Lesson video

Hello, Mr. Robson here.

Well done for joining me today.

What are we learning? We're improving our estimation of the gradient of a curve.

Well that sounds pretty exciting, so let's get stuck in.

Our learning outcome is that we'll be able to improve the estimate of the gradient by considering the gradient of the tangent at a fixed point.

You'll see what all that means by the end of this lesson.

Keywords today, tangent.

A tangent to a curve at a given point is a line that intersects the curve at that point, both the tangent and the curve have the same gradient at that given point.

Two parts to our learning today, we're gonna begin by improving the accuracy of our estimate.

For a non-linear graph, like Y equals X squared, we can choose two points to help us estimate the gradient between those two points.

If I choose these two points on the curve, I can work out that gradient.

I'll use a ratio table to do that.

I'll find the gradient is three, but it's an estimated gradient, so the exact gradient, it's an estimated gradient for Y equals X squared between X equals zero and X equals three, and that estimate is that the gradient is three.

The problem with this is there's a large gap between the hypotheses of the triangle and the curve.

That means that this estimate is not particularly accurate and you know we mathematicians, we love accuracy.

If we reduce the distance between the points on the curve, we reduce the gap between the hypotenuse and the curve, this improves the accuracy of our estimate.

We'll add a few more points on our curve.

Notice we've got multiple points now and they're quite close together.

When we add triangles to calculate the gradient, we find gradients of one, three and five respectively.

Our estimated gradients, Y equals X squared between zero and one is one, between one and two is three.

Between two and three is five.

Crucially, the proximity of our hypotenuse to the curve makes these estimates far closer to the true gradient of that curve at those given points.

Finding multiple estimates also reflects how the gradient of the curve is changing.

The gradient goes from one to three to five.

We can see that our gradient is very quickly increasing.

Quick check you've got this.

For the curve Y equals five minus two to the power of X, which you can see on that graph there.

An estimated gradient of negative seven over three lacks accuracy.

What I'd like you to do is estimate the gradients between the points given below to improve the accuracy of the estimate.

I'd like you to work out the gradient between X equals zero and X equals one, and another calculation for between X equals one and X equals two, and another calculation for between X equals two and X equals three.

Pause and do this now.

Welcome back, hopefully you did something like that with the graph and found estimated gradients of negative one, negative two and negative four.

They are far closer to the actual gradient in those moments, well done.

If we compare the two images when we found the gradient of negative 7/3 versus finding multiple estimates of negative one, negative two, negative four, you can see the proximity, the hypotenuse of each of those triangles on the right hand side to the curve makes these improved estimates for the gradient.

When estimating the gradient for more complex curves, the importance of reducing the distance between the points is greater still.

An estimated gradient for the graph Y equals negative X cubed plus 6X squared minus 8X between X equals zero and X equals four is zero.

You can see on the graph there, the gradient between those two points, if we estimate it, it's zero.

Why is this not a good estimate? Pause, tell the person next to you or say your answer aloud to me at the screen.

See you in a moment.

Welcome back, I wonder what you said.

Hopefully you said something along the lines of an estimated gradient of zero for this period does not reflect the changing gradients and turning points of the curve.

That would be absolutely right.

So what might we do about this problem? Reducing the distance between the points when estimating the gradient of this curve better reflects its nature.

Rather than leaving that distance all the way from X equals zero to X equals four, let's add these points, there's a far lesser distance between them.

If I work out multiple gradients between these closer points, I can see that the curve goes from having a gradient of approximately negative three to approximately three, approximately three and approximately negative three.

Armed with that information, I can tell you a lot more about the positive and negative gradient of this curve and any turning points between X equals zero and X equals four.

Let's check you've got that.

Write a sentence to explain why using an estimated gradient of zero between X equals negative four and x equals negative one is not helpful for this curve.

Pause and write me a sentence now.

Welcome back, I wonder what you wrote.

You might have said, the points are too far apart.

An estimated gradient of zero neglects to tell us the moments of positive gradient and negative gradient, so we can't identify if there are any turning points.

Practise time now, for question one, this distance time graph shows the movement of an animal.

