Edmond W. GCSE Maths tutor, A Level Maths tutor

Edmond W.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Birmingham University

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About me

Hi! My name is Ed Wong and I'm a third year Mathematics student at the University of Birmingham. I have a great passion for maths, and would love to tutor anyone who is finding it difficult. I would say I am great at explaining my solutions to anyone- no matter the mathematical ability!

I achieved A* in GCSE maths, and A* in both Mathematics and Further Mathematics at A level.

Whether if you want to just pass your A-Levels, or even go on to further study, I will ensure that you have an enjoyable time while you learn with me.

I look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
Further MathsA-LevelA*
MathsA-LevelA*
ArtA-LevelA
PhysicsA-LevelB
EPQA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

General Availability

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Ratings and reviews

5from 6 customer reviews

Lisa (Parent) July 21 2016

Ed, thanks for keeping Bryce so motivated during the summer holidays. You have cover so much in so little time and made it enjoyable. Thank you for putting so much time into preparing the lessons and homework. Thanks, you are a star! I would highly recommend this tutor to anyone needed to boost their maths understanding.

Bryce (Student) July 19 2016

Very well prepared for each lesson, plans in advance and sets homework to ensure that what has been taught has sunk in. Very good, would recommend him to anyone struggling with A level Mathematics.

Lisa (Parent) July 18 2016

Great job, Edmond. Bryce is enjoying your lessons. You are well planned and have adapted each session to his needs. Thanks for setting homework too. Will be booking many more sessions. Lisa

Bryce (Student) July 21 2016

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Questions Edmond has answered

At t seconds, the temp. of the water is θ°C. The rate of increase of the temp. of the water at any time t is modelled by the D.E. dθ/dt=λ(120-θ), θ<=100 where λ is a pos. const. Given θ=20 at t=0, solve this D.E. to show that θ=120-100e^(-λt)

When solving any differential equation, the first method to consider is the seperation of variables. This is the simplest method and, conveniently, it works in this case. To seperate variables: 1. Put all of the same type of variables on their own side. In this case, our two variables are t a...

When solving any differential equation, the first method to consider is the seperation of variables. This is the simplest method and, conveniently, it works in this case. To seperate variables:

1. Put all of the same type of variables on their own side. In this case, our two variables are t and θ. So in this case, divide everything by (120-θ) and multiply everything by dt. This leaves us with dθ/(120-θ)=λdt.

2. The next step is to integrate both sides.
              On the left side, we have a variable ^-1 so our answer must have a natural log of the denominator. So lets work backwards, let's assume our answer is ln(120-θ). If we differentiate this, this gives us -1/(120-θ), which is what we started with, but multiplied by -1. Therefore, we must have -ln(120-θ).
              On the right side, this is simply integrating a constant (as λ is a constant), so we have λt +c. REMEMBER TO ADD OUR CONSTANT OF INTEGRATION!
              This leaves us with: -ln(120-θ)=λt +c.

3. Next we find what c is. This is found by applying the initial conditions given in the question, i.e. θ=20, t=0. Plugging this in and rearranging for c, we have c=-ln(100), leaving us with: -ln(120-θ)=λt-ln(100).

4. We now have our final expression, but it isn't in the correct form. Therefore, we must use the rules of exponentials to manipulate it. This will leave us with the correct form of:
θ=120-100e^(-λt).

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5 months ago

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