Using Graphs to Describe Change
Using Graphs to Describe Change in NSW Year 10 Core maths reads stories from graphs (NESA outcomes MA5‑RAT‑P‑01 and MA5‑RAT‑P‑02). A straight line shows a constant rate of change, a horizontal line shows no change, and a curve shows a rate that is speeding up or slowing down.
You will match situations to graphs, find rates of change such as speed from a distance–time graph or the cost per litre of petrol, and describe curves in words, for example a temperature that rises quickly at first and then levels off. You will also solve direct and inverse variation problems, from filling bottles to painting a school, sketch a walk to the shops, see how water fills containers of different shapes, and find when one car catches another. It is part of Year 10 Core Non-Linear Relationships.
Theory
Graphs show how one quantity changes as another changes, usually over time. This Year 10 maths page shows how to find a rate of change from a straight-line graph, how to read distance-time graphs, how to describe curves using phrases such as increasing at a decreasing rate, and how to solve direct and inverse variation problems. This content comes up if you're heading for the Standard or Advanced path in Year 11.
A rate compares two different quantities, such as kilometres per hour or dollars per litre. The rate of change on a graph is its gradient: change in the vertical quantity \(\div\) change in the horizontal quantity.
A straight-line graph shows a constant rate of change. A horizontal line shows no change. A curved graph shows a variable rate: the gradient changes.
On a distance-time graph the gradient is the speed. A steeper line means a faster speed and a flat line means the object has stopped.
Two quantities are in direct variation if \(y = kx\): doubling one doubles the other, and the graph is a straight line through the origin. They are in inverse variation if \(y = \dfrac{k}{x}\): doubling one halves the other.
Rates, speed and variation:
| The graph | Words to describe it |
|---|---|
| straight, rising | increasing at a constant rate |
| rising, getting steeper | increasing at an increasing rate |
| rising, flattening out | increasing at a decreasing rate |
| falling, getting steeper | decreasing at an increasing rate |
| falling, flattening out | decreasing at a decreasing rate |
| horizontal | not changing |
How to find a rate from a straight-line graph
- Pick two points on the line that are easy to read.
- Divide the change in the vertical quantity by the change in the horizontal quantity. A falling line gives a negative rate.
- Write the units, such as km/h or \(\$\)/L. Convert minutes to hours when the speed is in km/h.
How to describe a curve
- Direction: is it going up (increasing) or down (decreasing)?
- Steepness: is it getting steeper (increasing rate) or flatter (decreasing rate)?
- Combine, for example: decreasing at a decreasing rate.
How to sketch a graph from a story
- Draw one section for each part of the story: sloping for steady movement, flat for a stop, sloping back down for returning.
- Make faster sections steeper, and use curves when the rate changes.
(i) Find the rate of change, in litres per minute.
(ii) Describe the change in words.
(iii) When is \(240\) L left?
(i) The graph is a straight line, so the rate is constant. Use two points: \((0, 600)\) and \((40, 0)\).
| \(\text{rate}\) | \(=\) | \(\dfrac{0 - 600}{40 - 0} = \mathbf{-15}\textbf{ L/min}\) |
(ii) The water decreases at a constant rate of \(15\) litres every minute.
(iii) \(600 - 240 = 360\) L has drained.
| \(t\) | \(=\) | \(\dfrac{360}{15} = \mathbf{24}\textbf{ minutes}\) |
(i) Describe the shape of the graph of her distance from home.
(ii) Find her speed on each ride, in km/h.
(i) A straight line up from \((0, 0)\) to \((30, 12)\), a flat section to \((45, 12)\), then a straight line down to \((85, 0)\). The last section is less steep, because she rides home more slowly.
(ii) Speed \(=\) distance \(\div\) time. Change minutes to hours first.
| \(\text{out}\) | \(:\) | \(12 \div \dfrac{30}{60} = 12 \div 0.5 = \mathbf{24}\textbf{ km/h}\) |
| \(\text{home}\) | \(:\) | \(12 \div \dfrac{40}{60} = 12 \times \dfrac{3}{2} = \mathbf{18}\textbf{ km/h}\) |
At the bottom the vase is narrow, so each litre makes the water rise a lot. Higher up it is wider, so each litre spreads out and the depth rises more slowly.
The depth increases at a decreasing rate: the graph is steep at first and then flattens out.
Compare: a cylinder fills at a constant rate (a straight line).
(i) Wages \(W\) vary directly with hours worked \(h\), so \(W = kh\). Working \(6\) hours pays \(\$171\). Find \(k\) and the pay for \(15\) hours.
(ii) A trip takes \(4\) hours at \(90\) km/h. How long does it take at \(120\) km/h?
(i) Substitute to find \(k\), the rate.
| \(171\) | \(=\) | \(6k\) |
| \(k\) | \(=\) | \(28.50\) |
| \(W\) | \(=\) | \(28.50 \times 15 = \mathbf{\$427.50}\) |
(ii) The distance is fixed, so time varies inversely with speed: faster means less time.
| \(\text{distance}\) | \(=\) | \(90 \times 4 = 360\text{ km}\) |
| \(\text{time}\) | \(=\) | \(\dfrac{360}{120} = \mathbf{3}\textbf{ hours}\) |
Common pitfalls
Frequently asked questions
What is a rate of change?
It is how fast one quantity changes compared with another, such as litres per minute. On a graph it is the gradient.
How do you find speed from a distance-time graph?
Find the gradient of the section: the change in distance divided by the change in time. Convert the time to hours for a speed in km/h.
What does increasing at a decreasing rate mean?
The quantity is still going up, but more slowly as time goes on. The graph rises but flattens out.
How does the shape of a container affect its filling graph?
Where the container is narrow, the depth rises quickly and the graph is steep. Where it is wide, the depth rises slowly and the graph is flatter. A cylinder gives a straight line.
What is the difference between direct and inverse variation?
In direct variation both quantities increase together, \(y = kx\), and the graph is a straight line through the origin. In inverse variation one increases as the other decreases, \(y = \dfrac{k}{x}\).