Home » Archives for April 2019

Month: April 2019

Find the distance from a Speed time graph

Today’s post is about an exam question about a speed time graph I  was prepping for my students yesterday. I was surprised to see this question because it covers two points I associate more with A-Level work.


What can we find from the area under a curve?


a)  Area under the curve as the distance
b) Using the ‘ trapezium rule’

Before I say more, here is the question








The question asks for the distance the car travels, but the graph is of speed over time. This is where the first of my ‘surprise facts’ comes in – You can find the distance travelled by measuring the ‘area under the curve’. I won’t here show why that is the case – that is for A-Level students (OK I might post on that later). For now though, I will just apply the rule.

The area ‘under the curve’  is the area of the shape made by the curve at the top and the line ‘speed = 0’ at the bottom, bound to the left and right by the times we are interested in : In this case t=0 and t=20.

The Trapezium Rule

We can only estimate the area of this shape, given how irregular it is – the top side anyway – and this is where the ‘Trapezium rule’ comes in. To apply this we have to divide the area into strips, each one a trapezium (OK, the first one is a triangle, but we can apply a similar rule. To be fair, the question does give a clue to what to do by saying we should use ‘4 strips of equal width’

We divide up like this.

The first area is a triange with area

1/2 x 5 x 22*

*The value of the speed at t=5


The other areas are trapeziums with area 5 x 1/2(Height at start + height at end).

The ‘height’ in each case is the speed values at t= 10, 15 and 20.

If we look at the calculation closely we can save time by noting the middle times are included twice each, 1/2 each time.

so the area of the trapeziums are 5 x (22 + 28 + 32 + 35 x 1/2)

This is a quicker way that fiddling about with all the 1/2s. The 35 – value at t=20 – still needs the 1/2 because it is a measurement only on one trapezium.

We get the value 5 x 99.5.   This is where I took into account that we are only estimating, so using a value of 100 instead of 99.5 is justifiable.    This gives us an estimate of 500m

There is a part(b) for this question on the next page.

Over and under estimates


“Is this an over-estimate or an under-estimate for the distance. Give a reason for you answer.’


The answer in this case is that it is an underestimate, because all those trapeziums still left a small space under the curve. I don’t think our rounding up to 500m stops this being the answer.


More about the trapezium rule can be found here

A simple algebraic proof

As we get close to the exams, my students mostly want to do previous exam questions with me, so today I am sharing one which I found rather interesting – Algebraic proof. Perhaps because it foreshadows the kind of Maths one does at A-Level and degree level but its aimed at GCSE students.

The skills required are not that different to solving equations, but the level of thinking required is perhaps one level further on.

The question is…

Prove that the square of any odd number is one more than a multiple of 4.

The idea of an algebraic proof

The first thing to notice are those words ‘prove that’..  for students who are used to Maths being about doing a sum and finding an answer, this could seem quite new.  We are told what to expect. We just need to show the statement is true.

One of my students said ‘does this mean I have to look at every odd number’.  That would be quite difficult because there is a never -ending list of them. But its not a bad way to start, to convince yourself it might be true.

52 = 25  – and that is 1 more than 24 which is 4 time 6.

112 = 121 and that is 1 more than 120 which is 4 times 30

I could keep going like that – and you should do maybe a couple of your own – but I will never prove its true for ALL numbers like that.


What is easier : prove its true or prove it isn’t?


Its easier in Maths – and in life – to prove something isn’t true. We only need to find ONE odd number where the square is not 1 more than a multiple of 4 to prove that statement is not true*. A person only needs to show he was in Birmingham on Tuesday evening to show he didn’t commit the crime in Paris.

(*You could try but don’t waste your time looking,  it is true!)

Proving things are true is harder.  But we can still do it. We are going to do this with algebra, looking at the ‘general case’. That is why we call this algebraic proof.

We need to think of numbers more generally.  What is the ‘general’ odd number.  Take a number, any number. Will it be odd? Well there is a 50-50 chance.  But if we double the number we know it will be even (131 doubled is 262), then add 1. We then know the answer we get will be odd.

And in fact, all odd numbers can be found using  2n + 1 – That is a general odd number!!

So take our general odd number, and square it.  (2n+1)(2n + 1).  I could have written this with a 2 but I find its easier to multiply brackets if you can see both.

We get

Algebraic proof equation Equation 4n

Factorise just part of this you get 4(n2 + n) + 1

And that is actually what we are looking for! Let me explain. In 2n + 1, our ‘general’ odd number, n is just any whole number, and so n2 + n is also just a  whole number.

So 4(n2 + n) must be a multiple of 4 [ 4 times (n2 + n)] and 4(n2 + n) + 1 is 1 more!

This completes our example of algebraic proof.

Finding a turning point

For today’s post I’m going to look at an exam question I saw posted on Facebook.

I confess this is the sort of stuff I get nerdy about! I love the way that equations can describe pictures, and vice versa.




Before we start, lets make sure we understand the terms used. What exactly is a ‘turning point’?   Well, it is the point where the line stops going down and starts going up (see diagram below).   That point at the bottom of the smile.

There are three different ways to find that but in all cases, we need to start by finding the equation – finding out the values of ‘a’  and ‘b’.

