Linear Models and Rate of Change

MATH161

Reference: Stewart 1.2 • Chapter: 1 • Section: 2

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Linear Models and Rate of Change

Why Does Constant Change Matter?

When a scientist measures temperature at different altitudes, or an economist tracks how prices change over time, they're looking for patterns. The simplest pattern? Constant change. If temperature drops by exactly 6 degrees for every kilometer you climb, or prices rise by $2 every year, then a straight line captures the entire relationship.

This constant-change behavior is everywhere: your car's speedometer reading when cruise control is on, the amount of medicine remaining in your body as it gets filtered out, or the cost of a phone plan based on data usage. Recognizing when a linear model applies—and knowing how to build one from data—is your first tool for making predictions.

The slope isn't just a number—it's the rate of change in context. A slope of $-10$ degrees per kilometer tells you the temperature drops 10°C for every km you ascend. That physical interpretation is what makes linear models powerful.

Prerequisite Map

Quick Reference

Property	Value
Concept	Essential Functions
Chapter	Chapter 1, Section 2
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Linear Model

A linear model describes situations where the dependent variable changes at a constant rate with respect to the independent variable:

$$y = mx + b$$

Symbol	Name	Meaning
$m$	Slope	Rate of change (how much $y$ changes per unit change in $x$)
$b$	$y$-intercept	Initial value (value of $y$ when $x = 0$)
$x$	Independent variable	The input you control or measure
$y$	Dependent variable	The output that responds to $x$

Building a Linear Model from Two Points

Given two data points $(x_1, y_1)$ and $(x_2, y_2)$:

Step 1: Calculate the slope: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}$$

Step 2: Find the $y$-intercept using one point: $$b = y_1 - mx_1$$

Step 3: Write the model: $$y = mx + b$$

Interpreting Slope in Context

The slope $m$ always has units: it's $\frac{\text{units of } y}{\text{units of } x}$.

Scenario	Slope Interpretation
Temperature vs. altitude	$m = -6$ °C/km means temperature drops 6°C per km
Cost vs. quantity	$m = 15$ $/item means each additional item costs $15
Distance vs. time	$m = 60$ mi/hr means traveling 60 miles per hour
Population vs. year	$m = 2500$ people/year means growth of 2500 per year

Key insight: A negative slope means the quantity is decreasing; a positive slope means it's increasing.

When Linear Models Work (and When They Don't)

Linear models are appropriate when:

The rate of change is (approximately) constant
Data points fall (roughly) along a straight line
Small deviations from linearity are acceptable for your purpose

Linear models fail when:

The rate of change itself is changing (use polynomial or exponential)
Data shows curves, peaks, or oscillations
The relationship has sudden jumps or thresholds

Practice Problems

Level 1 Identifying Slope and Intercept

A taxi company charges according to the model $C = 2.50d + 3.00$, where $C$ is the cost in dollars and $d$ is the distance in miles.

What is the slope, and what does it represent?
What is the $y$-intercept, and what does it represent?

Thought Process

Compare the given equation to the standard form $y = mx + b$. Here $C$ plays the role of $y$, and $d$ plays the role of $x$. Match coefficients to identify $m$ and $b$, then interpret each in the context of taxi fares.

Show Answer

(a) The slope is $m = 2.50$. This means the cost increases by \$2.50 for each additional mile traveled.

(b) The $y$-intercept is $b = 3.00$. This represents the base fare—the cost you pay just for getting in the taxi, before traveling any distance.

Level 2 Building a Model from Data

A spring stretches when weight is added. With 3 kg attached, the spring is 18 cm long. With 7 kg attached, it's 26 cm long. Find a linear model relating length $L$ (in cm) to weight $w$ (in kg).

Thought Process

We have two data points: $(w_1, L_1) = (3, 18)$ and $(w_2, L_2) = (7, 26)$.

Calculate slope: $m = \frac{26 - 18}{7 - 3}$
Use point-slope form or substitute one point to find $b$
Write the final equation

Show Answer

Step 1: Calculate the slope: $$m = \frac{26 - 18}{7 - 3} = \frac{8}{4} = 2 \text{ cm/kg}$$

Step 2: Find the $y$-intercept using $(3, 18)$: $$18 = 2(3) + b$$ $$b = 18 - 6 = 12$$

Step 3: The linear model is: $$\boxed{L = 2w + 12}$$

Interpretation: The spring stretches 2 cm for each additional kg, and has a natural (unstretched) length of 12 cm.

Level 3 Making Predictions

Atmospheric CO$_2$ concentration was measured at 354 ppm in 1990 and 384 ppm in 2008.

Find a linear model for CO$_2$ concentration $C$ (in ppm) as a function of year $t$.
Use your model to estimate the concentration in 2000.
According to this model, in what year will the concentration reach 420 ppm?

Thought Process

Let $t$ represent the year (or years since 1990 for simpler numbers). We have points $(1990, 354)$ and $(2008, 384)$.

For part (a): Find slope, then intercept. For part (b): Substitute $t = 2000$ into the model. For part (c): Set $C = 420$ and solve for $t$.

