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Video transcript
This unit of work is on Accuracy and Bounds.
Accuracy and Bounds are often difficult ideas to teach because students find it challenging to consider the least and greatest values which would round to a particular number.
We are going to look at how to identify the upper and lower bounds for data given to a specified accuracy.
We will look at identifying the bounds for data given to the nearest 10, 100, 1000 or to a specified number of decimal places or significant figures and use these results to find the solutions to simple problems.
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. -
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. -
Video transcript
This unit of work is on set notation and Venn diagrams
Students often struggle to recall the notation of Venn diagrams and to link it to common English terms as well as to represent this notation diagrammatically.
We are going to look at how to draw and use Venn diagrams to represent information for increasingly complex problems. Set notation will be explored and learners will use Venn diagrams for finding probabilities.
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. -
Video transcript
Suppose we want to sort these people into two groups:
Those with black hair and those wearing a tie.
In maths we call groups of this nature sets and the items within them, elements.
In this case the people we are sorting are the elements.
A Venn diagram can help us.
The Venn diagram consists of a rectangle that acts as a boundary into which all the elements must be placed.
This rectangle is referred to as the Universal Set and it contains the two further sets, each representing the characteristics with which we wish to sort our people.
We can take each element in turn and add them to the Venn diagram by placing them in the correct position.
The first person has both black hair and a tie so we place them in the Venn diagram where the two sets overlap.
This section of the Venn diagram is called the intersection of the two sets.
The second person has a tie, but does not have black hair, so we place them in the set for ‘having a tie’, making certain that they are not in the part of set that overlaps with having black hair.
The third person has black hair but no tie so is placed in the ‘black hair’ region of the Venn diagram, outside the region that overlaps with having a tie.
The fourth person has neither a tie or black hair, but every object needs to be placed somewhere inside the universal Set on a Venn diagram so this person is placed inside the universal set, but outside of the indicated sets.
Rather than placing the objects physically into the Venn diagram it is often more useful to simply represent the number of items in each region with a number.
In this case the number of people wearing a tie is equal to 4.
The number of people with black hair is equal to 5.
You will notice that if you add together the number of people with black hair to the number of people wearing a tie, you get a total of 9, which is more than the number of people placed in the two sets.
The reason is that people who have both black hair and a tie - have been counted twice.
We call this overlap the intersection.
And it represents the elements that are in both of the sets because they have both black hair and a tie.
The intersection of two sets is written as shown, and is an important concept in Venn diagrams.
We can see that there are 2 people who have both black hair and a tie.
If you want to know how many elements are in at least one of the two sets - In other words - the number of people who satisfy at least one of the conditions of having black hair or of wearing a tie, then we need to count everyone in each of the two sets.
We call this set the Union of the two sets and it is written as shown.
We can see that there are 7 people altogether who have black hair or who wear tie.
The union of two sets is often described by using the word OR, but this also includes both possibilities happening and at the same time. -
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. -
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. -
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. -
Video transcript
This unit of work is on straight line graphs
It can be difficult to prepare resources to teach straight line graphs and can be a challenge to teach because learners find it hard to understand the connections between a graph and its equation.
We are going to explore straight line graphs, investigating gradients as a measure of steepness and recognising them as rates of change.
We will also look at straight line equations and help learners to make connections between straight line graphs and their corresponding equations.
It is useful for learners to experience a range of graphs and learners will study a variety of scales and contexts.
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. -
Video transcript
Lesson 1: What is gradient?
Look at this road sign. It tells road users the gradient of the road ahead.
25% is the same as one quarter. So we say this gradient is ‘1 in 4’. This means you go up 1 unit vertically for every 4 units horizontally. Let’s look at this on a diagram.
This car travels 1 unit up for every 4 units along. The gradient is the change in height divided by the horizontal distance travelled.
Some people call this rise over run. Here, the gradient is the change in height, which is 1, divided by the horizontal distance, which is 4.That’s 1 divided by 4, or one-quarter.
We often need to find the gradient of a straight line on a graph.
The gradient tells us the slope of the line. The larger the gradient, the steeper the slope. When we work with gradients on graphs, instead of saying the change in height divided by the horizontal distance travelled, we usually say the change in y divided by the change in x.
The gradient is the change in y divided by the change in x.
For this graph, the change in y is 2 and the change in x is 1 so the gradient is 2 divided by 1 - which is 2. Another way to think of this is that the graph goes up two units for every one unit across so the gradient is 2.
But what if the graph slopes in the opposite direction? We still work out the change in y divided by the change in x, but this time the answer is – 2. The negative symbol shows us the line is sloping the opposite way.
Another way to think of this is that the graph goes down two units for every one unit across, so the gradient is −2.
