Chris has worked throughly with my son on his IB past papers. I am extremely impressed with his professional approach to teaching, his quick response to my messages and his helpfulness in finding a solution with regard to timing, days, etc. He is very knowledgeable about the workings of MyTutorWeb, which put me at ease with the whole process. Thank you Chris! :o)

Sheila, Parent from Genève

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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.

This is an exapmle of the addition rule as part of the logorithm syllabus.
The general rule is log_{a}b + log_{a}c = log_{a}(b*c)
When the base of the logarithm is the same and they are added the to numbers of the logs may be multiplied therefore the answer to the problem would be log_{2}(5*3) = log_{2}15

Strictly speaking, the derivative is the rate of change of a function, process or pattern.

So far, to calculate the gradient of a given function we always computed the change in the y coordinates divided by the change in the x coordinates. This gives the rate of change from a point A to a point B. However, if we require the rate of change at a specific point, our current method no longer works and has to be amended.

What we do is we introduce the concept of the limit. We introduce a step length called dx, and claim that a step length of the limit of dx tending towards 0 from point A will be a very small increase. Therefore, if we define point B to be at (f(x+dx),x+dx), the gradient function will be calculated as follows:

lim dx->0 (f(x+dx)-f(x))/(dx) This is basically calculating the gradient of the function at point A, and is the formal definition of the derivative of a function i.e the rate of change/ gradient/slop at a specific instance.

So far, to calculate the gradient of a given function we always computed the change in the y coordinates divided by the change in the x coordinates. This gives the rate of change from a point A to a point B. However, if we require the rate of change at a specific point, our current method no longer works and has to be amended.

What we do is we introduce the concept of the limit. We introduce a step length called dx, and claim that a step length of the limit of dx tending towards 0 from point A will be a very small increase. Therefore, if we define point B to be at (f(x+dx),x+dx), the gradient function will be calculated as follows:

lim dx->0 (f(x+dx)-f(x))/(dx) This is basically calculating the gradient of the function at point A, and is the formal definition of the derivative of a function i.e the rate of change/ gradient/slop at a specific instance.

a) f'(x)=uv'+vu' if f(x)= uv
u=2x u'=2 v=sin(x) v'=cos(x)
g'(x)=2x cos(x) +2sin(x)
b) g'(π) = 2π cos(π)+2sin(π) = 2 π (-1) + 2 (0)
g'(π) = -2π

To integrate the volume of revolution first imagine a thin disk around the x-axis which we want to know the volume of: Volume=area x height. The area of a circle is given by pi r^2 and in our case let us use the height of dx. Hence the volume= pi r^{2 } dx. Now we will use the radius at each point as the y-value at that point, hence volume = pi y^2 dx = pi sin^2(x) dx. We will integrate this between the limits using the identity sin^2(x)=1/2 (1 - cos(2x)). Hence the volume of integration is given by V=pi/2 integral{0->pi} (1-cos(2x))dx = pi/2*[x -1/2 sin (2x)]{x=0 -> pi} = pi/2*(pi - 0 - (0-0)) = pi^2/2

Answered by Luke C.

Studies Mechanical Engineering at Bristol

Polar and coordinase complex numbers are two different ways of represent the same complex numbers.
The polar way uses the following formula: M*e^(angle*j), where M is the modulus of the complex number and can be obtained by Pithagoras´Theorem from the vector coordinates of the number. On the other hand, the angle of the number is calculated by the arctan(vertical coordinate/horizontal coordinate).
The cartesian way uses also the modulus and that angle, by in a different way. It is determined by applying M*cos(angle)+j*M*sin(angle)
And then I would show to the student a numerical example and we would analysed together some different exceptions.

Answered by LORENZO M.

Studies MECHANICAL ENGINEERING at Bath

dy/dx = 3x+1/ -2y-3x-1

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