Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Today, we're estimating the grain of a curve.

This is a pretty cool piece of maths.

So what are we waiting for? Let's get learning.

Our learning outcome is that we'd be able to estimate the gradient of a curved part of the graph by considering the gradient of a straight line connecting two points.

Lots of keywords in today's lesson.

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.

Gradient, rate of change, positive x-direction.

Look out for those throughout our learning today.

Two halves to our learning.

We're gonna begin with estimating the gradient.

For a linear graph, we can find the exact gradient using a triangle to identify the rate of change in the y-direction with respect to the positive x-direction.

For this linear graph, we can identify two coordinate pairs.

Draw a triangle like so.

Look at the rate of change between the x-direction at the y-direction, and we can declare that the gradient is 3.

What does that mean? That means for every change of positive 1 in x, there is a change of positive 3 in y.

On such a linear graph, it does not matter which two coordinate pairs we choose to calculate our gradient.

If I take two coordinate pairs from the same line that are a little more distant, a bit further apart, I'll have a change in x of positive 2, a change in y of positive 6.

Same gradient? Different gradient? Of course, it's the same gradient.

If we put the change in x and the change in y to a ratio table, 2 to 6 cancels down to 1 to 3.

The gradient is still 3.

We can still say for every change of positive 1 in x, there is still a change of positive 3 in y.

This is because a linear graph represents a constant rate of change.

Well, here's a graph that doesn't have a constant rate of change.

This is a non-linear graph, the graph of y equals 1/2x squared.

For a non-linear graph, we can choose two coordinate pairs to help us estimate the gradient in a given range.

I pick those two coordinate pairs.

I draw a triangle awfully similar to the one you saw draw on the last graph.

Look at the change in x, the change in y.

Pop it into a ratio table, and it cancels down to 1 to 3.

I told you it was similar to the other triangle.

The hypotenuse of our triangle does not fit our curve exactly, but it gives us a very reasonable estimate for the gradient of the curve between x equals 2 and x equals 4.

In fact, we're able to say, for the graph of y equals 1/2x squared, the gradient of the curve between x equals 2 and x equals 4 is approximately 3.

It's not exact.

It's an approximation.

It's an estimate, but it's a pretty good one.

Quick check you've got that and you can repeat that skill.

I'd like you to use this graph to estimate the gradient of the curve.

y equals 1/2 to the power of x between x equals negative 3 and x equals negative 1.

You want to draw a triangle and let's see if you can estimate the gradient.

Pause.

Give this a go now.

Welcome back.

Hopefully you pointed out those two coordinate pairs and then drew such a triangle.

A change in x is positive 2.

A change in y is negative 6.

Pop that to a ratio table.

2 to negative 6 cancels down to 1 to negative 3.

So we can say, for the graph of y equals 1/2 to the power of x, the gradient of the curve between x equals negative 3 and x equals negative 1 is approximately negative 3.

Please tell me you wrote down approximately negative 3.

We're just approximating and estimating.

Not exact, but it's a very, very good estimate.

Another quick check.

Sam is estimating the gradient of the graph of y equals x cubed minus 4x squared plus 3x between x equals 1 and x equals 2 and draws this.

By this, I mean that triangle on the curve.

Sam uses the triangle to declare the gradient is 2.

Write a sentence or two of feedback to explain to Sam what they have done wrong.

Pause.

Give this a go now.

Welcome back.

I hope you spotted it.

Look at the triangle Sam drew and look what should have been drawn.

What's the difference? Well done, a positive x-direction.

So you might have written, when finding the gradient, always consider the positive x-direction first.

Also, check your answer by sight.

You can see that this gradient is negative.

When Sam declared the answer to be positive 2, Sam should have been thinking, hmm, that's a negative gradient.

Positive 2 can't be right.

Oh, you are welcome Sam.

Practise time now.

For question one, I'd like to use this graph to estimate the gradient of the parabola y equals 5 minus x squared between the given ranges.

For a, I'd like an estimate to the gradient between x equals negative 3 and x equals negative 1.

