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1.2 Rates of change

Lesson

Introduction

Learning objectives

  • 1.2.A Compare the rates of change at two points using average rates of change near the points.
  • 1.2.B Describe how two quantities vary together at different points and over different intervals of a function.

Rates of change

The average rate of change of a function is like finding the slope between two points on the graph. It helps us understand how the output values change when the input values change over a certain interval.

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The average rate of change of a function between two points can be found by drawing a straight line between the two points and finding its slope.

The average rate of change of the function shown on the graph over the interval [-1,2] is -1. This means that on average the y variable changes by -1 units for every unit change of the x variable.

The average rate of change for a function over the interval [a,b] is represented algebraically as:

\displaystyle \dfrac{f\left(b\right)-f\left(a\right)}{b-a}=\dfrac{\Delta y}{\Delta x}
\bm{a, b}
x-values of the function
\bm{\Delta y}
vertical change
\bm{\Delta x}
horizontal change

The rate of change at a single point, or instantaneous rate of change, can be approximated by finding the slope of the line that just touches the graph at that point. This is called a tangent line. We can estimate this rate of change by looking at the average rates of change over very small intervals around that point.

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Notice as the intervals around the point get smaller and closer to the point the slope of the line dashed segments gets closer to the slope of the tangent line.

Examples

Example 1

The water level of a water tank as it drains is given by the function W\left(t\right), measured in meters. Time, t, is measured in minutes. This is represented by the following table.

t \text{ minutes}W(t) \text{ meters}
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111.5
210.8
39.9
48.8
57.5
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a

How fast is the water tank draining, on average, over the six-minute interval?

Worked Solution
Create a strategy

To find the average rate of change of the water level over the six-minute interval, we will calculate the difference in height divided by the difference in time using the average rate of change formula \dfrac{f\left(b\right)-f\left(a\right)}{b-a}.

Apply the idea

Average rate of change =\dfrac{W\left(b\right)-W\left(a\right)}{b-a}

The starting point of the water level is represented by \left(0,12\right) and the end of the interval is \left(6,6\right).

We will substitute these points into the formula:

\displaystyle \dfrac{W\left(b\right)-W\left(a\right)}{b-a}\displaystyle =\displaystyle \dfrac{6-12}{6-0}Substitute into the formula
\displaystyle =\displaystyle \dfrac{-6}{6}Evaluate the subtraction
\displaystyle =\displaystyle -1Evaluate the division

The rate of change is -1 meters per minute which means, on average the water level is decreasing by 1 meter each minute during that time interval.

b

Estimate the rate at which the water level is changing at t=3.

Worked Solution
Create a strategy

To estimate the rate at which the height of the water is changing at t=3, we will find the average rate of change over small intervals around t=3.

Apply the idea

Let's calculate the average rates of change for some values approaching t=3 from the left side.

Between t=0 and t=1 the average rate of change is : \dfrac{11.5-12}{1-0}=-0.5

Between t=1 and t=2 the average rate of change is : \dfrac{10.8-11.5}{2-1}=-0.7

Between t=2 and t=3 the average rate of change is : \dfrac{9.9-10.8}{3-2}=-0.9

The average rates of change appear to be approaching -1 as x approaches 3 so we can approximate the instantaneous rate of change at x=3 as -1.

Reflect and check

To be more confident in our approximation we can check the average rates of change for values to the right of x=3.

Between t=5 and t=6 the average rate of change is : \dfrac{6-7.5}{6-5}=-1.5

Between t=4 and t=5 the average rate of change is : \dfrac{7.5-8.8}{5-4}=-1.3

Between t=3 and t=4 the average rate of change is : \dfrac{8.8-9.9}{4-3}=-1.1

We can see the rates of change to the right of x=3 are also approaching -1.

Example 2

Consider the function: g\left(x\right) = \sqrt{5x + 2}

a

Find the average rate of change of g\left(x\right) on the interval [0,\, 3].

Worked Solution
Create a strategy

We will use the formula for the average rate of change, but first we need to substitue the provided x-values into the function to find the output values we will need for the formula.

Apply the idea

Substituting the x-values of 0 and 3 into g\left(x\right) we get:

g\left(0\right)=\sqrt{5\left(0\right)+2}=\sqrt{0+2}=\sqrt{2}

g\left(3\right)=\sqrt{5\left(3\right)+2}=\sqrt{15+2}=\sqrt{17}

Substituting into the rate of change formula:

Rate of change =\dfrac{g\left(b\right)-g\left(a\right)}{b-a}=\dfrac{g\left(3\right)-g\left(0\right)}{3-0}=\dfrac{\sqrt{17}-\sqrt{2}}{3}\approx 2.71

The average rate of change of g\left(x\right) over the interval [0,\,3] is approximately 2.71.

b

Which has the smaller instantaneous rate of change: x=7 or x=9?

Worked Solution
Create a strategy

We can view the behavior of the graph to determine what happens to the rate of change as the x-values increase.

Apply the idea

Sketch a graph for g\left(x\right):

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We can see that the function gets flatter as the x-values increase. This means that the rate of change is getting smaller for values further to the right. So the instantaneous rate of change is smaller at x=9 because it is further to the right.

Idea summary

Average rate of change over an interval [a,\,b] is:

\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{\Delta y}{\Delta x}=\dfrac{f\left(b\right)-f\left(a\right)}{b-a}

Instantaneous rate of change at a point can be approximated by finding the average rates of change of small intervals near the point.

Outcomes

1.2.A

Compare the rates of change at two points using average rates of change near the points.

1.2.B

Describe how two quantities vary together at different points and over different intervals of a function.

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