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About me: I am a thirdÂ year undergraduate at Corpus Christi College Cambridge. I am studying Natural Sciences and achieved first class honours in both first and second year. In second year I did Maths and Physics andÂ inÂ first year I did Physics, Chemistry, Computer Science and Maths.Â I have wanted to study physics since about the age of 10 which means that I have invested large amounts of time in making sure that I know the subject as thoroughly as I can - and hopefully this will come across in tutorials. I also spent the past summer working on isaacphysics.org an online physics resource designed to help GCSE to A level students learn physics and maths - this experience will hopefully improve the quality of my tutoring. In terms of tutoring, I have done a reasonable amount of online tutoring as well as plenty of informal tutoring with peers. The session: With my tutorials - I will try to see where the gaps in your knowledge and understanding are and then help you to fill those. I tend, rather than teaching only what is required to pass an exam, to explain topics more thoroughly. This means that you should be able to tackle any problem and not just the specific types of question that appear in exams.Â Meet the tutor sessions: Firstly, feel free to schedule a meet the tutor session with me and I will see what I can do to try and help you! Do drop me a webmail first though to check that I will be free In my meet the tutor session I ask basic information such as what you're studying and what boards. I'll also ask if there are any particular areas that you're struggling with. We will then try to find a time that's good (if you want me to tutor you) and decide what material we will cover in the first few sessions. Other qualifications: As well as A levels, I participated in Olympiads in Maths, Physics, Chemistry and Computer Science. I also took an AEA in maths (achieving distinction) and STEP 2 & 3 (achieving a 2 and a 1 respectively). - Thus if you would like help with any of these I would be happy to help with these as well.

CRB/DBS Standard |
âœ“ |
05/09/2014 |

CRB/DBS Enhanced |
No |

4.9from 21 customer reviews

Paula (Parent) January 18 2017

Vey kindly helped Ben at short notice and very clearly explained the subject.

Alex (Student) December 8 2016

I threw a topic at him that he wasn't so familiar with but he still gave me a really clear explanation of all the areas I needed to go over!

Alex (Student) November 3 2016

Went over a theory I was finding hard in A level physics. Makes a lot more sense now!

Molly (Student) October 13 2016

Great 2nd lesson. Explanations were clear & Vandan was very thorough in making sure that I fully understood every concept before moving on. I would highly recommend!

Whilst it is very easy to just plug the numbers into the formula you are
given on your formula sheet there is a much faster and more intuitive method of
doing this.
Â
If we have n numbers beginning at a with an increment d then we can write our
sequence as follows:
Â
a, a+d, a+2d, a+3d,......

Whilst it is very easy to just plug the numbers into the formula you are given on your formula sheet there is a much faster and more intuitive method of doing this.

If we have n numbers beginning at a with an increment d then we can write our sequence as follows:

a, a+d, a+2d, a+3d,...., a+(n-1)d

If we wish to sum these, we can note that all of the terms pair up either side of the middle term giving pairs equal to 2a+(n-1)d.

We know that we must have n/2 of these pairs as we can only pair up values either side of the middle value (if it exists, which it does exclusively for odd n).

Thus we must just multiply our average term by the number of terms:

n/2 * (2a + (n-1) d)

Whilst this is just the formula you are given, if you know what it is doing then that enables you to take shortcuts, such as - if you are given the first and last term then you can simply replace 2a + (n-1)d with a+L.

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I'm going to show the answer to this question in two different ways. - The
first is perhaps more obvious but the second is much more elegant.
Taylor series expansion:
Â
Using Taylor expansion (or your trusty A level formula sheet) you can show
that sin(x) = x - x^3/3! + x^5/5! + Re ( (-i)^n ...

I'm going to show the answer to this question in two different ways. - The first is perhaps more obvious but the second is much more elegant.

Taylor series expansion:

Using Taylor expansion (or your trusty A level formula sheet) you can show that sin(x) = x - x^3/3! + x^5/5! + Re ( (-i)^n * x^n / n! )

Thus dividing through by x:

sin(x)/x = 1 - x^2/3! + x^4/5! +...

if we then replace x by 0:

sin(0)/0 = 1 - 0 + 0 +... where ... here is all 0.

thus sin(0) / 0 = 1.

The other much faster way of doing this is using l'Hopital's rule which states that for a limit lim (f(x)/g(x)) = lim (f'(x) / g'(x)) for the same limit.

Thus lim[x-> 0] (sin(x) / x) = lim[x->0] (cos x / 1) = 1.

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Whilst this proof is beyond the scope of A level physics, it is well within
the scope of A level Maths as it relies solely on the chain rule.
First let us note that F*x = W where F is force, x is distance and W is work
or energy.
However if we have a varying force, we must sum the Force in dx...

Whilst this proof is beyond the scope of A level physics, it is well within the scope of A level Maths as it relies solely on the chain rule.

First let us note that F*x = W where F is force, x is distance and W is work or energy.

However if we have a varying force, we must sum the Force in dx sized chunks, ie. W = int ( F dx) and it is from there we get our proof.

F = ma (from Newton II)

a = dv/dt

F = m*dv/dt

Putting this in our integral:

int(F*dx) = int (m * dv*dx/dt)

We also know that v = dx/dt.

Therefore int(m*dv*dx/dt) = int(m*dv*v) = int(mv dv).

Taking the integral we can see that W = mv^{2}/2 + C where C is dependant on the initial speed we take the work done from. If initial speed is 0 then C also becomes 0 and we get the well known formula.

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