This is something which comes up in the topic of surds, although the word denominator is a term which comes up in fractions and so this Maths skill combines both topics as well as a few skills from algebra thrown in. It is a fairly high level skill in Mathematics and also, at the highest level, combines the algebra concept known as the difference of two squares. In this post I shall attempt to untangle these terms and link them together.

TRIGONOMETRY AND PYTHAGORAS

Rational and irrational numbers

Some numbers are easy to write down. The answer to 3 + 5 is 8. Some would take forever to write down. The answer to 1 ÷ 3 is 0.333333….. and so on. You can get around this by saying “nought point three recurring” which is written by putting a little dot above a 3 as shown below:

I could also write this as ⅓ if I wanted. But, some numbers are impossible write down in a way that would tell the reader what digit comes next. In the example above we know we just keep writing 3s. An irrational number also goes on forever, but we have no idea what the next digit is. If we take a circle and measure its circumference and then divide it by its diameter we will get the irrational number pi, written as π.

When you square root many numbers you also get irrational numbers. The square root of 9 is 3, because when I multiply 3 by itself I get 9. What number would I need to multiply by itself to get 2? Or 3, 5, 6, 7 or 8? It is impossible to say. When you square root any of these numbers on your calculator you get a whole mess of digits. In other words, they are irrational numbers. Irrational numbers are usually written in different forms for this reason. The square root of 2 is simply left as: √2. But… I take √2 and I multiply it by √2 I get 2. I have now rationalised it. It started off irrational and now the answer is rational.

√2 x √2 = 2

WHAT ARE QUADRATIC EQUATIONS AND HOW DO I SOLVE THEM?

Let’s start rationalising some denominators!

The denominator of a fraction is the number at the bottom. The top number is called the numerator. 

So in the fraction ⅓, the numerator is 1 and the denominator is 3. We are essentially dividing 1 by 3. Let’s look at a fraction when we have a square root as the denominator:

We may be asked to rationalise the denominator here. If we were to multiply the denominator by root 3, we would get 6: 2√3 x √3 = 2 x 3 = 6. But, we would be changing the value of the expression if we did. When we are asked to rationalise the denominator, what is often unsaid is that the value of the expression must not change, it is simply rewritten in a different way.

One thing we can do to any number without changing it is multiply it by 1. Doesn’t sound much use really does it? Well, any number divided by itself is also 1, so I could do the following without changing the value of the expression:

When multiplying fractions you simply multiply the top by the top and the bottom by the bottom. You then simplify if necessary:

… and we have rationalised the denominator! Cool. It is still an irrational number of course, so it still has an irrational component, but it is now the numerator rather than the denominator.

WHY SHOULD YOU BOOK A MATHS TUTOR?

The difference of two squares

Let’s imagine we have two squares. One has lengths a and the second has lengths b. The first area would be a² and the second area would be b². The difference of the two areas would be a² – b². We can call this the difference of two squares.

Now let us imagine we factorised another expression:

(a+b)(a-b)

= a² + ab – ab – b²

… we notice the middle two terms cancel out.

= a² – b²

… and we have the original expression which I referred to as the difference of two squares! In other words:

(a+b)(a-b) = a² – b²

Rationalising more difficult surds

Imagine we were asked to rationalise this expression:

As discussed earlier we can multiply it by anything which is divided by itself as that would be the same as multiplying it by 1. Which one of the options below do you think makes the most sense? Cover the text below to test yourself!

Hopefully you said option… B! We are allowed to do all examples apart from option D, but only option B applies the difference of two squares which ends up rationalising the denominator because:

(1 + √2)(1 – √2) = 1 – 2 = -1

… and so

5(1 – √2)

-1

= 5√2 – 5

So, we have rationalised this denominator as well! That’s pretty much it, but if you think you would like more practice and teaching on this topic check out this video by Maths genie and this one by Cognito. You might also like to practice using this worksheet by Corbett Maths.

Good luck!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Rationalising the denominator appeared first on The Tutor Team.

Some terminology

I shall start by explaining some terminology that is useful for algebra and which is usually expected to be learnt by students. If any of this seems confusing for now, don’t worry, just read on and it might be less confusing as you continue.

• variable: This is the unknown value in the equation represented by a letter, such as x.
• operator: This refers to the symbol for adding, subtracting, multiplying, dividing and so forth.
• equation: This is like a statement using variables, numbers and operators to say something.

Eg. x + 3 = 5

This means “If I add the variable x together with 3, it would make 5.”
Equations must have equal signs otherwise it would not be saying anything.

• expression: These are often confused with equations, but do not have equal signs. They don’t tell you any information. x+5 is an expression, for example.
• product: This what you get when you multiply two values together. The product of 3 and 5 is 15. When you multiply a variable by a number, you can show the product by writing the number to the left of the variable. Eg. 3a means 3 multiplied by a. Two variables multiplied together are written next to each other in any order. Eg. xy or yx.
• term: This is anything that is being added or subtracted to something else within an expression. This could include products.
• solve: This means to find the value of the variable. It is not always possible to do this.
• rearrange: This means to change the equation in some way, which may help to use it more easily, and may help to solve it.

