__About Me__

I am an **enthusiastic **and **approachable **physics student at Oxford University. For as long as I can remember I have taken an active interest in **science **and **maths, **and I have always been keen to pass my enthusiasm onto others.

I am also a very sporty person, and have a wealth of teaching experience from coaching children as young as 4 to play tennis. I also spent a year tutoring maths A level to younger students whilst at sixth form college.

As a tutor I will always search for **imaginitive **and **intuitive** ways to help my students fully **understand **a topic which they are struggling with, with the aim that they should be able to explain the answer to me by the time we are finished. There will be a strong focus on **student interaction **in my tutorials to ensure that you are getting the most out of our time together.

__Subjects Covered__

I am currently offering tutoring for maths GCSE, as well as maths and further maths AS and A levels.

__Oxford Physics Entrance Exam__

As I recently sat and passed the 'PAT' test for admission into Oxford university I am able to offer tutorials on this subject. This could be particularly beneficial as there is a general lack of good online resources for preparing for this test.

__What Now?__

Please do get in touch through this website if you have any queries relating to my tutoring, and hopefully I will speak to you soon!

__About Me__

I am an **enthusiastic **and **approachable **physics student at Oxford University. For as long as I can remember I have taken an active interest in **science **and **maths, **and I have always been keen to pass my enthusiasm onto others.

I am also a very sporty person, and have a wealth of teaching experience from coaching children as young as 4 to play tennis. I also spent a year tutoring maths A level to younger students whilst at sixth form college.

As a tutor I will always search for **imaginitive **and **intuitive** ways to help my students fully **understand **a topic which they are struggling with, with the aim that they should be able to explain the answer to me by the time we are finished. There will be a strong focus on **student interaction **in my tutorials to ensure that you are getting the most out of our time together.

__Subjects Covered__

I am currently offering tutoring for maths GCSE, as well as maths and further maths AS and A levels.

__Oxford Physics Entrance Exam__

As I recently sat and passed the 'PAT' test for admission into Oxford university I am able to offer tutorials on this subject. This could be particularly beneficial as there is a general lack of good online resources for preparing for this test.

__What Now?__

Please do get in touch through this website if you have any queries relating to my tutoring, and hopefully I will speak to you soon!

No DBS Check

When solving a quadratic equation like this it is useful to write it in the form (x+a)(x+b)=0, as this is simply saying 'two numbers multiplied together equal zero'. A general rule in maths is that whenever two numbers multiply together to equal zero, either one or both of those numbers equal zero (try multiplying any number you can think of by zero and see what you get!)

The next question is how do I write x^2+4x+3 in the form (x+a)(x+b)? The answer to this is simple, you need to search for two numbers which multiply together to make three, and also add together to make 4. In this case the answer is 3 and 1, so we can re-write our original equation as (x+3)(x+1)=0. Now because this **product** is equal to zero, we can write that either x+3=0 or x+1=0 (because when a product equals zero either one or both of the numbers being multiplied must be zero).

Starting with x+3=0, subtracting 3 from both sides gives us our **first **solution: x=-3.

We can now subtract 1 from both sides of x+1=0, giving us our **second** solution: x=-1.

As you can see, there are two solutions to this equation! If you find this hard to believe, you can check that both solutions are correct by replacing all of the x's in the original equation with each solution in turn, and they should both equal zero!

When solving a quadratic equation like this it is useful to write it in the form (x+a)(x+b)=0, as this is simply saying 'two numbers multiplied together equal zero'. A general rule in maths is that whenever two numbers multiply together to equal zero, either one or both of those numbers equal zero (try multiplying any number you can think of by zero and see what you get!)

The next question is how do I write x^2+4x+3 in the form (x+a)(x+b)? The answer to this is simple, you need to search for two numbers which multiply together to make three, and also add together to make 4. In this case the answer is 3 and 1, so we can re-write our original equation as (x+3)(x+1)=0. Now because this **product** is equal to zero, we can write that either x+3=0 or x+1=0 (because when a product equals zero either one or both of the numbers being multiplied must be zero).

Starting with x+3=0, subtracting 3 from both sides gives us our **first **solution: x=-3.

We can now subtract 1 from both sides of x+1=0, giving us our **second** solution: x=-1.

As you can see, there are two solutions to this equation! If you find this hard to believe, you can check that both solutions are correct by replacing all of the x's in the original equation with each solution in turn, and they should both equal zero!

At first, you might think that it is possible to perform this integral simply by inspection, using the 'backwards chain rule'. This method would consist of adding one to the power, to get cos^{3}(x), then dividing by the new power and the derivative of the function, giving you -(1/3sin(x))cos^{3}(x). However, once you have performed an integration it is always wise to check your result by differentiating to see if you get your starting function back. In this case, it is clear that differentiating -(1/3sin(x))cos^{3}(x) does not give cos^{2}(x), because you have to use the quotient rule to differentiate cos^{3}(x)/sin(x).

This means that a different approach is required to perform the integration, and that is to use the trig identity cos^{2}x=1/2+(1/2)cos(2x) to change the integrand to something which can be integrated easily. It is then simple to integrate 1/2 +(1/2)cos(2x) using the familiar method, giving the correct answer of (1/2)x+(1/4)sin(2x)+c (not forgetting the constant of integration!).

Similarly, sin^{2}(x) can be integrated quickly using the trig identity sin^{2}(x)=1/2-(1/2)cos(2x), so these two identities are definitely worth memorizing!

At first, you might think that it is possible to perform this integral simply by inspection, using the 'backwards chain rule'. This method would consist of adding one to the power, to get cos^{3}(x), then dividing by the new power and the derivative of the function, giving you -(1/3sin(x))cos^{3}(x). However, once you have performed an integration it is always wise to check your result by differentiating to see if you get your starting function back. In this case, it is clear that differentiating -(1/3sin(x))cos^{3}(x) does not give cos^{2}(x), because you have to use the quotient rule to differentiate cos^{3}(x)/sin(x).

This means that a different approach is required to perform the integration, and that is to use the trig identity cos^{2}x=1/2+(1/2)cos(2x) to change the integrand to something which can be integrated easily. It is then simple to integrate 1/2 +(1/2)cos(2x) using the familiar method, giving the correct answer of (1/2)x+(1/4)sin(2x)+c (not forgetting the constant of integration!).

Similarly, sin^{2}(x) can be integrated quickly using the trig identity sin^{2}(x)=1/2-(1/2)cos(2x), so these two identities are definitely worth memorizing!