10.01 Average and instantaneous rates of change
We now venture into the topic of calculus - this is one of the fundamental studies in Mathematics. Here we will focus on the branch called differential calculus which is concerned with rates of change. A rate of change describes the speed at which one variable changes with respect to another variable. In calculating and interpreting the slope of linear functions we have encountered rates of change previously. Let's review rates of change for linear functions.
Constant rates of change
| Time, seconds ($x$x) | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|---|
| Distance, metres ($y$y) | $10$10 | $14$14 | $18$18 | $22$22 | $26$26 | $30$30 |
The table above shows a constant rate of change. As the time increases by $1$1 second the distance increases by $4$4 metres. The graph of distance against time would form a straight line, as it would for any two variables where the rate of change is constant. The rate of change, in this case the speed, is given by the change in distance divided by the change in time. This can be calculated using any two points from the table or graph.
| $\text{Rate of Change}$Rate of Change | $=$= | $\frac{\text{change in dependent variable}}{\text{change in independent variable}}$change in dependent variablechange in independent variable |
| $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
For our example, let's calculate the speed using the points $\left(1,14\right)$(1,14) and $\left(2,18\right)$(2,18):
|
Rate of Change (speed) |
$=$= | $\frac{\text{change distance}}{\text{change in time}}$change distancechange in time |
| $=$= | $\frac{\left(18-14\right)m}{\left(2-1\right)s}$(18−14)m(2−1)s | |
| $=$= | $4$4 m/s |
This should look familiar, this is the gradient formula. The rate of change of a linear function is given by the gradient of the line. Check you can use any two points and the result is the same constant speed. For examples set in context, don't forget to give the rate of change units. The units for a rate of change will be the dependent variable units divided by the independent variable units.
Average rates of change
In many situations, the rate of change may vary rather than stay constant. In exponential growth, we saw applications such as bacterial growth where the rate of growth increased with time and in reality cars are not likely to drive at a constant rate for very long. What would the distance-time graph of a typical drive around town look like? We could still calculate the change in distance divided by the change in time between two points for a journey with a variable speed, this calculation will now give us the average rate of change (average speed in the case of a distance-time graph).
To calculate the average rate of change between two points $A\left(x_1,y_1\right)$A(x1,y1) and $B\left(x_2,y_2\right)$B(x2,y2), we calculate the change in the dependent variable divided by the change in the independent variable. This is equivalent to calculating the gradient of the secant passing through points $A$A and $B$B. (A secant is a line that intersects a curve at two distinct points).
$\text{Average rate of change}=\frac{y_2-y_1}{x_2-x_1}$Average rate of change=y2−y1x2−x1
In calculus, function notation is very commonly used and being familiar with both $y=...$y=... notation and equivalent function notation will become an important skill. The equivalent statement for the average rate of change between $x=a$x=a and $x=b$x=b is:
$\text{Average rate of change}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$Average rate of change=f(b)−f(a)b−a
Worked examples
Example 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 $m^3$m3/week | ||
| (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 $m^3$m3/week | ||
| (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 $m^3$m3/week |
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.
Example 2
A population of rabbits is growing according to the function: $P(t)=200\times1.08^t$P(t)=200×1.08t, where $t$t is time in months.
(a) Find the average rate of change in the population between $2$2 and $4$4 months.
(b) Find the average rate of change in population between $4$4 and $7$7 months.
Think: Plot this relationship first (this can be done using technology), is the rate variable? Can you describe what is happening to the growth rate of the population over time?
The population is growing at an increasing rate, we can see this by the graph becoming steeper. We expect to see our calculation for part b) to be larger than that for part a).
To calculate the rate of change for the intervals we need to evaluate the function, that is find the population of rabbits at $t=2$t=2, $t=4$t=4 and $t=7$t=7.
