Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Well done for joining me today.

What are we learning today? We're going to be efficiently estimating journeys from non-linear graphs.

Now mathematicians like you and I, we love efficiency, so this lesson should be marvellous.

Let's take a look.

Our learning outcome is I'll be able to efficiently estimate the distance travelled for non-linear speed-time graphs.

Some keywords we'll encounter today, estimate, a quick estimate for a calculation is obtained from using approximate values, often rounded to one significant figure.

Generalisation.

We mathematicians love a generalisation.

A generalisation is a statement or rule that applies correctly to all relevant cases.

Two parts to our learning today, and we'll begin by efficiently making accurate estimations.

On a speed-time graph the area bound by the graph and the horizontal axis is the distance travelled.

We could find the area of one polygon to estimate the distance travelled in this model.

That's a trapezium with an area of 160.

In the context of a speed-time graph, that means we'll estimate the distance travelled to be 160 metres.

But the large gap between the graph and the polygon means our estimation lacks accuracy.

It's a comfortable overestimate, here.

We can improve the accuracy of our estimation by mapping more polygons onto the graph.

If I take it as two trapeziums mapped onto the graph, I work out the areas like so, I get 50 and 90.

Sum those together and I get an estimate now for the distance travelled of 140 metres.

The gap between the graph and the polygons is less, which makes this a far better estimate.

In fact, the more polygons we map onto the graph, the more accurate our estimation becomes.

Here I've mapped five trapezia onto the curve.

The gaps between the graph and the polygons are negligible, so this will provide a really accurate estimate.

I just have to work out these five area, a half of eight plus nine multiplied by two plus a half of nine plus 10.

5 multiplied by two, et cetera, trust me, this is right.

Those five area sum to 130 metres so we can estimate the distance travelled to be 130 metres.

I'd like to see if you can repeat that skill now.

I'd like you to use five strips of equal width to estimate the distance travelled in this speed-time graph, by five strips of equal width, I mean five trapeziums of the same height.

The Y coordinates are accurate to one decimal place, and those are in the table underneath the graph.

By five strips, I mean those five trapezium.

Work out the area of those, sum the area together and you'll have a really good estimation for the distance travelled on this speed-time graph.

Pause and work this out now.

Welcome back.

Let's see how you got on.

Your working out should have looked like that.

That's the area of those five trapezia.

Individually they were 5.

5, 6.

55, 7.

8, 9.

3, 11.

05.

Sum those together and you'll get an area of 40.

2, so, we can estimate the distance travelled to be 40.

2 metres.

Have you spotted a problem with this method? Is anything bothering you about it? Yes, you have, haven't you? Whilst it's fairly accurate, it's very time consuming.

I bet that problem took you a while to work out.

So what do we do about that? Can we make this calculation any more efficient? Look at the calculation.

Is there anything you see in there that could make this a little more simple? Pause, have a conversation with a person next to you or a good think to yourself, and we'll resolve this problem in a moment.

Welcome back.

I wonder what you said.

I wonder what ideas you came up with.

I wonder what you noticed about this calculation that enables us to make it a little more efficient, or a little more simple.

We can perform this calculation far more efficiently by factorising.

Factorising, i.

e.

what common factors are there that we can take out? Can you spot them? Did you spot them? A half.

Each of our trapeziums required us to multiply by a half.

Also the height of our trapeziums, or the width of our strips, they were all one.

So we can take out the common factor of a half and the common factor of one.

When we do that, our calculation looks like this.

We can make it more simple again.

What do you notice in this part of the calculation? Well done.

Two lots of six, two lots of 7.

1, two lots of 8.

5, two lots of 10.

1.

So let's write that moment like that.

What do you notice again? Oh, of course, two lots of six, two lots of 7.

1.

We can factorise out those twos.

We end up writing a half multiplied by one multiplied by bracket five plus 12 plus two lots of bracket six plus 7.

1 plus 8.

5 plus 10.

1.

By factorising, we just turned that awfully lengthy calculation into a much shorter one.

I know which one I'd prefer to type into my calculator.

This calculation gives us the exact same result just far more efficiently, and we mathematicians, we love efficiency.

Quick check you've got this.

The area of four trapezia of equal width can be used to accurately estimate the distance travelled in this speed-time graph.

What I'd like you to do is fully factorise the calculation at the bottom of the screen so that it can be more easily entered into a calculator.

I need the answer to that calculation but I don't wanna type it all into my calculator, so you're going to simplify it for me, and make my life a little easier.

Thank you.

Pause and do this now.

Welcome back.

See what we did? Hopefully you identified common factors of half, and two.

Factorise those out.

What do we notice next? Oh yes, two lots of 8.

8, two lots of 7.

5, two lots of 5.

2.

Can we go further? Absolutely.

