Lesson video

Hello, Mr. Robson here.

Welcome to "Maths." Today, we're problem-solving with real-life graphs.

I love problem-solving, you love problem-solving, so let's get problem-solving.

Our learning outcome is we'll be able to use our knowledge of interpreting real-life graphs to solve problems. Some key words that we're gonna hear throughout this lesson.

Gradient.

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

It is calculated by finding the rate of change in the y-direction with respect to the positive x-direction.

Another is tangent.

A tangent of a circle is a line that intercepts the circle exactly once.

Two parts to our learning today and I'm going to begin by comparing speed-time graphs.

Two cars are on a journey that can be modelled on a distance-time graph.

We can interpret a lot of information from this graph.

For example, we could find out how fast car (a) is travelling.

We'll do that by finding the gradient.

Why? Because the gradient will give us the rate of change in distance with respect to the rate of change in time.

Distance over time, that's speed.

So, we find the gradient of line (a), 70/2, that's 35.

So in the context of the distance being in the units of miles, and time being in the units of hours, we'll call that 35 miles per hour.

We can do the same thing for car (b).

How fast is car (b) travelling? We'll find the gradient, that will be 100/2.

And if you think about a car doing 100 miles in 2 hours, that's a speed of 50 miles per hour.

There's other stuff we can find out too.

For example, what distance headstart did car (a) have over car (b)? Well that would be that moment there.

The difference in the y intercepts is 60, so that means car (a) had a 60 mile headstart over car (b).

Is there anything else we can read from this graph? Absolutely.

We can see when car (b) overtakes car (a).

We saw earlier that car (b) has a steeper gradient, it has a faster speed.

And we can see the moment on the graph that car (b) overtakes.

It's that intersection there, it's the coordinate (4, 200).

So we could say car (b) overtakes car (a) after 4 hours, but we always like a little more detail, we mathematicians, so we'd also say, "after car has travelled 200 miles." Jun and Sam are discussing a journey by two cars modelled in this speed-time graph.

When does car (b) overtake car (a)? That's the question that was posed to them.

Jun says, "That's easy! It's after 5 seconds." You can see where the two graphs intersect, the coordinate (5, 20).

That is indeed when T, or seconds, equals 5.

Do you agree or disagree with Jun? What features of the diagram support your decision? I'd like you to pause and consider this question now.

Welcome back.

Did you agree with Jun? Did you disagree? If you disagreed, why? I wonder what you said.

Sam intervenes and points out to Jun, "The intersection is the moment car (b) is travelling faster than car (a).

It's not the distance travelled." And Jun realises, "Of course! It's a speed-time graph.

Thanks Sam!" It's really important to pay attention that this is a speed-time graph.

That intersection, as Sam tells us, is just the moment that the two cars are doing the same speed.

Therefore any moment after that, car (b) will be travelling faster.

However, we don't know the distance travelled.

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

The area under (b) is clearly less than the area under (a) after 5 seconds.

So how do we find the moment in the future where the two areas i.

e.

, the distances travelled, are equal? You might want to pause and see if you can answer that question now.

Or you might want to continue with me and I'll talk you through it.

What we need is an expression for the area under each graph at any given point.

Start with the equation of the line.

So car (b), we'd call that line representing car (b), y = 4x.

If you think about linear graphs in terms of y = mx + c, this is a gradient of 4, with an intercept at the origin.

So what we need to do now is figure out the area underneath that graph, at any value of x.

Well the area underneath line (b) is a triangle, so we need half of the base times the height.

The base is always going to be x, the x coordinate.

The height is always gonna be the y coordinate.

But because it's the line y = 4x, the y coordinate is going to be 4x.

So the area of that triangle at any given point will be x multiplied by 4x multiplied by 1/2.

We could write it in that form but we always like to simplify things, so we would turn that into 4x squared, over 2, which simplifies to 2x squared.

The area under line (b) at any point will be 2x squared.

What we need now is an expression for the area under line (a) at any given point.

Let's start with the equation of the line, (a) is the line y = 2x + 10.

That's a gradient of 2, and a y intercept of (0, 10).

So what about the area underneath this graph, the area bound by the line on the horizontal axis? Well it's a trapezium.

You might want to rotate your head 90 degrees so you can see clearly that it's a trapezium.

How do we find the area of a trapezium? Well, we need to know the upper base, the lower base, and the height of the trapezium.

In this case, the upper base of our trapezium is always going to be 10 because of the y intercept.

The height will always be the x coordinate.

And the lower base will always be the y coordinate.

The y coordinate will of course be 2x + 10.

This is gonna enable us to write an expression for the area in terms of x.

The area of the trapezium will therefore be 1/2 (10+(2x+10)) multiplied by the height, which is x.

We'll simplify that, of course.

10 + 2x + 10, that becomes 2x + 20.

