£30 /hr

Vandan P.

Degree: Physical Natural Sciences (Bachelors) - Cambridge University

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#### Subjects offered

SubjectLevelMy prices
Chemistry A Level £30 /hr
Further Mathematics A Level £30 /hr
Maths A Level £30 /hr
Physics A Level £30 /hr
Chemistry GCSE £30 /hr
Maths GCSE £30 /hr
Physics GCSE £30 /hr
Science GCSE £30 /hr
-Oxbridge Preparation- Mentoring £30 /hr
.MAT. Uni Admissions Test £30 /hr
.STEP. Uni Admissions Test £30 /hr

#### Qualifications

MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
 CRB/DBS Standard ✓ 05/09/2014 CRB/DBS Enhanced No

#### Ratings and reviews

4.9from 29 customer reviews

Molly (Student) May 21 2017

Another great lesson with very clear explanations. I would highly recommend!

Jamie (Student) April 18 2017

Really great session, helped a lot

Jamie (Student) April 6 2017

Really great, covered the basics and slowly moved onto the harder concepts. 10/10 would recommend

Paula (Parent) January 18 2017

Vey kindly helped Ben at short notice and very clearly explained the subject.
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### How to sum an arithmetic progression?

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,......

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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2 years ago

1069 views

### What is sin(x)/x for x =0?

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 ...

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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2 years ago

678 views

### Why is Kinetic Energy mv^2/2?

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 = mv2/2 + 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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2 years ago

636 views
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