# Lesson 7

Interpreting and Using Exponential Functions

## 7.1: Halving and Doubling (5 minutes)

### Warm-up

This warm-up reinforces the idea that finding the growth or decay factor for a fractional interval of input involves taking a root of (rather than dividing) the factor for a whole-number interval. Given a description of how an exponential function changes over some hours, students explain why the growth or decay factor for one hour is a certain number. In doing so, students practice reasoning quantitatively and abstractly (MP2).

### Student Facing

- A colony of microbes doubles in population every 6 hours. Explain why we could say that the population grows by a factor of \(\sqrt[6]{2}\) every hour.
- A bacteria population decreases by a factor of \(\frac{1}{2}\) every 4 hours. Explain why we could also say that the population decays by a factor of \(\sqrt[4]{\frac12}\) every hour.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Activity Synthesis

Invite students to share their explanations. If no students reason about the hourly factor in terms of what it must equal when raised to a particular exponent (6 for the first function and 4 for the second), as shown in the Student Responses, bring this up.

For example, for the second question, point out that if the substance decays by $\frac12$ in 4 hours, the decay factor for one hour, say $b$, must be such that $b \boldcdot b \boldcdot b \boldcdot b = \frac12$ or $b^4 = \frac12$.

## 7.2: Radiocarbon Dating (20 minutes)

### Activity

In this activity, students use fractional exponents to answer questions about amounts of radioactive isotopes in old artifacts. Unlike in previous activities in which the input values of the exponential functions were straightforward, more reasoning is needed here to determine the input values to use. Students are not expected to answer rigorous questions about finding the ages of artifacts at this point, as doing so requires finding unknown exponents. That work will occur later, when students are equipped with the knowledge of logarithms.

To answer the last question about the amount of carbon-14 in the fossil today, students may:

- Find the exact (or estimated) number of years since 20,000 BCE, substitute it for \(t\) in the equation of the function, and calculate the output.
- Make a new table counting by intervals of 5,730 in the years column and multiplying the value in the mass column by \(\frac12\) for each additional row until the years column is about 22,000. (This would include using a spreadsheet tool.)
- Estimate the number of half-lives between 20,000 BCE and today, and multiply the amount by \(\frac12\) that many times.

Monitor for students who use each strategy to share their reasoning during the discussion.

Making spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Give students a brief overview of radioactive dating. Explain that a radioactive element is an element that emits rays or particles of energy when its atoms break down. Some radioactive elements break down at an exponential rate, and their decay rate is measured in terms of half-lives. Carbon-14 is one of those radioactive elements whose decay rate is known: its half-life is 5,730 years. Because of this, scientists commonly use the amount of carbon-14 in ancient artifacts to estimate their age or date of origin.

Point out that the amount of carbon-14 found in most artifacts is incredibly small. Although carbon is one of the most abundant elements, the most common form of carbon is carbon-12 (which has the same number of protons but a different number of neutrons than carbon-14). For every one part of carbon-14, there are about 1,000,000,000,000 (1 trillion) parts of carbon-12!

*Reading: MLR6 Three Reads.*Use this routine to support reading comprehension, without solving, for students. Use the first read to orient students to the situation. Ask students to describe what the situation is about without using numbers (The amount of carbon-14 in a fossil is decreasing at an exponential rate). Use the second read to identify quantities and relationships. Ask students what can be counted or measured without focusing on the values. Listen for, and amplify, the important quantities that vary in relation to each other in this situation: amount of carbon-14, in picograms, and the number of years since the fossil had 3 picograms of carbon-14. After the third read, ask students to brainstorm possible strategies to complete the table. This helps students connect the language in the word problem and the reasoning needed to solve the problem while keeping the cognitive demand of the task.

*Design Principle: Support sense-making*

*Representation: Internalize Comprehension.*Differentiate the degree of difficulty or complexity by asking students to first complete the row with the number of years being 5,730 (one half-life) to begin with more accessible values. Highlight connections between representations by inviting students to write the correct expression for one half-life next to the corresponding row in the table. Invite students to refer to this as they process the questions.

