7.0 – Review of Rates of Change

A key topic in Calculus is rate of change. There are two kinds of rates of change:

Average Rate of Change: The change in the dependent variable divided by the change in the independent variable. We can calculate this in two ways:
 Algebraically: AvgRateCha nge

 y
 x
 f ( x
2
) x
2

 f x
1
( x
1
)
 Graphically: Find the slope of the secant line using m sec ant

 y
 x
 y
2 x
2

 y x
1
1

A positive rate of change means the quantity is increasing on the interval, a negative rate of change means the quantity is decreasing on the interval.

Non-linear relationships do not have a constant rate of change. Avg rate of change calculated over different intervals will give different (and possibly less accurate results).

Instantaneous Rate of Change: The instantaneous rate of change of the dependent variable is the rate at which the dependent variable changes at a specific value of the independent variable, x = a . We can calculate this in several ways:

Algebraically: Use the location where the instantaneous rate of change occurs
( a , f ( a )) and a general point ( a
 h , f ( a
 h )) , then using
InstRateCh ange small, such as

0 .

01 f ( or a

 h
0 h
)

.
001 f (
. a )
. You must choose values for h that are very
 Graphically: Find the slope of the tangent line using m tan gent

 y
 x
 y
2 x
2

 y x
1
1

The best estimate for the instantaneous rate of change occurs when the interval used to calculate the average rate of change is made as small as possible.
Example:
1.
A pebble is dropped from a cliff with a height of 100 m. After t seconds, it is s metres above the ground, where s ( t )

100

5 t
2
, 0
 t

4 . a) Find the average velocity of the pebble between the times t = 1s and t = 3s. (2 marks) b) Find the average velocity of the pebble between the times t = 1s and t = 1.5s. (2 marks) c) Explain which answer is closer to the instantaneous velocity of the pebble at t = 1s. (1 mark)
Now you try:
2.
The profit P(x) of a cosmetics company, in thousands of dollars, is given by
P ( x )
 
5 x
2 
400 x

2550 , where x is the amount spent on advertising in thousands of dollars. a) Calculate the average rate of change in profit on the interval 8
 x

10 . b) Estimate the instantaneous rate of change when x = 50. c) Discuss the significance of the signs in your answers to parts a) and b)

Rate of change can be calculated in the following ways:


Rate of change graphically:
Rate of change numerically:

Rate of change algebraically:

Although all methods should give the same answer, sometimes one type of rate of change calculation is easier or more appropriate than another for a particular situation.

Example 1: Calculate rate of change of polynomial function (OAME 1.2, 1.3)

Example 2: Calculate rate of change of a rational function
The electrical current in a circuit varies with time according to c
 3 s 3 s
 
3 s

2
10
5 s
, where the current, c , is in amperes, and time s is in seconds. Find the average rate of change from 0.75 seconds to 1.5 seconds, and find the instantaneous rate of change at 1.5 seconds. Identify any vertical asymptotes.

Example 3: Calculate rate of change of a logarithmic function
A culture of bacteria was serial diluted and spread plated to get the following titers once an hour:
Time
8 AM
9 AM
10 AM
11 AM
Noon
1 PM
2 PM
3 PM
4 PM
5 PM
Hours
0
1
2
3
4
5
6
7
8
9
Titer (cells/ml)
2 × 10 7
2.1 × 10 7
2.3 × 10 7
3.25 × 10 7
6.6 × 10 7
1.4 × 10 8
2.92 × 10 8
6.1 × 10 8
1.1 × 10 9
1.23 × 10 9
1.3 × 10 9 6 PM 10
In this activity, we are interested in determining some average rates of growth as well as the instantaneous rate of growth at a specific time during the experiment. We can calculate the average rate of growth by dividing the difference in titer for a given time period by the number of hours in the time period.
1. Calculate the average rate of growth over the 10 hours of the experiment.
 9   7
Solution:
  8
.
10
The average rate of growth over 10 hours is 1.28 x 10 8 cells/ml/h.
2. Calculate the average rate of growth over the first five hours of the experiment. Is it the same, less, or greater than the average for the entire experiment?