The key thing to remember is that the gradient tells us the slope of a line. You work out the gradient by calculating the change in y divided by the change in x. -
Video transcript
In real life we often need to share the details of real-life objects with people who can't see that object, so we need to be able to put them down on paper.
For example, manufacturers need to share the detail of their cars with owners and mechanics in their handbooks, so they know how they work and can maintain them. But it is often not possible to draw on papers the actual size of real-life objects such as the real size of a car or an airplane.
Let's look at an example.
The length of a typical car is about 5 metres.
But the length of an A4 piece of paper is only 297mm
How many pieces of paper would you need to be able to draw the length of the car full size?
Consider what we need to do first to answer this question.
First, we need to change both measurements into the same units.
The car is measured in metres and the paper in millimetres.
In this case we're going to change the length of the car into millimetres.
We are converting metres into millimetres.
The prefix 'milli' which means one thousandth can guide us.
At this point you might be asking yourself the question "Do I need to multiply or divide?"
To answer this - think about the question "Are you expecting more or less millimetres than metres?"
The answer to this question should be more.
For example, it would require a lot of tiny millimetre fleas to be the same height as the dog.
This should be telling you to multiply.
The scale factor to convert from metres to millimetres is therefore multiply by a thousand.
Now we need to work out how many pieces of paper we would need to put together to be able to draw the car full size.
Dividing the length of the car by the length of a piece of paper you can see it would take 17 pieces of paper end-to-end, just to be able to accommodate the full length of the car.
This should explain to you why we need scale drawings to represent large objects like this. -
Video transcript
Let's look at how to convert between reality and your representation.
A wall is one metre long in reality. How long will it be on a 1 to 100 scale drawing?
Remember your prefixes again - 'centi' means one hundredth.
So this means that the 1 metre wall is 100cm long.
The scale tells us that we need to draw 1 centimetre on the page for every 100cm in real life.
So the wall will be drawn as 1cm long on the scale drawing.
So how long will the same one metre long wall be on a 1:500 scale drawing?
Lets use the approach we used at the beginning of this unit.
On a scale of 1:500, doing the same to the left hand side we get 1 divided by 5 or 0.2.
This preserves the ratio of 1:500.
So on a scale of 1:500, 1m will be drawn as 0.2 centimetres which is the same as 2 millimetres - a very short line! -
Video transcript
Once you're happy converting from reality to your drawings, the next step is to convert between drawing scales.
This may seem more complicated but the technique is the same.
On a 1:50 scale drawing, one unit on the scale drawing represents 50 units in reality.
But on a scale drawing of 1:100 the same unit now represents 100 units in reality so you will need less space to draw the same object.
In fact you will need 50 divided by 100 or half the space to draw the same object on a 1:100 scale drawing.
Now consider the scale factor from a 1:500 to 1:200 scale drawing.
On a scale drawing of 1:500 one unit on the scale drawing will be equivalent to 500 units in reality.
On a scale drawing of 1:200 you will need more space to complete the drawing.
You will require 500 divided by 200 units to represent the same object on the scale drawing.
This means 2 and a half units for every one you drew previously.
So an object which is 1cm long on a scale of 1:500 will be 2.5cm long on a scale of 1:200. -
Video transcript
This unit of work is Unit conversions
Unit conversions can be a difficult topic to teach because students often learn rules without really understanding how they work and are then unable to apply these rules independently and in more complex, less familiar situations. Learners often think of different types of conversion as completely different topics and do not make the wider link to proportional reasoning.
We are going to look at how to use different units of mass, length, area, volume and capacity in practical situations and express quantities in terms of larger or smaller units.
We are also going to interpret and use graphs to inform our understanding of direct proportion linked to conversions, applying our understanding in practical situations
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. -
Video transcript
A typical example you are likely to meet is converting miles to kilometres. We are going to explore a way of representing this problem that will help you solve any problems like this and lots of other problems as well. Remember there is not necessarily a right or a wrong way of solving this type of problem just valid and invalid ways of solving them. We are going to demonstrate a simple way of setting out this type of problem that might help you.
We’re going to start by looking at how you solve this type of problem into steps. The first step in this example would be to find out how many kilometres there are in one mile. To do this on the left-hand side we need to divide by 10.
If we are going to maintain the equavalence of the left-hand side to the right-hand side of our visual representation, then we must divide the right-hand side by 10 as well. This will give us 1.6.
We want to work out how many kilometres there are in 56 miles, so our next step is to multiply 1 x 56.
What do you think we have to multiply the right-hand side by this time?
Yes, that’s right we need to multiply the right-hand side by 56 as well and this gives us an answer of 89.6.
This tells us that 56 miles is approximately equivalent to 89.6.
We use the word equivalent as our initial assumption that 10 miles is the same as 16 km is an approximation.
