Lesson 5
Representing Exponential Decay
5.1: Two Other Tables (5 minutes)
Warm-up
This warm-up asks students not only to identify that a quantity is changing linearly or exponentially, but also to identify a term when the preceding value is not given.
Student Facing
Use the patterns you notice to complete the tables. Show your reasoning.
Table A
\(x\) | 0 | 1 | 2 | 3 | 4 | 25 |
\(y\) | 2.5 | 10 | 17.5 | 25 |
Table B
\(x\) | 0 | 1 | 2 | 3 | 4 | 25 |
\(y\) | 2.5 | 10 | 40 | 160 |
Student Response
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Anticipated Misconceptions
For the last entry in each table, students may struggle to find the actual value to write in the table. Explain that an expression representing the value is all that’s needed.
Activity Synthesis
Focus the discussion on how students used the patterns in the tables to generate a \(y\)-value for a non-consecutive value of \(x\). Make sure students are able to express the pattern of repeated addition in the first table, using an expression like \(2.5 + 7.5x\), and that they can express the pattern of repeated multiplication in the second table using an exponential expression like \((2.5) \boldcdot 4^x\). Highlight the words linear and exponential to describe how the quantities are changing suggested by the tables.
5.2: The Algae Bloom (15 minutes)
Activity
In this activity, students focus on producing and interpreting a graph representing a quantity that decays exponentially. They can then use the graph to make some estimates about the context, which is about algae control. After students create their graph, use the activity synthesis to encourage students to interpret the meaning of input values that are not whole numbers.
Note that the axes on the given graph aren’t scaled. Students will need to make sure that the vertical intercept can be plotted and that it is not too close to 0. If students do the work to decide how to scale the axes, they are creating a coherent representation of the relationship (MP2).
Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
To introduce the context and keep students thinking about exponential growth and decay, consider presenting a riddle:
A certain area of the pond is covered in algae. Each day, the algae-covered area doubles. If it takes 24 days for the algae to completely cover the pond, how many days did it take to cover half of the pond? Given the word "half," students may be inclined to immediately say that it takes 12 days. But if they ponder the situation a bit, they should recognize that it takes 23 days to cover half the pond, because it only takes one more day or one more doubling to cover the whole pond.
Supports accessibility for: Visual-spatial processing; Organization
Student Facing
In order to control an algae bloom in a lake, scientists introduce some treatment products.
Once the treatment begins, the area covered by algae \(A\), in square yards, is given by the equation \(A = 240 \boldcdot \left(\frac{1}{3}\right)^t\). Time, \(t\), is measured in weeks.
- In the equation, what does the 240 tell us about the algae? What does the \(\frac13\) tell us?
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Create a graph to represent \(A = 240 \boldcdot \left(\frac{1}{3}\right)^t\) when \(t\) is 0, 1, 2, 3, and 4. Think carefully about how you choose the scale for the axes. If you get stuck, consider creating a table of values.
- About how many square yards will the algae cover after 2.5 weeks? Explain your reasoning.
Student Response
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Student Facing
Are you ready for more?
The scientists estimate that to keep the algae bloom from spreading after the treatment concludes, they will need to get the area covered under one square foot. How many weeks should they run the treatment in order to achieve this?
Student Response
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Anticipated Misconceptions
Students may find it challenging to choose a scale for the axes in a way that helps them plot the points and see a pattern. If they are still struggling to choose a scale for the axes after a few minutes, ask students to think about the greatest and least vertical coordinates they need to show on the graph and what the height of each rectangle on the grid should be to show these values. In addition, if they get stuck plotting points, suggest that they first make a table of values.
Activity Synthesis
Once students have created reasonable graphs, focus the discussion on the non-integer values of \(t\). Discuss questions such as:
- “What does a \(t\) value of 2.5 mean?” (a value \(t\) of 2.5 means 2.5 weeks or a little more than 17 days after treatment begins.)
- “How did you estimate the area covered by algae after a certain non-whole-number of weeks?” (We can use the graph to estimate the area of algae, which is between the area in week 2 and that in week 3, though not too much precision can be given. It is definitely less than 27 square yards and more than 9 square yards. Following the general trend of the points decreasing less and less rapidly as time goes on, an estimate of 15 is reasonable.)
Time permitting, consider discussing:
- “Someone says that after seven weeks, the algae will disappear. Do you agree?” (Sample responses: No, if we keep multiplying by \(\frac13\), the area will keep getting smaller, but never reach 0. No, after seven weeks there would be about \(\frac{1}{9}\) of a square yard or 1 square foot covered by algae. Yes, rounded to the nearest square yard, there are 0 square yards covered by the algae after seven weeks.)
5.3: Insulin in the Body (15 minutes)
Activity
Here students make sense of a graph representing a situation characterized by exponential decay. They justify why the amount of insulin in a patient's body could change exponentially and use the graph to answer questions about the situation. The numbers, chosen explicitly to provide a realistic model of the insulin context, are more complex than what students have seen so far. They will need to apply their understanding both to identify the growth factor and to write an expression showing how much insulin remains after \(m\) minutes.
In this activity, students are building skills that will help them in mathematical modeling (MP4). They don't decide which model to use, but students have an opportunity to figure out how to justify that an exponential model is a good model for the data. Also, they are prompted to construct an equation to represent the model in a scaffolded way.