For part A, I'd like you to use an estimate of the gradient to estimate the animal's speed between T equals zero and T equals three.

For part B, you're gonna estimate the speed using shorter intervals of T equals zero and T equals one, T equals one and T equals two, T equals two and T equals three.

You're gonna work out three different gradients there.

For part C, you're going to write a sentence or two to explain why your estimates in B are better than your estimate in A, pause and do all of that now.

Welcome back, feedback time.

See how we did, for one A, I asked you to use an estimate the gradient to estimate the animal speed between T equals zero and T equals three.

We should have done something like that the graph, put into a ratio table, three to seven cancels down to one to 7/3 or 7/3, so we can say there's an estimated speed of 7/3 metres per second, between T equals zero and T equals three.

You might have written 2.

3 to one decimal place.

That's perfectly fine.

For part B, I ask you to estimate the speed using shorter intervals.

These three short intervals, you would've found estimated speeds of four metres per second, two metres per second, one metre per second.

Then part C, I ask you to explain why those are better estimates than your one in A.

You might have said, these are more accurate estimates because the hypotenuse are closer to the curve.

If you wrote that, you'd be absolutely right.

You might also have said they reflect the fact that the animal is decelerating.

That would've been a nice comment also.

Onto the second half of our learning now, using a tangent to estimate, that sounds pretty interesting.

Let's take a closer look.

The gradient of a curve is constantly changing.

Would it make more sense to calculate the gradient at a given point on the curve instead of taking the gradient from two points? The answer of course is emphatic yes.

For example, we could try and read the gradient of Y equals X squared when X equals one.

We could take another point on the curve, draw a line through the two points and find the gradient.

If I find the gradient between these two points, I could use that line.

In this example, we have a distance between the X coordinates of one and the line has a gradient of one, but you know that's going to be quite a poor estimate for that gradient.

There's a large distance between that line and our curve, but what if we change it to this? Take another point on the curve, draw a line through those two points and find this gradient.

In this example, we have a distance between the X coordinates of just 0.

5 and the line has a gradient of 1.

5.

Can you see how that's a more accurate estimate than the previous one I showed you? What if we keep going with this idea? How about those two points even closer together? My second point's now even closer to X equals one.

In this example we have a distance between the X coordinates of just 0.

25 and the line has a gradient of 1.

75.

Can you see how it's really starting to closely reflect what the gradient must actually be at one? There's those three examples on the same screen.

As the distance between the X coordinates reduces, the intersections get closer together and the gradient of the line gives us a more accurate reflection of the gradient of the curve at X equals one.

If we make the distance between the X coordinates zero, the line will intersect the curve only once.

This line will be a tangent to the curve and both the line and the curve will have the same gradient at that moment, what might that look like? It look like this.

This tangent to the curve at Y equals X squared at X equals one tells us that the gradient at that point is two.

We can draw a tangent to the curve at any point to find the gradient.

It's the same curve, Y equals X squared, but now I'm gonna look at the gradient when X equals two.

I don't want a pair of points.

I need just one point and a tangent at that point.

A tangent to one equals X squared at X equals two shows us that the gradient at this moment is four.

Quick check you can do that.

This is the graph of Y equals X cubed minus X squared minus X.

What I'd like you to do is find the gradient of the tangent to the curve at X equals negative one and therefore find the gradient of the curve at this point, I've drawn the tangent for you find the gradient, therefore find the gradient of the curve at that point.

Pause and do this now.

Welcome back.

Hopefully you said the gradient at X equals negative one is positive four.

Same curve, different question.

I'd like you to find the gradient of the tangent to the curve at X equals zero.

Therefore, find the gradient of the curve at this point.

Once again, I've drawn that tangent for you.

Read the gradient and you'll find the gradient of both the tangent and the curve when X equals zero.

Pause and do this now.

Welcome back, I hope you said negative one.

The gradient of this curve at X equals zero is negative one.

One more check, how about the gradient of the tangent to the curve at X equals one? What's the gradient of that tangent and therefore what's the gradient of the curve at that point? Pause, have a think about that one.

Welcome back.

Well done, it is indeed a gradient of zero, so X equals one, the gradient zero, oh look, it's a turning point.

