Adam S. A Level Further Mathematics  tutor, A Level Maths tutor, A Le...

Adam S.

Currently unavailable: for regular students

Degree: Artificial Intelligence (Masters) - Edinburgh University

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About me

I love maths and physics. I loved them as a child, I loved them at university and I even loved them at school. I hope that during my tutorials you will grow to love these subjects too.

About me

I graduated from the University of York this year, where I studied Theoretical Physics. Right now, I'm doing Artificial Intelligence at the University of Edinburgh.

Physics

Physics is, in some ways, the most important of the sciences. It's an incredibly broad subject in which the goal is to try to explain why things behave in the way that they do. In physics tutorials, I will always try to encourage you to ask questions, but by the end, you should have enough confidence to begin tackling these questions yourself.

At school, physics exams have a lot to do with remembering which equation to apply to the problem that they give you. You can get good at this with practice, but if you learn to think in the right way and ask yourself the right questions, you'll find you can be half-way to the answer before you've finished reading the question!

Maths

Maths can often feel like learning recipes. "If they ask this, follow these steps. If they ask this, follow these." This can get you full marks in an exam, but it's absolutely no fun, and if you don't understand what you're doing, a slightly different question or a simple mistake will leave you in a mess. In maths tutorials, I will try to help you understand how everything fits together, to see why maths works, and also to see why maths is important, through examples and sample problems.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr

Qualifications

QualificationLevelGrade
PhysicsMasters Degree1*
MathematicsA-LevelA*
Further MathematicsA-LevelA
ChemistryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

Ratings and reviews

5from 5 customer reviews

Chloe (Student) January 19 2016

The tutorial was fun :D

Chloe (Student) January 12 2016

I have more of an understanding of Maths every tutorial.

Chloe (Student) January 5 2016

I always feel stupid at college, and always feel better after having a tutorial.

Chloe (Student) December 15 2015

Enjoyed the tutorial and already feeling more confident.
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Questions Adam has answered

Given that abc = -37 + 36i; b = -2 + 3i; c = 1 + 2i, what is a?

Substituting the given values for b and c into the equation for abc, a(-2 + 3i)(1 + 2i) = -37 + 36i Multiplying out the terms in brackets, a(-2 - 4i + 3i - 6) = -37 + 36i Collecting like terms and multiplying through by -1, a(8 + i) = 37 - 36i The complex number a can be represented as m + ...

Substituting the given values for b and c into the equation for abc,

a(-2 + 3i)(1 + 2i) = -37 + 36i

Multiplying out the terms in brackets,

a(-2 - 4i + 3i - 6) = -37 + 36i

Collecting like terms and multiplying through by -1,

a(8 + i) = 37 - 36i

The complex number a can be represented as m + ni, where m and n are constants we need to find.

(8 + i)(m + ni) = 37 - 36i

Multiplying out the terms in brackets,

8m + 8ni + mi - n = 37 - 36i

Collecting like terms and equating the real and imaginary parts, we end up with two simultaneous equations for m and n.

8m - n = 37 (from real part)

8n + m = -36 (from imaginary part)

Rearranging the first equation, we find that n = 8m - 37. Substituting this into the second equation,

8(8m - 37) + m = -36

64m - 296 + m = -36

65m - 296 = -36

65m = 260

m = 4

Subsituting this value for m back into the second equation,

8n + 4 = -36

8n = -40

n = -5

Putting it all together,

a = 4 - 5i

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1 year ago

272 views

Given that y = exp(2x) * (x^2 +1)^(5/2), what is dy/dx when x is 0?

y = e2x(x2+1)5/2 The first step is to calculate dy/dx. We can do this by splitting y into two parts and using the chain rule of differentiation: y = uv where u = e2x and v = (x2+1)5/2. We now differentiate u and v separately with respect to x. Here, remember that df(g(x))/dx is equal to df/...

y = e2x(x2+1)5/2

The first step is to calculate dy/dx. We can do this by splitting y into two parts and using the chain rule of differentiation:

y = uv

where u = e2x and v = (x2+1)5/2. We now differentiate u and v separately with respect to x.

Here, remember that df(g(x))/dx is equal to df/dg times dg/dx. So,

du/dx = 2e2x and dv/dx = 5/2 (x2+1)3/2 2x = 5x(x2+1)3/2

Using the chain rule,

dy/dx = u dv/dx + v du/dx

= e2x 5x(x2+1)3/2 + 2e2x (x2+1)5/2.

Now, when x=0, the first term disappears, since it's multipled by x. exp(0) is equal to 1, as is 15/2, so the second term reduces to 2 times 1. The answer is therefore 2.

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1 year ago

331 views
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