Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Super choice to join me.

Today, we're calculating the rate of change in real life graphs.

It's an important one, this one, so pay close attention.

A learning outcome is that we'll be able to calculate the gradient and interpret this as the rate of change for real life contexts.

Some key words we're going to see today, rate of change.

The rate of change is how one variable changes with respect to another.

If the change is constant, there is a linear relationship between the variables.

Gradient.

The gradient is a measure of how steep a line is.

It's calculated by finding the rate of change in the Y direction with respect to the positive X direction.

Look out for these words throughout our lesson.

Two parts to our learning today.

We're gonna begin with calculating gradient.

When graphing a linear equation, we can see the rate of change between the variables before we plot the graph.

If we populate this table for X and Y values, like so, there's a constant rate of change in the Y values.

Can you see what I mean by that? There, the Y values change by positive three every time.

The X values, of course, change by positive one in this table, that enables us to say that for every change of positive one in X, there is a change of positive three in Y.

Quick check you've got that.

I'd like you to fill in the blanks in the two sentences at the bottom of this screen.

Pause and do that now.

Welcome back.

Identifying these changes in the X and Y values in the table would enable you to fill in sentences like so.

There is a constant rate of change in the Y values and for every change of positive one in X, there is a change of, well done, negative four in Y.

The constant rate of change can be seen when the graph is drawn.

Let's look at this equation again.

Y equals three X plus two.

We know in the table that for every change of positive one in X, there's a change of positive three in Y.

We can see this manifest on the graph for every change of positive one in the X direction, there's a change of positive three in the Y direction.

It happens there and there and there and there and there and there and forever and ever.

That's what we call it, a constant rate of change, and you can see it's formed a linear graph.

The gradient of the line is three.

We define gradient as the rate of change in the Y direction with respect to the positive X direction.

So for this graph, we can say for every change of positive one in X, there's a change of positive three in Y.

That's what we mean when we say the gradient is three.

Quick check you've got that.

I'd like you to find the gradient of this line and describe the rate of change using a complete sentence.

To help you out, I've left you a definition of rate of change there.

Pause, give this a go now.

Welcome back.

Let's see how we got along.

Hopefully, you identified this on the graph.

The gradient of the line is negative two and it happens constantly.

It's a constant rate of change.

For every change of positive one in X, there was a change of negative two in Y.

Jun looked at the same graph and made a common error.

Jun says, I see the gradient of the line as positive two, and if you look at the graph, you can see how Jun has identified that.

Write a sentence or two to explain his error to him.

Pause and do that now.

Welcome back.

I wonder what you thought.

Hopefully, you said something along the lines of gradient is the rate of change in the Y direction with respect to the positive X direction.

Look at what Jun drew.

He was going in the negative X direction, that's not right.

You might have helped him out by drawing this on the graph, and identifying that the gradient is indeed negative two.

Jun says, "I see the gradient is negative two.

Thank you." You are welcome, Jun.

The gradient of the first graph here is easy to see.

We can identify those coordinate pairs and between them we can see that for every change of positive or X as a change of positive three in Y, so we can define it as a gradient of three.

The second graph is not so easy.

If I look at what's happening for every change of positive one in X, a little over a half, the gradient is a little over a half is not accurate enough.

We are mathematicians, we love accuracy, so how do we improve this statement? We'll often encounter gradients that are not integer values, but we can still accurately calculate them.

We need to identify any two coordinate pairs.

Once I've identified these two, zero zero and five three, I can see that for every change of positive five in X is a change of positive three and Y.

If I put this into a ratio table, I can then divide through both sides by five and I'll find out what the change in Y is for every change of one in X, three divided by five, what do you think we're going to write there? Well done, three fifths, so we can say here the gradient is three fifths or three over five.

Quick check you've got that.

I'd like you to use a ratio table to find the gradient of this line.

Pause and do this now.

Welcome back.

Hopefully, you identified for every change of positive four and X as a change of positive three in Y.

Lemme put, we put that into the table.

We can then cancel our ratio down to 1, 2, 3 quarters.

Well done.

The gradient is three quarters or three over four.

Another question for you, same graph.

I'm just doing something different this time.

The question is, do we get the same result if we use two different coordinate pairs? I'd like you to pause and have a think about that question for a moment.

Welcome back.

I hope you said an emphatic yes.

We do get the same result.

How? Well, if we divide both sides of our ratio by eight, we end up with six over eight or six eighths.

That's a fraction we can cancel down to, of course, three quarters or three over four.

The gradient is three quarters, that's the exact same result.

Practise time now, for question one, tables of values for linear equations, I'd like you to fill in the missing values and complete the statements about the rate of change.

Pause and do this now.

Question two, three graphs.

