Using Graphs to Describe Change
Graphs describe change. A straight line shows a constant rate of change, while a curve shows a rate that is speeding up or slowing down, described with phrases like growing more and more slowly. In NSW Year 10 Advanced maths you read, describe and sketch graphs of real situations and work out rates of change from them (NESA outcome MA5‑RAT‑P‑02).
You will describe the shape of graphs for a cooling coffee, a used car's value, a new app's users and a skydiver's speed, and match a graph to water filling a bowl. You will find rates from graphs of a draining tank, an electricity bill and a tide, and work out average speeds for a thrown ball and an accelerating car. Harder questions use a tangent to estimate a rate, and model a tank that fills and drains at once. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Theory
A graph shows not just whether a quantity is changing but how fast. A straight line is a constant rate; a curve is a variable rate, described as increasing or decreasing at an increasing or decreasing rate. This Year 10 Advanced maths page covers average rates, rates at an instant, filling containers and graphs whose rate changes.
The rate of change of one quantity with respect to another is how much the first changes for each unit of the second, such as litres per minute or dollars per year. On a graph it is the gradient, with units \(\dfrac{\text{vertical units}}{\text{horizontal units}}\).
A straight-line graph has a constant rate of change: positive if it rises, negative if it falls. A curved graph has a variable rate.
Describe a curve in two parts: increasing or decreasing (does it go up or down?), then at an increasing rate or at a decreasing rate (is it getting steeper or flatter?).
The average rate of change between two points is the gradient of the chord joining them. The rate at a single instant is the gradient of the tangent there, the line that just touches the curve.
It comes up if you are heading for Standard, Advanced or Extension maths in Years 11 and 12.
Average rate of change from \((x_1,\ y_1)\) to \((x_2,\ y_2)\):
For a constant rate, the quantity is a linear function of time:
How to describe change from a graph
- Direction: rising means increasing, falling means decreasing, flat means no change.
- Steepness: straight means a constant rate. Getting steeper means the rate is increasing in size; flattening means it is decreasing.
- Numbers: for a straight section, find the gradient with units. For a curve, use a chord (average) or a tangent (instant).
Problem types
- Matching a description to a graph: split the story into stages and check each stage's direction and steepness (steep then levelling off, sudden drop, constant).
- Average rates from a table or a rule: compute each interval's rate and compare them to describe the shape (Example 1). With a rule such as \(h = 25t - 5t^2\), substitute the end times first.
- Travel and money graphs: a distance-time gradient is a speed; a cost graph's gradient is the price per unit. A graph in sections has a different rate in each section; for a total, add the sections (e.g. the first \(250\) units at one rate, the rest at another).
- Containers: divide the flow rate by the cross-section area for each part (Example 2). A curved container gives a curved graph.
- Changing rates: write a rule for each stage, using the value at the end of one stage as the start of the next (Example 3).
- Tangents: read two points on the drawn tangent and find its gradient. Compare tangents at two times to say whether the rate is growing or shrinking.
- Exponential growth: average rates over equal later intervals are larger, so the graph is increasing at an increasing rate.
| Week \(t\) | \(0\) | \(2\) | \(4\) | \(6\) |
|---|---|---|---|---|
| Height (cm) | \(5\) | \(13\) | \(18\) | \(20\) |
(i) Find the average rate of growth in each \(2\)-week period.
(ii) Describe how the height is changing.
(i) Average rate \(=\) change in height \(\div\) change in time:
| \(t = 0 \text{ to } 2\) | \(:\) | \(\dfrac{13 - 5}{2} = \mathbf{4}\text{ cm/week}\) |
| \(t = 2 \text{ to } 4\) | \(:\) | \(\dfrac{18 - 13}{2} = \mathbf{2.5}\text{ cm/week}\) |
| \(t = 4 \text{ to } 6\) | \(:\) | \(\dfrac{20 - 18}{2} = \mathbf{1}\text{ cm/week}\) |
(ii) The rates are all positive, so the height is increasing; they get smaller, so it grows more slowly. The height is increasing at a decreasing rate, and its graph rises and flattens.
Rate of rise \(=\) flow rate \(\div\) base area. Time \(=\) height \(\div\) rate of rise.
| \(\text{lower}\) | \(:\) | \(\dfrac{1500}{300} = 5\text{ cm/min},\ \ \dfrac{15}{5} = \mathbf{3}\text{ min}\) |
| \(\text{upper}\) | \(:\) | \(\dfrac{1500}{150} = 10\text{ cm/min},\ \ \dfrac{20}{10} = \mathbf{2}\text{ min}\) |
Each part has a constant cross-section, so each section of the graph is straight. The narrower part fills faster, so its line is steeper. The container is full after \(5\) minutes, at a depth of \(35\) cm.
First \(12\) minutes: \(V = 800 - 25 \times 12 = 500\) L. After that the net rate is \(15 - 25 = -10\) L/min, so for \(t \ge 12\):
| \(V\) | \(=\) | \(500 - 10(t - 12)\) |
| \(500 - 10(t - 12)\) | \(=\) | \(200\) |
| \(t - 12\) | \(=\) | \(30\) |
| \(t\) | \(=\) | \(\mathbf{42}\text{ min}\) |
The graph has two straight sections with gradients \(-25\) and \(-10\): decreasing at a constant rate, then at a smaller constant rate.
(i) Find the average rate of growth from day \(2\) to day \(6\).
(ii) The tangent to the graph at \(t = 4\) passes through \((1,\ 84)\) and \((6,\ 164)\). Find the rate of growth at \(t = 4\).
(iii) Describe the growth.
(i) \(A(2) = 108\) and \(A(6) = 172\). The gradient of the chord is the average rate:
| \(\dfrac{172 - 108}{6 - 2}\) | \(=\) | \(\dfrac{64}{4} = \mathbf{16}\text{ cm}^2\text{/day}\) |
(ii) The gradient of the tangent is the rate at that instant:
| \(\dfrac{164 - 84}{6 - 1}\) | \(=\) | \(\dfrac{80}{5} = \mathbf{16}\text{ cm}^2\text{/day}\) |
(iii) The graph rises and gets steeper: the area is increasing at an increasing rate. A tangent drawn later, at \(t = 6\), would be steeper than \(16\).
Common pitfalls
Frequently asked questions
What does increasing at a decreasing rate mean?
The quantity keeps going up, but by less and less each time. On a graph the curve rises but becomes flatter, like a plant that grows quickly at first and then slows down.
How do you find the rate of change from a graph?
For a straight line, find the gradient: the change in the vertical quantity divided by the change in the horizontal quantity, with units. For a curve, use the gradient of a chord for an average rate or a tangent for the rate at one point.
What is the difference between average and instantaneous rate of change?
The average rate is the gradient of the chord joining two points on the graph. The instantaneous rate is the gradient of the tangent at a single point, the rate at that exact moment.
How does the shape of a container affect the water depth graph?
With a steady flow, depth rises fast where the container is narrow and slowly where it is wide. Straight-sided parts give straight-line sections; curved sides give curved graphs.
What does a negative rate of change mean?
The quantity is decreasing. For example, a rate of negative 25 litres per minute means 25 litres are lost every minute, and the graph slopes down.