	848	848
2	896	898.88
10	1{,}280	about 1{,}433

After one year the two agree exactly, because 6\% of 800 is 48. So a value such as “848 pairs after a year” would pass the test for both rules: a test checks a rule but cannot choose the family. The words do that. After the first year B pulls ahead, because its 6\% is taken of a larger count each time; by year 10, 800 \times 1.06^{10} \approx 1{,}432.7, about 153 pairs more than A.

Example 3

Day 1: silt removed by a percentage

A tank of pond water holds 400 grams of silt. Each pass through a filter removes 25\% of the silt still in the water. A test shows 225 grams of silt left after 2 passes. Write a rule for the amount of silt, and test it.

Solution

Let p be the number of passes through the filter and S the mass of silt left in the water, in grams. “25\% of the silt still in the water” is a percentage of the current amount, so each pass leaves 1 - 0.25 = 0.75 of what was there: exponential, with start 400 and factor 0.75.

S = 400(0.75)^p

Test: 400(0.75)^2 = 400 \times 0.5625 = 225. It matches the test. Passes are counted whole, so p = 0, 1, 2, \dots

Reading “25\%” as 100 grams every pass gives S = 400 - 100p, which predicts 200 at p = 2, and the test rejects it. The first pass removes 100 grams, but the second removes 25\% of 300, only 75.

Example 4

Day 2: a joining fee and a total

A climbing wall charges a joining fee, then the same charge every month. Liesel paid a joining fee of 40 dollars, and after 6 months of membership she had paid 190 dollars in total. After 12 months she had paid 340 dollars. Write a rule for her total and state its domain.

Solution

Let m be the number of whole months of membership and C the total Liesel has paid, in dollars. The same charge each month makes the rule linear. At m = 0 she has paid only the fee, so the start is 40. The 6 months added 190 - 40 = 150 dollars, so the rate is \displaystyle \frac{150}{6} = 25 dollars per month:

C = 40 + 25m

The 12-month figure was not used to build the rule, so it tests it: 40 + 25(12) = 340. It matches. Months are counted whole, so the domain is m = 0, 1, 2, 3, \dots, ending when the membership does. This is the build in Modeling with linear relationships; the new step is reading the rate out of a total.

Example 5

Day 2: working back to the start

A second gannet colony grows by 10\% a year. Three years after counts began there were 1{,}331 nesting pairs, and after two years there were 1{,}210. Write a rule for the number of pairs.

Solution

Let y be the number of years since counts began and P the number of nesting pairs. “By 10\% a year” makes the rule exponential with factor 1.1, but the start a is not given. The description gives P at y = 3, so

a(1.1)^3 = 1{,}331

Since 1.1^3 = 1.331, dividing both sides by 1.331 gives a = 1{,}000:

P = 1{,}000(1.1)^y

Test with the unused value: 1{,}000(1.1)^2 = 1{,}000 \times 1.21 = 1{,}210. It matches.

Taking 10\% off 1{,}331 gives 1{,}197.9, not 1{,}210. The growth was 10\% of the earlier, smaller count, so only dividing by the factor 1.1 undoes it.

Example 6

Day 2: halving every eight hours

The medicine and its numbers are made up for practice and are not medical advice. A 320-milligram dose of a medicine is taken, and the amount in the body halves every 8 hours. The next dose is due 24 hours later. Write a rule with its domain, and find the amount after 12 hours.

Solution

Let t be the number of hours since the dose was taken and A the amount in the body, in milligrams. “Halves” is a factor of \displaystyle \frac{1}{2}, so the rule is exponential, but the period is 8 hours. After t hours, \displaystyle \frac{t}{8} halvings have happened:

A = 320\left(\tfrac{1}{2}\right)^{t/8}, \quad 0 \le t \le 24

The domain stops at 24, where the next dose changes the story. Test: at t = 8, \displaystyle A = 320 \times \frac{1}{2} = 160, one halving; at t = 16, 80; at t = 24, 40.

After 12 hours, \displaystyle \frac{12}{8} = 1.5 halvings have happened: A = 320\left(\tfrac{1}{2}\right)^{1.5} \approx 320 \times 0.3536 \approx 113 milligrams. That is between 160 and 80, as it should be, and a little below 120, the value halfway between, because the curve falls faster early in each period than late.

Figure 1Example 6, a made-up medicine, not medical advice. Every 8 hours the amount halves, so each drop is half the one before. The curve stops at 24 hours, when the next dose is due.

Example 7

Day 2: a warden’s note with extra numbers

The note and all its numbers, the room on the stack included, are made up. A warden writes: “The sea stack is 90 meters high. When counts began there were 2{,}500 nesting pairs of gannets, and the colony has grown by 4\% a year since. I count it once a year, in June, with 3 volunteers. The stack has room for about 5{,}000 pairs.” Write a rule with a reasonable domain.