*Supports accessibility for: Conceptual processing*

### Student Facing

Carbon-14 is used to find the age of certain artifacts and fossils. It has a half-life of 5,730 years, so if an object has carbon-14, it loses half of it every 5,730 years.

- At a certain point in time, a fossil had 3 picograms (a trillionth of a gram) of carbon-14. Complete the table with the missing mass of carbon-14 and years.
number of years after fossil had

3 picograms of carbon-14mass of carbon-14

in picograms0 3 1,910 5,730 0.75 - A scientist uses the expression \((2.5) \boldcdot \left(\frac{1}{2}\right)^{\frac{t}{5,730}}\) to model the number of picograms of carbon-14 remaining in a different fossil \(t\) years after 20,000 BC.
- What do the 2.5, \(\frac{1}{2}\), and 5,730 mean in this situation?
- Would more or less than 0.1 picogram of carbon-14 remain in this fossil today? Explain how you know.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Anticipated Misconceptions

If students struggle figuring out how much carbon-14 remains after 1,910 years, consider asking them how 1,910 compares to 5,730. (It is one third.)

Some students may mistake the decay factor after 1,910 years as \(\frac13 \boldcdot \frac12\) or \(\frac16\) because 1,910 is a third of 5,730, and then calculate the amount of carbon-14 left as \(3 \boldcdot \frac 16\) or 0.5 picogram. Ask them if this amount makes sense given the amount of carbon-14 after 5,730 years (which is 1.5 picograms). Remind them of earlier activities in which we learned that if, for example, the hourly growth factor of an exponential function is 16, the factor for a quarter hour is \(\sqrt[4]{16}\), not \(\frac {16}{4}\).

### Activity Synthesis

Begin the discussion by inviting students to share how they completed the table, focusing on the second row (1,910 years). Make sure students recall that finding the amount of carbon-14 after 1,910 years means multiplying the original amount (3 picograms) by a factor of \(\left(\frac12\right)^\frac13\) or \(\sqrt [3]{\frac12}\) because the function at hand is exponential and 1,910 years is \(\frac13\) of 5,730 years.

Next, select previously identified students to share their strategies for the last question, having as many strategies as possible represented. If no one brings up the fact that the present time is less than four half-lives from 20,000 BCE, which means more than \(\left(\frac12\right)^4\) or \(\frac{1}{16}\) of the carbon-14 remains, point this out. Connect this to the idea of equal factors over equal intervals—the intervals being 5,730 years in this case.

## 7.3: Old Manuscripts (10 minutes)

### Activity

In this activity, students continue to reason about half-lives and exponential functions to solve problems. Both questions provide possible reasoning for a calculation and then prompt students to construct an argument to justify their position (MP3).

The context continues to be the decay of carbon-14 and students investigate two different situations. In each case, the amount of time that has passed since the carbon-14 began to decay is not a multiple of the half-life. This means that students need to make estimates, either in terms of a fraction of a half-life or a whole number of half-lives.

### Launch

In case students are unfamiliar with papyrus, explain that it is a material that was used as a writing surface in ancient times, similar to how paper is used today. Papyrus is made out of the papyrus plant and once it is harvested and turned into a type of paper, the amount of carbon-14 in the papyrus starts decreasing.

*Engagement: Develop Effort and Persistence.*Break the class into small discussion groups and then invite a representative from each group to report back to the whole class. While students work, provide tools for creating a visual display for them to record information for sharing. Encourage students to record their group’s thinking by writing out explanations using equations, words, pictures, or other symbolic representations such as timelines, that explain their approaches.

*Supports accessibility for: Language; Social-emotional skills; Attention*

### Student Facing

The half-life of carbon-14 is about 5,730 years.

- Pythagoras lived between 600 BCE and 500 BCE. Explain why the age of a papyrus from the time of Pythagoras is about half of a carbon-14 half-life.
- Someone claims they have a papyrus scroll written by Pythagoras. Testing shows the scroll has 85% of its original amount of carbon-14 remaining. Explain why the scroll is likely a fake.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Student Facing

#### Are you ready for more?

A copy of the Gutenberg Bible was made around 1450. Would more or less than 90% of the carbon-14 remain in the paper today? Explain how you know.