Launch
Explain that insulin is a hormone that is needed to process glucose (sugar) in the body. For people with diabetes, a disorder in which the body cannot process glucose properly or does not produce enough insulin, injections of insulin are sometimes needed. Once in the bloodstream, insulin breaks down fairly rapidly, so additional injections are needed over the course of a day to control the person’s glucose.
Display the graph for all to see. Ask students to observe the graph and be prepared to share one thing they notice and one thing they wonder. Give students a moment to share their observations and questions with a partner.
Design Principle: Support sense-making
Supports accessibility for: Memory; Conceptual processing
Student Facing
A patient who is diabetic receives 100 micrograms of insulin. The graph shows the amount of insulin, in micrograms, remaining in his bloodstream over time, in minutes.
- Scientists have found that the amount of insulin in a patient’s body changes exponentially. How can you check if the graph supports the scientists’ claim?
- How much insulin broke down in the first minute? What fraction of the original insulin is that?
- How much insulin broke down in the second minute? What fraction is that of the amount one minute earlier?
- What fraction of insulin remains in the bloodstream for each minute that passes? Explain your reasoning.
- Complete the table to show the predicted amount of insulin 4 and 5 minutes after injection.
time after injection (minutes) 0 1 2 3 4 5 insulin in the bloodstream (micrograms) 100 90 81 72.9 - Describe how you would find how many micrograms of insulin remain in his bloodstream after 10 minutes. After \(m\) minutes?
Student Response
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Anticipated Misconceptions
Students may notice that the insulin values decrease by \(\frac{1}{10}\) each minute and say, in response to the first two questions, that \(\frac{1}{10}\) of the original insulin remains after each minute. Ask them what \(\frac{1}{10}\) of 10 mg is. Then ask them if that is the amount of insulin that remains after 1 minute. If needed, ask them what fraction of 10 mg is the amount of insulin that remains after 1 minute. Is that the same as the fraction of 9 mg of insulin that remains after 2 minutes?
Activity Synthesis
Invite students to share their responses with the class. If not already mentioned in students’ explanations, highlight the connections between the equation, the graph, and the quantities in the situation. Ask students:
- “How can we use an equation to express the milligrams of insulin in the body (\(I\)) after \(m\) minutes using an equation?” (\(I = 100 \boldcdot \left(\frac{9}{10}\right)^{m}\))
- “Where can we see the 100 mg in the graph?” (the vertical intercept)
- “What about the \(\frac{9}{10}\)?” (The coordinates tell us that \(\frac{1}{10}\) of insulin gets removed each minute, so \(\frac {9}{10}\) of it, which is most of it, is left in the body every minute. The amount of insulin decays relatively slowly, which shows in the points dropping gradually.)
Also consider asking:
- “Will there be any insulin remaining in the bloodstream an hour after the injection?”
- “If so, when will the body completely run out of insulin?”
These are difficult questions to answer. The mathematical model will never reach 0 because a positive quantity multiplied by \(\frac{9}{10}\) is always positive. Practically speaking, however, the insulin given to the patient will eventually all disappear, unless a new dose is injected. The model also doesn't account for any insulin that the body produces naturally. This is a good opportunity to remind students that mathematical models are simplified descriptions of reality.
Lesson Synthesis
Lesson Synthesis
Use examples from the lesson or the new example presented here. Here is a graph showing the amount, in mg, of a medicine in a person’s body at some different times, measured in hours after taking some medicine. Every hour, \(\frac25\) of the medicine is broken down by the body.
Discuss questions such as:
- “What is the vertical intercept and what does it mean in this context?” (400. It is the milligrams of medicine in the person's body right after taking the pill.)
- “How can you tell from the graph that \(\frac25\) of the medicine is broken down after one hour?” (The amount of medicine after 1 hour dropped by 160 mg, and 160 is \(\frac25\) of 400.)
- “How can you tell from the graph that \(\frac35\) remained after each hour?” (240 is \(\frac35\) of 400, and 144 is \(\frac35\) of 240, and so on.)
- “What is an equation representing the amount of medicine in mg, \(m\), \(t\) hours after taking the medicine?” (\(m = 400 \left(\frac{3}{5}\right)^t\))
- “Why does it make sense to write \(\left(\frac35\right)^t\)?” (\(\frac35\) of the medicine remained with the passing every hour, so for \(t\) hours, we multiply the 400 by \(\frac35\), \(t\) times.)
5.4: Cool-down - Acetaminophen (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
Here is a graph showing the amount of caffeine in a person's body, measured in milligrams, over a period of time, measured in hours. We are told that the amount of caffeine in the person's body changes exponentially.
The graph includes the point \((0, 200)\). This means that there were 200 milligrams of caffeine in the person's body when it was initially measured. The point \((1, 180)\) tells us there were 180 milligrams of caffeine 1 hour later. Between 6 and 7 hours after the initial measurement, the amount of caffeine in the body fell below 100 milligrams.
We can use the graph to find out what fraction of caffeine remains in the body each hour. Notice that \(\frac {180}{200} = \frac{9}{10}\) and \(\frac {162}{180} = \frac{9}{10}\). As each hour passes, the amount of caffeine that stays in the body is multiplied by a factor of\(\frac{9}{10}\).
If \(y\) is the amount of caffeine, in milligrams, and \(t\) is time, in hours, then this situation is modeled by the equation:
\(\displaystyle y = 200 \boldcdot \left(\frac{9}{10}\right)^t\)