9  THINGS TO ASK A PRIVATE TUTOR – BEFORE YOU BOOK THEM

Solving an equation

Let’s start by looking at the equation below:

2x + 5 = 11

x, or whatever letter is there, represents an unknown value, or a variable. We call it a variable because the value of it could vary. If the equation has more than one variable it cannot (usually) be solved. If a number is written to the left of a variable, like ‘2x’ as above, it means 2 multiplied by x. This little fact is often not understood by some students early on and can cause huge confusion later on.

Think of x as a box of apples, labelled x, which cannot be opened. All of these x-boxes will have the same number of apples in. You can now tell your son or daughter that you’ve been learning about x-boxes! What the equation is saying is:

“Two boxes and 5 apples will add up to 11 apples”

Thinking of it like this, you might realise that 2 boxes and 4 apples would add up to 10 apples. In fact two boxes and no apples would add up to 6 apples. I’ve just taken 5 apples from the left and 5 from the right.

Remember that each x has the same number of apples, so it must be 3 each! I just divided both sides of the equation by 2.

WHY SHOULD YOU BOOK A MATHS TUTOR?

My Golden rule with rearranging equations

What we did there was rearrange and solve an algebraic equation. There were two steps.

Step 1: Take away 5 from both sides (of the equals sign).
Step 2: Divide both sides by 2.

Probably the most important thing to remember when rearranging equations is:

Whatever you do to one side, you must do the same thing to the other side.

If we were to solve the above equation without using my (hopefully) helpful boxes of apples analogy, it would look like this:

2x + 5 = 11

Step 1: Subtract 5 from both sides.
2x + 5 – 5 = 11 – 5

Adding 5 and then subtracting cancels out, so…
2x = 6

Step 2: divide both sides by 2
x = 3

When deciding on what step to take, you are trying to free the variable, so as it is finally on its on own on one side of the equals sign, with a number on the other side. To free it, you should ask “What is currently being done to the side it is on?” At first 5 was being added to the left, so I did the opposite. I subtracted 5. Next, x was being multiplied by 2, so I did the opposite. I divided both sides by 2.

This is, more or less, how you do algebra. Well… it should get you started anyway! Good luck!

3 REASONS WHY PARENTS BOOK PRIVATE TUTORS

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Algebra for Parents appeared first on The Tutor Team.

I want to focus in this post on the differences between trigonometry and pythagoras. When to use each one and so on. Let’s start by reminding ourselves of the equations you will hopefully be familiar with by now, presuming you have learnt a little bit about both.

Sin, Cos and Tan

This will always be involving three buttons on your scientific calculator:

Sin, cos and tan. These are not values, but functions. They do something to the numbers you put into them. They are like machines.

sin 30 is often written as sin(30) to emphasise this point. It does not mean ‘sin multiplied by 30’. That doesn’t make any sense. It would be like saying ‘plus multiplied by 30’. We’ll not worry too much about why sin(30) =  0.5 or sin(90) = 1 here. We’ll just trust that something is happening inside of our calculator that we don’t understand, but helps us to solve the problems.

Soh Cah Toa Vs Pythagoras

Let’s take a look at three problems…

In problem 1, we have an angle and a side, and we need to find the length of another side. In problem 2, we have 2 sides and we need to find the size of one of the angles.

But, in both problems we are working with 2 sides and one angle. This is the key. Don’t think “What do I have?” Think “What am I working with?” If we’re working with angles and sides it’s trigonometry.

In problem 3 we don’t have any angles and we don’t want to know any angles. So we are only working with 3 sides. If we are only working with sides, this means its pythagoras.

How to label the triangle when using Soh Cah Toa

When I mention trigonometry, I really mean what is often called soh cah toa. There is a more advanced type of trigonometry involving sin, cos and tan, but we will probably look at that in a different post. Let’s take another look at problem 1.

We need to label the triangle first with the following labels:

o for the side that is opposite the angle we are working with.

a for the side that is adjacent (next) to the angle we are working with.

h for the hypotenuse. This is the side that is opposite to the right angle and is always the longest side on a right angled triangle. 

We can also label the angle we are working with using a greek letter, θ, pronounced theta.

See the diagram below:

Applied to problem 1, it would look like this:

Problem 1 is asking us to calculate the adjacent side.

Using Soh Cah Toa

The trick to using this is to raise the middle letter each time you write out soh cah toa as below.

We can then draw a triangle around each one and we have instant formula triangles to work with. I’ve also colour coded the letters the same colours as in the above diagram.

s = sin(θ)

c = cos(θ)

t = tan(θ)

In our question we are working with o and a, that sounds like toa, so we will be using the 3rd formula triangle. 

We want the adjacent side, so covering that up we can see that:

So RT = 14cm ÷ tan (53°) = 10.5cm (1dp)

What if we need to find out an angle?

In problem 2 it’s an angle we need to find, not a side. Using the same technique as described above we can see that…

Sin (ACB) = 8cm ÷ 10.5cm = 0.762 (3sf)

But… This is not the angle ACB. This is just the sin of the angle. We need to put it through the sin machine backwards to ‘un-sin’ it! To do that we use an inverse-sin (sin-1). The same would be the case with tan and cos. You can access the inverse sin tapping ‘shift’ and then the pressing sin. 

So…

Angle ACB = Sin -1(8cm ÷ 10.5cm) = 49.63° (2dp)

And finally, problem 3!

Let’s briefly go over

In problem 3 we are not interested in any angles and so we’re going to use pythagoras. Let’s briefly look at how to label a triangle with pythagoras… Because it’s different!

