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Lesson plan Grade 8 Science

Havenport's Aging Water Mains: A Linear Replacement Plan

A 50-minute grade 8 lesson where students model a fictional town's cast-iron water main replacement with linear functions for miles remaining and cumulative cost.

Objectives

  1. Students will construct linear functions for miles of pipe remaining and cumulative cost.
  2. Students will interpret the rate of change and initial value of each function in context.

Vocabulary

Water main
A big underground pipe that carries drinking water beneath the streets to a neighborhood. Smaller service lines branch off it to reach each building.

Also: linear function, rate of change, initial value

Teacher overview

Many towns still carry drinking water through cast-iron mains that are close to a century old, and those old pipes break more often as they age. This lesson uses the fictional town of Havenport to turn a pipe-replacement plan into two linear functions, one for miles of old pipe left and one for money spent. A common misconception is that "linear" means the numbers stay the same each year; here students see that a constant rate of change is what makes a function linear, even as the running totals keep moving.

Procedure

Hook (5 min): Describe a main break: a street floods, crews dig, and water service stops for a block. Ask students why very old pipes might break more often than new ones. In the fictional town of Havenport, 60 miles of the water mains are old cast iron. Record a few student ideas without grading them.

Direct instruction (12 min): Explain that Havenport replaces 2 miles of old main each year. Model writing the function for miles remaining: start at 60, subtract 2 for each year, so M = 60 − 2y. Suppose the town wants to know the miles left after 4 years; substitute to get 60 − 8 = 52 miles. Name the rate of change (−2 miles per year) and the initial value (60 miles). Then introduce the cost function: each mile costs 1.2 million dollars, and the town replaces 2 miles per year, so it spends 2.4 million dollars per year and the cumulative cost is K = 2.4y million dollars.

Guided practice (13 min): Hand out the Student Handout. In pairs, students complete items 1 through 3, building and using the miles-remaining function. Circulate and check that pairs write M = 60 − 2y before substituting.

Independent practice (12 min): Students complete items 4 through 6 on their own, moving to the cost function and solving for a budget limit. Remind them to keep units (miles, years, million dollars) attached to their answers.

Closure and exit ticket (8 min): Students complete item 7 on the handout as an exit ticket, writing a function for a different pipe segment and stating when it is finished. Collect the tickets and scan for a correct rate of change before dismissal. If more than a few students miss the sign of the rate, reteach the subtraction step next class.

Havenport's Aging Water Mains: A Linear Replacement Plan — Grade 8 Science
Name: ____________________________ Date: ______________

Student Handout

Havenport is a fictional town. Answer each item with a function or a number, and keep the units attached.

  1. Havenport has 60 miles of old water mains and replaces 2 miles each year. Write a linear function for the miles M of old main remaining after y years.
  2. Using your function, how many miles of old main remain after 10 years?
  3. In how many years will all 60 miles be replaced? Solve 60 − 2y = 0.
  4. Each mile costs 1.2 million dollars, and Havenport replaces 2 miles per year, so it spends 2.4 million dollars per year. Write a linear function for the cumulative cost K, in millions of dollars, after y years.
  5. What is the cumulative cost after 10 years?
  6. Havenport's council can spend 12 million dollars total on this work. Using your cost function, how many years of replacement can they afford?
  7. Exit ticket: A different pipe segment is 8 miles long and is replaced at 2 miles per year. Write a function for the miles remaining, then state after how many years the segment is finished.

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: M = 60 − 2y.
    The initial value is 60 miles and the rate of change is −2 miles per year.
  2. Answer: 40 miles.
    M = 60 − 2(10) = 60 − 20 = 40 miles.
  3. Answer: 30 years.
    Solve 60 − 2y = 0, so 2y = 60 and y = 30 years.
  4. Answer: K = 2.4y.
    The town spends 2.4 million dollars per year (2 miles at 1.2 million dollars per mile), so cumulative cost is 2.4y million dollars, with rate of change 2.4 and initial value 0.
  5. Answer: 24 million dollars.
    K = 2.4(10) = 24 million dollars after 10 years.
  6. Answer: 5 years.
    Solve 2.4y = 12, so y = 5 years of replacement.
  7. Answer: M = 8 − 2y, finished after 4 years.
    Start at 8 miles and subtract 2 each year; solve 8 − 2y = 0 to get y = 4 years.

Differentiation

Support

Give students a two-column table (Year, Miles remaining) already started with years 0, 1, and 2 filled in, so they can see the constant drop before writing the function. Provide the frame "M = ___ − ___ y" with the blanks labeled initial value and rate of change.

Extension

Ask students to graph both functions on the same axes (miles remaining and cumulative cost against years) and describe where the lines cross the axes. Then have them find the year when Havenport has replaced half of its old mains and explain how they used the function to find it.

ELL supports

Pre-teach "main," "replace," and "cumulative" with a simple picture of a pipe under a street and a running total. Provide sentence frames such as "After ___ years, ___ miles remain." Allow students to show functions and substitutions rather than writing full explanations.