New Zealand
Level 6 - NCEA Level 1

# Interpret Rate of Change

Lesson

We have already learnt that a rate is a ratio between two measurements with different units.

When we graph these rates, the rate of change can be understood as the gradient, steepness or slope of a line. Further, we look at the equations in gradient-intercept form (that is, $y=mx+b$y=mx+b, where $m$m is the gradient), the larger the absolute value of $m$m, the steeper the slope of the line.

For example, a line with a slope of of $4$4 is steeper than a line with a slope of $\frac{2}{3}$23. Similarly, a line with a slope of $-2$2 is steeper than a line with a slope of $1$1, even though one is positive and one is negative.

## Increasing or Decreasing?

The rate of change in a graph can be increasing or decreasing.

The lines below have increasing slopes. Notice how as the values on the $x$x axis increase, the values on the $y$y axis also increase.

These next graphs have decreasing rates of change. Unlike graphs with a positive gradient, as the values on the $x$x axis increase, the values on the $y$y axis decrease.

## Rate of Change

The rate of change of a line is a measure of how steep it is.  In mathematics we also call this the gradient.

The rate of change is a single value that describes:

• if a line is increasing (has positive gradient)
• if a line is decreasing (has negative gradient)
• how far up or down the line moves (change in the $y$y value) with every unit step to the right (for every 1 unit increase in $x$x)

Take a look at this line for example. I've highlighted the horizontal and vertical steps.

We call the horizontal measurement the run and the vertical measurement the rise.

Here, for every $1$1 step across (run of $1$1), the line goes up $2$2 (rise of $2$2).  This line has a rate of change of $2$2.

Sometimes it is difficult to measure how far the line goes up or down (how much the $y$y value changes) in 1 horizontal unit. In this case we calculate the gradient by using a formula.

$\text{Gradient }=\frac{\text{rise }}{\text{run }}$Gradient =rise run

Where you take any two points on the line whose coordinates are known or can be easily found, and look for the rise and run between them.

## Finding the rate of change from a pair of coordinates

If you have a pair of coordinates, such as $\left(3,6\right)$(3,6) and $\left(7,-2\right)$(7,2) we can find the gradient of the line between these points also using the rule.

$m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1

Let's substitute our coordinates into this formula:

 $m$m $=$= $\frac{y_2-y_1}{x_2-x_1}$y2​−y1​x2​−x1​​ $=$= $\frac{6-\left(-2\right)}{3-7}$6−(−2)3−7​ $=$= $\frac{6+2}{3-7}$6+23−7​ $=$= $\frac{8}{-4}$8−4​ $=$= $-2$−2

So the rate of change between these coordinates is $-2$2.

## Gradient of horizontal and vertical lines

### Horizontal lines

horizontal lines have a rate of change of 0

Why?

Horizontal lines have NO rise value.  The $rise=0$rise=0.  So:

 $\text{Rate of change }$Rate of change $=$= $\frac{\text{Rise }}{\text{Run }}$Rise Run ​ $=$= $\frac{0}{\text{Run }}$0Run ​ $=$= $0$0

It doesn't matter what the run is, the gradient will be $0$0.

### Vertical Lines

##### The rate of change of vertical lines is undefined

Why?

Vertical lines have NO run value.  The $run=0$run=0.  So:

 $\text{Rate of Change }$Rate of Change $=$= $\frac{\text{Rise }}{\text{Run }}$Rise Run ​ $=$= $\frac{\text{Rise }}{0}$Rise 0​

It doesn't matter what the rise is, any division by $0$0 results in the value being undefined

Remember

Description of rate of change:     $\text{Rate of change }=\frac{\text{rise }}{\text{run }}$Rate of change =rise run

Gradient of Vertical Line = undefined

Gradient of Horizontal Line = $0$0

Rate of change can also be seen in everyday situations. Let's work through some worked examples.

#### Examples

##### Question 1

What kind of slope does the following line have?

1. Positive

A

Negative

B

Undefined

C

Zero

D

Positive

A

Negative

B

Undefined

C

Zero

D

##### Question 2

After Mae starts running, her heartbeat increases at a constant rate.

1. Complete the table.

 Number of minutes passed ($x$x) Heart rate ($y$y) $0$0 $2$2 $4$4 $6$6 $8$8 $10$10 $12$12 $49$49 $55$55 $61$61 $67$67 $73$73 $79$79 $\editable{}$
2. What is the unit change in $y$y for the above table?

3. Form an equation that describes the relationship between the number of minutes passed ($x$x) and Mae’s heartbeat ($y$y).

4. In the equation, $y=3x+49$y=3x+49, what does $3$3 represent?

The change in one minute of Mae’s heartbeat.

A

The total time Mae has run.

B

The total distance Mae has run.

C

The change in one minute of Mae’s heartbeat.

A

The total time Mae has run.

B

The total distance Mae has run.

C

##### Question 3

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol.

 Number of litres ($x$x) Cost of petrol ($y$y) $0$0 $10$10 $20$20 $30$30 $40$40 $0$0 $16.40$16.40 $32.80$32.80 $49.20$49.20 $65.60$65.60
1. Write an equation linking the number of litres of petrol pumped ($x$x) and the cost of the petrol ($y$y).

2. How much does petrol cost per litre?

3. How much would $47$47 litres of petrol cost at this unit price?

4. In the equation, $y=1.64x$y=1.64x, what does $1.64$1.64 represent?

The unit rate of cost of petrol per litre.

A

The number of litres of petrol pumped.

B

The total cost of petrol pumped.

C

The unit rate of cost of petrol per litre.

A

The number of litres of petrol pumped.

B

The total cost of petrol pumped.

C

### Outcomes

#### NA6-7

Relate graphs, tables, and equations to linear, quadratic, and simple exponential relationships found in number and spatial patterns

#### NA6-8

Relate rate of change to the gradient of a graph

#### 91028

Investigate relationships between tables, equations and graphs