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Extending the Definitions of Exponents, Variation 2


Alignments to Content Standards: F-LE.A N-RN.A.1

Task

A biology student is studying bacterial growth. She was surprised to find that the population of the bacteria doubled every hour.

  1. Complete the following table and plot the data.

    Hours into study 01234
    Population (thousands) 4
  2. Write an equation for $P$, the population of the bacteria, as a function of time, $t$, and verify that it produces correct populations for $t$ = 1, 2, 3, and 4.

  3. The student conducting the study wants to create a table with more entries; specifically, she wants to fill in the population at each half hour. However, she forgot to make these measurements so she wants to estimate the values.

    Instead she notes that the population increases by the same factor each hour, and reasons that this property should hold over each half-hour interval as well. Fill in the part of the table below that you've already computed, and decide what constant factor she should multiply the population by each half hour in order to produce consistent results. Use this multiplier to complete the table:

    Hours into study 0$\frac12$1$\frac32$2$\frac52$3
    Population (thousands) 4
  4. What if the student wanted to estimate the population every 20 minutes instead of every 30 minutes. What multiplier would be necessary to be consistent with the population doubling every hour? Use this multiplier to complete the following table:

    Hours into study 0$\frac13$$\frac23$1$\frac43$$\frac53$2
    Population (thousands) 4
  5. Use the population context to explain why it makes sense that we define $2^{\frac{1}{2}}$ to be $\sqrt{2}$ and $2^{\frac{1}{3}}$ as $\sqrt[3]{2}$.

  6. Another student working on the bacteria population problem makes the following claim:

    If the population doubles in 1 hour, then half that growth occurs in the first half-hour and the other half occurs in the second half-hour. So for example, we can find the population at $t=\frac{1}{2}$ by finding the average of the populations at $t=0$ and $t=1$.

    Comment on this idea. How does it compare to the multipliers generated in the previous problems? For what kind of function would this reasoning work?

IM Commentary

The goal of this task is to develop an understanding of why rational exponents are defined as they are (N-RN.1), however it also raises important issues about distinguishing between linear and exponential behavior (F-LE.1c) and it requires students to create an equation to model a context (A-CED.2)

Solution

  1. Hours into study 0 1 2 3 4
    Population (thousands) 4 8 16 32 64

    Students would be expected to find these values by repeatedly multiplying by 2. The plot below consists of the exponential function $P(t)=4\cdot 2^t$ which students will derive in the next part. The plot of the data alone would consist of the 5 plotted blue points.

    2_f1483994df84cea56acfd9dc44d60553
  2. One possible equation is $P(t)=4\cdot 2^t$, since as we tallied above via repeated multiplication, we have $4\cdot 2^1=8$, $4\cdot 2^2=16$, $4\cdot 2^3=32$, $4\cdot 2^4=64$, etc.  Others include $P(t)=2^{t+2}$, $P(t)=2\cdot 2^{t+1}$, etc.

  3. Hours into study 0 $\frac12$ 1 $\frac32$ 2 $\frac52$ 3
    Population (thousands) 4 5.657 8 11.314 16 22.627 32

    Let $x$ be the multiplier for the half-hour time interval. Then since going forward a full hour in time has the effect of multiplying the population by $x^2$, we must have $x^2=2$, and so the student needs to multiply by $\sqrt{2}$ every half hour.

  4. Hours into study 0 $\frac13$ $\frac23$ 1 $\frac43$ $\frac53$ 2
    Population (thousands) 4 5.010 6.350 8 10.079 12.699 16

    Similarly, since waiting three 20-minute intervals should double the population, the new multiplier has to satisfy $x\cdot x\cdot x=2$, which gives $x^3=2$. So you would need to multiply by $\sqrt[3]{2}$ every 20 minutes to have the effect of doubling every hour.

  5. We already know that the equation for population is $P=4(2)^t$ when $t$ is a natural number. Given this, it's reasonable to use the expression $P(\frac{1}{2})=4(2)^{\frac{1}{2}}$ to define $2^{\frac{1}{2}}.$ However, we reasoned above that $P(\frac{1}{2})=4\cdot\sqrt{2}$, and equating the two gives $2^{\frac{1}{2}}=\sqrt{2}$. Similarly, equating the expression $P(\frac{1}{3})=2^{1/3}$ with the calculation $P(\frac{1}{3})=\sqrt[3]{2}$ gives the reasonable definition $2^{1/3}=\sqrt[3]{2}$.

  6. The reasoning mistakenly assumes linear growth within each hour, i.e., that the amount of population growth is the same each half hour. We know instead that the percentage growth is constant, not the raw change in population. If we were to apply the faulty reasoning to the first hour, we would get the following values:

    Hours into study 0 $\frac12$ 1
    Population (thousands) 4 6 8

    However, this does not have constant percentage growth: from $t = 0$ to $t = \frac12$ this population grew by 50% ($\text{ratio} = 1.5$), but then from $t = \frac{1}{2}$ to $t = 1$ the ratio is only $1.33$. If you graphed this data, instead of seeing a smoothly increasing curve, you would see a series of connected line segments of increasing slopes.

Cam says:

about 2 years

Thank you very much for this insightful and detailed comment.

You are of course completely correct that there are multiple ways to write down an expression for $P(t)$, and that the alternate form you provided is absolutely correct. The solution implied a little too strongly that there was only one equation, and I've fixed that.

I don't believe this should cause much consternation over the language of "factor" -- from the table constructed in part (a), or really the definition of the growth rate, it is clear that a common factor of 2 occurs between successive hours. Parts (c) and (d) don't seem to involve your answer to part (b) at all, in fact. Indeed, the point of the problem -- the part (e) you reference -- is to compare your formula from (b) with your tables in (c) and (d).

Given your formula of $P(t)=2^{t+2}$, I would reason as follows: Since from the table there were $4\sqrt{2}$ thousand at $t=1/2$, and since $P(1/2)=2^{2+1/2}$, then it should be the case that $4\sqrt{2}=2^22^{1/2}$. Dividing by 4 gives $2^{1/2}=\sqrt{2}$, precisely the punchline of that part of the task.

I agree that plotting the data is somewhat tangential, though might help develop a familiarity of the growth pattern, and plausible estimates for half-integer values.

I would worry that it's unclear what exactly students would be being asked to interpret in part (a).

Again, our thanks for carefully working through the task and coming back with feedback!

Brent says:

about 2 years

Our math education class was asked to complete and analyze this task to gain a better understanding of how we would use math tasks in our future classrooms. Upon the completion of our analysis of this task we came up with some comments and questions.

As our group progressed through the task we noticed multiple equations that could model the data in part A. For example P(t)=2^(t+2) is also a valid equation for the data. This led to confusion in parts C and D referring to a constant factor in the equation. We were still able to complete the tables either by guess and check or plugging the values into our various equations. However we struggled to naturally progress to the conclusion in part E. We continue to think about ways to modify this task to address the problems that we encountered.

We wondered how plotting the data in part A contributes to the understanding required to complete later prompts in the task.

We also wondered if adding "interpret your results" to part A would assist in connecting the goal of part F to the plotted data.