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Sam G.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Durham University

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About me

Hi, my name's Sam and I'm a mathematics (and physics) student at Durham.

I know for some people the most important thing is just to get through their exams; I can and will help you with this. I can't stress enough, however, how much easier this will become if you thoroughly understand the maths and physics you're doing. I will do my utmost to ensure you do. The main benefit that a comprehensive understanding gives is that you will not need to memorise a huge number of equations as they’ll become obvious or easily derivable. As a result complicated concepts will become clear as you will gain familiarity and develop a natural intuition for the subject.

I have done a fair amount of tutoring before, when I was taking my GCSEs I’d often help members of my class by going through things on the blackboard during lunches and breaks. One friend I tutored over the course of a day and after getting a U in her mock exam, I worked with her to achieve an A grade in the actual exam by explaining concepts according to her learning style -there are few things better than watching the “click” one someone’s face as everything falls into place. During my A levels, I also volunteered to help those doing GCSEs in my school and gave frequent advice and help to A-level friends.

My experience tutoring a range of friends and students alike has given me the skills to relate to tutees and approach problems from their mind-set in a relaxed environment. I’ve also learnt there is no such thing as a silly question, if something doesn’t make sense and you think it should, I urge you to tell me so we can address the issue as soon as possible: please don’t be embarrassed by it. I’d rather be helping you with division of fractions than partial fraction decomposition if you have trouble with the former!

I am always always up for a challenge, so if there’s a particularly tough question, or you want to understand something beyond the syllabus (I have attended a few lectures on the nature of infinity and done an EPQ on the Riemann Hypothesis for example), or perhaps you’re looking into STEP or MAT problems, I’ll be more than happy to help.

See you soon!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £24 /hr
Maths A Level £24 /hr
Chemistry GCSE £22 /hr
Maths GCSE £22 /hr
Physics GCSE £22 /hr
Science GCSE £22 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA
ChemistryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

05/11/2013

CRB/DBS Enhanced

No

Currently unavailable: for regular students

Ratings and reviews

4.9from 18 customer reviews

Sarita (Parent) November 5 2016

First session went well and my daughter found that she could actually make sense of Chemistry! Thanks Sam, keep up the good work.

Maggy (Parent) October 1 2015

Good session, able to implement different teaching styles to explain maths topic.

Rachel (Parent) May 29 2015

Thanks Sam for helping Toby so much. He has gained a great deal in terms of knowledge and confidence. You have been constructive and flexible with the arrangements even taking into account the time differences and our dodgy internet. We will hope to give you good news come August!

Rachel (Parent) April 12 2015

v successful
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Questions Sam has answered

How do you take the derivative of a^x ?

There are two ways you can take the derivative of a^x. 1)  Let y = a^x    now we're trying to find dy/dx 2) ln(y) = xln(a)    take logs of both sides and use log rules 3) (dy/dx)*(1/y) = ln(a)   take the derivative of both sides using the chain rule                                         ...

There are two ways you can take the derivative of a^x.

1) 

Let y = a^x    now we're trying to find dy/dx

2)

ln(y) = xln(a)    take logs of both sides and use log rules

3)

(dy/dx)*(1/y) = ln(a)   take the derivative of both sides using the chain rule                                           on the left hand side.

4)

dy/dx = ln(a)*y              multiply both sides by y

5)

dy/dx =  ln(a) *a^x        realise y= a^x and replace it

Now we're done!

 

Alternatively we could realise that any exponent can be written as e to the power of something with a log in it.

So

1)

y = a^x = (e^ln(a))^x    just rewritting 'a'

2)

y = e^xln(a)                multiplying exponent rule

3)

dy/dx = ln(a)*e^xln(a)   take the derivative of both sides using the chain                                                   rule for the right hand side

4)

dy/dx  = ln(a)*a^x                         substitute back to get desired result

 

 

 

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2 years ago

295 views

How does integration work?

There are a few different types of integration, the most common being Riemann integration. Riemann integration allows you to find the area under the graph of a function between two points. Its definition uses something called a limit and it basically says we can approximate the area underneat...

There are a few different types of integration, the most common being Riemann integration.

Riemann integration allows you to find the area under the graph of a function between two points. Its definition uses something called a limit and it basically says we can approximate the area underneath the graph by adding up areas of rectangles (which is trivially "base times height") so that the width of all the rectangles together go from the first point to the second point and the varying heights of the rectangles goes from the bottom up and "hugs" the curve as best as possible.

Imagine you have a curve and approximate the area underneath it between a and b by finding the area of two rectangles each of width 

(b-a)/2, we realise it's quite a poor approximation, but if you make the approximation with 10 rectangles each of width (b-a)/10 and lots of varying heights we realise the approximation is better. Newton and Leibnitz (the independent founders of calculus) then said "what if we take the width of the rectangles to be really really small, so small that the width of the rectangles approaches zero!" then we realise we'd have an infinite number of rectangles to add up all of varying heights and all of width "essentially zero". 

What is really going on is that we've said let the width be dx and let us add up all the areas of rectangles as the limit of dx approaches zero. This can be seen in the standard lay out of integrals:

f(x) dx is simply the height of the function multiplied with a width (giving rise to the area of a rectangle) the integral sign at the beginning is an elongated S for "sum", hence you sum up all the areas of rectangles.

 
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2 years ago

345 views

Some videos I've made

Prerequisite for addition formula: https://www.youtube.com/watch?v=iHZmi2MHqWc   Proof of formula for sin(A+B):  https://www.youtube.com/watch?v=BEETggUVMrY   Proof of formula for cos(A+B):  https://www.youtube.com/watch?v=gTpNnlRxnIY   Proof of formula for tan(A+B):  https://www.youtube.co...

Prerequisite for addition formula: https://www.youtube.com/watch?v=iHZmi2MHqWc

 

Proof of formula for sin(A+B):  https://www.youtube.com/watch?v=BEETggUVMrY

 

Proof of formula for cos(A+B):  https://www.youtube.com/watch?v=gTpNnlRxnIY

 

Proof of formula for tan(A+B):  https://www.youtube.com/watch?v=D5VJiL0DMWU

 

Some C3 Trig stuff: https://www.youtube.com/watch?v=Gcby6WYAVKg

 

The product rule proof with limits:https://www.youtube.com/watch?v=9FRFRVTUthk

 

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1 year ago

312 views
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