Factorise out the twos.

Instead of two lots of 8.

8 and two lots of 7.

5 and two lots of 5.

2, we just say two lots of bracket, 8.

8 plus 7.

5 plus 5.

2.

Now that we're at this stage, we're fully factorised.

So do I type that into my calculator? I hope you noticed a half of two, oh that's one.

So we just end up with one lot of 9.

5 plus four, 1.

4 et cetera.

That's further simplified again.

I want to type that into my calculator, which I can do in a matter of seconds.

You'll get an area of 53.

9, in the context of a speed-time graph, we say that's 53.

9 metres travelled.

What a lovely bit of simplifying.

Well done.

Another check.

Now that we know we can simplify those calculations, can you do the whole thing, from start to finish? If I say to you, here's a speed-time graph and I'd like you to use three strips, three trapezia, of equal width to find a good estimate for the distance travelled in this speed-time graph.

The values you'll need are in the table at the bottom of the screen.

Once you've got the area of your trapeziums, I want you to factorise and simplify your calculation before you enter it into your calculator.

Pause and give this problem a go now, I'll be back in a moment with the solution.

Welcome back.

Hopefully your trapezia look like so, the area of which look like that, but we don't wanna enter that into our calculators, so we factorise out the common factors half and three.

We further simplify that, to say we have two lots of 4.

5 and 2.

8.

We tap that into our calculators.

We find an estimate for the distance travelled of 43.

8 metres.

Well done.

Alex is looking at the same speed-time graph, and he says, "I can make a more accurate fit to the graph if I use three polygons instead".

Whilst the estimation is more accurate, there's a problem in Alex's calculation.

Can you see it? Pause, have a look at that calculation, have a think, why wouldn't we do this? Welcome back.

I wonder what you spotted.

Did you notice this? The trapeziums don't have an equal width in Alex's model, so we don't have a common factor to simplify our calculation.

This is why it's really important that we use strips of equal width, trapeziums of equal height.

True or false? If we want to efficiently estimate, we need strips, trapeziums of equal width.

Is that true? Is it false? I'd like you to justify your answer with one of the two statements at the bottom of the screen.

Pause, answer this one now.

Welcome back.

Well done, you were paying attention.

You said true, and you justified that by saying strips of equal width give us a common factor which enables us to operate efficiently.

Well done.

Practise time now.

Question one, I'd like you to use four strips of equal width, trapezia of equal height, to estimate the distance travelled in this speed-time graph from zero to eight seconds.

Pause, and do this now.

Question two, multiple parts of this question.

For part A you're going to use two strips of equal width to estimate the distance travelled in this speed-time graph from zero seconds to 16 seconds.

What I'd like you to do before calculating is factorise, simplify your calculation before you type it in.

For part B, you're going to estimate the distance using four strips of equal width.

Again, I'd like you to factorise your calculation before you type it into your calculator.

For part C, you're then gonna comment on the difference between the two estimates that you got for part A and for part B.

You'll want to write at least one sentence for that part.

Pause and do this now.

Welcome back, feedback time, see how we did.

Question one we were using four strips of equal width to estimate the distance travelled on this speed-time graph from zero to eight seconds.

The four strips should have looked like so, equal width of course.

The area of those trapeziums looks like this, but we don't type that into our calculator.

We simplified down like so, and we get 30.

4 metres as our estimate for the distance travelled.

For question two, there were three parts to this, for part A I asked you to use two strips of equal width to estimate the distance travelled.

Your two strips would've look like so, your calculation like so, we'll simplify that of course, and we'll get 143.

6 metres as our estimate.

That's the estimate for part A, two strips.

Part B, I asked you to use four strips.

Your four strips should have looked like so.

Your calculation like so, obviously we're not typing that into our calculator, factorising and simplifying to so, typing that in, gives us 135 metres.

Part C, I asked you to comment on the difference between the two estimates.

You might have said the estimate using four strips is less of an overestimate.

It is more accurate.

The more strips, the more trapezia we use, the better our estimate.

Onto the second part of our learning now, where we're generalising a rule.

In mathematics when calculations are as lengthy as the ones we've seen in this lesson, we seek to generalise a rule which works for all cases to help improve our efficiency and accuracy.

It's reassuring to know that we don't have to type that calculation at the bottom into our calculators.

Generalisation is going to make this far simpler.

We know the formula for the area of one trapezium.

We need to define these variables in the context of a Cartesian coordinate grid.

We can define them by labelling the trapezium like so.

In this context we know the value of A.

The value is the value of the Y coordinate at this point.

We also know the value of B.

It's the value of the Y coordinate at this point.

Will they be the same Y values? Well done, not necessarily.

So we label them as different Y values.

We label them as Y0 and Y1.

