About me Hi, I'm Andrew. I'm currently going into my second year studying maths at the University of Warwick. I find maths incredibly fascinating, as my passion stems from the idea that maths is everywhere and is fundamental to understanding the infrastructure of the universe. At heart I prefer pure maths, but I enjoy pursuing the applied side too. I also have a great passion for science, having been to CERN on a college trip completely opened my eyes as to how exciting cutting edge science and maths really is! At uni, I try to balance the maths by taking some physics modules on the side. I also enjoy playing the French horn, reading and watching films in my free time. How I can help you I am a big fan of tailoring sessions towards the student (which is a massive advantage of 1 on 1 tuition). Everyone's different, and the way people learn is no exception! So if something doesn't sink in right away with the first explanation (be that in class or with me), I'll try to come up with one which will hopefully shed some more light on the area! To aid me, I will also use past exam questions or examples that I myself have prepared prior to the lesson My main goal is for you to understand why a method works, rather than to just learn 'that it just does' and then regurgitate it, otherwise maths turns into a very dull memory game, which is not exactly fun to play... I truly believe that anyone and everyone can find maths fun, but if you don't agree with me, why not give it a go?About me Hi, I'm Andrew. I'm currently going into my second year studying maths at the University of Warwick. I find maths incredibly fascinating, as my passion stems from the idea that maths is everywhere and is fundamental to understanding the infrastructure of the universe. At heart I prefer pure maths, but I enjoy pursuing the applied side too. I also have a great passion for science, having been to CERN on a college trip completely opened my eyes as to how exciting cutting edge science and maths really is! At uni, I try to balance the maths by taking some physics modules on the side. I also enjoy playing the French horn, reading and watching films in my free time. How I can help you I am a big fan of tailoring sessions towards the student (which is a massive advantage of 1 on 1 tuition). Everyone's different, and the way people learn is no exception! So if something doesn't sink in right away with the first explanation (be that in class or with me), I'll try to come up with one which will hopefully shed some more light on the area! To aid me, I will also use past exam questions or examples that I myself have prepared prior to the lesson My main goal is for you to understand why a method works, rather than to just learn 'that it just does' and then regurgitate it, otherwise maths turns into a very dull memory game, which is not exactly fun to play... I truly believe that anyone and everyone can find maths fun, but if you don't agree with me, why not give it a go?

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

Standard DBS Check

12/06/2016Here you will deal with all problems of the same type in a very similar way.

Intuition first tells us that we can’t really find the limit, as both the top (4x^{2}+5), and the bottom (x^{2}-6) parts of the fraction tend to infinity as x does likewise. However, as is sometimes the case, this is the wrong approach as playing with infinities is a dangerous game!

Note that we’re comfortable at guessing the limit of, say 1/x^{2} as x tends to infinity (if you’re not, then notice that as x gets very big, 1/x gets very small – the bigger x is, the smaller 1/x is, and since x > 0, 1/x must always be larger than 0 as it can’t be negative, so 1/x gets closer and closer to 0 as x increases. This is another way of saying that “1/x tends to 0 as x tends to infinity”. The limit of 1/x^{2} is calculated in a very similar way). So in order to get the expression into a format that we’re happy dealing with, it is favourable to divide it by x^{2}/x^{2} so we’re not changing the value of the expression, but this leaves us with (4+(5/x^{2}))/(1-(6/x^{2})). Notice now that we can compute the definite limit of each ‘part’ of the fraction: (4+(5/x^{2})) tends to 4 as x tends to infinity (as 5/x^{2} tends to 0), and (1-(6/x^{2})) tends to 1 as x tends to infinity (as -6/x^{2} tends to 0). So we’re left up with 4/1 = 4. So (4x^{2}+5)/(x^{2}-6) tends to 4 as x tends to infinity.

Note that if you divide by the highest power of x in the expression (over itself) , it will whittle down into something that you’re probably more able to deal with

Here you will deal with all problems of the same type in a very similar way.

Intuition first tells us that we can’t really find the limit, as both the top (4x^{2}+5), and the bottom (x^{2}-6) parts of the fraction tend to infinity as x does likewise. However, as is sometimes the case, this is the wrong approach as playing with infinities is a dangerous game!