This time the hypotenuse is labelled as c and the other two sides are a and b. It doesn’t matter which way around you label a and b, but c is always the longest side, opposite the right angle. You can also see the famous Pythagoras theorem as well.

Notice we have another a, but this does not mean adjacent. What would it be adjacent to? We are not working with angles this time! It’s just a.

Ok, in our case it is the hypotenuse we are trying to find out. So…

AB2 = 152 + 82 = 225 + 64 = 289.

But… This is AB2, not AB. So I need to square root it.

So there you have it. 3 problems. Problem 1 using soh cah toa with a missing side. In Problem 2, we had to use an inverse trigonometry function. Problem 3 was pythagoras. I hope you found that helpful and I’d suggest searching the internet for a few practice questions with answers in these topics to see if you are able to do them now.

Thanks!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Trigonometry and Pythagoras appeared first on The Tutor Team.

When you first start studying algebra the equations you will be looking at will no doubt be all linear equations. I’ll write some examples below:

1. 1. 1. x + 5 = 8
      2. 3a -10 = 26
      3. 6w – 3 = 7 – 4w

Feel free to try and solve these equations as a little warm up. They get gradually more difficult, but are a lot easier, and less weird, than quadratic equations. One thing they all have in common is that the variable, which is the unknown value that has been represented by a letter in bold, are all to the power of 1. That is that nothing has been squared, cubed, square rooted or anything like that. In quadratic equations the variable could also be squared, but not cubed or any other power. I have put some examples below:

IMPLICIT DIFFERENTIATION AND ITS USE IN DERIVATIVES

What makes quadratic equations special?

The strange thing about quadratic equations is that they can have more than one solution! What?? I know right… How could that possibly be?? Let’s imagine a very simple equation:

x² = 9

How would you solve this equation? Well, we want to find what x is, and we currently know what x² is. Like all equations, we can split this equation into two sides: Left of equal sign and right of the equal sign. I shall refer to these sides as the left and right sides.

The golden rule for all equations is:

“Whatever you do to one side, you must do to the other side of the equation.”

 So to change x² to x I just need to square root it.

However, following the above rule, I would have to also square root 9. That’s easy, it’s 3 right? 3² = 9 and so the square root of 9 is 3. Ta da!

Really? Why not? Well… 3²  does in fact equal 9, but (-3)² also equals 9! So there are two solutions to x² = 9:

 x = 3 or x = -3

I know right?? Mind blown!

Both solutions satisfy the equation, so there really are two solutions.

INTEGRATION IN MATHS: THE PADDINGTON BEAR APPROACH

The trick to solving quadratic equations

Let’s expand the equation below:

(x-3)(x-5) = 0

 I’m going to presume you know how to expand double brackets here, but if not you might want to brush up. Below is the equation expanded and simplified:

x² – 8x +15 = 0

You may wonder why I’m having zero on the right hand side of the equation. All I shall say for now is that it is very useful to do so, but read on and you shall see why.

Because all I have done is expand the equation, they are in fact the same equation, but arranged differently. Once we expand out the brackets we see the equation is a quadratic because it has a maximum power of 2.

Now in the first rearrangement we have two algebraic expressions being multiplied by each other and equaling zero. The first expression: x-3, and the second expression: x-5. At the risk of making this more complicated, let’s say:

a = x – 3    and   b = x – 5

So…

ab = 0

The only way this could be true is if a equals zero, b = zero or they both equal zero.

a = 0

or

b = 0

 Which means that:

x – 3 = 0

or

x – 5 = 0

In other words…

x = 3

or

x = 5

 Two solutions! Mind blown… No wait, I’ve done that bit…

Ok, let’s see if this actually works, or have I just broken a load of Maths rules?? We’re going to put the values for x back into the left side of the equation (x² – 8x +15) to see if it does in fact equal zero:

x = 3: 3² – (8×3) + 15

= 9 – 24 +15 =  0

x = 5: 5² – (8×5) + 15

= 25 – 40 +15 = 0

  No way!!

Hang on though… That’s all well and good, but we started with: (x-3)(x-5) = 0, which made it a lot easier. What if we started with: x² – 8x +15 = 0?

IMPROVING NUMBER CONFIDENCE – A FRACTION AT A TIME

Factorising quadratics

Without going into too much of an explanation, I shall teach you the method of how to factorise, which is the opposite of expanding.

Factorising: x² – 8x +15 → (x-3)(x-5)

Expanding: (x-3)(x-5) → x² – 8x +15

We always want our quadratic equation to be in the form of:

ax² + bx + c = 0

In our equation: x² – 8x +15, a = 1, b = -8 and c = 15.

We are looking for two numbers, let’s call them m and n, where:

m + n = b

and

mn = c

In our case:

m + n = -8

and

mn = 15

 With a little bit of thinking we see the numbers are: -3 and -5. It doesn’t matter which one is n and which one m. We then put that into the form of

(x+n)(x+m)

In our case: (x-3)(x-5)

So, there you have it. That’s how to factorise and you already know why to factorise.

Important note… This only works when a = 1. We shall look at other situations in a future post.

One more thing before I sign off… Here is an AWESOME resource which helps you solve any quadratic equation. Check it out. I shall see you next post!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post What are quadratic equations and how do I solve them? appeared first on The Tutor Team.