For part b, an estimate to the gradient between x equals negative 1 and x equals 0.

For part c, an estimate of the gradient between x equals 0 and x equals 2.

Pause and give this a go.

Question two.

This distance time graph shows the metres travelled by a car in the first five seconds of its journey.

For part a, I would like to estimate the gradient and therefore the car's mean average speed between s equals 0 and s equals 2.

For part b, I'd like you to estimate the gradients and therefore the cars mean average speed between s equals 3 and s equals 5.

Pause.

Give this one a go now.

Feedback time.

Let's see how we got on with question one.

For part a, I asked you to estimate the gradient of this parabola between x equals negative 3 and x equals negative 1.

Hopefully you identified those two coordinate pairs, drew a triangle like so for change in x of positive 2, a change in y of positive 8.

2 to 8, what does that cancel to? 1 to 4.

I hope you said it's an estimated gradient of 4.

Please tell me, use the word estimated or approximate gradient.

It's not exact, just an estimation, an estimated gradient of 4.

How about part b? Between x equals negative 1 and x equals 0.

We should have identified those two coordinate pairs, drawn a triangle like so, and estimated the gradient to be positive 1.

How about part c? Estimating the gradient between x equals 0 and x equals 2.

Hopefully you have those two coordinate pairs, a triangle like so.

Your triangle must move in the positive x-direction.

First, I've got a change of positive 2 in the x-direction, a change of negative 4 in the y-direction.

Now positive 2 to negative 4, cancel down to 1 to negative 2.

We've got an estimated gradient of negative 2 between x equals 0 and x equals 2.

Well done.

For question two, we were looking at the distance time graph that shows the metres travelled by a car in the first 5 seconds of its journey.

Asked you to estimate the gradient and therefore the speed in the first two seconds of its journey between S equals 0 and S equals 2.

You should have identified those two coordinate pairs, drawn a triangle like so.

3 over 2 is 1.

5 is a decimal, so we can say it's an estimated gradient of 1.

5, and therefore approximate the speed to 1.

5 metres per second.

For part b, I ask you to estimate the gradient between S equals 3 and S equals 5.

The triangle would've looked like so.

2 to 24 would cancel down to 1 to 12, so you can say it's an estimated gradient of 12, so we can estimate that speed to be 12 metres per second between S equals 3 and S equals 5.

Onto the second half of our learning now, problems with estimation.

Sometimes an estimation of the gradient does not give us enough information about the graph.

Consider this example.

A curve passes through the points negative 2,1 and 2,1 and has an estimated gradient of 0 between the two points.

Sketch the curve.

There's the two coordinate pairs.

There's an estimated gradient of 0 between them.

And I said sketch the curve.

What would you draw? Pause.

Have a think about that now.

Welcome back.

Exactly.

We haven't got anywhere near enough information to be able to sketch this curve.

You're telling me it's a curve, but it could be any curve between those two points.

It might be the parabola y equals 5 minus x squared.

It could be that.

We're still an estimated gradient of 0 between those two coordinate pairs.

It could even be a cubic curve such as y equals x cubed minus 4x plus 1.

We've still got the same estimated gradient of 0 between those two points.

So what are we to do about this problem? There's many curves that could be drawn between these two points and therefore automatically have an estimated gradient of 0.

The problem with this estimate is the distance between the points.

If the problem with the estimate is the distance between the two points, how do you think we're going to resolve that problem? Well done.

We're going to reduce the distance between the two points.

Reducing the distance between the x values allows a more accurate sketch to be drawn.

I've now got x values at negative 2, negative 1, 0, 1, and 2.

If we'd estimated the gradients between those moments, we would've got positive 3, negative 3, negative 3, and positive 3.

What's the benefit of this? If we'd had this level of information initially, we would've been better informed to make assumptions about the graph.

If I told you those gradients between those intervals, you'd have been far more likely to sketch something like our actual graph.

Quick check you've got that.