Of course the gradient is zero.

All of the tangents that I've shown you so far have been entirely accurate.

I driven with graphic technology so they were perfect tangents to the curves at those moments.

However, in many cases we'll be drawing a tangent by hand to a curve and in these moments, a gradient will only be an estimate.

We can do some very good estimates, but we have to be aware when drawing a tangent by hand we're only estimating the gradient.

Let's look at this example.

We can estimate the speed of the particle modelled in this distance time graph when T equals three.

That's the moment when T equals three.

I drew that tangent by hand, not with graphing technology.

Can you spot the gradient of that tangent? I've got a change of positive three in T and a change of positive four in D.

When I do the vertical change divided by the horizontal change, I get four divided by three.

My calculator tells me that's 1.

3 to one decimal place.

That's my estimated gradient.

I'll use that gradient to estimate the speed.

The speed of this particle when T equals three, I estimate to be 1.

3 metres per second to one decimal place.

Quick check you've got that.

I'd like you to use this hand drawn tangent to estimate the speed of the particle modelled in this time distance graph when T equals 6.

5.

Pause, have a go at this now.

Welcome back, hopefully you picked coordinate pairs like so where we can see a change in T of positive four and a change in T of positive six.

Six divided by four gives an estimated gradient of 1.

5 so we can estimate the speed to be 1.

5 metres per second to one decimal place.

Practise time now, for question one, I'd like you to use these tangents to the curve Y equals six minus X squared to find the gradient at these points.

For A, I'd like to know the gradient at X equals negative one and for B, I'd like to know the gradient at X equals two.

Once again, I used graphing technology to draw these tangents.

They are perfect, you won't be estimating here.

These will be accurate.

Pause, find those gradients now.

Question two, I'd like you to use these tangents to the curve Y equals X cubed minus X squared minus 2X to find the gradient at these points, that's for A, X equals negative one and for B when X equals two, pause and find those gradients now.

Question three, this distance time graph model is the acceleration of a particle, units are in metres and seconds.

I'd like you to draw a tangent to the curve by hand, find the gradient and thus estimate the speed when T equals 1.

5.

I'd like you to give your answer to one decimal place, just an estimate this time.

Be as accurate as you possibly can with that tangent.

Wonder if you'll get the same as me.

Let's see, pause, give it a go.

Feedback time now, let's see how we did.

Question one I asked you to use these tangents to the curve Y equals six minus X squared to find the gradient at these two points, at X equals negative one we should have found the gradient to be positive two.

At x equals two, we should have found the gradient to be negative four.

Question two, this curve was Y equals X cubed minus X squared minus 2X and asked you to find the gradient at these two points.

X equals negative one and X equals two.

At X equals negative one, the gradient of a tangent at that point is two.

Therefore the gradient on the curve at that point is two.

At X equals two, we've got a gradient of positive six.

Question three, we were drawing a tangent to the curve when T equals 1.

5 to estimate the gradient, therefore estimate the speed of this accelerating particle.

That's what my tangent looks like.

Hand drawn of course, so yours shouldn't be exactly the same as mine, but your result to one decimal place should be pretty similar.

When I drew my tangent, I found two coordinate pairs with a change in T of positive 0.

3 between them and a change in D of positive 0.

8 between them.

Sticking that into my calculator, 0.

8 divided by 0.

3 gave me 2.

7 to one decimal place.

That's my estimated gradient at that time.

Therefore, that's my estimated speed at that time, 2.

7 metres per second to one decimal place.

I did go on and check that my tangent was perfect with my graphing technology and it was, so 2.

7 to one decimal place is accurate.

Sadly, at the end of the lesson now, shame, I was enjoying that, but what have we learned? We've learned that the gradient between two points on a curve can be used to estimate the gradient of the curve.

The closer the points are, the better the estimate.

The gradient of a curve is constantly changing and at any given point on the curve, a tangent can be drawn.

The tangent to the curve has the same gradient as the curve itself at that point.

Tangents can be used to estimate the gradient at any given point on a curve.

Hope you enjoyed this lesson as much as I did and I look forward to seeing you again soon for more maths.

Goodbye for now.