What is Implicit Differentiation in calculus?

Before starting the introduction of implicit differentiation, we must have a sound knowledge about implicit function as in implicit differential we have to calculate the derivative of implicit functions. The function becomes an implicit function when the dependent variable is not explicitly isolated on either side of the equation.

A function defined in terms of both dependent and independent variables, such as,

X² + 3Y = 0

 is an implicit function. An explicit function, on the other hand, is expressed in terms of an independent variable, such as y = f(x). An implicit function can be easily differentiated without rearranging the function and instead distinguishing each word in the case of differentiation. Because y is a function of x, we’ll use the chain rule, as well as the product and quotient rules.

The technique of obtaining the derivative of an implicit function is known as implicit differentiation. Explicit and implicit functions are the two types of functions. The dependent variable “y” is on one of the sides of the equation in an explicit function of the type y = f(x). However, having ‘y’ on one side of the equation is not necessarily required. Consider the following functions, for example:

• X³ + 3Y = 5
• xy² + cos(xy) = 0

Even though ‘y’ is not one of the sides of the equation in the first case, we can still solve it to write it as y = 1/3 (5 – x³), and it is an explicit function. However, in the second situation, we cannot readily solve the equation for ‘y,’ and this sort of function is known as an implicit function. Implicit differentiation calculator can be used for the accurate results for this type of derivative.

7 WAYS TO GET THE BEST RESULTS FROM PRIVATE TUTORING

Method to solve this type of differentiation

We can’t start with dy/dx in the process of implicit differentiation since an implicit function isn’t of the type y = f(x), but rather f (x, y) = 0. It’s important to be aware of derivative rules like the power rule, addition rule, product rule, exponent rule, quotient rule, chain rule, and so on. The step-by-step method to find the implicit differentiation is mentioned below.

• In the given equation, first, take the derivative of the given equation with respect to x, differentiate each term of f (x, y) = 0.
• Sort the terms that have dy/dx on one side and the ones that don’t have dy/dx on the other.
• Make dy/dx a function of either x or y, or both.

How to evaluate the problems of implicit differentiation?

The steps for performing implicit differentiation have been demonstrated. Is there a certain formula we came across along the way? No!! There is no specific formula for implicit differentiation; instead, to obtain the implicit derivative, we follow the processes mentioned above.

Important Points to Remember About Implicit Differentiation:

• When the function is of the form f(x, y) = 0, implicit differentiation is the process of determining dy/dx.
• Simply differentiate on both sides and solve for dy/dx to discover the implicit derivative dy/dx. However, whenever we are distinguishing y, we should write dy/dx.
• In the process of implicit differentiation, all derivative formulas and techniques must be applied as well.

To understand this concept let’s take some examples.

Example 1

Find the implicit differentiation of x³ + y² = 9

Solution

Step 1: write the given function.

x³ + y² = 9

Step 2: Apply d/dx on both sides of the given equation.

d/dx (x³ + y²) = d/dx (9)

Step 3: Apply the rule of the differentiation.

d/dx (x³) +d/dx (y²) = d/dx (9)

Step 4: Solve the derivative.

3x² + 2y dy/dx = 0

Step 5: Arrange the above equation to get dy/dx.

3x² + 2y dy/dx = 0

2y dy/dx = -3x²

dy/dx = -3x²/2y

Thus, the implicit differentiation of the given function is dy/dx = -3x²/2y.

Example 2

Find the implicit differentiation of 4x² + 2y² = 6y

Solution

Step 1: write the given function.

4x² + 2y² = 6y

Step 2: Apply d/dx on both sides of the given equation.

d/dx (4x² + 2y²) = d/dx (6y)

Step 3: Apply the rule of the differentiation.

d/dx (4x²) +d/dx (2y²) = d/dx (6y)

Step 4: Solve the derivative.

8x + 4y dy/dx = 6dy/dx

Step 5: Arrange the above equation to get dy/dx.

8x + 4y dy/dx = 6dy/dx

4y dy/dx = 6 dy/dx – 8x

so 4y dy/dx – 6 dy/dx = – 8x

(4y -6) dy/dx = -8x

dy/dx = -8x / (4y – 6)

so dy/dx = -8x / 2(2y – 3)

dy/dx = -4x / (2y – 3)

Thus, the implicit differentiation of the given function is dy/dx = -4x / (2y – 3). You can also find the antiderivative or integral of a function using antiderivative calculator.

Example 3

Find the implicit differentiation of x² + y² = 7y² + 7x

Solution

Step 1: Write the given function.

x² + y² = 7y² + 7x

Step 2: Apply d/dx on both sides of the given equation.

d/dx (x² + y²) = d/dx (7y² + 7x)

Step 3: Apply the rule of the differentiation.

d/dx (x²) +d/dx (y²) = d/dx (7y²) + dy/dx (7x)

Step 4: Solve the derivative.

2x + 2y dy/dx = 14y dy/dx +7

Step 5: Arrange the above equation to get dy/dx.

2x + 2y dy/dx = 14y dy/dx +7

2y dy/dx = 14y dy/dx +7 – 2x

so 2y dy/dx – 14y dy/dx = 7 – 2x

(2y -14y) dy/dx = 7 – 2x

dy/dx = (7 – 2x) / (2y -14y)

Thus, the implicit differentiation of the given function is dy/dx = (7 – 2x) / (2y -14y).

