Lesson 1 ~40 min Rates of Change · Path +85 XP

Constant Rates of Change

A graph made of straight segments is telling you that something is changing steadily. The gradient of each segment is the rate at which it changes, and reading the units off the axes turns a bare number into a statement about the world.

Today's hook: A graph rises steeply, then flattens, then falls gently. Without any numbers you can already say a great deal: something increased quickly, then stopped changing, then decreased slowly. Attaching numbers is just reading gradients, and attaching meaning is just reading the axis labels.

0/5QUESTS

Think First

warm-up

A tank is filling. The graph of volume against time is a straight line from $(0,0)$ to $(5, 90)$, with volume in litres and time in minutes. Work out the gradient. Now say the answer as a sentence about the tank, including the units. Which of those two answers would a marker call complete?

Record your answer in your workbook.

The Big Idea

+5 XP to read

A straight-line segment on a graph means the quantity is changing at a constant rate. That rate is the gradient, and its units are the vertical axis unit divided by the horizontal axis unit.

$$\text{rate} = \text{gradient} = \frac{\text{change in the vertical}}{\text{change in the horizontal}}$$

Sign and size say different things. The sign says whether the quantity is rising or falling; the size says how fast. A steep downward segment is a large rate of decrease, not a small rate.

$\text{rate} = \text{gradient}$

Units from the axes

Vertical unit over horizontal unit. Litres up and minutes across gives litres per minute.

Straight means steady

Any straight segment has one rate throughout. Curved means the rate is changing.

Flat is not stopped

A horizontal segment means the quantity is not changing, which is not the same as nothing happening.

What You'll Master

objectives

Know

That a straight-line segment represents a constant rate of change
That the gradient of the segment is that rate
That the units of the rate come from the two axis labels

Understand

Why the sign and the size of a gradient carry different information
Why a direct variation graph is the special case of a constant rate starting from zero

Can Do

Calculate a rate of change from a straight-line graph, with units
Describe a piecewise-linear graph in words
Compare rates by comparing steepness

Words You Need

vocabulary

Rate of changeHow much one quantity changes for each unit change in another.

GradientRise divided by run. On a graph of two quantities, it is the rate of change.

Constant rateA rate that does not change, shown by a straight-line segment.

Piecewise linearA graph made of straight segments joined end to end.

IntervalA stretch of the horizontal axis over which one segment applies.

Straight Means Steady

+5 XP to read

On any graph of one quantity against another, a straight-line segment means the first quantity is changing at a constant rate with respect to the second.

The reason is what a gradient measures. Between any two points on a straight line, the ratio

$$\frac{\text{change in the vertical}}{\text{change in the horizontal}}$$

is the same, whichever two points are chosen. So equal steps across produce equal steps up, throughout the segment. That is exactly what "changing at a constant rate" means.

A curve tells you the opposite: the ratio changes as you move along it, so the rate is not constant. Deciding whether a graph is straight is therefore the first question to ask about it, and the whole of Lesson 3 is about what to do when the answer is no.

A graph made of several straight segments is describing a situation with several phases, each at its own steady rate, and the joins are the moments when something changed.

The Gradient Is the Rate, With Units

+5 XP to read

Calculating the gradient is the same operation you already know:

$$\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}$$

What is new is the interpretation. The number that comes out is a rate, and its units are the vertical axis unit divided by the horizontal axis unit.

A tank graph with litres up and minutes across gives litres per minute. A journey graph with kilometres up and hours across gives kilometres per hour, which is a speed. A cost graph with dollars up and items across gives dollars per item, which is a price.

From the hook: a line from $(0,0)$ to $(5,90)$ has gradient $\dfrac{90-0}{5-0} = 18$, so the tank fills at $18$ litres per minute.

"The gradient is $18$" is a half answer. "The tank fills at $18$ litres per minute" is a complete one, and the difference is a mark. Read the axis labels before you write the sentence.

The link to the previous area

A direct variation graph is exactly a constant-rate graph that starts at the origin, and its constant of variation is its rate of change. So $y = kx$ and "changing at $k$ units per unit" are the same statement, seen from two directions.

