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Video transcriptThis unit of work is Differentiation.
Differentiation can be a challenging topic to teach because it requires learners to have a conceptual understanding of the association between graphs, functions and equation solving.
Learners will often learn rules for finding a derivative without really understanding how they work and are then unable to apply these rules independently and in more complex, less familiar situations.
We start the unit by reviewing graphs that are essential prior knowledge.
Differentiation is introduced and learners will explore the language and notation of the topic.
By the end of the unit learners will locate the coordinates of turning points and be able to classify them as maximum points or minimum points.
This unit of work is just one of several approaches that you could take when teaching this topic and you should aim to adapt the resources to match ability level of your learners as well as your school context.
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Lesson 1 video
Video transcript
In this video we will give you an overview of what calculus is. We will then identify and recap some of the areas of mathematics that you will need to be familiar with before we start studying this topic.
Calculus is a branch of mathematics which can be divided into two parts – integral calculus and differential calculus.
Integral calculus (or integration) can be used to find the area under curves and the volumes of solids.
Integration has developed over a very long time. In fact, the beginnings of integration can be found in the work of the ancient Greeks including Archimedes. Archimedes was one of the early contributors when he developed a method to find an approximation for the areas bounded by curves. He went on to extend this to finding the volumes of certain solids.
Differential calculus (or differentiation) can be used to find rates of change. When we illustrate a function on a graph the rate of change is the gradient, so when we differentiate we can find the gradient at specific points on a graph. One of the most practical uses of differentiation is in finding the maximum or minimum value for a real world function.
The development of differentiation has taken place over hundreds of years. Differentiation developed from work on tangents to curves in the early 17th century. You might be interested to know that there was a bitter dispute about who had discovered the method first as two mathematicians – Newton and Leibniz – made the discovery independently. It is now known that Newton made the discovery first, but that Leibniz published first.
In the late 17th century it was recognised that integration and differentiation are the inverses of each other.
So why would you be interested in learning about calculus? Well, calculus is used extensively in mathematics and in a lot of real world contexts. For example, in physics, calculus is used in the topics of motion, electricity, acoustics, astronomy and others. In chemistry it is used in reaction rates and radioactive decay and in biology it can be used to model birth and death rates.
If you watch the next two videos then you will discover how to find the derivatives of a function and apply differentiation to finding the gradient of a graph and to sketching the graph of a function. Before you start this there are three topics that you should be familiar with.
The first topic is substitution of both positive and negative numbers.
The second topic is finding the gradient of straight lines and finding the gradient at a point on a curve by drawing a tangent to a curve.
The third topic is knowing the shapes of the graphs of quadratic and cubic functions.
Let’s have a look at an example of each of these so that you can assess your existing knowledge before moving on to learn about differentiation.
Substitution of both positive and negative numbers
Let’s have a look at a couple of examples. Let’s start with the equation on the screen, y= 3x-2.
What is the value of y if x=5? We can find this by substituting x=5 into the equation, remember that 3x means three times x. x=5
Following the steps shown on the screen we found that y=13.
What about if x=-4?
Following the steps shown on the screen we found that y=-14.
Now let’s have a look at a slightly more complicated example. Let’s consider y=3x2-5x.
What is the value of y if x=4? Remember the order of operations.
So we found that y=28.
What about if x=-2? Remember – when you square a negative number you get a positive answer.
So we found that y=22.
If you found any of these tricky then it would be worth spending some time reviewing this topic before you move on to the next video.
Finding the gradients of straight lines
Try to remember how you would find the gradient of a straight line. You might want to pause the video now to do this.
Let’s have a look at an example of how to find the gradient of a straight line. On the screen we have a straight line graph. If you want to have a go at this calculation before we do then pause the video now.
In order to find the gradient of a straight line we start by choosing two points and finding the difference in the x values for these two points and the difference in the y values.
I’ve added two arrows to show the differences that I will be working with.
We can see that the difference in the x values is 2 and the difference in the y values is 6. Let’s use these to calculate the gradient of the line.
The gradient of the line is given by the difference in the y values divided by the difference in the x values. Substituting our differences in we find that the gradient of this line is 3.
You will also need to be able to find gradients where the graph has a negative gradient. That is where the line slopes downwards as we move from left to right.
Finding an approximation of the gradient at a point on a curve by drawing a tangent
Let’s build on from our work on the gradient of straight line graphs to considering the gradient of points on curves. We can’t find a single value for the gradient of a curve as the gradient changes along the length of the curve. We can find an approximate value for the gradient of a point on a curve by drawing a tangent to the curve at that point.
Here is an example of a curve, this is the graph of y=x2-3. We will try to find an approximation of the gradient at the point (2,1).
Firstly, we need to add a tangent to the curve at (2,1). An example of this is shown on the screen now.
Next, we find the gradient of the tangent Here we can see that, for the two points on the tangent I have chosen, the difference in x is 2 and the difference in y is 8. So the gradient is 8 divided by 2 which gives us 4.