This process we have just looked at is a way of finding the multiplier for any conversion. First you calculate how much a single unit is when converted. In this case what one mile is when converted to kilometres. This is why it’s called unitary method. You then use the multiplier to complete the conversion. In this case we multiplied 1.6 km by a multiplier of 56 to get our conversion.
If you are feeling confident you can solve this problem in one step.
Look at the image what would you have to divide and then multiply 10 by in order to get to 56? Remember what you did in the starter for this lesson.
Yes, that’s right you need to divide by 10 and then multiply by 56.
As a fraction this can be written as 56 over 10. This can be simplified to 28 over five or it could be written as a decimal 5.6. However, at the moment it will be better to leave it in its original form as this will help us to see more clearly what is happening.
As you can see we now need to multiply the right-hand side by the same fraction and as you can see we arrive at the same answer that we got when completing the calculation into steps. Which is of course what we would expect.
There is still another way that we can calculate this conversion using the functional relationship between miles and kilometres.
What would you have to divide 10 by and then multiply your answer by to get 16?
Yes that’s right you would need to divide by 10 and then multiply by 16. Once again you could simplify this fraction or even turn into a decimal but is more useful to keep it as it is as it helps you to see clearly what is happening when you do this calculation.
As you can see if you then divide 56 by 10 and multiply it by 16 you get 89.6 which is the same answer we got using the two previous methods.
As well as using this method as an alternative you can also use it to doublecheck what you’ve already done as both methods, using a multiplier or a functional relationship, will arrive at the same answer and if they don’t you know you’ve done something wrong.
You can use this framework to solve any proportional reasoning problem including conversion problems. -
Video transcript
Lets consider a bearings problem about the journey of a ship.
A ship sails 22km on a bearing of zero four degrees from point A, and a further 30km on a bearing of zero nine degrees to arrive at point B.
What is the bearing of B from A?
The first thing you need to do with any problem like this is to draw a diagram.
Spend a few moments drawing the diagram yourself.
Here is our diagram for this problem.
To find the bearing of B from A we will need to work out the size of angle alpha.
The triangle that we need to work with to solve this problem is a non-right-angled triangle so this tells us that we are likely to have to use either the sine or cosine rule to help solve this problem.
But at the moment we have very little information about this non-right-angled triangle.
First lets concentrate on finding more information about the right-angled triangle.
Angle theta sits within a right-angled triangle.
We use the fact that the sum of angles in a triangle is 180 degrees to calculate theta as we know the sizes of the other angles.
Spend a few moments working out the size of angle theta yourself.
Theta is 48 degrees.
Let's mark this angle on our diagram.
Now we can find the size of angle beta as both these angles lie on a straight line and angles around a point on a straight line sum to 180 degrees.
Work out the size of angle beta yourself.
Angle beta is 132 degrees.
Having put this information into our diagram. Let's look again.
If we remove all the other information from this diagram we can see that we have a non-right-angled triangle for which we know two sides and the angle between them.
In this situation the cosine rule can be used to find the opposite length AB.
Spend a few moments calculating the length of AB yourself.
AB is 48km to the nearest km.
We now have nearly all the information that we need to calculate the solution to this problem.
We can use our non-right-angled triangle for which we have two sides and an opposite angle.
To find angle alpha we can use a rearrangement of the sine rule.
Spend a few moments calculating the size of angle alpha yourself.
Angle alpha is 28 degrees to the nearest degree.
We haven't quite finished yet and it is very easy to give our final answer as 028 degrees to the nearest degree but we still need to work out the bearing of B from A measured from North.
We must add out new value for alpha to our original bearing 042 degrees.
Don't forget the final bearing needs to be written using 3 figures.
So the bearing of B from A is 070 degrees.
Well done! -
Video transcript
As we have already seen a vector has magnitude (size) and direction. When looking at a vector diagram the length of the line shows its magnitude and the arrow represents the direction of the vector.
We're going to start by looking at how you add vectors. We can add two vectors by joining them end on end.
Here you can see two vectors a and b.
To add vector b to vector a you join the tail of vector b to the head of vector a.
The result is a vector that joins the tail of vector a to the head of vector b. This is called the resultant vector a + b.
Think about two non-parallel vectors, a and b. Then a plus b is the translation of a followed by the translation of b.
You should be able to see from the diagram that is doesn't matter which order you add them, you will get the same result.
So a plus b is exactly the same as b plus a.
This tells us that vector addition, like ordinary numerical addition is commutative, in other words the order doesn't matter.
You can add vectors with or without the use of a diagram for example if you want to add two vectors p and q you can add the horizontal components and then the vertical components.
This will give you the vector five-two. Five to the right and two up.
You can always check your results using a diagram.
You can also subtract vectors.
a - b is defined as the translation of a followed by the translation of minus b.
To subtract vector b from vector a you join the tail of vector -b to the head of vector a.
