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Xavier D.

Currently unavailable: until 23/05/2016

Degree: Computer Science (Masters) - University College London University

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About me

My background

I study postgraduate Computer Science at University College London, and before that I graduated from the University of Edinburgh reading Philosophy & Economics.

I truly believe everybody has the potential to excel in mathematical disciplines, since we all speak and understand language. We combine symbols into new sentences according to grammar, we perform logical deductions, and we seek to invoke pictures in each others' minds. This is precisely what mathematics is about too.

I have volunteered as a teaching assistant in a GCSE maths class, and I have tutored various students for GCSE and A-level maths, physics and economics.

My approach to teaching

But I understand mathematical subjects can be daunting, to say the least! So I am all too happy to explain a concept from several angles, until it clicks with the student and they are able to describe it in their own words.

I think it's often good to begin with a casual discussion of the topic in question, to get a sense of why it's important, placing it in the wider body of human knowledge. Then it comes alive, beyond just being an examinable topic, making it much easier to work through concrete problems.

We can work from a particular exam syllabus, textbook, or simply a chosen topic.

I really get a kick out of maths/science and I enjoy sharing that with others. Most importantly, I want sessions to be fun because this makes it so much easier to learn effectively.

Next steps

Please feel free to get in touch with any queries. You can reach me via the MTW webmail system or arrange a free "Meet The Tutor" session to find out if I'm the right tutor for you!

Subjects offered

SubjectLevelMy prices
Economics A Level £20 /hr
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Philosophy A Level £20 /hr
Philosophy and Ethics A Level £20 /hr
Physics A Level £20 /hr
Chemistry GCSE £18 /hr
Economics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Economics IB £20 /hr
Maths IB £20 /hr
Physics IB £20 /hr
Maths 13 Plus £18 /hr
Science 13 Plus £18 /hr

Qualifications

QualificationLevelGrade
IB SL ChemistryBaccalaureate6
IB SL EnglishBaccalaureate7
IB SL ItalianBaccalaureate5
IB HL MathsBaccalaureate6
IB HL PhysicsBaccalaureate6
IB HL EconomicsBaccalaureate6
Economics & PhilosophyBachelors Degree1st Class
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

23/05/2016

Questions Xavier has answered

What are differential equations, and why are they important?

Let's begin by reviewing a fairly simple (linear) equation: (1) y = 2x + 3 (1) describes the relationship between two variables, x and y. Given a value for one, we can use(1) to work out the value of the unknown. For instance, when x=4, y=11. Differential equations are more general. They inc...

Let's begin by reviewing a fairly simple (linear) equation:

(1) y = 2x + 3

(1) describes the relationship between two variables, x and y. Given a value for one, we can use (1) to work out the value of the unknown. For instance, when x=4, y=11.

Differential equations are more general. They include derivative terms, denoted dy/dx or y'. Let us consider an example:

(2) y' = dy/dx = ky

While (1) described the relation between two variables, (2) describes the relation between y's slope (with respect to x) and y itself. In English, it says that "y's slope is directly proportional to y." Unlike (1), the solution to (2) is a function y, that satisfies the above description. But what kind of function has these properties? Let us solve (2) to find out:

* Take all the 'x' terms to one side, and all the ‘y’ terms to the other side:

dy/dx = ky

dy = ky dx

dy/y = k dx

(1/y) dy = k dx

* Take the integral of both sides (we omit some details about definite integrals and boundary conditions, for simplicity):

INTEGRAL( (1/y) dy) = INTEGRAL(k dx)

* Evaluate the above integral:

ln(y) = kx

* Which is true if and only if:

ekx = y

Thus we have shown that the exponential function y(x) = ekx satisfies the differential equation (2). This is what we should expect, since (2) said "y's slope is directly proportional to y" and this is exactly how the number e is defined.

In further study, you will find that differential equations like (2) have applications in diverse fields such as: in Economics, to model growth rates and continuously compounded interest; Physics, to model radioactive decay or damped oscillations; in Biology to model bacteria populations. Now let us consider a more difficult example:

(3)2ψ/∂x2 = (1/v2)(∂2ψ/∂t2)

(3) looks a little daunting but it is not much more complicated than (2). The solution to (3) needs to be a function ψ(x, t). x stands for a spatial direction, and t is time. v stands for ‘speed.’ We are given details about how ψ’s second (partial) derivatives, with respect to x and t, relate to each other. (3) is known to Physicists as the 'Wave Equation', because it can be derived by carefully studying springs in oscillation and sine/cosine waves like (4) are solutions to it.

(4) ψ = Asin(kx - ωt)

Let us verify this.

* Calculate (4)’s second partial derivative with respect to x (we will need the Chain Rule):

(5)2ψ/∂x2= -k2Asin(kx - ωt)

* Calculate (4)’s second partial derivative with respect to t:

(6)2ψ/∂t2= -ω2Asin(kx - ωt)

* We also need the following result (known as a dispersion relation). In physics, It can be shown that:

(7) ω = kv

where ω is ‘angular frequency’, k is ‘wavenumber’ and v is the speed that a wave propagates at.

* Substituting (5) and (6) into (3) gives:

-k2Asin(kx - ωt) = -(1/v2)ω2Asin(kx - ωt)

-k2 = -(1/v2)ω2

k2/ω2 = 1/v2

v2 = ω2/k2

kv = ω

* Which we know to be correct because of result (7), which follows from the definitions of ‘angular frequency’, ‘wavenumber’ and ‘wave speed’.

Hence the function ψ(x, t) = Asin(kx - ωt) is a solution to the differential equation (3). In fact, Schrödinger’s equation is a famous wave equation, incorporating complex numbers, whose solution is the ‘wavefunction’ of a system of particles. So I hope you can see, from this brief introduction, that we can get a lot of mileage out of this concept of ‘differential equations’!

 

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