It is important to remember that using this method to find the gradient of the curve at the point gives us an approximation of the gradient. This is because we have drawn the tangent by eye. In the next video you will see how to find the gradient of a tangent at a point more accurately by using differentiation.
The shapes of the graphs of quadratic and cubic functions
Before we start, try to remember what shape of graphs you would expect to see for quadratic and cubic functions. Pause the video now and sketch these if you can.
So, what do the graphs of quadratic functions look like? Well, quadratic graphs have a parabolic shape.
When the coefficient of x squared (the constant multiplying the x squared part of the function) is positive then the graph of the function would look something like this:
When the coefficient of x squared is negative, then the graph of the function would look something like this:
So, what about cubic graphs? Well, again we have two possibilities.
When the coefficient of x cubed is positive, then the graph of the function would look like this:
When the coefficient of x cubed is negative, then the graph of the function would look like this:
Conclusion
In this video we have introduced calculus and the two areas of integration and differentiation. We have identified some uses of integration and differentiation.
Finally, we reviewed some topics which will be helpful when you watch the next two videos which will introduce the differentiation content for the Cambridge iGCSE in Mathematics. -
Lesson 2 video
Video transcript
We encounter rounded numbers every day, whether it’s telling us that 10% of people own 90% of global wealth, or that there are 85 million pet owners in the US.
The accuracy with which the number has been rounded tells us how large or small the real value could have been.
Let us consider an example. A newspaper reports that a local cheese factory has produced 24000 kg this year.
The newspaper is likely to have rounded the actual figure
If the newspaper had rounded to the nearest 1000 kg what is the smallest amount of cheese that could have been produced?
What is the largest amount of cheese?
This leads us to 2 questions
If the newspaper had rounded to the nearest 1000 what is the smallest amount of cheese that could have been produced?
What is the largest amount of cheese?
Let’s look at this on a number line.
If we are rounding to the nearest 1000, the smallest amount of cheese that could have been produced is 23500 kilograms.
If we are rounding to the nearest 1000, the largest amount of cheese that could have been produced is 24500 kilograms 23500 is the lower bound, the smallest value 24500 is the upper bound, the largest value.
But what if the newspaper had instead rounded to the nearest 100?
What would the smallest and largest possible values of the cheese be?
If we are rounding to the nearest 100, the smallest amount of cheese that could have been produced is 23950 kilograms
If we are rounding to the nearest 100, the largest amount of cheese that could have been produced is 24050 kilograms
In this case 23950 is the lower bound, the smallest value 24050 is the upper bound, the largest value
So what does this tell us?
When a number is rounded the accuracy of the rounding is important to tell us what the smallest and largest possible values of the number are.
These values are called the lower and upper bounds of the number. -
Lesson 3 video
Video transcript
We encounter rounded numbers every day, whether it’s telling us that 10% of people own 90% of global wealth, or that there are 85 million pet owners in the US.
The accuracy with which the number has been rounded tells us how large or small the real value could have been.
Let us consider an example. A newspaper reports that a local cheese factory has produced 24000 kg this year.
The newspaper is likely to have rounded the actual figure
If the newspaper had rounded to the nearest 1000 kg what is the smallest amount of cheese that could have been produced?
What is the largest amount of cheese?
This leads us to 2 questions
If the newspaper had rounded to the nearest 1000 what is the smallest amount of cheese that could have been produced?
What is the largest amount of cheese?
Let’s look at this on a number line.
If we are rounding to the nearest 1000, the smallest amount of cheese that could have been produced is 23500 kilograms.
If we are rounding to the nearest 1000, the largest amount of cheese that could have been produced is 24500 kilograms 23500 is the lower bound, the smallest value 24500 is the upper bound, the largest value.
But what if the newspaper had instead rounded to the nearest 100?
What would the smallest and largest possible values of the cheese be?
If we are rounding to the nearest 100, the smallest amount of cheese that could have been produced is 23950 kilograms
If we are rounding to the nearest 100, the largest amount of cheese that could have been produced is 24050 kilograms
In this case 23950 is the lower bound, the smallest value 24050 is the upper bound, the largest value
So what does this tell us?
When a number is rounded the accuracy of the rounding is important to tell us what the smallest and largest possible values of the number are.
These values are called the lower and upper bounds of the number. -
Lesson 4 video
Video transcript
In this video we are going to look at an exemplar question taken from specimen paper 6. We will recap the key knowledge required to answer the question and then work through the solution.
Let’s start by having a look at the whole question. As you can see the question is split into two parts. In part (a) we are asked to find the co-ordinates of the two turning points and in part (b) we are asked to determine whether each of the turning points is a maximum or minimum.
Main content
Let’s start by looking at how we would answer part (a) of this question. In this part of the question we are asked to find the co-ordinates of the two turning points. The first thing to notice is that we are being asked to find the co-ordinates of the two turning points. How would we know how many turning points to expect if we were not told this in the question?