So we can now define the area of this trapezium as a half of bracket Y0 plus Y1 multiplied by H.

That's the area of our first trapezium, but we're gonna use multiple trapezia, so we need to find the area of multiple trapeziums, what's the area of this second trapezium? We need strips of equal width so we have the same H in our formula.

And the Y value might be different again, so we label it differently.

We label it as Y2.

So the area of this trapezium is going to be a half of bracket Y1 plus Y2 multiplied by H.

So we have a formula for the area of the first two trapeziums. What do you think we'll use for the area of the third trapezium? Pause, see if you can write that formula.

See you in a moment.

Welcome back.

Hopefully you identified the next Y coordinate as Y3 and of course you used H in your formula because we've got strips of equal width.

Your formula you should have read, area equals a half bracket Y2 plus Y3 multiplied by H.

Well done.

What we now need to do is sum the area of these three trapezia.

When summing the areas we can simplify by factorising.

Can you spot the common factors? Well done.

And again, let's take out the common factors of half and H.

We'll end up writing this.

Can you see anything else now? Well spotted.

Two lots of Y1 and two lots of Y2.

How might we write that differently? Why don't we write two lots of Y1 and two lots of Y2? What do you notice now? Two.

Let's factorise that out.

We're now at the point where we've got the simplest form for the total area of the first three trapezia.

Quick check you've got this.

The area of the first four trapezia would look like so.

What I'd like you to do is write an expression for the sum of their areas, and then simplify it just like we did for the three trapezia, I'd like a simplified form for the four trapezia please.

See you in a few moments.

Welcome back.

Let's see how we did.

The area of the first four trapezia, we'll start by writing that, and then we'll identify a half and H are the common factors we can take out, and we wouldn't write all of that out because we've got two lots of Y1 and two lots of Y2 and two lots of Y3, so we'd write it like so.

And we can also factorise out that two, and there we go.

It's the simplest form for the total area of the first four trapezia.

We can write a general rule that works for any number of strips, any number of trapezia.

Half H will always factorise out.

If we tried this activity for five trapezia, six trapezia, seven trapezia, a half H will always be common factors.

Y0 is always going to be the first Y value, and in the context of four trapezia, Y4 was the last Y value.

And then we ended up with two lots of the sum of the Y values in between our first and last Y value.

So what would that look like for any number of trapezia? Not three, not four, not five, any number? Well we're still labelling the first Y values as Y0, Y1, Y2.

How do you think we're gonna label the last Y value? Well done.

YN.

N will be the number of strips we use, the number of trapezia.

If YN is the last value, what's coming before that? Well done.

YN minus one.

That'll be the penultimate Y value.

So we'll use this formula.

The first Y value is still Y0, but the last Y value is now YN.

And then we've got two lots of the sum of all the Y values in between.

We need to define now how we're gonna find H strips of equal width.

What's the value of H gonna be? Well N is the number of trapezia that we're using, and the distance on the horizontal axis goes from point A to point B.

So how do we find H? Well, we'd have to subtract A from B and then divide by N, the number of trapezia.

If we do that, we'll find the value of H.

This is the trapezium rule.

It's a generalisation that's going to work for any number of trapezia.

It's really important, so you are gonna wanna pause and write this down.

We can use this general rule to estimate the area under this curve between T equals zero and T equals 16.

We use four strips of equal width or four trapezium of equal height.

What would H the height be, in the context of these four strips? Well we do 16, our last T value, minus zero, our first T value, and then divide that by four.

We find H equals four.

These strips have a width of four.

Let's plug all this into trapezium rule and it looks like so.

Half multiplied by H is half multiplied by four in this case and we multiply that by 16, our first Y value, plus 6.

4, our last Y value, plus two lots of the sum of the Y values in between.

When I type that into my calculator, I get 115.

2.

So we say the area or the distance travelled is approximately 115.

2 metres.

How much quicker and simpler was that? Quick check you can repeat that skill.

I'd like you to use this general rule, the trapezium rule, to estimate the area under this curve between T equals zero and T equals 20.

I'd like you to use five strips of equal width or five trapezium of equal height.

Pause, give this a go now.

Welcome back.

There's our five strips of equal width, five trapezia of equal height.

What is that height that we're going to plug into a trapezium rule? It'll be 20 minus zero divided by five, four, the trapeziums have a height of four, or the strips have a width of four.

Let's substitute that in, and it looks like so, half multiplied by four multiplied by bracket 16, that's the first Y value, plus 12, that's the last Y value, plus two lots of the sum of all the Y values in between.

Type that into your calculator and you should have found 152 metres as your estimate for the distance travelled.

Well done.

We can use this rule to estimate the area of any time period.

The two examples we just saw, we started at T equals zero.

We won't always want to know the distance travelled from T equals zero.