I'd like you to find the gradients of those three lines.

Pause and do it now.

Question three, Jun is asked to find the gradient and describe the rate of change of this line and Jun says the gradient is positive two.

For every change of positive one an X, there's a change of positive two in Y.

I'd like you to write at least two sentences of feedback for Jun explaining why he's wrong.

Pause and do this now.

Question four.

I'd like to find the gradients of these lines.

These are a little bit trickier.

You might find that ratio tables help.

Pause and do this now.

Welcome back, feedback time.

Question one asked to fill in the missing values in the tables and complete the statements about the rate of change.

Hopefully in table A, you inserted those values and completed the sentence so that it reads for every change of positive one in X, there was a change of positive five in Y.

For B, hopefully you identified those missing values and completed the statement.

For every change of positive one X, there was a change of negative four fifths in Y.

You might have been caught out and put negative four in Y for part B, but look at the X values.

They're moving up in steps of five.

Hence it's negative four fifths and not negative four.

You might wanna pause, just check that your tables of values match mine and your statements match mine.

Welcome back.

Question two, find the gradient of these lines.

Hopefully, you identified a gradient of negative two for the line on graph A, a gradient of positive four for B, and a gradient of positive seven for C.

Question three, we were writing a couple of sentences to explain Jun's error to him.

You might have written pay attention to the scale on your diagram.

You've not got a change of positive one in X.

Jun's X values go from one to 1.

5, so that means our gradient is not two because that's not a change of positive one in X.

If you wanted a change of positive one in X, your triangle would have to look something like that to reveal a gradient of four.

Question four, find the gradients of these lines and I advise you that ratio tables might help.

So hopefully for A, you identified a change of positive three and X for a change of positive five and Y, put that into a table, divide both sides by three, and we find that the gradient is five thirds or five over three.

For B, we should have identified for every change of positive two and X, it's a change of negative five in Y.

Put that into a table.

We find the gradient is negative five halves or negative five overdue.

For C, we've got a change of positive six in the X direction, positive seven in the Y direction.

That gives us a gradient of seven over six or seven sixths.

Onto the second half of our lesson now, where we're going to look at gradient in real life contexts.

Whilst gradient is typically defined as the rate of change in the Y direction with respect to the positive X direction, we frequently see axes labelled with variables other than X and Y.

In a graph of direct proportion, we might see variables such as U and R representing units sold and revenue on this graph.

We find the gradient in the exact same way.

Any two coordinate pairs.

How does X change? How does Y change or how does U change? How does R change in this context? Pop into a ratio table and we've calculated the gradient in the exact same manner.

It's 15.

However, our interpretation of this rate of change is different.

For every change of positive one in U, there is a change of positive 15 in R.

What does that mean? In this context, this means that every one unit sold is worth 15 in revenue.

Quick check you've got this.

This direct proportion graph relates P, British pounds, to B Thai baht, it's a currency conversion graph.

I'd like you to calculate and interpret the rate of change in this context.

Pause and do this now.

Welcome back.

I hope you calculated the gradient like so and found it to be 40.

We've got for every change of positive one in P, there's a change of positive 40 in B.

Let's interpret that in this context.

We can put that into a sentence.

This tells us that for every one pound, we'll get 40 Thai baht or one pound is worth 40 Thai baht.

This distance time graph shows a journey of M, miles, versus T, hours.

When we pick these two coordinate pairs, we can see a change of four in the time and a change of 150 in the miles.

From there, we could calculate the gradient.

It's a gradient of 37.

5.

Therefore, for every one hour, we travel 37.

5 miles.

What's wrong here? I'd like you to pause this video and think, "What's wrong with that bit of maths?" Welcome back.

I hope you spotted it.

Our gradient reading did not come from a triangle.

Look at the hypotenuse of our triangle versus the line of the graph.

They're not one and the same.

So the 37.

5 means that we have a mean average speed for the first four hours of travel, but at no point will we actually travelling at 37.

5 miles per hour.

So how do we get round this problem? For this graph, we have two gradients.

They need separate calculations.

The first section has a change of positive two in time, positive a hundred in miles, so you can find a gradient of 50 there.

And the second section, for every change of positive two in time, we've got a change of positive 50 in miles, so we can say it's a gradient of 25 after that change of gradient.

Interpreting this in context, we could say for the first two hours of travel, the rate of change is 50 miles per hour.

After two hours, the rate of change is 25 miles per hour.

That's a comment on the rate of change.

Given this context of speed, time, distance, you might see it more commonly phrased as for the first two hours of travel, the speed is 50 miles per hour, after two hours, the speed is 25 miles per hour.

Both are valid comments about the rate of change in this graph.

Quick check you've got that.