Do: Calculate population at the specified times:
| Population at $t=2$t=2: | Population at $t=4$t=4: | Population at $t=7$t=7: | ||||||||
| $P(2)$P(2) | $=$= | $200\times1.08^2$200×1.082 | $P(4)$P(4) | $=$= | $200\times1.08^4$200×1.084 | $P(7)$P(7) | $=$= | $200\times1.08^7$200×1.087 | ||
| $\approx$≈ | $233$233 rabbits | $\approx$≈ | $272$272 rabbits | $\approx$≈ | $343$343 rabbits |
(a) Average rate of change between $t=2$t=2 and $t=4$t=4:
| $\text{Average rate of change}$Average rate of change | $=$= | $\frac{\text{Change in population}}{\text{Change in time}}$Change in populationChange in time |
| $=$= | $\frac{P(4)-P(2)}{4-2}$P(4)−P(2)4−2 | |
| $\approx$≈ | $\frac{272-233}{2}$272−2332 | |
| $=$= | $19.5$19.5 rabbits per month |
(b) Average rate of change between $t=4$t=4 and $t=7$t=7:
| $\text{Average rate of change}$Average rate of change | $=$= | $\frac{\text{Change in population}}{\text{Change in time}}$Change in populationChange in time |
| $=$= | $\frac{P(7)-P(4)}{7-4}$P(7)−P(4)7−4 | |
| $\approx$≈ | $\frac{343-272}{3}$343−2723 | |
| $=$= | $23\frac{2}{3}$2323 rabbits per month |
Reflect: Did we get a larger growth rate for part b) as expected? Notice that although it does not make sense to state a population with a fraction of a rabbit a rate of change does not need to be an integer. A rate of $0.5$0.5 rabbits per week would be equivalent to $1$1 rabbit per fortnight.
The average rate of change of a function, $f(x)$f(x), from $x=a$x=a to $x=b$x=b, is given by:
$\text{Average rate of change}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$Average rate of change=f(b)−f(a)b−a
If $A$A is the point on $f(x)$f(x) at $x=a$x=a and $B$B is the point of $f(x)$f(x) at $x=b$x=b, then the average rate of change is the gradient of the secant through points $A$A and $B$B.
Practice questions
Question 1
Does the graphed function have a constant or a variable rate of change?
Constant
AVariable
B
Question 2
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 3
The value of a particular coin can be modelled by the equation $y=80\left(2^x\right)$y=80(2x), where $x$x is the number of years from now.
What is the average rate of change of its value over the interval $\left[0,4\right]$[0,4]?
Instantaneous rates of change
Having just looked at the 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).
Take for example a trip from Canberra to Sydney, we can calculate the average speed for the journey, which is the total distance travelled divided by the time taken. If I drove from Canberra to Sydney in $5$5hrs, and the distance travelled is $340$340km, then my average speed is $\frac{340}{5}$3405 km/h = $68$68km/h. However, it would be impossible to have achieved this route with a constant rate, imagine that from the moment I started my car, to the moment I arrived in Sydney that I travelled at $68$68km/h the whole time. It's more likely that at some stages I was travelling $100$100km/h and others $40$40km/h through roadworks and other variations in the speed including stopping briefly at traffic lights. It is 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 speed of the vehicle at that point in time. 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.
How do we calculate instantaneous rates of change? If we think about the formula for average rates of change, for our example above we have: $\text{Average speed}=\frac{\text{change distance}}{\text{change in time}}$Average speed=change distancechange in time. However, in any particular instant, no distance has been travelled and no time has elapsed so we need a new method.
We will formally look at some algebraic techniques to calculate the instantaneous rate of change next in this chapter. For now, let's look at the fundamental ideas behind these techniques and some ways to approximate the instantaneous rate of change.
Worked example
Example 3
An object is moving according to the distance-time graph shown below. Find an estimate for the instantaneous rate of change at point A.
Think: How can we estimate the speed at this point? Can you imagine if at point $A$A the object then continued from this point at the speed it had at point $A$A rather than continuing along the curve. The object would follow the straight line shown below.
