8.2 Linear (Algebraic) Growth

Marco is a collector of antique soda bottles. His collection currently contains 437 bottles. Every year, he budgets enough money to buy 32 new bottles. Can we determine how many bottles he will have in 5 years, and how long it will take for his collection to reach 1000 bottles?

# Ch 8.2 - linear growth: Marco's antique soda bottle collection.
P0 = 437        # bottles he owns now          TRY IT: change any of these three
D = 32          # bottles he buys every year
TARGET = 1000   # the size he is aiming for

print(f"recursive form:  P0 = {P0},   Pn = P(n-1) + {D}")
print(f"explicit  form:  Pn = {P0} + {D}n\n")
print(" n    recursive        explicit")
running = P0
for n in range(9):
    if n:
        running += D
    print(f"{n:>2}    {running:>9}    {P0} + {D}({n}) = {P0 + D * n}")

if D == 0:
    print(f"\nwith d = 0 the collection never changes, so {TARGET} never arrives")
else:
    n_star = (TARGET - P0) / D
    print(f"\nreaching {TARGET}:  {TARGET} = {P0} + {D}n")
    print(f"   n = ({TARGET} - {P0}) / {D} = {n_star:.2f}")
    print(f"   n has to be a whole year, so it happens in year {-(-n_star // 1):.0f}")

# The same equation from TWO measurements instead of a stated rate: the elk
# population was 12,000 in 2003 and 15,000 in 2007.
Y1, V1, Y2, V2 = 2003, 12000, 2007, 15000   # TRY IT: use 2003 and 2014 data
d = (V2 - V1) / (Y2 - Y1)
print(f"\nelk: d = ({V2} - {V1}) / ({Y2} - {Y1}) = {d:g} elk per year")
print(f"     Pn = {V1} + {d:g}n  with n = 0 at {Y1}")
for year in (2014, 2020):
    print(f"     {year}: n = {year - Y1}, Pn = {V1 + d * (year - Y1):,.0f} elk")
print("\nConstant CHANGE per step is what makes growth linear. Compare this with")
print("section 8.4, where the growth is a constant PERCENT instead.")

While both of these questions you could probably solve without an equation or formal mathematics, we are going to formalize our approach to this problem to provide a means to answer more complicated questions.

Suppose that $P_{n}$ represents the number, or population, of bottles Marco has after $n$ years. So $P_{0}$ would represent the number of bottles now, $P_{1}$ would represent the number of bottles after 1 year, $P_{2}$ would represent the number of bottles after 2 years, and so on. We could describe how Marco’s bottle collection is changing using:

$P_{0}=437$

$P_{n}=P_{n-1}+32$

This is called a recursive relationship. A recursive relationship is a formula which relates the next value in a sequence to the previous values. Here, the number of bottles in year $n$ can be found by adding 32 to the number of bottles in the previous year, $P_{n-1}$ . Using this relationship, we could calculate:

$P_{1}=P_{0}+32=437+32=469$

$P_{2}=P_{1}+32=469+32=501$

$P_{3}=P_{2}+32=501+32=533$

$P_{4}=P_{3}+32=533+32=565$

$P_{5}=P_{4}+32=565+32=597$

We have answered the question of how many bottles Marco will have in 5 years. However, solving how long it will take for his collection to reach 1000 bottles would require a lot more calculations.

While recursive relationships are excellent for describing simply and cleanly how a quantity is changing, they are not convenient for making predictions or solving problems that stretch far into the future. For that, a closed or explicit form for the relationship is preferred. An explicit equation allows us to calculate P_n directly, without needing to know P_n_-1. While you may already be able to guess the explicit equation, let us derive it from the recursive formula. We can do so by selectively not simplifying as we go:

$\begin{array}{lr}P_{I}=437+32 & =437+1(32) \\ P_{2}=P_{1}+32=437+32+32 & =437+2(32) \\ P_{3}=P_{2}+32=(437+2(32))+32 & =437+3(32) \\ P_{4}=P_{3}+32=(437+3(32))+32 & =437+4(32)\end{array}$

You can probably see the pattern now, and generalize that

$P_{n}=437+n(32)=437+32 n$

Using this equation, we can calculate how many bottles he’ll have after 5 years:

$P_{5}=437+32(5)=437+160=597$

We can now also solve for when the collection will reach 1000 bottles by substituting in 1000 for $P_{n}$ and solving for $n$ .

$1000=437+32 n$

Scatter plot with the horizontal axis Years from now marked 0 to 5 and the vertical axis Bottles marked 0 to 700. Six points joined by a straight line climb evenly from 437 bottles at year 0 to 597 bottles at year 5, a gain of 32 bottles a year. $563=32 n$

$n=563 / 32=17.59$

So Marco will reach 1000 bottles in 18 years.

