Hi there! My name's Catherine. I am a second year student at the University of Bath studying Physics and Astrophysics. I spent a year at UPMC, a university in Paris between my A levels and coming to Bath, studying Physics, Chemistry, Mechanics and Maths. I have A* in Further Maths and Maths A levels and A in French, Physics and Chemistry A levels.

I have mentored quite a few different students of different ages and levels, for Physics, Maths, French and Chemistry before. I prefer tutoring Maths but I can also tutor Physics. I have done some Chemistry at university level, both in Paris and one semester at Bath. I have a lot of experience for French as well, as I grew up going to France a lot and learning it outside of school as well as living in Paris.

I'm really passionate about all four of these subjects and love to help people to understand things. I find Physics fascinating, hence why I chose it as my degree course, and I love speaking French as it is such a beautiful language. I am patient and I always find different ways of explaining things if you don't understand first time. You will choose what you want to cover and when, and obviously you will choose the pace too. You are the tutee, so you get to decide!

Hi there! My name's Catherine. I am a second year student at the University of Bath studying Physics and Astrophysics. I spent a year at UPMC, a university in Paris between my A levels and coming to Bath, studying Physics, Chemistry, Mechanics and Maths. I have A* in Further Maths and Maths A levels and A in French, Physics and Chemistry A levels.

I have mentored quite a few different students of different ages and levels, for Physics, Maths, French and Chemistry before. I prefer tutoring Maths but I can also tutor Physics. I have done some Chemistry at university level, both in Paris and one semester at Bath. I have a lot of experience for French as well, as I grew up going to France a lot and learning it outside of school as well as living in Paris.

I'm really passionate about all four of these subjects and love to help people to understand things. I find Physics fascinating, hence why I chose it as my degree course, and I love speaking French as it is such a beautiful language. I am patient and I always find different ways of explaining things if you don't understand first time. You will choose what you want to cover and when, and obviously you will choose the pace too. You are the tutee, so you get to decide!

Standard DBS Check

06/06/20155from 5 customer reviews

Simon (Parent from Trowbridge)

September 18 2017

Very good lesson

Juliet (Parent from London)

April 5 2016

amazing!

Simon (Parent from Trowbridge)

October 16 2017

Simon (Parent from Trowbridge)

September 20 2017

This is a past paper question for an A level OCR MEI paper for Maths.

We need to find the gradient of the curve so we know right away that we need to use differentiation.

The equation y = ln(1-cos2x) is difficult to differentiate by itself so we use the chain rule and a substitution. We say y = ln(u) where u = 1-cos2x.

dy/du = 1/u. This is something you just have to learn, that the differential of y= lnx is 1/x.

If you're stuck with finding du/dx, you can use another substitution if you want where u = 1-cosv and v = 2x, but most people know (after a lot of practice) that the differential of y=coskx is -ksinkx. I believe it's also in most formula books.

So here du/dx = -2sin2x (remember the change of sign when going from cos to sin when differentiating).

So from the chain rule, we know that dy/dx = dy/du x du/dx so dy/dx = 1/u x -2sin2x = -2sin2x/u and u = 1-cos2x so dy/dx = -2sin2x/(1-cos2x)

Now we need to find the value of this gradient when x = π/6

dy/dx = -2sin2x/(1-cos2x) = -2sin(2 x π/6)/(1-cos(2 x π/6)) = -2sin(π/3)/(1-cos(π/3))

Remember that we're working in radians here so for this, you need to put your calculator in radians too

We know that cos(π/3) = 1/2 (or 0.5, but 1/2 is easier for the moment) and sin(π/3) = root(3)/2

So dy/dx at x=π/3 gives (-2 x root(3)/2)/(1-1/2) => multiply top and bottom by 2 => -2root(3)/(2-1) = -2root(3)/1 = -2root(3)

This is a past paper question for an A level OCR MEI paper for Maths.

We need to find the gradient of the curve so we know right away that we need to use differentiation.

The equation y = ln(1-cos2x) is difficult to differentiate by itself so we use the chain rule and a substitution. We say y = ln(u) where u = 1-cos2x.

dy/du = 1/u. This is something you just have to learn, that the differential of y= lnx is 1/x.

If you're stuck with finding du/dx, you can use another substitution if you want where u = 1-cosv and v = 2x, but most people know (after a lot of practice) that the differential of y=coskx is -ksinkx. I believe it's also in most formula books.

So here du/dx = -2sin2x (remember the change of sign when going from cos to sin when differentiating).

So from the chain rule, we know that dy/dx = dy/du x du/dx so dy/dx = 1/u x -2sin2x = -2sin2x/u and u = 1-cos2x so dy/dx = -2sin2x/(1-cos2x)

Now we need to find the value of this gradient when x = π/6

dy/dx = -2sin2x/(1-cos2x) = -2sin(2 x π/6)/(1-cos(2 x π/6)) = -2sin(π/3)/(1-cos(π/3))

Remember that we're working in radians here so for this, you need to put your calculator in radians too

We know that cos(π/3) = 1/2 (or 0.5, but 1/2 is easier for the moment) and sin(π/3) = root(3)/2

So dy/dx at x=π/3 gives (-2 x root(3)/2)/(1-1/2) => multiply top and bottom by 2 => -2root(3)/(2-1) = -2root(3)/1 = -2root(3)

This is a past exam question from an A level paper for OCR Physics B.

We know that in one second, a volume of water (V), travelling at 2.5 m/s is passing through an area of 500 metres squared in one second. This volume can be represented as a column, with the cross section (area at the front) equal to the area the water is passing through, so 500 metres squared. Since we know that v=s/t, we can rearrange this to get s=vt meaning that in one second, all the water molecules travel v.t metres of 2.5 x 1 = 2.5 metres. This gives us our bottom side for our column, giving us a total volume of V=Al = 500 x 2.5 = 1250 metres cubed.

So we now have the volume (V) and the density (ρ) but want to find the mass (m) which are all linked in the equation ρ=m/V which when rearranged gives m=ρV giving us an answer of m=1030 x 1250 = 1,287,500 kg which we can say is roughly equal to 1 x 10^6 kg

This is a past exam question from an A level paper for OCR Physics B.

We know that in one second, a volume of water (V), travelling at 2.5 m/s is passing through an area of 500 metres squared in one second. This volume can be represented as a column, with the cross section (area at the front) equal to the area the water is passing through, so 500 metres squared. Since we know that v=s/t, we can rearrange this to get s=vt meaning that in one second, all the water molecules travel v.t metres of 2.5 x 1 = 2.5 metres. This gives us our bottom side for our column, giving us a total volume of V=Al = 500 x 2.5 = 1250 metres cubed.

So we now have the volume (V) and the density (ρ) but want to find the mass (m) which are all linked in the equation ρ=m/V which when rearranged gives m=ρV giving us an answer of m=1030 x 1250 = 1,287,500 kg which we can say is roughly equal to 1 x 10^6 kg