This graph shows the cost pounds that a broadband company charge new customers over T, time, in months, find and interpret the gradients.

Notice that's gradients plural.

Pause and do this now, I'd like a sentence for each of the gradients.

Welcome back.

Hopefully, you found those gradients and then wrote something along the lines of for the first three months, the of change is 20 pounds per month.

After three months, the rate of change is 40 pounds per month.

Or given this context you might have said for the first three months, the cost is 20 pounds per month.

After three months, the cost is 40 pounds per month.

Either of those statements is a valid one about the rate of change in this graph.

This graph shows the charges of a taxi firm in pounds for every mile travelled.

The gradient of the line is two, so Jun makes this statement.

Journeys with this taxi firm cost two pounds per mile.

What's wrong about this statement? Pause, have a conversation with a person next to you or a good thing to yourself where's the error in this? Welcome back.

Lots we could say here.

One thing we might identify is that if Jun's statement were correct, a five mile journey would cost 10 pounds, but it doesn't.

A five mile journey costs 15 pounds, so it can't be right to say that journeys with this taxi firm cost two pounds per mile.

The reason for this disconnect is because this is not a direct proportion graph.

It does not intercept the axes at the origin.

Jun spotted this now and says, "I understand.

Instead, I should say the rate of change of the cost is two pounds per additional mile travelled." Well done, Jun.

That is a better application of language.

Quick check you've got that.

Now, which statements are accurate about the rate of change for this taxi firm's charges? Four statements to choose from, some are accurate and some aren't.

You decide, see you in a moment.

Welcome back.

I hope you said statement A is not accurate.

It's not direct proportion, it's not true.

If a four-mile journey costs 16 pounds as you can read from this graph, then it's not fair to say that journeys cost three pounds per mile.

Even if the gradient of that line is three.

B is a slightly contentious one.

It's accepted, but it's a little bit ambiguous, so I'd ask you to avoid it.

C is much better.

The rate of change of the cost is three pounds per mile.

Those two words at the end there per mile.

I'm relating the two variables and relating the change in pounds cost to the change in miles travelled.

It's better if your sentence relates the two variables.

D is perfectly acceptable.

Practise time now.

Question one.

This time distance graph shows a walk taken between two friends, interpret the gradient.

I'd like you to use a complete sentence and the units are kilometres and hours.

Pause and do that now.

For question two, this graph shows the cost in pounds that a manufacturer pays for the number of units, U, of rubber valves that they purchase from a supplier.

Write at least one sentence interpreting the rate of change.

Pause and do that now.

Question three, this graph shows the charge in pounds of a mountain bike company for the number of hours, T, time, that a bike is hired.

For part A, I'd like you to write a sentence explaining why it is not correct to say a bike costs eight pound an hour to hire, despite the fact that the gradient of the line is eight.

For part B, I'd like you to write a sentence giving a better interpretation of this rate of change.

Pause and do those now.

Feedback time now.

Let's see how we did.

Question one, we were looking at time distance graph of a walk between two friends and we were interpreting the gradient.

You might have written, "The rate of change is four kilometres per hour" or given the context, you might have said, "The friend's speed is four kilometres per hour." Either is good.

Question two, this was the graph of the cost in pounds that a manufacturer pays for the number of units, rubber valves that they purchase from their supplier.

Hopefully, you identified a gradient of 0.

2 in the first section and identified that the gradient changes to 0.

15.

So we've got a gradient of no 0.

2 at the start and then after 500 units changes to 0.

15.

So you might have written as your interpretation of these gradients the rate of change is 20 pence per unit for the first 500 units and then 15 pence per unit for the next 500 units.

Or given this context, you might have said the cost is 20 pence per unit for the 500 units, then 15 pence per unit for the next 500 units.

Either of those two are perfectly acceptable.

Question three, part A, we would write in a sentence to explain why it's not correct to say a bike costs eight pounds an hour to hire, despite the fact the gradient of a line is eight.

You might have written the statement is incorrect because a five-hour hire does not cost 40 pounds.

It's useful once you make that statement to identify a coordinate pair which contradicts the statement.

You might then have written, this is not a direct proportion graph, it does not intercept the axes at the origin.

For part B, I ask you to write a sentence giving a better interpretation of this rate of change.

You might have written, "The rate of change of the cost is eight pounds per hour of higher," or "Every extra hour of higher adds eight pounds to the cost." Well, well done, you.

We're at the end of the lesson now.

We've learned that the gradient of a line can be calculated and interpreted as the rate of change in real life contexts.

It's important to communicate clearly when commenting on a rate of change.

Hope you've enjoyed this lesson as much as I have.

I look forward to seeing you again soon for more mathematics.

Goodbye for now.