Solution

Let y be the number of years since counts began and P the number of nesting pairs. Go through every number. 2{,}500 is the start. 4\% a year is a percentage of the current count, so the factor is 1.04 per year. The 90 meters and the 3 volunteers measure the rock and the counting team, not the colony, so they stay out of the rule. “Once a year” and “room for about 5{,}000 pairs” stay out of the rule too, but they set the domain.

P = 2{,}500(1.04)^y

Counts are yearly, so y is a whole number. The rule stops being believable once the stack is full, since a colony with nowhere to nest cannot keep growing by 4\%. Trying years: 2{,}500(1.04)^{17} \approx 4{,}870 and 2{,}500(1.04)^{18} \approx 5{,}065. So the domain is y = 0, 1, 2, \dots, 17.

Boundaries

The rule describes the words, not the world. “A steady speed” or “by 6\% each year” gives a line or an exponential because the description says so; how long the real lift or colony keeps to it is the domain question, taken further in Where a model holds: domain, interpolation and extrapolation.
Linear needs a fixed amount per period; exponential needs a fixed percentage of the current amount, which is a fixed factor. A percentage of a fixed amount, such as “of the original silt”, is a fixed amount, so it is linear.
A description must give enough: a start, or a later value to work back from, and a rate or factor with its period. “The colony grows every year” alone fixes no rule.
One rule needs one way of changing throughout. A description whose rate changes partway needs a piecewise model.
An exponential decay never reaches 0 in the rule, though the real amount soon becomes too small to measure. A falling linear rule reaches 0 and then goes negative, so its domain ends there.

Where this goes next

Interpreting the numbers in a model reads the start, rate and factor of a finished model back into the situation. A reasonable domain for a real situation says more about the domain stated with a rule, and Where a model holds: domain, interpolation and extrapolation asks how far past the given information a model can be trusted. When the information arrives as a table or a graph instead of words, Writing an exponential function from a table or graph builds the rule from that.

Exercises

Day 1: naming the quantities

A lift in a mine shaft leaves the surface and descends at a steady 3 meters per second. Mattias writes “d = distance, t = time”. Rewrite both names in full, with units, write the rule, and find the depth 50 seconds after the lift leaves.

Day 1: choosing the family

For Exercises 2 to 6, name the quantities in full, say whether the words call for a linear or an exponential rule, write the rule, and find the output when the input is 4.

A gannet colony has 600 nesting pairs and gains 45 pairs each year.
A gannet colony has 600 nesting pairs and grows by 15\% each year.
A tank of water holds 250 grams of silt, and each pass through a filter removes 10\% of the silt still in the water.
A newly settled sea stack has 12 pairs of gannets, and the number doubles every year.
A mine lift starts 20 meters below the surface and rises at a steady 2 meters per second.

Day 1: testing a rule

A tank of water holds 600 grams of silt, each pass through a filter removes 20\% of the silt still in the water, and 480 grams remain after 1 pass. Temperance writes S = 600(0.2)^p. Test her rule and correct it.
A climbing wall charges a joining fee of 30 dollars and 35 dollars a month, and 3 months of membership cost 135 dollars in total. Cormac writes C = 30m + 35. Test his rule and correct it.
A colony of 400 pairs grows by 5\% each year, and there were 441 pairs after 2 years. Perpetua writes P = 400(1.05)^y and Briallen writes P = 400 + 20y. Test both.

Day 2: two facts to combine

A climbing wall charges a joining fee of 55 dollars and a fixed monthly charge. After 8 months, Mattias had paid 255 dollars in total. Write the rule.
A gannet colony grows by 20\% a year, and after 2 years there were 432 pairs. Write the rule.
Each pass through a filter removes 20\% of the silt still in a tank of water, and after 2 passes 256 grams remain. How much silt was there at the start? Write the rule.

Day 2: a period longer than one

Made-up numbers, not medical advice. A 240-milligram dose of a medicine halves every 3 hours. Write the rule and find the amount after 9 hours.
A gannet colony of 150 pairs doubles every 5 years. Write the rule and find the number of pairs after 15 years.
Made-up numbers, not medical advice. A 500-milligram dose of a medicine halves every 8 hours. Find the amount after 20 hours, to one decimal place.

