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Lesson plan Grade 11 Math

Pelsworth's Dry Spell: Fitting Ludworth Brook's Recession Curve

A grade 11 lesson fitting an exponential decay model to six fictional streamflow readings, then predicting when the brook drops below an intake.

Objectives

  1. Students will test whether a set of streamflow readings follows exponential decay by comparing successive ratios.
  2. Students will write an exponential model for baseflow recession and use it to predict a later flow.
  3. Students will compute and compare average rates of change over two windows of the same recession.

Vocabulary

Baseflow
Baseflow is the water that keeps a stream flowing even when it has not rained for a while. It comes from groundwater slowly seeping into the streambed.

Also: recession curve, decay constant, average rate of change, semi-log plot

Teacher overview

When rain stops, a stream keeps flowing on water that drains slowly out of the soil and shallow groundwater around it, and that draining flow falls off in a pattern close to exponential decay rather than in a straight line. This lesson gives students six gauge readings taken over twenty rainless days on the fictional Ludworth Brook, which supplies the fictional town of Pelsworth, and asks them to decide what kind of function fits, build it, and use it to predict when flow reaches the intake's minimum. The misconception the lesson addresses is the assumption that a falling quantity drops by a constant amount per day: the readings here fall by about 28 percent every four days, so the daily loss shrinks as the flow shrinks, and a linear model would run the brook dry well before the data supports.

Procedure

Framing the question (6 min): Display the six gauge readings without comment and tell students that the last rain fell on day 0 and no rain has fallen since. Ask what shape they expect a graph of these numbers to have, and record two or three predictions on the board without resolving them. Name the intake constraint: the fictional Pelsworth intake needs at least 8 cfs to operate its pumps, and the class question for the day is what date the operator should start planning for.

Testing for a constant ratio (10 min): Have students compute the ratio of each reading to the one before it. Circulate while they work and watch for students who subtract instead of divide, since the differences here shrink steadily while the ratios stay near 0.72. Bring the class back together long enough to establish that successive ratios agree to two decimal places, which is the evidence that an exponential model fits better than a linear one.

Teacher-modeled example (8 min): Work a separate example at the board so students see the method before they apply it. Suppose a different fictional gauge read 40.0 cfs on the day the rain stopped and 28.8 cfs four days later. The four-day ratio is 28.8 divided by 40.0, or 0.72, and the daily constant is the fourth root of 0.72, which is 0.921. Write the model as Q of t equals 40.0 times 0.921 raised to the t, and evaluate it at day 10 to get about 17.6 cfs. Do not work any handout number at the board.

Independent work on the handout (20 min): Students complete the nine handout items, working alone for the first ten minutes and then checking items 1 through 5 with a partner. Require that item 4 be written as a function before anyone attempts item 5, since students who skip the model tend to extend a table by hand and lose the point of the lesson.

Formative check (5 min): Collect a one-question exit response: give a flow of 30.0 cfs on day 0 with a daily constant of 0.921 and ask for the flow on day 5, which is about 19.9 cfs. Sort responses into correct, off by a factor of ten, and no attempt. If more than a third of the class misses it, reteach the exponent entry on the calculator at the start of the next period before moving on to logarithms.

Closing discussion (6 min): Return to the two predictions recorded at the start and ask which one the data supported. Close by asking why an operator would rather have the day-25 estimate from item 8 than a simple statement that flow is falling, and connect the answer to the lead time a utility needs to arrange an alternate source.

Pelsworth's Dry Spell: Fitting Ludworth Brook's Recession Curve — Grade 11 Math
Name: ____________________________ Date: ______________

Student Handout

Ludworth Brook, the Pelsworth intake, and these gauge readings are fictional. The last rain fell on day 0, and no rain fell during the twenty days recorded here.

Day Flow (cfs)
0 62.0
4 44.6
8 32.1
12 23.1
16 16.7
20 12.0

The Pelsworth intake needs at least 8 cfs to run its pumps.