The resultant vector a - b joins the tail of vector a to the head of vector b.
You can see that the resultant vector a minus b can be considered either as a minus b or as minus b plus a.
You can also subtract vectors with or without the use of a diagram by subtracting components.
First subtract the horizontal components and then the vertical components.
In this example the resultant vector is one eight - one to the right and eight up.
This can be pictured on a diagram as follows. -
Video transcript
This unit of work is on the probability of combined events
Students often struggle with combined event problems although calculating probabilities for these is similar process to that of single events in that it amounts to counting up the number of equally likely outcomes that fit a particular situation.
In this unit we will use possibility space diagrams and explore how probabilities can be calculated using fraction arithmetic.
We will explore the different ways of representing combined probability situations diagrammatically, introducing tree diagrams as a way of finding the probabilities of combined events.
A separate lesson covers the language associated with conditional probability.
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. -
Video transcript
Life is full of uncertainties. Take for example, the likelihood that when playing a game you correctly guess the first number drawn from 10 tokens consecutively numbered from 1 to 10 in a bag.
We can calculate this probability easily enough because there are only 10 outcomes, one of which matches our guess. We therefore have a 1 out of 10 chance of guessing correctly or a probability of one tenth
However, selecting a single token would be a very short and dull game. It is more interesting to continue to draw tokens from the bag and to keep guessing. This makes listing the likely outcomes and calculating the required probabilities much more complicated.
When we are calculating probabilities based on two more events needing to be matched we call this calculating combined probabilities.
Let us consider another example.
At a fairground, visitors play a game that involves using a spinner.
They spin the spinner twice and win a prize if the spinner lands on the same colour for both spins.
What is the probability of winning a prize?
To calculate the probability of winning we need to consider the number of equally likely outcomes.
We can do this by methodically listing all of the possible outcomes.
Suppose on the first spin the spinner lands at the blue shown.
Then on the second spin the spinner could land on either the same blue section
Or the first green section
Or the second blue section
And so on
Altogether there are 8 possible outcomes for both spins with the first spin landing on the top blue sector.
Now lets consider how many possible outcomes are there for both spins if the first spin lands on the first green sector instead?
There are 8 of these as well.
With this information we should now be able to work out how many outcomes are there altogether for both spins.
We can illustrate all the possible outcomes by putting the results into a table.
A table like this is called a possibility space diagram
We can see from the diagram that there are 64 equally likely outcomes
And we look to see how many of these have the same colour on both spins
Altogether there are 24 of these.
This means that the probability of getting the same colour when the spinner is spun twice is 24 out of 64.
Or a probability of twenty four, sixty fourths. Which simplifies to three eighths.
Possibility spaces are great way of visualising simple combined probability problems where the probability of the second event does not depend on the first.
For example a student plays a game where they draw a ticket from the bag which contains six tickets numbered from 1 to 6.
If they pick a number less than 3, they eat a chocolate.
If they put the ticket back in the bag every time, the probability of picking a winning ticket from the bag each time remains the same.
This makes it easy to find the probability they win 2 chocolates if they play the game twice.
Using the possibility space diagram it is still relatively easy to see that the probability of winning in both games is 4 out of 36 or one ninth.
It is also possible to calculate this probability, without the need to draw a possibility space diagram by considering the probability of the student winning each of the games separately and multiplying them together.
The probability that he draws a winning ticket in the first game is 2/6 and if he replaces the ticket the probability of him drawing a winning ticket in the second game is still 2/6, meaning that the probability that he wins two consecutive games is the same as 1/3 times 1/3, which is equal to 1/9.
If the student fails to return the first ticket to the bag after the first game, the possibility space diagram becomes difficult to draw, as the results of the second game would depend on the first result.
However, we can still calculate the probability that the student wins both games by multiplying together the probability of him winning each game.
In this case, the probability that he wins the first game is 1/3 as before but the probability that he wins the second game is reduced to 1/5, as there are only 5 tickets left in the bag and only one of them is a winning ticket. This means that the probability that he wins in both games is 1/3 multiplied by 1/5 which is 1/15. We can illustrate this method using a tree diagram. The first part of the tree represents the outcomes for the first game. And we can extend the branches to reflect the outcome of the second game If the student wins the first game, there will still only be five tickets left, but only 1 winning ticket. If you put these together you can find the probability of winning both games by multiplying along the branches. From the tree diagram it is possible to calculate the probability of any outcome from the two games by multiplying along the branches and adding up the probabilities of the outcomes required. For example you can find the probability of getting exactly 1 winning ticket by multiplying along two sets of branches and adding the probabilities together. The probability that when you play the game you win the first time and lose the second is 8/30 and the probability you lose the first time, but win the second is also 8/30. This means that the probability of winning exactly one chocolate is 16/30 which is the same as 8/15