Summary

A function defined in terms of both dependent and independent variables, such as X² + 3Y = 0, is an implicit function. Implicit differentiation is a process of finding dy/dx of the given implicit function equation such as f (x, y) = 0. Implicit differentiation has no specific formula to solve the problems rather it has some steps to solve the problems of implicit differentiation.

By following those steps we can easily solve any problem related to this type of differentiation.

The post Implicit differentiation and its use in derivatives appeared first on The Tutor Team.

To start with, you will need to be confident in two other skills, namely basic rearranging of formulae and solving basic linear equations.

In its most basic form, it is how one amount changes as another changes. So, for example, (and it is a classic example!) if you want to make twice as many cakes, you need twice as many ingredients. On that level, it seems to be obvious. However, there are many ways in which the difficulty can be ramped up in GCSE maths.

Direct Proportion

Here is an example of a question at the first level of difficulty:

A recipe requires 3 eggs to make 12 cupcakes. How many eggs do I need to make 36 cupcakes?

Here we are looking at how many times more cupcakes we are looking at making :

So, 3612 = 3

This means we need 3 times more eggs, 3 x 3 = 9

OR to put it another way :

12 Cupcakes 36 Cupcakes
Needs Needs
3 Eggs 9 Eggs

This is an example of DIRECT PROPORTION – as one increases, so does the other at the same rate, or in the same proportion – think how many times more?

HOW TO FACTORISE AND SOLVE QUADRATIC EQUATIONS

Pancakes!

Next up, we get the less obvious increases, again, in direct proportion, but with slightly more difficult numbers involved, but only because the number of times each is increasing isn’t so obvious. It is easier to go through this using a worked example:

A recipe states that you need 200g of flour to make 8 pancakes. How much flour do you need to make 30 pancakes?

This is about adopting a slightly different strategy, based on working out how much flour one pancake needs in order to then multiply it back up by 30. Ask yourself, “How much do I need for one?”

From the first part of the question, we can see that 200 ÷ 8 = 25g of flour per pancake

So, it follows that 25 x 30 is how much we need to make 30 pancakes = 750g.

Decreases in Direct Proportion

Now let’s have a look at an example of a decrease, remaining in direct proportion. In this case, we are looking at as one decreases, so does the other at the same rate, or in the same proportion – think how many times less?

Using the first question we looked at, let’s see how this might work:

A recipe requires 3 eggs to make 12 cupcakes. How many cupcakes can I make if I only have 1 egg?

We have 3 times less in terms of the number of eggs, so we can make three times fewer cupcakes

3 Eggs 1 Egg
Makes Makes
12 Cupcakes 4 Cupcakes

3 REASONS WHY PARENTS BOOK PRIVATE TUTORS

How Much For One?

Once again, let’s do the same with some less obvious numbers, using the “How much for one?” approach.

A recipe for making 16 pancakes uses 800g of flour. How much flour do we need to make 10 pancakes?

So for this – let’s look at the “How much for one?” approach :

If 16 pancakes need 800g of flour, it stands that 1 pancake will need (800÷16)g = 50g

We now know we need 50g per pancake, so 10 pancakes will need 10 x 50 = 500g.

Once you have got to grips with the “how many times more / less?” and “how much for one?” ideas then we can finish with a look at the most common way these questions are framed in a GCSE assessment – the ‘Best Buy” question.

Proportion in Daily Life

Go to any supermarket and you will see all kinds of offers and similar products from different brands. Best Buy questions are all about value for money and being able to compare two very similar products or offers.

There are two main types of question at level 4 / 5 and are as follows:

What we need to ask ourselves here is “how much for one?” – but that leads to “one what?”

We need to identify a common unit of volume in this question that can be applied to both buying options – given the amounts in question, I would suggest finding out how much one-litre costs in each case:

Multipack 6 x 330ml = 1980ml = 1.98 litres
Cost of this is £1.70
So cost per litre = £1.70 ÷ 1.98 L = £0.85858585……, so rounded to 86p per litre.

For the bottle, we have 1.5 litres
The cost of this is £1.50
So cost per litre = £1.50 ÷ 1.50 L = £1.00 per litre.

From the two calculations, we can now compare the costs and can see that the multipack offers better value for money as the cost per litre is less.

INTEGRATION IN MATHS: THE PADDINGTON BEAR APPROACH

Special Offers

Another classic example is the “special offer” scenario and involves the use of percentages and other skills:

Laundry liquid normally costs £6 per bottle.

Two different supermarkets have competing offers as follows:

Tesdi supermarket – 30% off all laundry liquid.

Snozbury’s supermarket – buy 3 get one free on laundry liquid.

Which offers the best value?

For Tesdi, we now have a new price of £6 – 30% = £4.20 per bottle.

For Snozbury’s we are now buying 4 bottles, but only paying for 3.
Total cost = 3 x £6 = £18
This is now spread over the 4 bottles, so cost per bottle is £18 ÷ 4 = £4.50 per bottle

So from this, we can see the 30% off offer from Tesdi is the better value.

For these types of question, the two questions that we need to consider are: “how many times more/less?” and/or “how much for one?” in order to do a direct comparison and are all linked directly to proportion.