Solution:
 8   7
  7
5
The average rate of growth over the first 5 hours of the experiment is 2.4 x 10 7 cells/ml/h. It is less than average for the entire experiment.
3. Calculate the average rate of growth over the last five hours of the experiment. Is it the same, less, or greater than the average for the first half of the experiment? Can you suggest reasons for any differences?
 9   8
Solution:
5
  8
The average rate of growth over the last 5 hours is 2.
32 x 10 8 cells/ml/h. It is greater than the average for the experiment. It makes sense that the growth would be faster in the second half of the experiment because as more bacteria grow, more are available to multiply. This suggests an exponential growth rate.
4. How could you calculate the instantaneous rate of growth at the exact half-way point in the experiment (1 pm)?
The answer to the last question is that there is more than one way to estimate the instantaneous rate of growth at exactly 1pm. We will take two different approaches today; first, a numerical approach, and second, a graphical approach. In both cases, we will work with graphing calculators.

Example 4: Calculate rate of change of a trigonometric function
Given the function: f
1. Sketch on an interval








,
6 6






6



and the points:

3
. Sol’n
and
2

3
2. Is the function increasing or decreasing on the interval
3. Draw the line through the points f

 
and f




2

3

to
3




. Sol’n:
2

3
? Sol’n: Increasing
4. Find the average rate of change of the function f
 this mean? f



3



 f
 2
3 3




2

3



 
3

 
3

0.025
6. Find the instantaneous rate of change at
3

This is the slope of the line through the points

3
.
,1.5
and


 

6


from

3
2

3
, 3
to
2

3
. What does

To find instantaneous rate of change at
At
At
At
At
At

2
5

12
7

18
13

36
61

180 f f f f f

 
 f

 
2 3


5

12


 f

5
 
12 3

7

18

 f

 
7
 
18 3


13

36


 f

13
 
36 3

61

180

 f

 
61
 
180 3
2.5981 1.5


1.0981

6
2.1213 1.5

12
1.9284 1.5

18

3
, choose values for θ which move closer to


1.7207 1.5

36
0.6213
0.4284

6

12

18
0.2207

36

1.5451 1.5

0.0451
1 1
0.0366



0.0414
0.0428

0.0441
0.0451

3
Approaches 0.05. This means that the slope of the line tangent to f
 


from


6

3
is 0.05
2

3

 at
.
WORK :

Rate of Change Review Questions
Rates of Change with Exponential and Logarithmic Functions
The data in the table below shows the number of bacteria grown over a 10 hour period in a laboratory. It assumes a zero death rate.
Time (hours) Number of bacteria
0 1
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
2
4
8
16
32
64
128
256
512
1 024
5.5
6
6.5
7
7.5
8
2 048
4 096
8 192
16 384
32 768
65 536
8.5
9
131 072
262 144
9.5
10
524 288
1 048 576
1. Calculate the average rate of growth for the following periods: a) The entire 10 hours of the experiment b) The last 5 hours of the experiment c) The first 5 hours d) The middle 5 hours (2.5 to 7.5 hours)
2. Predict the highest instantaneous rate of growth and when it occurred.
3. Determine the instantaneous rate of growth for the following times: a) At time = 3.5 hours b) At time = 7 hours c) At time = 8.5 hours d) At time = 10 hours
Without recounting the actual steps, provide an overview of the differences in calculating average and instantaneous rates of change in exponential applications.
Rates of Change with Rational Functions
As you get farther from Earth’s surface, gravity has less effect on you. For this reason, you actually weigh less at higher altitudes. A person who weighs 55kg can use the function’s
 6400(55) h

64000
to find their weight, W in kgs, at a specific height, h in feet above sea

level, above the Earth’s surface. Find the average rate of change from heights of 750 ft to
1200 ft above sea level, and find the instantaneous rate of change at 1200 ft above sea level.
After you eat something that contains sugar, the pH or acid level in your mouth changes.
This can be modeled by the function ( )

 m
20.4
2  m
36

6.5
, where L is the pH level and m is the number of minutes that have elapsed since eating. Find the average rate of change from 1.5 minutes to 3 minutes, and find the instantaneous rate of change at 3 minutes.
Rates of Change in Trigonometry
For each of the following functions, sketch the graph on the indicated interval. Find the average rate of change using the identified points, then find the instantaneous rate of change at the indicated point.
1. In a simple arc for an alternating current circuit, the current at any instant t is given by the function f (t)=15sin(60t). Graph the function on the interval 0 ≤ t ≤ 5. Find the average rate of change as t goes from 2 to 3. Find the instantaneous rate of change at t = 2.
2. The weight at the end of a spring is observed to be undergoing simple harmonic motion which can be modeled by the function D(t)=12sin(60π t). Graph the function on the interval 0 ≤ t ≤ 1. Find the average rate of change as t goes from 0.05 to 0.40.
Find the instantaneous rate of change at t = 0.40.