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Juliet (Parent) April 5 2016
Juliet (Parent) April 8 2016
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)see more
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 kgsee more