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)

3.40 minutes

PLAN:

1.Select roles, if you have not already done so, and decide how you will carry out steps 2
first cycle of the graph are hand 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, but for any length of 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 moths 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 and don't have to mess with 5 - h in some cases, 5+h in others]
We get an expression in which h - the time to the other end of the interval - shows up - indicating that the average rate of change changes for different lengths of time.
Using this formula, we would find that the average rate from x=5months to x = 8months (that is, with h = 3) would give a rate of change 10 + 3 = 13 hundred aphids/mo. (as in #1) and the average rate from x = 2mo to x = 5mo (h = -3) would give an average rate 10 + (-3) = 7 hundred aphids/mo (also as in #1)
If we take shorter and shorter intervals (h closer and loser to 0) we average over shorter times – and get closer to the growth rate at time x = 5 mo. – so the growth rate at 5 mo is = = 10 + h = 10 hundred aphids/mo.

3.General (formula) form of the difference quotient for average rate at any time:
The next step is to look at average rate of change not only for any length of time (h) but focused at any time (x), so instead of always using intervals starting/ending at 5 months (or any other particular value) we allow use of any value x. Then our General form of the difference quotient, for average rate of change on an interval of length h at time x is = . For the function we have been using, f(x) = x2 - 2, this gives
= = = = 2x + h hundreds of aphids/mo.
If we put x = 5 into this formula, we get the results in #2 – what is different here is that we get a formula that works for all x values.
If we take the limit as h approaches 0, we get a formula for the growth rate that applies for any time x mo.
(Growth rate at time x) = = = 2x + h = 2x

EXERCISES:

In all of these, we are considering a bacterial culture growing so that its mass (M - in grams) at any time (t - in hours after an experiment begins) is given by M(t) = t2 - t + 5

1.
a. Calculate the average growth rate = (g/hr) [this is the average rate of change of size] for the time interval t = 3 hr to t = 5 hr and the average growth rate for the time interval t =1 hr to t = 3 hr .
b. Calculate the average growth rate from t = 3 hr to t = 4 hr and from t = 2 hr to t = 3 hr.

2.
a.) Calculate the average growth rate for time intervals beginning/ending at time 3 hr
b.) Use the result from a.) to calculate the average growth rate for t = 1 hr to t = 3 hr (h = -2), for t=3 hr to t = 5 hr (h = 2) for t=2 hr to t = 3 hr (h = -1) and for t=3 hr to t=4 hr (h = 1). Your results should match the results in #1 a & b
c.) find the limit as h approaches 0 of the difference quotient you found in a.) [that is , find ] – this is the actual growth rate at t=3 hr.

3.
a.) Calculate the average growth rate for time intervals beginning and ending at variable time t (you will get a formula containing both t and h – not a number). Substituting t = 3 and the appropriate h values in your formula should give the same results as in #1 and #2
b.) Take the limit as h approaches 0 – this gives a formula (the variable t is still in the result) for the actual growth rate at any time t hr.

CRITICAL THINKING QUESTIONS:(answer individually in your journal)

1.What is the problem if we try to calculate a rate of change at a particular time (the instantaneous rate of change) directly, without using the difference quotient and a limit?

2.The units on “rate of change” are always “per” (miles per hour, grams per week, pounds/inch). What part of the work done here explains getting a “fraction – type” unit?

3.How well is your team working with learning the new material as we move beyond algebra and into the calculus?

SKILL EXER CISES(assignment)

Text p. 169 #1-3, #9-10 (instantaneous velocity is actual rate of change of position – given by the process we have just worked on) #23, 25, 27, 28