So in this example, let's find out the distance travelled between T equals eight and T equals 20.

The model will look like that.

The strips have a height of four.

20 minus eight divided by three, strips have got a height of four, a width of four.

Substitute that into our trapezium rule and we get a half multiplied by four multiplied by 4.

8 is now our first Y value, 12, our last Y value, two lots of the Y values in between, and we find an estimate of 75.

2 metres for the distance travelled between T equals eight and T equals 20.

Your turn.

I'd like you to use the rule to estimate the area between T equals four and T equals 20.

I'd like you to use four strips of equal width, four trapeziums of equal height, pause and do that now.

Welcome back.

Let's see how we did.

The model would look like that.

H in this case is 20 minus four divided by four, that's four, plug that into our trapezium rule.

Half of four multiplied by 8.

8 was your first Y value, 12, your last Y value, two lots of the sum of the Y values in between, giving you 102.

4 metres as your estimate for the distance travelled.

Practise time now.

For question one, I'd like you to use the trapezium rule with four strips to estimate the distance travelled between T equals zero and T equals 12.

For part B, I'd like you to use the trapezium rule with three strips to estimate the distance travelled between T equals six and T equals 15.

For part C, you're gonna use the trapezium rule with six strips to estimate the distance travelled between T equals zero and T equals 18.

All of the values you need are in the table at the bottom of the screen.

Pause and do these now.

For question two, a particle's movement in a six second period can be modelled by the equation S equals 0.

9 plus 0.

3 T squared.

For part A, I'd like to use a trapezium rule with three strips to approximate the distance travelled over the whole six seconds.

For part B, I'd like you to use a trapezium rule with five strips to estimate the distance travelled between T equals two and T equals six.

Your obvious question is where is our table of values? It's not there I'm afraid.

My hint is that you're going to use the equation S equals 0.

9 plus 0.

3 T squared to find S values to one decimal place to substitute in.

You've got to do that bit of work yourself this time, but you've got this, don't worry.

Pause, give this problem a go.

I'll see you in a moment for the solutions.

Feedback time.

Question one, part A, trapezium rule with four strips to estimate the distance travelled between T equals zero and T equals 12.

That's those values within the table there.

Substituting it the trapezium rule should look like half multiplied by three, multiplied by two being your first Y value, 9.

6 being your last Y value, and two lots of the sum of the Y values in between.

You should have found an estimated distance of 81.

6 metres.

For part B, trapezium rule with three strips to estimate the distance travelled between T equals six and T equals 15.

That's those values from the table.

Your H on this occasion, well you should have done 15 minus six divided by three to get a H value of three, we're plugging in 7.

4 as our first Y value, 10.

1 is our last Y value, two lots of the Y values in between, and you should have got 81.

15.

So an estimated distance travelled of 81.

15 metres.

For part C, we're using the trapezium rule with six strips to estimate the distance travelled between T equals zero and T equals 18.

This is the joy of the trapezium rule, it makes this one a lot quicker.

It's all the values in that table that we're using now.

Can you imagine having to write out the sum of the area of those six trapezia? No thank you, I'll use a trapezium rule, and you should get 141.

9.

Therefore an estimated distance of 141.

9 metres.

Question two, part A, trapezium rule with three strips to estimate the distance travelled over the whole six seconds on the graph that would look like so.

Three strips of equal width from T equals zero to T equals six, H on this occasion is therefore two.

We need to populate a table of values, we wanna find those S values to one decimal place before substituting into the trapezium rule.

We need to read from T equals zero to T equals two, T equals four, T equals six, and you substitute in those T values into the equation, you get these S values.

Plug those into our trapezium rule and it looks like so.

We've got an estimated distance travelled of 28.

2 metres from T equals zero to T equals six seconds.

Question two, part B, five strips to estimate the distance travelled between T equals two and T equals six.

Slightly trickier this but nothing to be afraid of.

T equals two to T equals six, five strips, we're no longer dealing in integer values of T.

Not a problem though, H is gonna be 0.

8.

That's six minus two divided by five.

So we need to find the S values when T equals two, 2.

8, 3.

6, 4.

4, et cetera.

Plug those T values in to find those S values, they're accurate to one decimal place.

Plug all that into our trapezium rule, and we get an estimated distance travelled of 24.

6 metres.

Sadly, we're at the end of the lesson now, but we've learned that the distance travelled for non-linear speed-time graphs can be estimated using trapeziums. The more trapezia used, the better the estimation.

Using trapezia of equal width means the sum can be written more simply.

The trapezium rule can be used to efficiently carry out estimations of the distance travelled in non-linear speed-time graphs.

I enjoyed this lesson, I hope you did too, and I look forward to seeing you again soon for more mathematics.

Goodbye for now.