Day 2: extra numbers and a domain

“The cage is 3 meters tall and carries up to 10 workers. It leaves the surface at seven o’clock and descends at a steady 4 meters per second to the bottom of the 320-meter shaft.” Write the rule with its domain, and say which numbers are left out.
“Membership at the climbing wall costs a 40-dollar joining fee and 30 dollars a month, for at most 24 months. A day pass costs 12 dollars, and the wall is 15 meters high.” Write the rule for a member’s total with its domain, and give the largest total.
“The colony had 1{,}200 nesting pairs when counts began and grows by 3\% a year. The warden counts it each summer on 4 visits. The stack has room for about 1{,}500 pairs.” Write the rule with a reasonable domain.

Reasoning

A description says a colony went from 500 to 550 pairs in its first year. Explain why this alone cannot decide between a linear and an exponential rule, and what the second year would give under each.
A tank of water holds 400 grams of silt, and each pass through a coarse filter removes “5\% of the original silt”. Which family is this, and why? Write the rule, and say when it reaches 0.
Made-up numbers, not medical advice. Briallen writes the rule for a dose of a medicine that halves every 8 hours as a\left(\tfrac{1}{2}\right)^{8t}. Test it at t = 8 and explain which exponent is right.

Answers

t = seconds since the lift left the surface; d = depth of the lift below the surface, in meters. A steady speed from depth 0: d = 3t. After 50 seconds, d = 150 meters.
y = years since the count, P = nesting pairs. A fixed amount each year, so linear: P = 600 + 45y. At y = 4, P = 780 pairs.
Same names. 15\% of the current count, so exponential: P = 600(1.15)^y. At y = 4, P \approx 1{,}049 pairs.
p = passes through the filter, S = grams of silt left in the water. A percentage of the current amount, so exponential: S = 250(0.9)^p. At p = 4, S = 250 \times 0.6561 \approx 164 grams.
y = years since settling, P = pairs. Doubling multiplies by 2, so exponential: P = 12 \cdot 2^y. At y = 4, P = 12 \times 16 = 192 pairs.
t = seconds since it started rising, d = depth below the surface in meters. A steady speed, so linear, the depth falling by 2 each second: d = 20 - 2t. At t = 4, d = 12 meters.
Her rule gives 600(0.2) = 120, not 480. Removing 20\% leaves 0.8: S = 600(0.8)^p, and 600(0.8) = 480.
His rule gives 30(3) + 35 = 125, not 135: the fee and the monthly charge are swapped. C = 30 + 35m gives 30 + 105 = 135.
400(1.05)^2 = 400 \times 1.1025 = 441 matches; 400 + 20(2) = 440 does not. “By 5\% each year” is exponential; Briallen took 5\% of the first count every year.
m = months of membership, C = dollars paid. Rate \displaystyle \frac{255 - 55}{8} = 25, so C = 55 + 25m.
a(1.2)^2 = 432, so \displaystyle a = \frac{432}{1.44} = 300. P = 300(1.2)^y, with y = years since counts began.
a(0.8)^2 = 256, so \displaystyle a = \frac{256}{0.64} = 400. The tank held 400 grams; S = 400(0.8)^p, with p = passes through the filter.
A = 240\left(\tfrac{1}{2}\right)^{t/3}. After 9 hours, 3 halvings: \displaystyle A = 240 \times \frac{1}{8} = 30 milligrams.
P = 150 \cdot 2^{y/5}. After 15 years, 3 doublings: P = 150 \times 8 = 1{,}200 pairs.
A = 500\left(\tfrac{1}{2}\right)^{t/8}, and \displaystyle \frac{20}{8} = 2.5 halvings: A = 500\left(\tfrac{1}{2}\right)^{2.5} \approx 88.4 milligrams.
t = seconds since leaving the surface, d = depth in meters: d = 4t. It reaches 320 at t = 80, so 0 \le t \le 80. Left out: the 3 meters and the 10 workers. Seven o’clock says when t = 0 is, not a number in the rule.
C = 40 + 30m for m = 0, 1, 2, \dots, 24. Largest total: 40 + 30(24) = 760 dollars. The 12 and the 15 stay out.
P = 1{,}200(1.03)^y, y whole years. 1{,}200(1.03)^7 \approx 1{,}476 and 1{,}200(1.03)^8 \approx 1{,}520, so y = 0, 1, \dots, 7. The 4 visits stay out.
550 = 500 + 50 and 550 = 500 \times 1.1, so “50 pairs a year” and “10\% a year” both fit. In year 2 the linear rule gives 600 and the exponential 605. Only the words, or a second value, decide.
Linear: 5\% of the original 400 is a fixed 20 grams each pass. S = 400 - 20p, reaching 0 at p = 20 passes, which ends the domain.
At t = 8 her rule gives a\left(\tfrac{1}{2}\right)^{64}: sixty-four halvings in 8 hours, leaving almost nothing, where there should be exactly one. \displaystyle \frac{t}{8} counts periods of 8 hours, giving one halving at t = 8; 8t would count eight halvings every hour.