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.

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

Integration by parts is not only a very useful techinque for integrating, but it is also one of the techinques that appears more in the IB Mathematics exams. The most important step is to choose the part of the integrand that works better for u or f(x), as a right choice will save you a lot of time (and time is everything in the IB!). Whenever you are in doubt, choose u or f(x) to be the following:
1. The LOgarithm. If there is none,
2. The ROot. If there is none,
3. The POlynomial. If there is none,
4. The Trigonometric function. If there is none,
5. The Exponential function.
To remember this rule, think about the sentence "LOve ROasted POtatoes Till the End", and its initials will tell you the order of the functions you need to look for to be u or f(x)!!

Answered by Xell B.

Studies Mathematics and Biology at Edinburgh

If the two graphs intersect, it means that they will share the same y and x coordinates at one particular point. (I will draw diagram to show point).
Therefore, you can set f(x)=g(x) so that x^2 -ax +a -1 = x-5
Then, x^2 -x(a+1) +a + 4 = 0
If they only intersect at one particular point, this means that the previous quadratic equation has only one solution. This is translated into an equation in terms of the determinant so that the determinant must be 0.
(If necessary, I will explain the difference number of solutions that one gets for different determinants).
Then, one requires b^2 - 4ac = 0 , where b is the coefficient multiplying the x, a is the coefficient multiplying the x^2 and c is the coefficient with no x in the previous equation.
This leads to (a+1)^2-4(a+4)=0 which is a quadratic equation in a:
a^2 -2a -15 =0 , which, using the quadratic formula, has solutions a= 5 and a=-3.

Answered by Andres O.

Studies PhD in Theoretical Particle Physics at Durham

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