In the previous example, Marco’s collection grew by the same number of bottles every year. This constant change is the defining characteristic of linear growth. Plotting the values we calculated for Marco’s collection, we can see the values form a straight line, the shape of linear growth.

Note

Linear Growth

If a quantity starts at size $P_{0}$ and grows by $d$ every time period, then the quantity after $n$ time periods can be determined using either of these relations:

Recursive form:

$P_{n}=P_{n-1}+d$

Explicit form:

$P_{n}=P_{0}+d n$

In this equation, represents the common difference – the amount that the population changes each time $n$ increases by 1

Example 1 Connection to prior learning – slope and intercept

You may recognize the common difference, $d$ , in our linear equation as slope. In fact, the entire explicit equation should look familiar – it is the same linear equation you learned in algebra, probably stated as $y=m x+b$ .

In the standard algebraic equation $y=m x+b$ , $b$ was the $y$ -intercept, or the y value when $x$ was zero. In the form of the equation we’re using, we are using P0 to represent that initial amount.

In the $y=m x+b$ equation, recall that m was the slope. You might remember this as “rise over run”, or the change in $y$ divided by the change in $x$ . Either way, it represents the same thing as the common difference, $d$ , we are using – the amount the output Pn changes when the input $n$ increases by 1.

The equations $y=m x+b$ and $P_{n}=P_{0}+d n$ mean the same thing and can be used the same ways, we’re just writing it somewhat differently.

Example 2

The population of elk in a national forest was measured to be 12,000 in 2003, and was measured again to be 15,000 in 2007. If the population continues to grow linearly at this rate, what will the elk population be in 2014?

Show solution

To begin, we need to define how we’re going to measure $n$ . Remember that $P_0$ is the population when $n = 0$ , so we probably don’t want to literally use the year 0. Since we already know the population in 2003, let us define $n = 0$ to be the year 2003. Then

$P_0 = 12,000$ .

Next we need to find $d$ . Remember d is the growth per time period, in this case growth per year. Between the two measurements, the population grew by $15,000-12,000 = 3,000$ , but it took $2007-2003 = 4$ years to grow that much. To find the growth per year, we can divide: 3000 elk / 4 years = 750 elk in 1 year.

Alternatively, you can use the slope formula from algebra to determine the common difference, noting that the population is the output of the formula, and time is the input.

$d=$ slope $=\frac{\text { change in output }}{\text { change in input }}=\frac{15,000-12,000}{2007-2003}=\frac{3000}{4}=750$

We can now write our equation in whichever form is preferred.

Recursive form:

$P_{0}=12,000$

$P_{n}=P_{n-1}+750$

Explicit form:

$P_{n}=12,000+750 n$

To answer the question, we need to first note that the year 2014 will be $n = 11$ , since 2014 is 11 years after 2003. The explicit form will be easier to use for this calculation:

$P_{11}=12,000+750(11)=20,250$ elk

Example 3

Gasoline consumption in the US has been increasing steadily. Consumption data from 1992 to 2004 is shown below[1]. Find a model for this data, and use it to predict consumption in 2016. If the trend continues, when will consumption reach 200 billion gallons?

# Ch 8.2 - build a linear model from real data by choosing TWO of the points.
# Which two you choose changes the prediction. Try it and see how much.
import matplotlib.pyplot as plt

YEAR = list(range(1992, 2005))
GAS = [110, 111, 113, 116, 118, 119, 123, 125, 126, 128, 131, 133, 136]
A, B = 1993, 2003        # TRY IT: 1992 and 2004; or 1992 and 1996; or 1999 and 2000
i, j = YEAR.index(A), YEAR.index(B)
P0 = GAS[i]
d = (GAS[j] - GAS[i]) / (B - A)

print(f"anchoring on {A} and {B}:")
print(f"   P0 = {P0}   (consumption in {A}, so n = 0 means {A})")
print(f"   d  = ({GAS[j]} - {GAS[i]}) / ({B} - {A}) = {d:.4g} billion gallons per year")
print(f"   model:  Pn = {P0} + {d:.4g}n\n")
for year in (2010, 2016, 2030):
    n = year - A
    print(f"   {year}: n = {n:>2},  Pn = {P0} + {d:.4g}({n}) = {P0 + d * n:.1f} billion gallons")
if d > 0:
    n200 = (200 - P0) / d
    print(f"   hits 200 billion at n = {n200:.4g}, i.e. during {A + n200:.0f}")

miss = [GAS[k] - (P0 + d * (YEAR[k] - A)) for k in range(len(YEAR))]
worst = max(range(len(YEAR)), key=lambda k: abs(miss[k]))
print(f"\n   biggest miss on the 1992-2004 data: {abs(miss[worst]):.2f} billion gallons"
      f" (in {YEAR[worst]})")
print("   Two points is a crude way to fit a line, but it is enough to see the trend.")