A bit about the author of this article, David M:

You can enquire about tutoring with David here

The post Direct and Inverse Proportion appeared first on The Tutor Team.

I often stated that multiplying fractions together is one of the easiest things to do when working with maths.  By simply multiplying the numerators together and multiply the denominators together we do not have to find the lowest common multiple.  It is understood at an early age that the word ‘of’ means times and that a ‘half of a half’ of a pizza is the same as a half times a half, which is a quarter:

½   x   ½    =  ¼

PROBABILITY – A MASTER CLASS

Cross Canceling a Fraction

Cross canceling is often taught as a way of saving time for more challenging fraction work:

Can be cross-canceled in the diagonal going from 18 to 36 by dividing

each number by 18 – striking through 18 and writing 1 and striking through 36 and writing 2.

Sim

ilarly, for the diagonal going from bottom left to top right, we can divide through by 11 to obtain 3 on the bottom and 1 at the top. We then have:

which is simply equal to  1/6

HOW TO REVISE: 5 STUDY TIPS THAT REALLY WORK

Dividing a fraction

Dividing fractions is also very easy if we remember the strategy of  ‘Flip, Kiss, Keep’ which involves inverting the second fraction, replacing the divide with a times (the kiss), and keeping the first fraction the same. Therefore:

2/3 divided by 4/5  becomes

2/3 times 5/4 which is 10/12

which simplifies to 5/6.

Fraction multiplication and division can also successfully achieved by first changing the mixed number into a top-heavy fraction.  For example,

2 ½   x  3 ¾    can be calculated by first converting each mixed number;

I have found that using highlighter pens makes this process a lot easier for younger pupils if they are asked to shade the whole number and the denominator of each fraction first to remind them that the process involves multiplying the big number by the denominator and adding the top number to obtain:

5/2  x   15/4

which becomes  75/8

The most memorable breakthroughs with teaching fractions have happened when large class sizes have considered how best to calculate:

1/3  +  ¼

I will always remember the looks on the faces of my first ever adult students in Warminster who gave up two hours on a Monday evening for an academic year to achieve a cornerstone qualification with maths.  Students told me that during their school days, they got lost with fractions, and the feelings of failure carried on from there with other maths topics.

WHY SHOULD YOU BOOK A MATHS TUTOR?

Fractions in daily life

I instinctively remembered that I had a twenty dollar bill in my wallet as well as a ten pound note.  With both notes on the table, I asked my adult students to say how much money was on the table.  Inevitably, the students quickly responded that we needed to first use the exchange rate between pounds and dollars and suddenly there was a fantastic lightbulb moment for the entire class.  This was especially helped by the exchange rate being one pound equals two dollars at the time and the answer of twenty pounds in total was quickly understood.

We then decided to write the first few times table for 3 and the first few times tables for 4 to find the first number in both lists (12)  – hence finding the lowest common denominator became easy.  I also mentioned the gifted Classical Greeks had found a ‘Golden Rule’ for fractions even before the birth of Christ:

“A fraction remains unchanged if we multiply (or divide) the numerator and denominator by the same number.”

Hence  1/3  became 4/12 and ¼ became 3/12.   Using curvy arrows between the numerators and the denominators really helped.  As we now had a common currency of twelfths, the answer of 7/12 quickly became understood.  After that, it was a fantastic feeling to see every student in the group achieve their target grade. In addition, this led to career opportunities for the nurses and army trainees in this group; this even matched my feeling of seeing lots of ‘A’ level success in future years as a subject leader.

I wonder how the adult evening course students are doing now….

A bit about the author of this article, Chris C:

Chris is a freelance tutor who has 28 years of full-time teaching experience. He was Head of Department in a large Somerset college from 1992 to 2014.

He holds a B.Sc. in Maths & Science and a B.Ed. in Mathematics.

With an enthusiasm and interest in mathematics, Chris can raise self-confidence and belief by making maths fun. He concentrates on setting the language in context and building clear scaffolding in topics based on his insight and understanding of the subject.

You can enquire about tutoring with Chris here

The post Improving number confidence – A fraction at a time appeared first on The Tutor Team.

The fascinating topic of probability is more relent now than ever.  We take probability choices every day of our lives and make critical decisions to enhance our safely and our well-being.  For example, “The chances are if I wear a facial mask, I have a reduced risk of contracting covid”…..”the chances are if I am respectful to the people I meet in life, they will be respectful to me in return”.

Learning Probability

Students learn at an early age that probability values range from zero (impossible) to 1 (certain) with a myriad of events in between.

If we consider probability work in year 7, 8, 9, 10 or 11, it is fascinating to hear the range of responses when you pose the question “If I roll two dice, what is the chance that both dice come down as a 6”.  The most popular incorrect answers include 2 in 12, 1 in 12 and 2 in 6!

The dream solution to finding a way in to probability came unexpectedly and fortuitously when I accompanied a group of Frome College students to a mathematics Master Class at the University of Bath which was run as a series of Saturday mornings.  This was the first time I had seen the impact of having six dice available and a score sheet for the game of ‘ZILCH’.

WHY SHOULD YOU BOOK A MATHS TUTOR?

ZILCH – The Probability Game

Two students rolled one dice each and the highest score would go first.