Note that we’re comfortable at guessing the limit of, say 1/x^{2} as x tends to infinity (if you’re not, then notice that as x gets very big, 1/x gets very small – the bigger x is, the smaller 1/x is, and since x > 0, 1/x must always be larger than 0 as it can’t be negative, so 1/x gets closer and closer to 0 as x increases. This is another way of saying that “1/x tends to 0 as x tends to infinity”. The limit of 1/x^{2} is calculated in a very similar way). So in order to get the expression into a format that we’re happy dealing with, it is favourable to divide it by x^{2}/x^{2} so we’re not changing the value of the expression, but this leaves us with (4+(5/x^{2}))/(1-(6/x^{2})). Notice now that we can compute the definite limit of each ‘part’ of the fraction: (4+(5/x^{2})) tends to 4 as x tends to infinity (as 5/x^{2} tends to 0), and (1-(6/x^{2})) tends to 1 as x tends to infinity (as -6/x^{2} tends to 0). So we’re left up with 4/1 = 4. So (4x^{2}+5)/(x^{2}-6) tends to 4 as x tends to infinity.

Note that if you divide by the highest power of x in the expression (over itself) , it will whittle down into something that you’re probably more able to deal with

I am going to assume you know what an irrational number is, given that that should have been taught right at the beginning the topic of surds. When I talk about a “rational” number, here I mean “whole number” or “integer”, although a rational number is a number that is not irrational.

There are two types of problem that you could come across when being asked to rationalise the denominator of a fraction:

The first type is when you are asked to rationalise a fraction which has its whole denominator under a square root. For example 1/sqrt(2) or 5/sqrt(8) etc. To rationalise the denominator here, we somehow have to square the denominator to get a whole number. However, we can’t change the value of the fraction! A great way of doing this is therefore by multiplying the fraction with denominator/denominator as you’ll notice that this is just equal to 1: Let’s say we had the fraction n/d (n for numerator, and d for denominator), if we multiply this by d/d, we get nd/d^{2} (which is still equal to our original fraction). If ‘d’ was irrational, e.g. d=sqrt(2), then d^{2} would be rational, e.g. (sqrt(2))^{2}=2, and therefore we have successfully rationalised the denominator! For example, we could multiply 1/sqrt(2) by sqrt(2)/sqrt(2) to get sqrt(2)/( sqrt(2))^{2} = sqrt(2)/2, or 5/sqrt(8) by sqrt(8)/ sqrt(8) to get 5sqrt(8)/(sqrt(8))^{2} = 5sqrt(8)/8.

The second type is a little trickier as it relies on a concept we call “the difference of two squares”. First, let’s consider (x+y)(x-y) = x^{2}-y^{2}; this result is known as “the difference of two squares”, and it is incredibly useful because if either ‘x’ or ‘y’ were irrational, then we now know that (x+y)(x-y) is rational. For example, if x is irrational (e.g. x=sqrt(2)) and y is rational (e.g. y=3), then (x+y)(x-y)= x^{2}-y^{2} =(sqrt(2))^{2}-3^{2}=2-9=-7, or if x is rational (e.g. x=4) and y is irrational (e.g. y=sqrt(5)), then (x+y)(x-y)= x^{2}-y^{2}=4^{2}-(sqrt(5))^{2}=15-5=11. Now let’s return to the second type of problem: you could be asked to rationalise the denominator of a fraction whose denominator comprises of a rational number added to a whole number, for example 1/(1+sqrt(2)) or 3/(sqrt(3)-5) etc. This is definitely more tricky than the first type of problem, but here is where we use what we’ve just learned about “the difference of two squares” to make life a lot easier for us: The denominators of these fractions look a lot like the examples I gave for (x+y) or (x-y) (so 1+sqrt(2) is just (x+y) where x=1 and y=sqrt(2) etc.) so, we probably want to find a way to multiply the denominator of the fraction by the ‘opposite form’ in order to rationalise it. For example, multiplying (1+sqrt(2)) by (1-sqrt(2)) gives us the difference of two squares: 1^{2}-(sqrt(2))^{2}=1-2=-1, and multiplying (sqrt(3)-5) by (sqrt(3)+5) = (sqrt(3))^{2}-5^{2}=3-25=-22. Now we have a way of turning the denominator into a whole number, we are back to thinking about how we can do this without changing the value of the fraction. Using our trick from the first type, you can see that, if we were given a fraction a/(b+sqrt(c)), then multiplying it by (b-sqrt(c))/ (b-sqrt(c)) wouldn’t change the value ( as (b-sqrt(c))/ (b-sqrt(c))=1) but gives us b^{2}-(sqrt(c))^{2} on the bottom, which is b^{2}-c, which is a whole number! Going back to my earlier examples, 1/(1+sqrt(2)) x(1-sqrt(2))/ (1-sqrt(2)) = (1-sqrt(2))/-1 = sqrt(2)-1, and 3/(sqrt(3)-5) x(sqrt(3)+5)/ (sqrt(3)+5)=3(sqrt(3)+5)/-22 = (3sqrt(3)+15)/-22 =(-3sqrt(3)-15)/22.

