26.01 Rates of change (Extended)
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.
Practice questions
Question 1
Does the graphed function have a constant or a variable rate of change?
Constant
AVariable
B
Instantaneous rates of change
Instantaneous rate of change is the rate of change at that particular point.
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.
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 1
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.
Practice questions
question 2
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