MATH 115 ACTIVITY 5: Rate of change: Introduction, average and exact (instantaneous) rate of change
WHY: We will take advantage of the limit ideas to define a notion of rate of change (represented graphically by slope of
the graph) for functions which are not linear - this is the first big idea of calculus, leading to the derivative of a
function. Because calculating rate of change (or slope) directly requires two points, we will have to take an
indirect approach to produce an idea that works and makes sense.
LEARNING OBJECTIVES:
1. Recognize the ideas of average rate of change over an interval and of actual (instantaneous) rate of change at a point in the
context of functions.
2. Be able to calculate average rate of change and work with the difference quotient in its specific (numerical) and general
(formula) forms
3. Work as a team, using the team roles.
CRITERIA:
1. Success in completing the exercises.
2. Success in working as a team and in filling the team roles.
RESOURCES:
1. Your text section 3.3
2. The team role desk markers (handed out in class for use during the semester)
3. Your calculators
4. 40 minutes
PLAN:
1. Select roles, if you have not already done so, and decide how you will carry out steps 2 and 3
2. Work through the models given here – be sure all parts are clearly understood
3. Complete the exercises given here - be sure all members of the team understand and agree with all the results in the
recorder's report.
4. Assess the team's work and roles performances and prepare the Reflector's and Recorder's reports including team grade .
MODELS:
1. Average rate of change for a specific interval. The average rate of change for a function f over an interval x = a to x = b is
given by the difference quotient = , since f (change in f) = f(b) - f(a) and x (change in x) = b - a.
For example, if f(x) = x2 - 2 represents the size of a population (hundreds of aphids, perhaps) at time x months, then over
the interval x = 3 months to x=5 months, the change in f (that is, f) is f(5) - f(3) = [ 52 - 2] - [32 - 2] = 23 - 7 = 16
hundred aphids, corresponding to an elapsed time (x) of 5 - 3 = 2 months , so the average rate of change is 16/2 = 8
hundred aphids per month.
Over the interval x = 5 months to x = 8 months, the average rate of change is given by = =13 hundred aphids per
month [much faster growth than from 3 to 5 months].
For a linear function, this is exactly the same as the calculation of slope. For a non-linear function, different intervals give
different [average] rates of change - so the rate of change must be different for different points - but we can't calculate
directly with a slope-like calculation.
2. General (formula) form of the difference quotient for average rate for a particular start or end time:
We will want to focus on "average rate of change for intervals ending at 5 months" and "average rate of change for
intervals starting at 5 months" - so we want to represent the length of the interval in our formula.
For "intervals starting at 5 months" ("from x = 5 months to a bit later") we use x = 5 as the starting time and 5+h as the
finish time, so f = f(5+h) - f(5) and x = h [For our example of x = 5 months to x = 8 months, we would have had h =
3] so that = which can be simplified to = = 10 + h
For "intervals ending at 5 months (such as 3 months to 5 months) we use 5 as the ending time and 5+h (with h a negative
number) as the starting time [for our example of x = 3 months to x = 5 months, we would have x = 5 as the ending time
and h = -2 giving x = 5+h = 5-2 = 3 as the starting time] so that t = which can be simplified to = = 10 + h . Notice
that is exactly equal to [Here you see the reason for allowing negative h values - we get the same expression both ways