Increasing, Decreasing and Flat

+5 XP to read

Three cases, distinguished by the sign of the gradient.

Positive gradient: increasing at a constant rate. The graph rises from left to right. In context, the quantity is growing steadily: a tank filling, a savings balance growing by a fixed amount each week, a distance increasing at a steady speed.

Negative gradient: decreasing at a constant rate. The graph falls from left to right. A tank draining, a fuel level dropping, a debt being repaid at a fixed amount per month.

Zero gradient: no change. The graph is horizontal. The quantity is neither rising nor falling.

Two points of care. First, a horizontal segment does not mean nothing is happening: on a distance-time graph it means the object is stationary, but time is still passing. Second, when describing a decrease, the rate is usually quoted as a positive number with the word "decreasing" doing the work:

a gradient of $-4$ litres per minute is normally read as "draining at $4$ litres per minute", not "filling at $-4$". Either is acceptable if it is clear, but mixing them, as in "decreasing at $-4$ litres per minute", says the opposite of what is meant.

Comparing and Describing

+5 XP to read

When a graph has several segments, describing it well means saying three things about each: whether it rises, falls or is flat; how steeply; and over what interval.

Take the graph in the diagram, with quantity up and time across. A complete description:

it increases at a constant rate for the first phase, and that rate is the largest on the graph;
it then stays constant for a shorter interval, neither rising nor falling;
it then decreases at a constant rate until the end, more slowly than it rose.

Comparing steepness compares rates directly, because the gradient is the rate. Steeper means faster, in whichever direction the segment goes. When comparing a rise with a fall, compare the sizes of the gradients and describe the directions separately:

a segment of gradient $6$ and one of gradient $-9$ mean an increase at $6$ per unit and a decrease at $9$ per unit, so the second is the faster change even though it is the smaller number.

That last point is the one most often mishandled, and saying "faster" rather than "greater" avoids it.

Common Pitfalls

+5 XP to read

Quoting a gradient as a bare number with no units.

Fix: read the two axis labels and divide one by the other. "The gradient is $18$" is incomplete; "$18$ litres per minute" is the answer.

Saying a horizontal segment means the graph has stopped or the situation has ended.

Fix: it means the quantity is not changing while the horizontal variable continues. On a distance-time graph, the object is at rest but time still passes.

Calling a gradient of $-9$ a smaller rate of change than a gradient of $6$.

Fix: the sign gives the direction and the size gives the speed. A gradient of $-9$ is a faster change than one of $6$.

Writing "decreasing at $-4$ litres per minute".

Fix: the word and the sign both express the decrease, so together they cancel. Write either "the rate is $-4$ litres per minute" or "decreasing at $4$ litres per minute".

Watch Me Solve It · 3 examples

Watch Me Solve It · Reading a rate with units

+15 XP per step

PROBLEM

A graph shows the volume of water in a bath, in litres, against time in minutes. It is a straight line from $(0, 12)$ to $(8, 140)$. Find the rate at which the bath is filling, and explain what the value $12$ represents.

1

Calculate the gradient

$\text{gradient} = \frac{140 - 12}{8 - 0} = \frac{128}{8} = 16$

Rise over run, using the two given points.
2

Attach the units from the axes

Litres are on the vertical axis and minutes on the horizontal, so the units are litres per minute.
3

State the rate as a sentence

The bath fills at $16$ litres per minute. The rate is constant, because the graph is a straight line.
4

Interpret the starting value

The value $12$ is the volume at time zero, so there were already $12$ litres in the bath when timing began. Note that this means the relationship is linear but not direct variation, since the graph does not pass through the origin.

Answer$16$ litres per minute, with $12$ litres already present at the start

Watch Me Solve It · A graph with several phases

+15 XP per step

PROBLEM

A graph shows the fuel in a tank, in litres, against time in hours. It falls in a straight line from $(0, 60)$ to $(3, 24)$, is horizontal from $(3, 24)$ to $(4, 24)$, then rises in a straight line from $(4, 24)$ to $(4.5, 60)$. Describe each phase with its rate.