I'd like you to explain why you wouldn't use the points x equals negative 4 and x equals 6 to estimate the gradient of this curve and suggest a better choice of x values to use to describe the nature of this curve.

Pause and try this now.

Welcome back.

Hopefully you said something, along the lines of the points are too far apart.

Using those x values gives us an estimated gradient of 0 and neglects to tell us the moments of positive gradient and negative gradient, therefore we can't identify the turning points.

You might also have said, it would be better to reduce the distance between the points and estimate the gradients between x equals negative 4 and x equal negative 2, negative 2 and 0, 0 and 2, 2 and 4, 4 and 6.

We had the estimated gradients of negative 4, 4, 3, negative 1, negative 2, you've got a lot more information about this graph.

You'd be able to point out those turning points to me.

A different example shows us another benefit to using points which are closer together.

This distance time graph shows a particle's movement in the first few seconds of motion.

We could estimate the gradient like this.

It's two coordinate pairs, triangle, change of positive 3 in t, change of positive 7 in d.

Pop it into a ratio table, we found a gradient of 7 over 3.

Good estimate? What do you think? There's a problem here.

In this example, there's a large gap between the hypotenuse of our triangle and the curve that makes this, 7 over 3, a weak estimate.

If we reduce the distance between the points, we reduce the gap between the hypotenuse and the curve, which makes these better estimates for the particle speed as it moves.

If I have a gradient of positive 1 changing to a gradient of positive 2, changing to a gradient of positive 4, these estimates are actually really close to the actual speed of that particle in these given intervals.

Quick check you've got that.

True or false? These two points will give us an accurate estimates of the gradient or speed in this distance time graph.

Is that true? Is it false? Once you've decided, use one of the two statements at the bottom of the screen to justify your answer.

Pause and do this now.

Welcome back.

I do hope you said false and I hope you justified that with statement b.

The gap between the hypotenuse and the curve is large making this a weak estimate.

Practise time now.

Question one.

This is the quadratic graph, y equals x squared minus 2x minus 3.

For part a, I'd like to estimate the gradient between the coordinate pairs negative 1.

0 and 3,0 and explain why this is not a good estimate.

For part b, I'd like you to estimate the gradient between the coordinate pairs, 1, negative 4 and 4,5 and explain why this is also not a good estimate.

Part c, you'd be glad to hear we're gonna do some good estimating in this part.

I'd like you to estimate the gradients between the coordinate pairs 1, negative 4, 2, negative 3, 3,0, and 4,5.

You'll have three estimates for the gradient between those pairs and explain why these are better estimates.

Pause and do these now.

Welcome back.

Feedback time.

Let's find out how we got on question one, part a.

Estimate the gradient between negative 1,0 and 3,0 and explain why it's not a good estimate.

There's the two coordinate pairs.

We get an estimated gradient of 0.

Why is that bad? A gradient of 0 between these two points does not reflect the fact we have a negative gradient switching to a positive gradient and therefore a turning point between these points.

Your explanation should have read a little like that.

Part b, estimating the gradient between 1, negative 4, and 4,5 would've looked like so.

We'd get an estimated gradient of 3.

What's wrong with this estimated gradient? I hope you said something along the lines of, there's a large gap between the hypotenuse and the curve, which makes this a weak estimate.

How about part c? Asked to estimate the gradients between these four coordinate pairs.

When you do so, you draw something like that on the curve and get gradients, or sorry, estimated gradients of positive 1, positive 3, and positive 5 respectively.

Asked you to explain why these are better estimates.

You should have said something along the lines of, there is very little gap between the hypotenuses of these triangles and the curve, which makes these more accurate estimates.

We're at the end of the lesson now.

Well done for all your efforts today.

What have we learned? We've learned that the gradient of a curved part of the graph can be estimated by considering the gradient of a straight line connecting two points.

Crucially, we learned that using points that are closer together gives a more accurate estimate of the gradient of the curve between those two points, and you know we mathematicians love our accuracy.

Well, I hope you've enjoyed today's lesson as much as I have and I look forward to seeing you again soon for more maths.

Goodbye for now.