The points scoring was quickly understood by the students:

• • • • • 3 dice scoring 2 in one go = 200
        • 3 dice scoring 3 in one go = 300
        • 3 dice scoring 4 in one go = 400
        • 3 dice scoring 5 in one go = 500
        • 3 dice scoring 6 in one go = 600
        • 3 dice scoring 1 in one go = 1000
        • 1,2,3,4,5,6 in one go = 1500
        • 1 Five = 50
        • 1 One =  100

The first player to reach 2000 or more points was the winner of ZILCH.

Firstly, each player rolls all six dice at once and places aside the scoring dice.  For example, a score of  2,2,2,  3, 4 and 5 scores 200 points (three twos) plus 50 points (one five) to give 250 points.

Then the player makes a decision to stick with this score or to carry on and ‘gamble’. To gamble, the player would throw again the non-scoring dice, in this case the 3 and the 4.

However, if a player decides to gamble and then does not score in the rolling of the other dice, they score ‘ZILCH’ or as they say in America –  Zilch means nothing.

After that, Zilch became a beautiful way in to my probability work for the next twenty years. ZILCH World Cups were a joy to observe with winners coming up to the whiteboard to put their names on the board before playing someone else.

INTEGRATION IN MATHS: THE PADDINGTON BEAR APPROACH

Learning the probabilities

After a few games, we would discuss the ZILCH dilemma using a ‘Sample Space Diagram’ where a player has two dice remaining and has a score or three hundred or so points. Should I gamble or should I stick?  Because of the points given for One 1 or One 5, students could see that there were 20 possibilities out of 36 for improving their score if a risk was taken.

The Odds

Students could also clearly see that the chances of rolling a six and a six with two dice were one in thirty six! In additions, we would give a round of applause to the first 1,2,3,4,5,6 in a lesson that was witnessed.

I even, on just one occasion in twenty years, witnessed a win on the first go. 1, 1, 1,   1, 1, 1!

In conclusion, ZILCH was instantly one of my favourite ways into teaching a topic in mathematics. After all, it simplified a topic that many students did not at first find easy to understand.  I will always be grateful to the University of Bath for running such sparkling sessions for pupils on an inspired Saturday morning workshop.

A bit about the author of this article, Chris C:

Chris is a freelance tutor who has 28 years of full-time teaching experience. He was Head of Department in a large Somerset college from 1992 to 2014.

He holds a B.Sc. in Maths & Science and a B.Ed. in Mathematics.

With an enthusiasm and interest in mathematics, Chris can raise self-confidence and belief by making maths fun. He concentrates on setting the language in context and building clear scaffolding in topics based on his insight and understanding of the subject.

You can enquire about tutoring with Chris here

The post Probability – A Master Class appeared first on The Tutor Team.

Standard Form is a way to write really large or really small numbers. It is a way to write these numbers in a format that everyone agrees to.

What’s in a name?

Although the idea is to write the numbers into a set format, the name of the format is not always agreed on. You may hear terms such as Standard Index Form, Scientific Notation, Standard Form and SIF. However, they are all the same. The one to watch is Engineering notation which is similar but does not follow the same rules.

INTEGRATION IN MATHS: THE PADDINGTON BEAR APPROACH

Where did they come from?

A famous mathematician and Greek philosopher called Archimedes was thought to have used them initially. He used them to help him try to calculate the size of the universe.

Why do we need it?

As we get more and more advanced we are often dealing with very large numbers. This is not just bank balances! Science is also a key area where large numbers are used. For instance, when talking about the distance between planets or even galaxies we often need to use large numbers.

Equally, as science explores the smaller world such as atoms, particles and quantum physics, then we are often using ridiculously small measurements and we need a way of writing these numbers down.

If we just write these large or small numbers normally (often referred to as ordinary form) then it is often difficult to count or quickly see the number of 0’s. If we miscount or accidentally miss one-off, then it could make a dramatic difference to any calculations we are trying to do. Just imagine if you were expecting £100 but were given £10, you might not be too happy. When numbers are 30 or 40 digits off then it is easy to make a mistake.

What does a number in Standard Form look like?

A number is Standard Form has to follow a set of 3 rules which everyone uses.

You can see that from this that the first part is a number between 1 and 10. This number can be 1 and go up to but not include 10. We sometimes write this as 1≤n<10. The number can be an integer (whole number) or a decimal.

Fun Fact – A little known fact is that this number has a special name. It is called the mantissa! Not a lot of people know that.

The second part is always the “x10” and this is always the same. This is because our place value system is based on 10 so by multiplying by 10 we are moving the digits one place left. For example, 8 becomes 80 and 1.23 becomes 12.3 .

The last part of the number is the power. This is sometimes called the index or exponent. This is placed on the “x10”. This has to be an integer but can be positive or negative. If it is positive, it means we are multiplying by 10 lots of times. So, for example, a power of 3 would multiply the number by 10 three times. If the power is 24, then we are multiplying by 10 24 times (which will make a very large number).

If the power is negative, then we are effectively dividing by 10 that number of times. For example, a power of -6 would suggest we are dividing by 10 six times.

So in the example above we are starting with the number 4.2 and multiplying it by 10 nine times. This means we are really doing

4.2 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x10

 or

4.2 x 1,000,000,000 which is 4,200,000,000

HOW TO FACTORISE AND SOLVE QUADRATIC EQUATIONS

Skills needed for GCSE

For GCSEs, you need to be able to convert numbers to and from Standard Form. If you can multiply and divide numbers by 10 then you should be able to do this without a calculator.