We can half that, so multiply the bracket by a half, we get x multiplied by x + 10.

When we expand that last bracket, we get x squared + 10x, a quadratic expression, for the area of our trapezium at any given x value.

So we've got an expression for the area under (b) at any point and we've got an expression for the area under (a) at any point.

What we do now is form this equation because it will tell us the exact moment that the two areas are equal.

That will be the point that the two distances travelled by (a) and (b) are equal.

So the equation is 2x squared + x squared + 10x.

Let's add negative x squared to both sides.

Let's put all the variable terms on one side.

Oh look, x squared - 10x = 0.

What can we do from here? Well done! We can factorise, x(x - 10) = 0.

So, if the expression is going to equal 0, either x is 0, or x - 10 is 0.

So we get these two solutions.

No surprise to see a quadratic equation with two solutions.

So x = 0, x = 10 are our two solutions.

The x = 0 solution tells us the areas are equal when x = 0.

Well of course because there's no distance travelled yet.

We're not interested in that solution.

We're interested in the other solution, when x = 10.

The areas are equal again when x = 10.

That is the moment when (b) overtakes.

Quick check you've got this.

Which of these equations gives you the area underneath line (b) at any given x value? Three to choose from.

Pause, and you figure out which one's right.

Welcome back.

Hopefully, you said it was option b, the Area = 4x squared at any point.

Why is that? Well the line is y = 8x.

The base of our triangle will always be x.

The height will always be the y coordinate which is of course 8x.

So we need a half of x multiplied by 8x.

That simplifies to 4x squared.

Next, which of these equations gives you the area under line (a) at any given point? This one's a little trickier because the area bound by the graph of (a) and the horizontal axis is a trapezium.

This one will take a bit more figuring but I'm sure you've got this.

Pause.

Take your pick from those three options.

Welcome back.

Hopefully you spotted it was a, the Area = 5/2 x squared + 12x.

Why is that? Well, line (a) is the line y = 5x + 12.

So, we have a trapezium with an upper base of 12, low base of 5x + 12, that's the y coordinate, a height of x.

Put that into the formula for area of a trapezium and simplify, and simplify further, and multiply out that bracket.

There we are, we get a 5/2 x squared + 12x.

That will give us the area under line (a) at any given point.

What I'd like you to do next is solve the equation to find the moment when the two distances travelled are equal.

I want to know the point when 4x squared = 5/2 x squared + 12x.

So you're solving that equation.

Pause and do this now.

Welcome back, let's see how we did.

There's lots of ways you can solve this equation.

For me, I think the most efficient way is to multiply every term by 2, so we've got no fractional coefficients.

From here, we can add -5x squared to both sides and -24x to both sides.

We'll have all the variable terms on one side, the right-hand side, being equal to 0.

Wonderful.

We can factorise again, 3x(x - 8) = 0.

Therefore, x = 0, or x = 8.

We're not interested in the x = 0 solution, we're interested in the x = 8 one.

That tells us the areas, therefore distances travelled, are equal when x = 8.

i.

e.

, when time = 8 seconds.

Practise time now.

Question 1, I'd like you to form and solve an equation to find the time when car (b) overtakes car (a).

Pause and do that now.

Question 2: These graphs model the movement of three particles.

Alex says, "Particle 'c' has travelled furthest after 8 seconds.

It's the highest line on the graph at that point." Part a, I'd like you to consider, is Alex correct? I'd like you to then write a sentence to justify your response, making reference to the graphs.

For part b, I'd like you to then calculate which particle has travelled furthest after 8 seconds.

You'll want to calculate the distance travelled by all three particles after 8 seconds.

Pause and do this now.

Question 3: In this race, car 'a' gets a 4 second head start.

I'd like you to form an equation to calculate at what time car 'b' catches car 'a'.

And give your answer to two decimal places.

Pause and do this now.

Feedback time, let's see how we got on.

Question 1, we were forming and solving an equation to find the time when car (b) overtakes car (a).

Line a, is the line y = 9x + 15.

Line b, is y = 12x.

The area bound by the graph of (a) and the horizontal axis at any point will be 9/2x squared + 15x.

That's the area of our trapezium.

The area underneath line (b) at any point will be 6x squared.

We formed that equation, we want to find the moment when the two areas are equal.

From there, I'd multiply 3 by 2 so we've got no fractional coefficients, rearrange, factorise, and solve.

Again, we're interested in the solution that's not x = 0.

We're interested in the x = 10 solution.

The distances are equal after 10 seconds.

This is when (b) overtakes.

For question 2, part a, I asked you to consider if Alex is correct, and then write a sentence to justify your response, making reference to the graphs.

An example of what you might have written is: Alex is wrong because the graphs show us that (c) is travelling at the fastest speed after 8 seconds.

However, we need the area under the graph, not the point, to tell us the distance travelled.

