15.01 Average and instantaneous rates of change
Average rate of change
We can find the rate of change of a non-linear function at a point by looking at the slope of the tangent. However, this will only tell us about the rate of change at one point and it requires knowing the tangent first. We’ll look at a method to deal with the first issue, and in the process we’ll deal with the second issue as well.
Exploration
Suppose that the relationship between the population ($n$n) of a colony of bacteria and the time ($t$t, in days) is described by the relationship $n=2^t$n=2t. We want to find the growth rate, which is the rate of change of the population. We can plot this relationship:
Instead of a tangent, we will use a secant. A secant is a line which touches a curve at two specific points. We can choose two points on this curve, $\left(1,2\right)$(1,2) and $\left(5,32\right)$(5,32), and draw the line connecting them, $y=7.5x-5.5$y=7.5x−5.5:
This secant has a slope of $7.5$7.5 so we say that the average rate of change from day $1$1 to day $5$5 is $8$8 per day.
We can draw secants through any two points of a graph. Here we have drawn the secants connecting $\left(1,2\right)$(1,2) to $\left(3,8\right)$(3,8), $\left(2,4\right)$(2,4) to $\left(4,16\right)$(4,16), and $\left(3,8\right)$(3,8) to $\left(5,32\right)$(5,32). These secants have slopes of $3$3, $6$6, and $12$12 respectively.
Notice that the average rate of change is variable since this is a non-linear function. This is true even when we take secants around the same point. $\left[1,5\right]$[1,5] and $\left[2,4\right]$[2,4] are both intervals around $3$3, but in the first case the average rate of change is $7.5$7.5 and in the second it is $6$6.
Worked example
Question 1
The volume of a lake over five weeks has been recorded below:
| Week | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|---|
| Volume (m3) | $123000$123000 | $142000$142000 | $135000$135000 | $111000$111000 | $104000$104000 | $123000$123000 |
(a) Find the average rate of change of volume in the first week
(b) Find the average rate of change of volume over the whole five weeks
(c) Find the average rate of change of volume in the last three weeks
Think: We don't know the function mapping the week to the volume. However, the average rate of change only requires two points on the graph. So we can find the average rate of change from just the data points.
Do: For each period, the average rate of change will be the change in volume divided by the number of weeks:
| (a) | average rate of change of volume in the first week | $=$= | $\frac{\text{change in volume}}{\text{number of weeks}}$change in volumenumber of weeks |
| $=$= | $\frac{142000-123000}{1-0}$142000−1230001−0 | ||
| $=$= | $\frac{19000}{1}$190001 | ||
| $=$= | $19000$19000 | ||
| (b) | average rate of change of volume over the whole five weeks | $=$= | $\frac{\text{change in volume}}{\text{number of weeks}}$change in volumenumber of weeks |
| $=$= | $\frac{123000-123000}{5-0}$123000−1230005−0 | ||
| $=$= | $\frac{0}{1}$01 | ||
| $=$= | $0$0 | ||
| (c) | average rate of change of volume in the last three weeks | $=$= | $\frac{\text{change in volume}}{\text{number of weeks}}$change in volumenumber of weeks |
| $=$= | $\frac{123000-135000}{5-2}$123000−1350005−2 | ||
| $=$= | $\frac{-12000}{3}$−120003 | ||
| $=$= | $-4000$−4000 |
Reflect: We can tell that the function is non-linear because the average rate of change is variable. Also notice that the average rate of change can be positive, negative or zero depending on the interval we choose. This is a significant limitation of average rates of change.
A secant is a line which intersects with a curve at two points
The average rate of change of a function over an interval is the slope of the secant on the function between the endpoints of the interval
Practice questions
Question 2
Does the graphed function have a constant or a variable rate of change?
Constant
AVariable
B
Question 3
Consider a function which takes certain values, as shown in the table below.
| $x$x | $3$3 | $6$6 | $8$8 | $13$13 |
|---|---|---|---|---|
| $y$y | $-12$−12 | $-15$−15 | $-17$−17 | $-22$−22 |
Find the average rate of change between $x=3$x=3 and $x=6$x=6.
Find the average rate of change between $x=6$x=6 and $x=8$x=8.
Find the average rate of change between $x=8$x=8 and $x=13$x=13.
Do the set of points satisfy a linear or non-linear function?
Linear
ANon-linear
B
Question 4
Does the function that is satisfied by the ordered pairs: $\left\{\left(-2,-5\right),\left(1,-20\right),\left(2,-25\right),\left(7,-50\right),\left(9,-60\right)\right\}${(−2,−5),(1,−20),(2,−25),(7,−50),(9,−60)} have a constant or a variable rate of change?
Constant
AVariable
B
Average and instantaneous rate of change
Having just looked at average rate of change, it's time now to look at another type of rate of change, called instantaneous rate of change.
Instantaneous rate of change is the rate of change at that particular point, (not the average over a range of points).
At this stage we won't need to calculate the instantaneous rate of change (although that comes next), but for now, let's really get our head around on contextually, what the difference is.
Worked example
Question 5
We can have average speed, which is the average speed taken over a journey. For example if I drove from Canberra to Sydney in $5$5hrs, and the distance traveled is $340$340km, then my average speed is $\frac{340}{5}$3405 km/h = $68$68km/h. Now it's unlikely (and nearly impossible) to imagine that from the moment I started my car, to the moment I arrived in Sydney that I traveled at $68$68km/h the whole time. It's more likely that at some stages I was traveling $100$100km/h and others $40$40km/h through roadworks or any speed in between. It is these specific occasions, like the $40$40km/h past the road work sign, or the $100$100km/h through the police speed checkpoint that are the instantaneous rate of change. So we have the average rate of change, being a measure of the average speed for the trip, and the instantaneous rate of change being a measure of the speed at a point in the journey.
Practice questions
question 6
Consider the function $f\left(x\right)=2x+3$f(x)=2x+3 drawn below.
Calculate the average rate of change between $x=-2$x=−2 and $x=2$x=2.
State the slope of the tangent at the point $x=-2$x=−2.
question 7
Consider the function $f\left(x\right)=x^3-3x^2+2x-1$f(x)=x3−3x2+2x−1
Calculate the average rate of change between $x=2$x=2 and $x=3$x=3.
A tangent line has been drawn at $x=2$x=2. Use the tangent to calculate the instantaneous rate of change at $x=2$x=2.
question 8
Consider the function $f\left(x\right)=2\left(3\right)^{-x}$f(x)=2(3)−x
Calculate the average rate of change between $x=-1$x=−1 and $x=0$x=0
The tangent line at $x=-1$x=−1 has been graphed, and its equation can be approximated by $y=-6.6x+6$y=−6.6x+6
Use the tangent to calculate the instantaneous rate of change at $x=-1$x=−1.