As this is a cubic equation we know that the graph will have up to two turning points.
Try to identify the steps you will take in answering this part of the question.
Now let’s find the co-ordinates of the two turning points.
In order to find the turning points of a curve we want to find the points where the gradient is 0.
The gradient function for a curve is found by differentiating the equation of the curve. So if we differentiate y=x3-6x2+16 we will obtain the gradient function of this curve. Let’s do this now.
Remember - if we have functions of the form y=axn where a is a constant and n is a positive whole number then these differentiate to give dy/dx=anxn-1. The expression is multiplied by n and the power reduces by one.
If we have y equal to a constant then this gives dy/dx=0.
As our equation is made up of three terms we differentiate each one in turn. Let’s do this now.
We differentiate the x3 term and obtain 3x2, differentiating -6x2 we obtain -12x1, and differentiating 16 we obtain 0.
We know that the gradient at the turning points will be 0. If we set dy by dx equal to 0 this will give us an equation which we can solve to find the x-coordinates for the turning point. Let’s do this now.
We need to solve this equation to find the x-coordinates of the turning points. We can do this by factorising.
We can see that either 3x is equal to 0 or x-4 is equal to 0. This means that either x is 0 or x is 4.
Once we have found the x-coordinates of the turning points we need to find the corresponding y-coordinates. We can do this by substituting the x values into the original equation. Let’s do this now. We can see that when x is 0, y is 16. Similarly, when x is 4, y is -16.
So we have found the two turning points for the curve we were given. The turning points are (0,16) and (4,-16).
We have now answered part (a) of the question. Let’s just summarise the key steps in finding the turning points. The first step was to differentiate the equation to find dy by dx. Once we had dy by dx we put this equal to 0 as we knew that the gradient is 0 at turning points. The next step was to solve the equation that we obtained to find our x-coordinates. Finally, we substituted our x values into the original equation to find the corresponding y-coordinates.
Now let’s look at how we can answer part (b) of the question. In this part of the question we are asked to determine whether each of the turning points that we have found is a maximum or a minimum and to provide reasons.
We will look at two methods which can be used to answer this part of the question.
We can use what we know about the shape of cubic graphs to give us an idea of what we are expecting to find. We know the general shape of the cubic curve when the coefficient of x3 is positive which is on the screen now. We can see from this that we are expecting (0,16) to be a maximum point and (4,-16) to be a minimum point.
Let’s look at how we use differentiation to show whether each of the two points that we have found is a maximum or a minimum. We will look at two alternative methods for this.
Method 1 – checking the gradient on either side of the turning point
Let’s find out the gradient of the curve on either side of the turning points that we have identified. If the gradient goes from positive to 0 to negative as x increases then the point will be a maximum.
If the gradient goes from negative to 0 to positive as x increases then the point will be a minimum.
Let’s start by considering the point (0, 16). We will find the gradient of the curve at x=-0.1 and x=0.1 as these points are on either side of the turning point. We can find the gradient by substituting these values into our equation for dy by dx. Substituting x=-0.1 we find that dy by dx is 1.23 and substituting x=0.1 we find that dy by dx is -1.17. We can see from the values we have obtained that the gradient goes from positive to 0 to negative and so we have a maximum at (0,16).
Now let’s consider the point (4,-16). We will find the gradient of the curve at x=3.9 and x=4.1 as these points are on either side of the turning point. We can find the gradient by substituting these values into our equation for dy by dx. Substituting x=3.9 we find that dy by dx is -1.17 and substituting x=4.1 we find that dy by dx is 1.23. We can see from the values we have obtained that the gradient goes from negative to 0 to positive and so we have a minimum at (4,-16).
So we have determined whether each of the turning points is a maximum or a minimum.
Method 2 – using the second derivative.
Let’s have a look at an alternative method for determining whether each turning point is a maximum or minimum. This method uses the second derivative.
We can find the second derivative by differentiating the equation we have for dy by dx. When we do this we obtain d2y by dx2. For the curve in the question we have already found that dy by dx is 3x2 – 12x. Differentiating this we obtain d2y by dx2 equals 6x -12.
If the second derivative at a point is positive then the point is a minimum, and if the second derivative at a point is negative then the point is a maximum.
Let’s calculate the value of d2y by dx2 when x is 0.
Since d2y by dx2 is negative when x is 0 the point (0,16) must be a maximum.
Now let’s calculate the value of d2y by dx2 when x is 4.
Since d2y by dx2 is positive when x is 4 the point (4,-16) must be a minimum.
So we have determined whether each of our turning points is a maximum or minimum.
Conclusion/Summary
In this video you have seen how we can use differentiation to find the co-ordinates of the turning points for a curve. We have also seen two methods for determining whether each of the turning points is a maximum or minimum.
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