Show Answer

(a) Calculate slope: $$m = \frac{384 - 354}{2008 - 1990} = \frac{30}{18} = \frac{5}{3} \approx 1.67 \text{ ppm/year}$$

Find intercept using $(1990, 354)$: $$354 = \frac{5}{3}(1990) + b$$ $$b = 354 - \frac{9950}{3} = 354 - 3316.67 = -2962.67$$

Model: $C = \frac{5}{3}t - 2962.67$ (or equivalently, $C = 1.67t - 2962.67$)

(b) For $t = 2000$: $$C = \frac{5}{3}(2000) - 2962.67 = 3333.33 - 2962.67 \approx \boxed{371 \text{ ppm}}$$

(c) Set $C = 420$: $$420 = \frac{5}{3}t - 2962.67$$ $$\frac{5}{3}t = 3382.67$$ $$t = \frac{3 \times 3382.67}{5} \approx \boxed{2030}$$

Level 4 Comparing Two Linear Models

Two phone plans are available:

Plan A: \$25 per month plus \$0.10 per text message
Plan B: \$40 per month with unlimited texting

Write a linear model for the monthly cost of Plan A as a function of the number of text messages $n$.
For what number of texts are both plans equally expensive?
A customer sends about 200 texts per month. Which plan should they choose?

Thought Process

Plan A has variable cost (depends on usage), so it's linear with positive slope. Plan B has constant cost regardless of usage—this is a horizontal line.

To find where they're equal, set the two cost expressions equal and solve. Then compare at $n = 200$ to make the recommendation.

Show Answer

(a) Plan A: $C_A = 0.10n + 25$

Plan B: $C_B = 40$ (constant, regardless of $n$)

(b) Set costs equal: $$0.10n + 25 = 40$$ $$0.10n = 15$$ $$n = \boxed{150 \text{ texts}}$$

(c) At $n = 200$:

Plan A: $C_A = 0.10(200) + 25 = 20 + 25 = \$45$
Plan B: $C_B = \$40$

Plan B is cheaper by \$5 per month for this customer.

Level 5 Limitations of Linear Models

A researcher models bacterial population $P$ (in thousands) as a linear function of time $t$ (in hours), using data from $t = 0$ to $t = 3$:

$t$ (hours)	0	1	2	3
$P$ (thousands)	2.0	2.5	3.1	3.9

Find the best-fit linear model through points $(0, 2.0)$ and $(3, 3.9)$.
Use the model to predict the population at $t = 10$ hours.
Explain why this prediction might be unreliable. What type of model would better describe bacterial growth?

Thought Process

For part (a), use the two-point method to find slope and intercept.

For part (b), substitute $t = 10$ into the model.

For part (c), think about what actually happens with bacteria: they reproduce by division, so the rate of growth depends on how many bacteria exist. More bacteria means faster growth—this is not constant growth.

Show Answer

(a) Using $(0, 2.0)$ and $(3, 3.9)$: $$m = \frac{3.9 - 2.0}{3 - 0} = \frac{1.9}{3} \approx 0.633$$

Since the point $(0, 2.0)$ gives us $b = 2.0$ directly: $$\boxed{P = 0.633t + 2.0}$$

(b) At $t = 10$: $$P = 0.633(10) + 2.0 = 6.33 + 2.0 = \boxed{8.33 \text{ thousand}}$$

(c) This prediction is likely unreliable because:

Bacterial growth is exponential, not linear. Bacteria reproduce by cell division—each bacterium can produce two, those two produce four, etc. The growth rate is proportional to the current population.

Extrapolating far beyond the data is risky. Our data covers only 3 hours, but we're predicting at $t = 10$. Any slight inaccuracy in the model gets magnified.

An exponential model $P = P_0 \cdot b^t$ would be more appropriate. If we fit $P = 2.0 \cdot (1.25)^t$, the $t = 10$ prediction would be $P = 2.0 \cdot (1.25)^{10} \approx 18.6$ thousand—much higher than the linear prediction.

Key lesson: Linear models assume constant change. Recognize when the underlying phenomenon has a different structure.

Mastery Checklist

[ ] I can identify the slope and $y$-intercept from a linear equation
[ ] I can build a linear model from two data points
[ ] I can interpret slope as a rate of change with appropriate units
[ ] I can use a linear model to make predictions (interpolation and extrapolation)
[ ] I can recognize when a linear model is appropriate and when it isn't

Mental Model

Think of slope as a conversion rate:

Just as "\$1.50 per gallon" tells you how to convert gallons to dollars, the slope tells you how to convert changes in $x$ to changes in $y$.

If $m = -6$ °C/km, then climbing 2 km means: $2 \text{ km} \times (-6 \text{ °C/km}) = -12$ °C change.

The units cancel, leaving you with the change in the dependent variable. This "dimensional analysis" view of slope makes interpretation automatic.

Connections

Looking back:

The slope formula from algebra is the foundation for calculating $m$
Graphing skills help you visualize why slope measures steepness

Looking ahead:

Polynomial functions model situations where the rate of change itself changes
The concept of rate of change leads directly to the derivative in Chapter 2

Real-world connections:

Economists use linear models for supply/demand curves (as approximations)
Engineers use linear models for springs (Hooke's Law: $F = kx$)
Climate scientists model short-term CO$_2$ trends linearly

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Last updated: 2026-01-22