Use a geometric model in context

A ball that bounces lower each time

A ball falls 5 meters and bounces. After each bounce it rises to \(\frac35\) of the height of its last fall, so the rises are 3, 1.8 and 1.08 meters.

Each bar shows the distance in one round, starting with the first fall of 5 meters. Each later round is a rise and an equal fall, so the next bars are \(2(3)=6\) and \(2(1.8)=3.6\).

Drag the slider to add landings. The teal column stacks the bars into the distance traveled, which is \(5+6+3.6=14.6\) meters at the third landing.

From the second bar on, each bar is \(\frac35\) of the one before it, so the total levels off. It approaches 20 meters, and the next steps show why.

Turn a repeated change into a series

Remove the numbers. The ball falls a first height \(a\), and each rise is \(r\) times the height of the last fall.

The first fall counts once. After that, each round is a rise and a fall, so the rounds add \(2ar\), \(2ar^2\) and so on.

Drag the slider for \(r\), the fraction that carries over from one bounce to the next. The bars change, but the rounds after the first fall still form a geometric series with ratio \(r\).

To model a repeated change, find the first amount and the fraction that carries over. Then decide what each round adds.

Notation

1. Name the first fall \(\cy{a}\) and the fraction \(r\) that carries over to each rise.

   \[\cy{a}=5,\qquad r=\dfrac35\]

2. Write the total distance as a sum of the rounds. The first fall counts once, and each later height counts twice.

   \[D=\cy{a}+2ar+2ar^2+2ar^3+\cdots\]

3. If \(0\lt r<1\), the geometric series after \(\cy{a}\) has the sum \(\frac{2ar}{1-r}\). Then combine the two terms over \(1-r\).

   \[D=\cy{a}+\dfrac{2ar}{1-r}=\cy{a}\cdot\dfrac{1+r}{1-r}\]

4. Substitute the numbers. Give the answer in meters, because \(D\) is a distance.

   \[D=5\cdot\dfrac{1+\frac35}{1-\frac35}=20\ \text{m}\]

When it fails

If r is 1 or more, the rounds do not shrink, because each rise is at least as high as the fall before it.

Then the total distance grows without limit, and the red arrow shows the column leaving the figure. The model needs \(0\lt r<1\).

Example 1: A dose that partly stays

A patient takes 200 mg of a drug each morning. By the next morning, 25% of the amount in the body remains. Find the amount immediately after the third dose, and the long-run amount.

1. Each morning, the body keeps 0.25 of the old amount, and the new dose adds 200, so the first amounts are 200, 250 and 262.5 mg.

   \[\begin{aligned}A_2&=200+0.25(200)=250\\ A_3&=200+0.25(250)=262.5\end{aligned}\]

2. Write the amount after n doses as a sum. The newest dose counts in full, and each older dose is multiplied by 0.25 once for each day since it was taken. The column stacks these parts, 200, 50 and 12.5.

   \[\begin{aligned}A_n&=200+200(0.25)\\ &\quad+200(0.25)^2+\cdots\\ &\quad+200(0.25)^{n-1}\end{aligned}\]

3. This is a geometric sum with \(a=200\), \(r=0.25\) and \(n\) terms.

   \[A_n=200\cdot\frac{1-0.25^n}{1-0.25}\]

4. Since \(0.25<1\), the power \(0.25^n\) approaches 0, so the amount after each dose approaches \(\frac{800}{3}\approx266.7\) mg.

   \[\lim_{n\to\infty}A_n=\frac{200}{1-0.25}=\frac{800}{3}\]

Example 2: Money that is spent again

A town receives a grant of 40 million dollars. Each person who receives money spends 75% of it in the town. How much is spent in all?

1. The rounds of spending are 40, then 75% of 40, then 75% of that.

   \[\begin{aligned}&40\\ 40(0.75)&=30\\ 30(0.75)&=22.5\end{aligned}\]

2. A tempting wrong move: The ratio is the fraction that is not spent, so \(r=0.25\).

   \[\begin{aligned}\text{total}&=\frac{40}{1-0.25}\\ &=\frac{160}{3}\approx53.3\end{aligned}\]

   With \(r=0.25\), the second round would be 10 million dollars, the red outline. The figure shows 30, because each round is 0.75 of the one before.

3. The ratio is the fraction that carries over to the next round, which is the spent fraction.

   \[r=\frac34\]

4. Since \(\frac34<1\), the series converges, and 160 million dollars are spent in all.

   \[\text{total}=\frac{40}{1-\frac34}=160\]

Practice

A pendulum swings 40 cm on its first swing. Each swing is 0.9 times as long as the swing before it. How far does the bob travel in all?

The top bars show each swing, and the dots below show the distance so far. Decide whether each swing counts once or twice.

Questions

Questions 1 and 2 use the figure. A ball is dropped from 12 meters and rises to half of its last fall after each bounce. Each bar is the distance in one round: the first fall, then a rise and an equal fall.

1. How far has the ball traveled when it hits the ground for the third time?

   • 30 m
   • 21 m
   • 24 m
   Answer

   30 m

2. If the ball bounces forever, what total distance does it travel?

   • 36 m
   • 24 m
   • 48 m
   Answer

   36 m

3. A worker packs 90 boxes on day 1. Each later day, the worker packs \(\frac23\) as many boxes as the day before. What total does the worker approach?

   • 270
   • 135
   • 180
   Answer

   270

Summary

You model a repeated change by writing each round as a term of a series, with the ratio equal to the fraction that carries over.

This works because carrying the same fraction to every round makes the terms geometric.

You use the sum to find the long-run total, and you state what it measures, such as distance, drug in the body or money spent.

This is one of four skills in topic 10.2 of Which & Why: AP Calculus.