Lesson

We can find the rate of change of a non-linear function at a point by looking at the gradient 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.

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`=2`t`. 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`=7.5`x`−5.5:

This secant has a gradient 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 gradients 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.

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 (m^{3}) |
$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.

Secants and 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 **gradient** of the **secant** on the function between the endpoints of the interval

Does the graphed function have a constant or a variable rate of change?

Loading Graph...

Constant

AVariable

BConstant

AVariable

B

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

BLinear

ANon-linear

B

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

BConstant

AVariable

B