Edmond W. GCSE Maths tutor, A Level Maths tutor

Edmond W.

Unavailable

Mathematics (Bachelors) - Birmingham University

5.0
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24 reviews

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

34 completed lessons

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!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!

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Ratings & Reviews

5from 24 customer reviews
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Thea (Student)

February 16 2017

Explained very well in detail, and answered all my questions. :)

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Larry (Parent from Woodbridge)

February 2 2017

he likes maths

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Lisa (Parent from Berkhamsted)

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.

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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.

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Qualifications

SubjectQualificationGrade
Further MathsA-level (A2)A*
MathsA-level (A2)A*
ArtA-level (A2)A
PhysicsA-level (A2)B
EPQA-level (A2)A

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr

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 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).

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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2 years ago

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