  1. Divide the day 4 flow by the day 0 flow. Round to two decimal places.
  2. Divide the day 20 flow by the day 16 flow. Round to two decimal places.
  3. Your answers to items 1 and 2 give the factor by which flow changes across four days. Find the daily factor by taking the fourth root of that number. Round to three decimal places.
  4. Write an exponential model Q(t) for the flow in cfs on day t, using the day 0 flow and your daily factor.
  5. Use your model to predict the flow on day 28. Round to one decimal place.
  6. Find the average rate of change of flow from day 0 to day 8, in cfs per day. Round to two decimal places.
  7. Find the average rate of change of flow from day 12 to day 20, in cfs per day. Round to two decimal places.
  8. Use your model to find the day on which flow first falls to 8 cfs. Round up to a whole day.
  9. Explain why plotting the logarithm of flow against the day number produces a straight line, while plotting flow itself against the day number produces a curve.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 0.72.
    44.6 divided by 62.0 = 0.7194, which rounds to 0.72.
  2. Answer: 0.72.
    12.0 divided by 16.7 = 0.7186, which rounds to 0.72. The agreement with item 1 across the far end of the record is the evidence that one constant factor describes the whole recession.
  3. Answer: 0.921 per day.
    The fourth root of 0.72 is 0.72 raised to the 0.25 power, which equals 0.9212, or 0.921 to three decimal places.
  4. Answer: Q(t) = 62.0(0.921)^t.
    The day 0 flow is the initial value and the daily factor is the base. Checking at day 8 gives 62.0 x 0.921^8 = 32.1 cfs, which matches the record.
  5. Answer: 6.2 cfs.
    Q(28) = 62.0 x 0.921^28. Since 0.921^28 = 0.0998, the predicted flow is 62.0 x 0.0998 = 6.19 cfs, which rounds to 6.2 cfs.
  6. Answer: -3.74 cfs per day.
    The change is 32.1 - 62.0 = -29.9 cfs across 8 days, so the average rate of change is -29.9 divided by 8 = -3.7375, or -3.74 cfs per day.
  7. Answer: -1.39 cfs per day.
    The change is 12.0 - 23.1 = -11.1 cfs across 8 days, so the average rate of change is -11.1 divided by 8 = -1.3875, or -1.39 cfs per day. The later window falls at roughly a third the rate of the earlier one even though the factor per day never changed.
  8. Answer: day 25.
    Set 62.0 x 0.921^t = 8, so 0.921^t = 8 divided by 62.0 = 0.1290. Taking logarithms of both sides gives t = log(0.1290) divided by log(0.921) = -0.8892 divided by -0.0357 = 24.9 days, so flow first falls to 8 cfs during day 25.
  9. Answer: The logarithm of an exponential function is linear in t.
    Writing the model as Q(t) = 62.0 x 0.921^t and taking the base-ten logarithm of both sides gives log Q = log 62.0 + t log 0.921. The right side is a linear expression in t with slope log 0.921, which is about -0.0357, and intercept log 62.0. Because a constant factor per day becomes a constant difference per day once logarithms are taken, the semi-log plot straightens out, while the flow itself keeps changing by a shrinking amount and stays curved.

Differentiation

Support

Give students the ratio column already computed for items 1 and 2 so the lesson starts at the modeling step rather than at the arithmetic. Provide a filled-in calculator sequence for taking a fourth root and for evaluating a decimal base raised to a power, and let these students answer item 8 by extending a table in one-day steps rather than by using logarithms.

Extension

Ask students to find the half-life of this fictional recession, meaning the number of days for flow to fall by half, and to explain why that number does not depend on which day the count starts. A further step is to add a small rain event on day 10 that raises flow by 15 cfs, then have students fit a new recession from the new peak and describe what the combined graph looks like.

ELL supports

Pre-teach "baseflow," "recession," "gauge," "ratio," and "intake" with a labeled cross-section drawing showing water draining from soil into a fictional stream channel. Provide a two-column card pairing "constant difference" with "straight line" and "constant ratio" with "curve," and allow item 9 to be answered by completing the frame "Taking the logarithm turns a constant ___ into a constant ___, so the points line up."

Extensions

Have students download a real daily-values hydrograph for a nearby gauge from the United States Geological Survey, find a stretch of at least two rainless weeks, and repeat items 1 through 4 on that record. Real recessions rarely hold one constant ratio for the whole period, because different parts of the watershed drain at different rates, and asking students to describe where their fitted model breaks down is a strong introduction to residuals.