‘b’ is the easier. take the case where x=0 and the two terms with x can both be disregarded and so b = -5 = the point where the curve crosses the y-axis (also known as the line ‘X = 0’.  That is similar to the reason why the term ‘c’ is often called ‘the y intercept’ for straight lines, equation y = mx+c

The other point we know is (5,0)  so we can create the equation

25 + 5a – 5 = 0 (By substituting the value of 5 in for x)

We can solve this for a giving a=-4


The full equation is  y = x2 – 4x – 5

I usually check my work at this stage  52 – 4 x 5 – 5 = 0 – as required

Now, I said there were 3 ways to find the turning point. I will give all three here bu, be warned, the third does require some A-Level maths. This diary is more aimed at GCSE students but method 3 is actually the way I would usually find the turning point, so I will give a brief description here.

Method 1 – The ‘parabola’ is symmetrical.

A Parabola is the name of the shape formed by an x2 formula


On this version of the graph

I’ve marked the turning point with an X and the line of symmetry in green. 

This tells us the value of x on the turning point lies halfway between the two places where y=0 (These are solutions, or roots, of x2 – 4x – 5 = 0

We know one of these is is x=5. We can get the other by factorising  to give (x-5)(x+1) = 0. So x = -1 is the other solution. Halfway between x = -1 and x = 5 is x = 2.

when x = 2, y = 22 – 4 x 2 – 5 = -9.  So the turning point is (2,-9).

Method 2  Complete the Square

If we ‘complete the square’ on this equation we get

x2 – 4x + 4 – 4 – 5  –  I’ve added in 4 and taken 4 away which looks like an eccentric thing to do, but this means we can factorise the first part into a square

(x – 2)2 – 9 (also combined the -4 and – 5 to make -9)

This where we use our knowledge that a ‘square’ is never less than 0, but it can be 0. So its minimum value is when that square = 0 – so x – 2 = , so x = 2. And we can also see the value of the whole thing there is 0 – 9 = -9

Method 3 – By differentiation

This is A-Level stuff really, so I’ll only give an overview. This is the way I would usually do this, but then I have studied Maths through A-level (and beyond!)

Differentiation is one branch of Calculus, the mathematics of measuring change.

By a rule you will learn if and when you first study calculus, the equation of how much x2 – 4x – 5 is changing is given by 2x – 4.

The turning point is where the line isn’t changing, so 2x – 4 = 0 (Zero change) so 2x = 4 and x = 2. y = 9 can be as above.



I’ve given three methods here. Fortunately they all give the same answer! (They wouldn’t be good alternatives otherwise!)

Choose which of the first two you feel most comfortable with.  The third method is the easiest to extend to where we have equations of than an a x2

Some tricky algebra – and how to make it easier

Let’s consider this question from a Higher GCSE paper.




First thing I think is – Wow, that looks complicated.

[Actually, the first thing I should think is, what sign is that. It is a divide sum in the middle, not an add sign as I first thought]

Algebra fractions can stump the brighter students, but its worth remembering to rules are just the same as fractions with numbers.

We could do this divide sum by flipping the second and multiplying the numerators together and then the denominators. But given that we are told the final answer is simple, there is certain to be a lot of simplifying we can we first.

The first thing I noitices was that the bottom of the second fraction was going to factorise. It looks so close to the ‘difference of 2 squares’ rules.

Indeed it does factorise to


And immediately we see there is some cancelling to be done with the top to give 1/x(x-5) for the second fraction

What about the bottom of the first fraction. Looks like there could be some factorising to be done there. And indeed, taking my cue from the (x-5) we have already seen, this factorises to

(x – 5)(x + 2)

Something I missed before because I was so busy factorising the higher powers of x, but the top can be factorised as 3(x+2) and so we can cancel to


NOW we can do the divide by flipping the 2nd fraction, and the operation is now much easier than before

3/(x – 5) *  x(x – 5)   [Using an * for multiply here to stop confusion with the x’s]

which is then 3x = so a = 3.





Circles and Squares – The Answer

In my last post I gave the following exam question which I had looked at with two of my students

And now, as promised, the answer.

I’ve added a few coloured lines to help me talk this through.

The width of the square is the width of two of the circles. I have marked this with the two parallel grey lines of different shades. As we are told each of the circles has a radius of 24cm, that makes the width of the square 4 x this – 96cm!

The height of the rectangle is a bit less, because there is an overlap of the circle diameters in that direction.

We do know the height includes two radiuses* – shown as the parallel red/pink lines. But there is also the height of the green triangle, shown in blue on the height of the rectangle.

We can find that using Pythagoras Theorem – By splitting the green triangle in half, we have two right angled triangle with sides 48cm (the hypotenuse) and 24cm…   so h2 + 242 = 482.

h = sqrt(1728) = 41.57cm (to 4 s.f.)**

It can be easy after doing the most complex part of a calculation to forget we haven’t yet finished the question. Let’s not fall into that trap. We still need to find the total height of the rectangle, and thus the area of the rectangle.

Total height is 41.57 + 2 x 24 = 89.57cm

Area = 89.57 x 96 = 8600cm2 (To 3 s.f)

A couple of notes about my solution
* Yes, the official plural is radii – from the Latin!  I always think that looks a bit old fashioned so I tend to say radiuses!
**Its often a judgement call how one rounds ones answer. Exam questions will sometimes say (Round to 3 s.f. or 1d.p) but if they don’t, show what rounding you have done. It is not appropriate to give more figures than any rounding given in the data provided….  BUT  if you know you are going to round to 3 s.f. at the end, keep one extra figure in the intermediate numbers. That’s why I use 4.s.f for h, even though I know I am going to give a less precise answer at the end