If we extend this line in both directions, obtaining the line shown below, we can estimate the instantaneous rate of change at $A$A by calculating the gradient of this line. The line that touches a curve and has a gradient matching the rate of change of the curve at the point of contact is called a tangent.
Do: Find two convenient points the tangent passes through to calculate the gradient. For our example, the line passes through point $A$A at $\left(1.5,7\right)$(1.5,7) and point $B$B at $\left(4.5,15\right)$(4.5,15), so our estimate for the instantaneous rate of change at point $A$A is:
| $\text{Instantaneous rate of change at A}$Instantaneous rate of change at A | $\approx$≈ | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
| $=$= | $\frac{15-7}{4.5-1.5}$15−74.5−1.5 | |
| $=$= | $\frac{8}{3}$83 m/s |
So to estimate the instantaneous rate of change at a point we can sketch a tangent to the curve and calculate the gradient of the tangent. We could obtain an accurate instantaneous rate of change by using technology to draw the tangent.
An alternative to sketching a tangent at point $A$A is finding the average rate of change between $A$A and a point, $B$B, very close to $A$A. The closer we make point $B$B to $A$A, the closer the secant through $A$A and $B$B will approximate the tangent at $A$A. If we calculate the average rate of change for points as they get closer to $A$A we may see that these rates are becoming closer to a particular value. We could make our estimate for the instantaneous rate of change the apparent limiting value. This diagram below shows the tangent at point $A$A and three secants at progressively closer points $B$B. We can see the secants become a closer estimate for the tangent at $A$A as $B$B draws closer.
Let's look at an example of this process.
Worked example
Example 4
Find an estimate for the instantaneous rate of change of the function $f\left(x\right)=3x^2$f(x)=3x2 at $x=2$x=2 by calculating the average rate of change between $x=2$x=2 and each of the following values $2.5,2.1,2.01,2.001$2.5,2.1,2.01,2.001.
Setting up the calculations in a table we obtain:
| $a$a | $b$b | $b-a$b−a | $\frac{f\left(b\right)-f\left(a\right)}{b-a}$f(b)−f(a)b−a |
|---|---|---|---|
| 2 | 2.5 | 0.5 | $13.5$13.5 |
| 2 | 2.1 | 0.1 | $12.3$12.3 |
| 2 | 2.01 | 0.01 | $12.03$12.03 |
| 2 | 2.001 | 0.001 | $12.003$12.003 |
We can see that as $b$b comes closer to $a$a the average rate of change appears to be getting closer to $12$12. We could try a very close point to see if this pattern continues. We could also try points that get closer from below $2$2, such as $1.9,1.99,1.999,...$1.9,1.99,1.999,... and see if the average rate of change again appears to approach this limit. This can be a tedious process and can we be sure the limit is in fact $12$12? In our next lesson we will look into how to evaluate limits, for now we will "estimate" the instantaneous rate of change of the function as the apparent limiting value. For this example, our estimate for the instantaneous rate of change of $f\left(x\right)$f(x) at $x=2$x=2 is $12$12.
Practice questions
question 4
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 5
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.
question 6
Consider the function $f\left(x\right)=x^3$f(x)=x3
By filling in the table of values, complete the limiting chord process for $f\left(x\right)=x^3$f(x)=x3 at the point $x=1$x=1.
Answer up to 4 decimal places if required.
$a$a $b$b $h=b-a$h=b−a $\frac{f\left(b\right)-f\left(a\right)}{b-a}$f(b)−f(a)b−a
$1$1 $2$2 $1$1 $\editable{}$ $1$1 $1.5$1.5 $\editable{}$ $\editable{}$ $1$1 $1.1$1.1 $\editable{}$ $\editable{}$ $1$1 $1.05$1.05 $\editable{}$ $\editable{}$ $1$1 $1.01$1.01 $\editable{}$ $\editable{}$ $1$1 $1.001$1.001 $\editable{}$ $\editable{}$ $1$1 $1.0001$1.0001 $\editable{}$ $\editable{}$ The instantaneous rate of change of $f\left(x\right)$f(x) at $x=1$x=1 is