1

First phase

$\frac{24 - 60}{3 - 0} = \frac{-36}{3} = -12$

The fuel is decreasing at $12$ litres per hour for the first three hours: the vehicle is being driven.
2

Second phase

$\frac{24 - 24}{4 - 3} = 0$

The gradient is zero, so the fuel level is not changing for one hour. The vehicle is stopped, or at least not using fuel. Time is still passing.
3

Third phase

$\frac{60 - 24}{4.5 - 4} = \frac{36}{0.5} = 72$

The fuel increases at $72$ litres per hour for half an hour: the tank is being refilled.
4

Compare the rates

The refuelling rate of $72$ litres per hour is six times the size of the consumption rate of $12$ litres per hour, which is why the refill takes so much less time than the driving. Comparing the sizes of the gradients compares the speeds of the two changes.

AnswerDecreasing at $12$ L/h for $3$ h; unchanged for $1$ h; increasing at $72$ L/h for $0.5$ h

Watch Me Solve It · Comparing two constant rates

+15 XP per step

PROBLEM

Two savings accounts are graphed with balance in dollars against time in months. Account A is a straight line from $(0, 200)$ to $(12, 920)$. Account B is a straight line from $(0, 500)$ to $(12, 980)$. (a) Find each rate. (b) State which grows faster. (c) Find when the balances are equal.

1

(a) Find both gradients

$A: \ \frac{920-200}{12} = 60, \qquad B: \ \frac{980-500}{12} = 40$

So A grows at $\$60$ per month and B at $\$40$ per month.
2

(b) Compare the rates

Account A grows faster, at $\$60$ per month against $\$40$. Its line is steeper, which is the same fact seen on the graph. Note that A grows faster despite starting lower.
3

(c) Write both equations

$A: \ b = 200 + 60t, \qquad B: \ b = 500 + 40t$

Each has the starting balance as its vertical intercept and the rate as its gradient.
4

(c) Set them equal and solve

$200 + 60t = 500 + 40t$

$20t = 300 \ \Rightarrow \ t = 15$

The balances are equal after $15$ months, at $200 + 900 = \$1100$ each. Note this is beyond the $12$ months graphed, so the answer relies on extending both lines, which assumes the rates continue unchanged.

Answer(a) $\$60$ and $\$40$ per month; (b) A; (c) after $15$ months, at $\$1100$

Brain Trainer · 5 problems

Brain Trainer · Read the rate

5 problems

Five items on constant rates. Work each one, then reveal the answer.

1 A line goes from $(0,0)$ to $(4, 100)$, with litres up and minutes across. State the rate.

Gradient is $\tfrac{100}{4}$, and the units come from the axes.$25$ litres per minute
2 A line goes from $(0, 50)$ to $(10, 20)$. State the gradient.

$\tfrac{20-50}{10-0}$.$-3$
3 What does a horizontal segment on a graph mean?

The gradient is zero.The quantity is not changing
4 Which is the faster change, a gradient of $5$ or a gradient of $-8$?

Size gives the speed, sign gives the direction.The gradient of $-8$
5 A graph has dollars up and hours across. What are the units of its gradient?

Vertical unit divided by horizontal unit.Dollars per hour

Complete in your workbook.

Multiple Choice · 5 questions

MC1

What straight means

+10 XP

A straight-line segment on a graph of one quantity against another shows that:

MC2

Units

+10 XP

On a graph of distance in metres against time in seconds, the gradient has units of:

MC3

Flat segments

+10 XP

A horizontal segment on a graph of volume against time means:

MC4

Comparing

+10 XP

Segment P has gradient $7$ and segment Q has gradient $-11$. Which changes faster?

MC5

Wording

+10 XP

A tank drains with a gradient of $-5$ litres per minute. The best description is:

Short Answer · 3 questions

One rate, fully described

+15 XP

SHORT ANSWER

A straight-line graph shows the height of a candle, in centimetres, against burning time in hours. It passes through $(0, 24)$ and $(6, 9)$.
(a) Find the gradient.
(b) State the rate of change as a sentence, with units.
(c) State what the value $24$ represents.
(d) Find when the candle would burn out, and state one assumption you have made.