I am going to assume you know what an irrational number is, given that that should have been taught right at the beginning the topic of surds. When I talk about a “rational” number, here I mean “whole number” or “integer”, although a rational number is a number that is not irrational.

There are two types of problem that you could come across when being asked to rationalise the denominator of a fraction:

The first type is when you are asked to rationalise a fraction which has its whole denominator under a square root. For example 1/sqrt(2) or 5/sqrt(8) etc. To rationalise the denominator here, we somehow have to square the denominator to get a whole number. However, we can’t change the value of the fraction! A great way of doing this is therefore by multiplying the fraction with denominator/denominator as you’ll notice that this is just equal to 1: Let’s say we had the fraction n/d (n for numerator, and d for denominator), if we multiply this by d/d, we get nd/d^{2} (which is still equal to our original fraction). If ‘d’ was irrational, e.g. d=sqrt(2), then d^{2} would be rational, e.g. (sqrt(2))^{2}=2, and therefore we have successfully rationalised the denominator! For example, we could multiply 1/sqrt(2) by sqrt(2)/sqrt(2) to get sqrt(2)/( sqrt(2))^{2} = sqrt(2)/2, or 5/sqrt(8) by sqrt(8)/ sqrt(8) to get 5sqrt(8)/(sqrt(8))^{2} = 5sqrt(8)/8.

The second type is a little trickier as it relies on a concept we call “the difference of two squares”. First, let’s consider (x+y)(x-y) = x^{2}-y^{2}; this result is known as “the difference of two squares”, and it is incredibly useful because if either ‘x’ or ‘y’ were irrational, then we now know that (x+y)(x-y) is rational. For example, if x is irrational (e.g. x=sqrt(2)) and y is rational (e.g. y=3), then (x+y)(x-y)= x^{2}-y^{2} =(sqrt(2))^{2}-3^{2}=2-9=-7, or if x is rational (e.g. x=4) and y is irrational (e.g. y=sqrt(5)), then (x+y)(x-y)= x^{2}-y^{2}=4^{2}-(sqrt(5))^{2}=15-5=11. Now let’s return to the second type of problem: you could be asked to rationalise the denominator of a fraction whose denominator comprises of a rational number added to a whole number, for example 1/(1+sqrt(2)) or 3/(sqrt(3)-5) etc. This is definitely more tricky than the first type of problem, but here is where we use what we’ve just learned about “the difference of two squares” to make life a lot easier for us: The denominators of these fractions look a lot like the examples I gave for (x+y) or (x-y) (so 1+sqrt(2) is just (x+y) where x=1 and y=sqrt(2) etc.) so, we probably want to find a way to multiply the denominator of the fraction by the ‘opposite form’ in order to rationalise it. For example, multiplying (1+sqrt(2)) by (1-sqrt(2)) gives us the difference of two squares: 1^{2}-(sqrt(2))^{2}=1-2=-1, and multiplying (sqrt(3)-5) by (sqrt(3)+5) = (sqrt(3))^{2}-5^{2}=3-25=-22. Now we have a way of turning the denominator into a whole number, we are back to thinking about how we can do this without changing the value of the fraction. Using our trick from the first type, you can see that, if we were given a fraction a/(b+sqrt(c)), then multiplying it by (b-sqrt(c))/ (b-sqrt(c)) wouldn’t change the value ( as (b-sqrt(c))/ (b-sqrt(c))=1) but gives us b^{2}-(sqrt(c))^{2} on the bottom, which is b^{2}-c, which is a whole number! Going back to my earlier examples, 1/(1+sqrt(2)) x(1-sqrt(2))/ (1-sqrt(2)) = (1-sqrt(2))/-1 = sqrt(2)-1, and 3/(sqrt(3)-5) x(sqrt(3)+5)/ (sqrt(3)+5)=3(sqrt(3)+5)/-22 = (3sqrt(3)+15)/-22 =(-3sqrt(3)-15)/22.