For part b, I asked you to find the distances travelled by each particle after 8 seconds.

We want the area under the graphs.

The area under (a) at that point is a trapezium.

We find the area using a 1/2(10 + 18) x 8 = 112, so the area in the context of the speed-time graph is 112 metres travelled.

B was also the area of trapezium, 1/2(5 + 12) x 8, giving us 104 metres travelled.

C was a triangle, base times height, multiplied by half, giving us 96 metres travelled for c.

So particle 'a' has travelled furthest after 8 seconds.

For question 3, we've got a race where car (a) is getting a 4 second head start and I asked you to form an equation to calculate at what time car (b) catches car (a).

Start with the equation of lines.

A was lovely, y = x.

B, a little trickier, but we still call that line y = 2x - 8.

It's a gradient of 2 and an interceptor of (0, -8).

Imagine if you could see that fourth quadrant, you'd see that y intercept at (0, -8).

So we now need an expression for the area underneath those graphs at any point.

The triangle underneath line (a) would be expressed by x squared over 2.

The triangle underneath line (b) is a little trickier because the height is gonna be the y coordinate, 2x - 8, but the base is not gonna be x.

Remember, (b) started 4 seconds later, so the base won't be x, it'll be x - 4.

So we get this expression for the area of the triangle under (b) at any point, (2x - 8)(x - 4) x 1/2.

Multiply out those brackets, multiply by a half, and you get expression x squared - 8x + 16.

We now need to find out when those two areas are equal.

We need to solve this equation.

In order to solve that, I'm going to multiply every term by 2 so I've got no fractional coefficients.

Rearrange from there.

Oh lovely, we've got a quadratic.

Lots of ways we could solve this.

I solved it by completing the square.

So x = 8 +/- the root 32.

We're gonna get two solutions.

We get x = 2.

34, to two decimal places; and x = 13.

66, to two decimal places.

X = 2.

34 is not a valid solution.

Remember, car (b) has not started until 4 seconds have passed.

So x = 2.

34 can't be a valid solution.

X = 13.

66, however, is a very valid solution.

It is the point at which car (b) catches car (a).

Onto the second half of our learning now.

We're gonna look at a very different type of problem-solving with real-life graphs.

We're going to look at circle theorems and coordinate geometry.

We can use coordinate geometry to demonstrate circle theorems such as, the tangents to a circle from an external point are equal in length.

The circle x squared + y squared = 5 has two radii, OA and OB.

You can see that in this diagram.

We can find the equations of the tangents to the circle at A and B.

If you wanted to find the tangent at A, a couple of useful things we can see from our diagram.

The point A is coordinate pair (1,2).

We can also see the gradient of the radius OA.

The gradient of that radius is 2.

Then the gradient of tangent must be the negative reciprocal of 2.

The gradient of the tangent must be -1/2.

So the tangent must be the line y = -1/2x + a constant.

We can find the constant by substituting in the one point that we know on this tangent, x = 1, y = 2.

Substitute that in to find the constant of 5/2.

The equation of the tangent at A is y = -1/2x + 5/2.

We can now set about finding the tangent at point B.

We know that point B is (1,-2).

And we can see the radius has a gradient of -2.

The negative reciprocal of -2 will be 1/2.

Y = 1/2x + c.

Let's find that constant.

We know one point on this tangent, x = one, y = -2.

Substitute in those values, we find the constant is -5/2.

So the equation of the tangent at B is y = 1/2x - 5/2.

Now that we know the equation of the lines of both tangents, we can find out where the two lines intersect, i.

e.

, when there's -1/2x + 5/2 = 1/2x - 5/2.

We form and solve that equation.

If I want to solve that equation, one thing I'm gonna do is multiply every turn by 2 so I don't have any fractional coefficients or fractional constants.

From there, we can rearrange and see that 2x = 10.

So we know that they intersect at x = 5.

But it's not enough just to say x = 5.

I need to know the y coordinate as well.

Coordinates come in pairs.

So I'm gonna substitute x = 5 to the equation of E, the tangent, and I'm gonna find the y coordinate to be 0.

The two tangents intersect at (5,0), like so.

We can now show that lengths AC and BC are equal.

How might we do that? Well consider the length AC, and think about that right triangle.

If we want to find AC, we could use Pythagoras theorem.

4 squared + 2 squared is 20.

Square root of 20 will tell us the length of that hypotenuse.

Oh look, AC has the length of root 20.

We're gonna leave it in that form for now and we're going to try and find the length BC.

We'll consider this right triangle.

And oh, look, the length of BC is again the square root of 4 squared + 2 squared, it's the root of 20.

Both of those lengths are the same, AC = BC.

We've used coordinate geometry to demonstrate that the tangents to the circle from an external point are equal in length.

Quick check you've got this.

Which of these lines is tangent to the circle at point A? Three to choose from.

