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Preparation for Calculus

P.2 Linear Models and Rates of Change

Focus on slope and line relationships, average rate of change, build a linear model.

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Ideas in this lesson

Idea 1 of 3

Slope and line relationships

Slope is vertical change divided by horizontal change. Parallel lines have equal slopes; perpendicular nonvertical slopes multiply to −1-1.

Worked example 1

Find the slope through (1,2)(1, 2) and (3,6)(3, 6).
  1. Δy=6−2=4\Delta y = 6 - 2 = 4
  2. m=42=2m = \dfrac{4}{2} = 2

Answer: 22

Worked example 2

A line perpendicular to y=3x+1y = 3x + 1 has what slope?
  1. m1m2=−1m_{1}m_{2} = -1
  2. m2=−13=−13≈−0.33m_{2} = -\dfrac{1}{3} = \dfrac{-1}{3} \approx -0.33

Answer: −13≈−0.33\dfrac{-1}{3} \approx -0.33

Idea 2 of 3

Average rate of change

Divide the change in function value by the change in input. This is the slope of a secant line.

Worked example 1

For f(x)=x2f(x) = x^{2}, find the average rate of change on [0,2][0, 2].
  1. xx0022
    f(x)f(x)0044
  2. m=4−02−0m = \dfrac{4 - 0}{2 - 0}
  3. m=2m = 2

Answer: 22

Worked example 2

For f(x)=x2f(x) = x^{2}, find the average rate of change on [1,3][1, 3].
  1. xx1133
    f(x)f(x)1199
  2. m=9−13−1m = \dfrac{9 - 1}{3 - 1}
  3. m=4m = 4

Answer: 44

Idea 3 of 3

Build a linear model

A constant rate produces Q(t)=Q(t) = mt +b+ b. Interpret mm as the rate and bb as the initial value.

Worked example 1

A quantity begins at 11 and increases by 22 per unit. Find its value after 44 units.
  1. Q(t)=2t+1Q(t) = 2t + 1
  2. Q(4)=2(4)+1=9Q(4) = 2(4) + 1 = 9

Answer: 99

Worked example 2

A quantity begins at 22 and increases by 33 per unit. Its value is 1414. How many units have elapsed?
  1. Q(t)=3t+2Q(t) = 3t + 2
  2. t=14−23=4t = \dfrac{14 - 2}{3} = 4

Answer: 44

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §P.2,pP.2, p. 1010. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.