I think it’s best I work through an example with you as these problems can vary quite a lot, but the general methods used are the same.

Example: Use the substitution u=x^{2}+5 to find: The integral of (x^{3}/sqrt(x^{2}+5)).dx between the limits of 2 and 1.

So the whole idea of using a substitution here is to simplify the integration for us. The first thing we must do is substitute the given substitution in, otherwise there wouldn’t be much point! In doing this, we also need to replace the .dx in the integration, we do this by finding du/dx and then rearranging for dx; in this particular example, du/dx = 2x, so dx = (1/2x)du, so the integral becomes (x^{3}/sqrt(x^{2}+5)). (1/2x).du =( x^{2}/2sqrt(x^{2}+5)).du. We then use the substitution to get the x’s in terms of u: the numerator, x^{2} becomes u-5 (as u=x^{2}+5), the denominator, 2sqrt(x^{2}+5) becomes 2sqrt(u). Finally, modifying the limits in terms of u: since the top limit is x=2, then this is equivalent to u=2^{2}+5=9, and the bottom limit becomes 1^{2}+5=6.

We now have our rephrased integration problem: Integrate ((u-5)/2sqrt(u)).du between the limits of 9 and 6. Notice we can split up the integrand (the thing we’re integrating): =(u/2sqrt(u))-(5/2sqrt(u)). You know that this is equivalent to 0.5u^{1/2}-5.5u^{-1/2}, which when integrated is (1/3)u^{3/2}-5u^{1/2}. The only thing left to do now is apply the limits: [(1/3)(9^{3/2})-5(9^{1/2})]-[ [(1/3)(6^{3/2})-5(6^{1/2})] = 9-15-2sqrt(6)+5sqrt(6) = __7sqrt(6)-6__

The steps to solving other substitution problems are very similar to the ones detailed above, obviously the manipulations will be different, but the ideas are the same.

I think it’s best I work through an example with you as these problems can vary quite a lot, but the general methods used are the same.

Example: Use the substitution u=x^{2}+5 to find: The integral of (x^{3}/sqrt(x^{2}+5)).dx between the limits of 2 and 1.

So the whole idea of using a substitution here is to simplify the integration for us. The first thing we must do is substitute the given substitution in, otherwise there wouldn’t be much point! In doing this, we also need to replace the .dx in the integration, we do this by finding du/dx and then rearranging for dx; in this particular example, du/dx = 2x, so dx = (1/2x)du, so the integral becomes (x^{3}/sqrt(x^{2}+5)). (1/2x).du =( x^{2}/2sqrt(x^{2}+5)).du. We then use the substitution to get the x’s in terms of u: the numerator, x^{2} becomes u-5 (as u=x^{2}+5), the denominator, 2sqrt(x^{2}+5) becomes 2sqrt(u). Finally, modifying the limits in terms of u: since the top limit is x=2, then this is equivalent to u=2^{2}+5=9, and the bottom limit becomes 1^{2}+5=6.

We now have our rephrased integration problem: Integrate ((u-5)/2sqrt(u)).du between the limits of 9 and 6. Notice we can split up the integrand (the thing we’re integrating): =(u/2sqrt(u))-(5/2sqrt(u)). You know that this is equivalent to 0.5u^{1/2}-5.5u^{-1/2}, which when integrated is (1/3)u^{3/2}-5u^{1/2}. The only thing left to do now is apply the limits: [(1/3)(9^{3/2})-5(9^{1/2})]-[ [(1/3)(6^{3/2})-5(6^{1/2})] = 9-15-2sqrt(6)+5sqrt(6) = __7sqrt(6)-6__

The steps to solving other substitution problems are very similar to the ones detailed above, obviously the manipulations will be different, but the ideas are the same.