I cannot praise this tutor enough. Highly skilled in mathematics, teaches at the student's pace, patient and super efficient. A conscientious person who responds immediately to messages. The lesson was tailor made to suit my own needs. Mayur is excellent at pin pointing math problems quickly - taking things back to basic principals where needed. A most enjoyable tutorial - I would highly recommend this very gifted tutor to students of every ability.

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Joseph has been the most fantastic help with GCSE revision this Easter. He's given up a huge amount of time, is encouraging, flexible, easy to deal with, focussed, targeted - the all round perfect tutor. His greatest achievement is to have taught my son how to enjoy maths. A lifetime's terror of the subject has turned into enjoyment. My son said he'll miss maths and he thinks he has a good chance of getting a high grade. Unthinkable before Joseph started helping him a few months ago.

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Extremely good attitude and commitment to the job! Tutor has been very consistent in trying to improve my son’s understanding of the subject. Really patient and understanding of my son’s needs in the subject, explains everything accordingly. Exam went well and my son greatly appreciates the help he has received. Highly recommend –

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Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your tutorials are recorded. Make the most out of your live session, then play it back after

Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

Let x = 0.2(54) = 0.2545454...
Then 100x = 25.454545... = 25.4(54)
So 99x = 100x - x = 25.4(54) - 0.2(54) = 25.2
Hence 990x =252
Therefore x = 252/990 = 14/55

Answered by Matt H.

Studies Mathematics at Warwick

This question requires integration by parts, using the formula:
Integral(u dv) = u v - integral(v du)
This is applied to find the integral of Ln(x) by writing Ln(x) as 1 * Ln(x), u is then Ln(x) and dv is 1.
Differentiating u=Ln(x) gives you du=1/x. Integrating dv=1 gives you v=x.
Then substituting into formula gives you: Integral(Ln(x)) = xLn(x) - Integral(x*1/x) = xLn(x) - Integral(1)
Therefore Intergral(Ln(x)) = xLn(x) - x + C, Where C is the integration constant

Answered by Isaac D.

Studies MSc in Physics at Birmingham

Common exam question: fully expand and simplify (x+3)(x-2)
Student asks: how should I best answer the question? How do I do it step by step?
My answer:

First expand each individual part of the brackets separately.

I.e. (x+3)(x-2) --> x^{2}+3x-2x-6

Then simplify the answer by working out if any 'x' components with the same power can be cancelled down

I.e. x^{2}+x-6

I'd explain to the student that all these steps are needed to get full marks, and that if you write the final answer straight away, you won't get the marks. I hope this is ok, I wasn't sure what you were asking, whether it's a question a student would ask or a question a student would face?? I'm happy to provide more information if needed.

First expand each individual part of the brackets separately.

I.e. (x+3)(x-2) --> x

Then simplify the answer by working out if any 'x' components with the same power can be cancelled down

I.e. x

I'd explain to the student that all these steps are needed to get full marks, and that if you write the final answer straight away, you won't get the marks. I hope this is ok, I wasn't sure what you were asking, whether it's a question a student would ask or a question a student would face?? I'm happy to provide more information if needed.

In short, **a radian is an angular measurement**, very much like a degree, except one radian is a lot bigger than one degree, and it has some special properties. A radian can often seem like an unnecessarily complicated alternative to the degree measurement, and it's easy to forget why it's so useful!
While **a circle is 360 degrees, a circle is also 2π radians** (that is 2pi, or about 6.283). This means one radian is approximately 57.30 degrees. Not a whole number, or even a rational number! Indeed, this is one of the confusing things about radians! But this conversion is the most important thing to remember.
Two properties of the radian are particularly important;
1) A simple law of circles states that the circumference equals the radius multiplied by 2pi. Because 2pi is also the number of radians in a circle, this leads to the equation for circle segments **L=r*θ**, where **r** is the radius, **θ **is the angle (__in radians__) subtended by the circle segment (the angle of the segment at the centre), and **L** is the arc length (the circumference of circular part of the segment).
Let's imagine a circular cake (yum!), with radius** 10cm**. If you cut the cake into six pieces, each piece subtends 60 degrees, or pi/3 radians, at the centre. The length of the curved edge of any slice (**L**) is then equal to **10*(pi/3)**, or about **10.47cm**. If instead you cut a slightly smaller slice which subtends an angle of one radian (about 57.30 degrees), then the length of the curved edge will be exactly the same as the radius, that is, **10cm**.
2) If you've had to differentiate trigonometric functions, such as *sin*(x), you should know for example that the derivative of *sin*(x) is *cos*(x). Nice and simple, right? This is however only true when **x** is measured in radians.
Why is this the case? It seems almost coincidental. But by considering the graph of *sin*(x) this might seem more believable! If you try drawing the graph **y=***sin*(x) with a 1:1 axis ratio, in the radian system, going through (0, 0) and (pi/2, 1), try and measure the gradient at x=0. It should give a gradient of about 1 (depending how accurate the drawing is!). Since the derivative of *sin*(x) is *cos*(x), and *cos*(0) = 1, you would predict the gradient at x=0 to be 1, so this makes sense!
If you tried the same with the degree system, you would need a very wide piece of paper for 1:1 axes, but it should be clear that the gradient at x=0 is much less than 1. Hopefully this makes it clear that differentiating *sin*(x) to *cos*(x) doesn't work so smoothly in the degree system.

When differentiating a function that is the sum of three different parts we can differentiate each part separately:
a) 2x^{3} is easy to differentiate. We remember the rule d/dx[ax^{b}] = abx^{b-1}. So
2x^{3} --> 6x^{2}
b) (cos(x))^{2} is a bit harder. We can use the chain rule, as we have a function raised to a power. The chain rule is:
d/dx[(g(x))^{n}] = n(g(x))^{n-1} * d/dx[g(x)]
Also we need to remember that cos(x) differentiates to -sin(x)
So we have that
(cos(x))^{2} --> -2cos(x)sin(x).
c) e^{x} is the easiest of the lot: it doesnt change when differentiated.
e^{x} --> e^{x}
Therefore the final answer is:
d/dx[f(x)] = 6x^{2} - 2cos(x)sin(x) + e^{x}

Answered by Seth P.

Studies Theoretical Physics at Durham

We must first use trigonometric identities to simplify cos^{2}(x). We can use the formula** cos(A+B) = cos(A)cos(B) - sin(A)sin(B)** , where *A=x* and *B=x*, so that we get **cos(2x) = cos**^{2}(x) - sin^{2}(x) = 2cos^{2}(x) - 1. Rearranging this we find that **cos**^{2}(x) = 1/2 + (cos(2x))/2 This gives us **(⌠(x+xcos(2x))dx)/2 = 1/2(⌠( x)dx + ⌠( xcos(2x))dx) = x**^{2} /4 + ⌠( xcos(2x))dx) /2 We can then use the integration by parts formula, **⌠(udv)dx = uv - f(vdu)dx **, where *u=x *and d*v=cos(2x)*, so that we get **x**^{2} /4 + ⌠( xcos(2x))dx) /2 = x^{2} /4 + xsin(2x)/4 - ⌠(sin(2x))dx)/4 = x^{2} /4 + xsin(2x)/4 + cos(2x)/8 Hence,the final answer is **x**^{2} /4 + xsin(2x)/4 + cos(2x)/8 + c

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