You also need to be able to compare numbers in Standard Form. This is done by looking at the power. The bigger the power, the bigger the number. Remember that small numbers will have a negative power.

Complete calculations with Standard Form. You need to be able to complete simple calculations with Standard Form. This could be both with and without a calculator. Your calculator has a special button on it for Standard Form.

Want your child to succeed? Why not try one of our expert tutors. You can read why this helps here: Why Should I book a Maths Tutor

A bit about Phil:

Phil offers tutoring up to secondary school aged students, up to GCSE, in Maths. Having taught GCSE maths for many years, he can help both Foundation Level and Higher-Level students improve their grades.

He is currently a teacher in a secondary school in Colchester and has been teaching maths for over 22 years. He is a fully qualified teacher with a degree (BSc) in Maths and Computer Science.

You can enquire about tutoring with Phil here

The post Standard Form appeared first on The Tutor Team.

A quick count of the marks in a recent (November 2020) A Level paper showed that 99 out of the 200 marks came from them. That’s nearly 50% of all the marks. Now whilst most pupils can cope with differentiation, nearly everyone finds integration in maths tricky. But why?

What’s the Difference?

The big difference is that differentiation is really the starting point and there are fixed ways of doing it. If you have a function in a particular form there is really only one way to differentiate it. These are nicely learnable. A product uses the product rule, a compound function uses the chain rule etc… There are even nice easy ways to remember them.

Product rule: “Keep Differentiate + Differentiate Keep” what could be simpler?

Integration, on the other hand, has lots of different methods and, whilst each method on its own is very learnable and not at all difficult, not one of them ALWAYS works – there is no silver bullet.

This wouldn’t, on its own, be a problem.

After all, just learning a few different methods isn’t the end of the world. The issue is that spotting which method to use can be very hard. Two very similar looking functions can need very different ways of approaching them. Changing  to , just by sticking in an innocent looking little “squared” changes it from a very simple, just write it down, integration problem to one that requires some serious university level maths.

To make it worse, there can often be more than one method that does work. This sounds great, but the different methods can lead to different forms of the answer achieved. Many an A Level pupil will have spent some considerable time trying to work out where they have gone wrong when their answer doesn’t match the one in the back of the book or the “show that” in the question, only to discover that their answer was correct, just written in a different way.

HOW TO FACTORISE AND SOLVE QUADRATIC EQUATIONS

So how can an A Level pupil “get there”?

The main thing, as in any A Level maths question, is to remember that you are doing an A Level maths question. This means that it will definitely work and will only need the methods that

you have been taught. So DON’T PANIC!

Next, start the question by employing the Paddington Method – give it a really hard stare!  Yes, seriously….

If you look really hard at the function and think “if I differentiated something and ended up here, what must I have started with?” this can often get you most of the way there.

You are used to doing this. No one, probably(!), actually divides numbers – if asked “What is 42 divided by7?” we all think “what do I have to multiply 7 by to get 42”

So can you see where you must have started from?

• Is there a “base function” that must have been involved? Sin must have come from cos, e^x stays the same etc..
• Might I have used the chain rule to get this product of two functions, or maybe the product rule produced this sum.
• Is it a fraction where the top is the differential of the bottom, or the differential of the inside of the brackets of the bottom?
• If the answer to the above is yes then you have almost certainly done most of the work already and just need to think about the multipliers..

Now I am not saying that this method is easy, but it is very useful and it can be practiced. You can get a set of ones that work like this from here and work your way through, seeing how many you can do in 5 minutes. Keep coming back to it and trying again so that you get better. The time spent on this now can save you lots of time and grief in an actual exam.

DEMYSTIFYING FORMULAS IN ALGEBRA

But what if even the hardest Paddington stare doesn’t work?

You’ve eaten all the marmalade sandwiches, tried again and it still doesn’t work. Well, in this case the reason why it doesn’t work will tell you what to do next.

A product that didn’t come from chain rule will need differentiation by parts or a fraction that isn’t from ln or chain rule needs to be flipped with a substitution or separated with partial fractions.

And don’t forget, rule number 1 – it does work, you are doing A Level questions! It’s not like real life where there are times when we can’t solve things. So, if it seems to be going wrong and taking too long you have either made a mistake or are doing it the wrong way…

So the key to integration in maths is…

Practice your Paddington stare and then just apply some common sense.  And like all good Paddington stories, it will work out in the end…

Practice Paddington questions                       Practice Paddington answers

Want your child to succeed? Why not try one of our expert tutors. You can read why this helps here: Why Should I book a Maths Tutor

A bit about Andy:

Andy is incredibly good at breaking down maths into easy to understand chunks, whether it be number bonds with primary pupils, the dreaded fractions with KS3 or the scarily hard Further Maths and undergraduate topics.

With a firm belief that you can teach anyone anything as long as it is approached in the right way, Andy has led a class of “sink” pupils in an inner city comprehensive through a lesson where they all “discovered” Pythagoras’ Theorem (and still remembered it 3 months later!) and has had year 8 pupils working through Year 13 algebraic proofs using multilink cubes.

You can enquire about tutoring with Andy here

The post Integration in Maths: The Paddington Bear Approach appeared first on The Tutor Team.