span = list(range(1992, 2036))
plt.figure(figsize=(7.5, 4.2))
plt.plot(YEAR, GAS, "o", color="tab:blue", label="measured consumption")
plt.plot(span, [P0 + d * (y - A) for y in span], color="tab:red",
         label=f"model  Pn = {P0} + {d:.3g}n")
plt.plot([A, B], [GAS[i], GAS[j]], "s", ms=11, mfc="none", mec="tab:red",
         label=f"the two anchor points ({A}, {B})")
plt.axhline(200, ls=":", color="grey")
plt.text(1993, 202, "200 billion gallons", color="grey", fontsize=9)
plt.xlabel("year")
plt.ylabel("billions of gallons")
plt.title("US gasoline consumption, and a two-point linear model")
plt.legend(loc="upper left", fontsize=9)
plt.tight_layout()
plt.show()

Year	$\cdot 92$	$\cdot 93$	$\cdot 94$	$\cdot 95$	$\cdot 96$	$\cdot 97$	$\cdot 98$	$\cdot 99$	$\cdot 00$	$\cdot 01$	$\cdot 02$	$\cdot 03$	$\cdot 04$
Consumption (billion of gallons)	110	111	113	116	118	119	123	125	126	128	131	133	136

Show solution

Scatter plot with the horizontal axis Year running from 1992 to 2004 and the vertical axis Gas Consumption running from 100 to 140 billions of gallons. Thirteen points, one for each year, rise in a nearly straight run from 110 in 1992 to 136 in 2004, taking the values 110, 111, 113, 116, 118, 119, 123, 125, 126, 128, 131, 133 and 136. Plotting this data, it appears to have an approximately linear relationship:

While there are more advanced statistical techniques that can be used to find an equation to model the data, to get an idea of what is happening, we can find an equation by using two pieces of the data – perhaps the data from 1993 and 2003.

Letting $n = 0$ correspond with 1993 would give $P_{0}=111$ billion gallons.

To find d, we need to know how much the gas consumption increased each year, on average. From 1993 to 2003 the gas consumption increased from 111 billion gallons to 133 billion gallons, a total change of $133 – 111 = 22$ billion gallons, over 10 years. This gives us an average change of 22 billion gallons / 10 year = 2.2 billion gallons per year.

Equivalently,

$d=\text {slope}=\frac{\text { change in output }}{\text { change in input }}=\frac{133-111}{10-0}=\frac{22}{10}=2.2$ billion gallons per year

We can now write our equation in whichever form is preferred.

Recursive form:

$P_{0}=111$

$P_{n}=P_{n-1}+2.2$

Explicit form:

$P_{n}=111+2.2 n$

Calculating values using the explicit form and plotting them with the original data shows how well our model fits the data.

We can now use our model to make predictions about the future, assuming that the previous trend continues unchanged. To predict the gasoline consumption in 2016:

$n=23 (2016 – 1993 = 23$ years later)

$P_{23}=111+2.2(23)=161.6$

Our model predicts that the US will consume 161.6 billion gallons of gasoline in 2016 if the current trend continues.

To find when the consumption will reach 200 billion gallons, we would set $P_{n}=200$ , and solve for $n$ :

$\begin{array}{ll} P_{n}=200 & \text{Replace } P_n \text{ with our model} \\ 111+2.2 n=200 & \text{Subtract 111 from both sides} \\ 2.2n = 89 & \text{Divide both sides by 2.2} \\ n = 40.4545 \end{array}$

This tells us that consumption will reach 200 billion about 40 years after 1993, which would be in the year 2033.

Your Turn

Try it Now 1

The number of stay-at-home fathers in Canada has been growing steadily[2]. While the trend is not perfectly linear, it is fairly linear. Use the data from 1976 and 2010 to find an explicit formula for the number of stay-at-home fathers, then use it to predict the number if 2020.

Year	1976	1984	1991	2000	2010
Number of stay-at-home fathers	20,610	28,725	43,530	47,665	53,555

Answer

Letting $n=0$ correspond with 1976, then $P_{0}=20,610$ .

From 1976 to 2010 the number of stay-at-home fathers increased by

$53,555-20,610=32,945$

This happened over 34 years, giving a common different d of $32,945 / 34=969$ .

$P_{n}=20,610+969 n$

Predicting for 2020, we use $n=44$

$P_{44}=20,610+969(44)=63,246$ stay-at-home fathers in 2020.

[1] www.bts.gov/publications/nati...ble_04_10.html

[2] www.fira.ca/article.php?id=140

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.