### Student Response

Student responses to this activity are available at one of our IM Certified Partners

### Anticipated Misconceptions

As in the previous activity, students may mistakenly think that after about \(\frac{1}{2}\) of a half-life, only about \(\frac{1}{4}\) of the carbon-14 remains. Ask them if more or less than \(\frac{1}{2}\) of the carbon-14 remains after 2,600 years. (Less since this is less than a half-life.)

### Activity Synthesis

Focus the discussion on the estimates students made and the properties of exponential functions that they used. Here are some questions for discussion:

- “About how old is the papyrus? About how many half-lives of carbon-14 is that?” (2,600 years, \(\frac{1}{2}\) half-life)
- “By what factor does the amount of carbon-14 decrease in one half-life? What about in half of a half-life?” (\(\frac{1}{2}\) and \(\frac{1}{\sqrt{2}}\))

One thing to note to students is that carbon-14 dating, while improving, is not an exact calculation but rather a good approximation for dating items from long ago. Consider telling students that carbon-14 dating is only used for objects that are less than about 50,000 years old. Ask,

- “About how many half lives is that?” (9)
- “About what fraction of the carbon-14 would be left in a 50,000-year-old artifact?” (\(\frac{1}{2^9}\) or \(\frac{1}{512}\))

*Speaking, Listening: MLR8 Discussion Supports.*Use this routine to support whole-class discussion. For each observation that is shared, ask students to restate what they heard using precise mathematical language. Consider providing students time to restate what they heard to a partner before selecting one or two students to share with the class. Call students' attention to any words or phrases that helped to clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.

*Design Principle(s): Support sense-making*

## Lesson Synthesis

### Lesson Synthesis

Highlight some problem-solving strategies that students have used in the lesson, which may include:

- writing an expression to represent a situation and evaluating the expression
- writing an equation and solving for an unknown value
- using a table
- extending a pattern (including using a spreadsheet)
- reasoning backward

Invite students to reflect on the merits of the different strategies when dealing with exponential functions. Discuss questions such as:

- “What kinds of questions were easier to answer using equations or expressions? Can you give an example from the lesson?”
- “What kinds of questions were easier to reason about without equations or expressions? Can you give an example?”

## 7.4: Cool-down - Half Gone, Again and Again (5 minutes)

### Cool-Down

Cool-downs for this lesson are available at one of our IM Certified Partners

## Student Lesson Summary

### Student Facing

Some substances change over time through a process called radioactive decay, and their rate of decay can be measured or estimated. Let’s take sodium-22 as an example.

Suppose a scientist finds 4 nanograms of sodium-22 in a sample of an artifact. (One nanogram is 1 billionth, or \(10^{\text-9}\), of a gram.) Approximately every 3 years, half of the sodium-22 decays. We can represent this change with a table.

number of years after first being measured |
mass of sodium-22 in nanograms |
---|---|

0 | 4 |

3 | 2 |

6 | 1 |

9 | 0.5 |

This can also be represented by an equation. If the function \(f\) gives the number of nanograms of sodium remaining after \(t\) years then \(\displaystyle f(t) = 4 \boldcdot \left(\frac{1}{2}\right)^{\frac{t}{3}}\)

The 4 represents the number of nanograms in the sample when it was first measured, while the \(\frac{1}{2}\) and 3 show that the amount of sodium is cut in half every 3 years, because if you increase \(t\) by 3, you increase the exponent by 1.

How much of the sodium remains after one year? Using the equation, we find \(f(1) = 4 \boldcdot \left(\frac{1}{2}\right)^{\frac{1}{3}}\). This is about 3.2 nanograms.

About how many years after the first measurement will there be about 0.015 nanogram of sodium-22? One way to find out is by extending the table and multiplying the mass of sodium-22 by \(\frac12\) each time. If we multiply 0.5 nanogram (the mass of sodium-22 9 years after first being measured) by \(\frac12\) five more times, the mass is about 0.016 nanogram. For sodium-22, five half-lives means 15 years, so 24 years after the initial measurement, the amount of sodium-22 will be about 0.015 nanogram.

Archaeologists and scientists use exponential functions to help estimate the ages of ancient things.