Write your working in your book.

A multi-phase graph

+15 XP

SHORT ANSWER

A graph shows the water in a pool, in kilolitres, against time in hours. It rises in a straight line from $(0, 0)$ to $(4, 30)$, is horizontal until $t = 7$, then falls in a straight line to $(9, 18)$.
(a) Find the rate of change in each of the three phases, with units.
(b) Describe the situation in words.
(c) State which phase involves the fastest change, and justify your answer.
(d) Explain why the first phase, but not the whole graph, shows direct variation.

Write your working in your book.

Comparing two rates

+15 XP

SHORT ANSWER

Two candles are lit at the same time. Candle P is graphed as a straight line from $(0, 30)$ to $(10, 10)$, and candle Q from $(0, 20)$ to $(10, 12)$, with height in centimetres and time in hours.
(a) Find the rate of change for each candle.
(b) State which burns faster, and by how much.
(c) Write an equation for each height in terms of time.
(d) Find when the two candles are the same height, and state whether that moment is inside the interval graphed.

Write your working in your book.

Stretch Challenge · What a rate does and does not tell you

+25 XP

CHALLENGE

(a) Two graphs of distance against time both have gradient $60$ km/h, but one passes through the origin and the other does not. Describe the difference in the two journeys.
(b) A quantity decreases at a constant rate of $5$ units per hour. Explain why its graph must eventually reach zero, and why that is not true of a quantity that halves every hour.
(c) A graph of cost against number of items has gradient $\$4$ per item and vertical intercept $\$30$. Explain why the average cost per item is not $\$4$, calculate it for $10$ and for $100$ items, and describe what happens to it as the number grows.

(a) Both travel at the same steady speed of $60$ km/h, since the gradient is the speed and the two gradients are equal. The difference is where they start. The graph through the origin describes a journey whose distance is measured from the starting point, so at time zero the distance is zero. The other has a non-zero vertical intercept, which means that when timing began the object was already some distance from the reference point: perhaps the distance is measured from a town the vehicle had already passed, or the timing started partway through the journey. The two graphs are parallel lines, so the gap between them stays constant: the second object is always the same distance ahead of, or behind, the first.

(b) A constant rate of decrease removes the same amount each hour. Starting from any value $V$, after $t$ hours the quantity is $V - 5t$, and choosing $t = \dfrac{V}{5}$ makes this exactly zero. Since that value of $t$ is a finite number, the quantity reaches zero in finite time. A quantity that halves every hour behaves quite differently: after $t$ hours it is $V \times \left(\tfrac{1}{2}\right)^t$, and a positive number multiplied by a positive number is never zero, however many times it is done. It gets arbitrarily close to zero without arriving. The underlying difference is that the first removes a fixed amount each hour while the second removes a fixed proportion, and only the first has a constant rate. The graph of the second is a curve, not a line, which is the subject of Lesson 3.

(c) The gradient of $\$4$ per item is the extra cost of one more item, not the average cost of the items ordered. The intercept of $\$30$ is a fixed charge incurred regardless of the number, and it is shared across however many items are bought. The total cost is $C = 30 + 4n$, so the average per item is $\dfrac{30 + 4n}{n} = 4 + \dfrac{30}{n}$. For $10$ items: $\dfrac{70}{10} = \$7$ per item. For $100$ items: $\dfrac{430}{100} = \$4.30$ per item. As $n$ grows the term $\dfrac{30}{n}$ shrinks towards zero, so the average falls steadily towards $\$4$ without ever reaching it: the fixed charge is spread ever more thinly. This is the same structure as the printing charges in the previous focus area, and it is why a fixed cost makes a relationship linear without making it proportional.

Quick Review

recap

Straight means steady

One rate for the whole segment

Gradient is the rate

With units from the two axes

Sign and size

Sign gives direction, size gives speed

Flat is a rate too

Zero rate, not zero value

Your Badges

0 of 6

First Steps

3-Day Streak

3 in a Row

Lesson Ace

Stretch Seeker

Daily Warrior

Mark lesson as complete

Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.

Rates of Change (Path) index · Maths subject page