Pause now, see if you can work out which one's right.

Welcome back.

Did you spot that it was c, y = 2/3x + 14/3? Why would it be that, the tangent to the circle at A? Well, we know A is the coordinate pair (2,6), and the centre of the circle is the coordinate pair (4,3).

The radius you can see has a gradient of -3/2, which means our tangent must have a gradient of 2/3.

There we go, option a was immediately ruled out.

We just need to find that constant, c, and we'll know whether it's option b or c.

So, we substitute in the one known point on that tangent, x = 2, y = 6, and we find the constant to be 14/3.

Therefore, it was option c.

Next check, which of these lines is tangent to the circle at B? Three to pick from.

Pause and do that now.

Welcome back.

Did we spot that it's option (a), y = 3/2x - 19/2? How do we know it's that one? Well we know the centre of the circle, we know B has the coordinates (7,1).

We can find out the gradient of the radius, that's -2/3.

The gradient of the tangent is gonna be the negative reciprocal of that.

What's the negative reciprocal of -2/3? Positive 3/2.

So if the tangent has got a gradient of positive 3/2, option b was immediately ruled out.

How do we find the exact equation of the line? Well we know one coordinate pair on this line, that's x = 7, y = 1.

Substitute that in, we find the constant -19/2, hence the line of the tangent is y = 3/2x - 19/2.

Next, we're gonna form and solve an equation to find the intersection of the tangents.

I want to know when 2/3x + 14/3 is equal to 3/2x - 19/2.

Pause.

See if you can do that now.

Welcome back.

Hopefully, you formed that equation.

There's all sorts of ways we can solve this equation.

If you know that 3 and 2 have a common multiple of 6, the useful thing to do is multiply through by 6.

Once we've done that, we've got no fractional coefficients, no fractional constants, we've got an easier equation to work with.

A little bit of rearrangement from there, and we find that 5x = 85, so x = 17.

And you know we're not done there.

We need to substitute 17 back into the equation of E, the tangent, to find the y coordinate.

We found the intersection, the coordinate pair (17,16).

When we put that on the diagram, it looks right.

They look like accurate tangents, absolutely.

Intersection point x = 17, y = 16.

Final step.

We're now gonna show that AC = BC.

Pause and see if you can demonstrate that fact now.

Welcome back.

I wonder what you wrote.

Hopefully, you spot it.

We can find the length AC by using that right triangle.

15 squared + 10 squared.

Take the root of that, it's root 325.

And we can just leave it in that form for now.

When we come to find the length of BC, the right triangle would be fairly similar.

It's the root of 10 squared + 15 squared, which is the root of 325.

Therefore, we found that AC = BC.

And we've shown that the tangents to a circle from an external point are equal in length.

Practise time now.

Question 1, the circle, (x - 2 squared) + (y + 4 squared) = 10, has radii OA, OB and OC.

You can see all that on the visual representation here.

For part (a), I'd like you to find the intersection of the tangents to the circle at A and B.

I'd like you to label that point as D.

From there, I'd like you to do part (b), which is to show that AD = BD; that those two lengths are equal.

For part (c), you're gonna find the intersection of the tangents to the circle at B and C.

Once you found that intersection, label it E.

And finally, part (d), show that the lengths BE and CE are equal.

Pause and give all this a go now.

Welcome back.

Feedback time, let's see how we did.

For part (a), we'll find the intersection of the tangents to the circle at A and B.

The tangent at A is the line y = 1/3x - 4/3.

The tangent at B, the line y = 3x - 20.

Form and solve an equation.

You'll find the intersect when x = 7.

Substitute x = 7 back into either line, you'll find that y = 1.

That's the intersection of the tangents to the circle at A and B.

Next, we were showing that AD = BD, that those two lengths are equal.

You can find the length AD by using Pythagoras theorem, the square root of 6 squared + 2 squared, that's the square root of 40.

Length BD, similar.

The square root of 2 squared + 6 squared gives us the square root of 40.

Therefore, we've shown that AD = BD.

Part (c), find the intersection of the tangents to the circle at B and C.

The tangent at B is the line y = 3x - 20.

The tangent at C is the line y = -1/3x - 20/3.

Where do those two lines intersect? You would form and solve an equation and find that x = 4.

Substitute that back into either line to find the y coordinate of -8, showing that BE = CE.

The length of BE is a square root of 3 squared + 1 squared.

That's a square root of 10.

And funnily enough, CE is also the square root of 10.

Therefore BE = CE.

We're at the end of the lesson now, sadly.

We have learned that our knowledge of interpreting real-life graphs can be useful when solving problems with speed-time graphs.

We've also learned that our knowledge of coordinate geometry can be used to demonstrate circle theorems. Hope you've enjoyed this lesson as much as I have and I look forward to seeing you again soon for more mathematics.

Goodbye for now.