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# OCR MEI C4 [June 14th] Discussion

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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1. Re: OCR MEI C4 [June 14th] Discussion
(Original post by Amy-Rose)
So gutted about what happened in January, it was cruel. Wouldn't surprise me if they do it again though, punish us, persay.
2. Re: OCR MEI C4 [June 14th] Discussion
(Original post by JumpingFrog)
The numerator of the fraction is always an order lower than the denominator so if you have just a linear term on the bottom you just have an A on the top. If you have a quadratic on the bottom e.g. (px+q)^2 then you have Ax+B on the top.
Cheers, I've been wondering this for a while.

Hmm, wouldn't you just do your example as (px+q)(px+q)? Hence you would get A/(px+q) + B/(px+q)?

Thanks!
Last edited by Benniboi1; 13-06-2012 at 13:36.
3. Re: OCR MEI C4 [June 14th] Discussion
(Original post by hhh123)
Are differential functions like sinkx ----> kcoskx in the formula book?
For tan kx, sec x, cot x and cosec x yes they are, for sin x and cos x, they're not but they're pretty straight forward to remember.
4. Re: OCR MEI C4 [June 14th] Discussion
(Original post by Benniboi1)
For tan kx, sec x, cot x and cosec x yes they are, for sin x and cos x, they're not but they're pretty straight forward to remember.
Ok thanks Benniboi.
5. Re: OCR MEI C4 [June 14th] Discussion
For anyone else freaking out about the comprehension paper, this is what you need to know : http://www.mei.org.uk/files/pdf/09co...9HandoutB1.pdf

6. Re: OCR MEI C4 [June 14th] Discussion
(Original post by Benniboi1)
Cheers, I've been wondering this for a while.

Hmm, wouldn't you just do your example as (px+q)(px+q)? Hence you would get A/(px+q) + B/(px+q)?

Thanks!
Yes, you could do that. Sometimes you lose marks for not simplifying enough, they seem to want the simplest form possible unless they give you the form they want in the question.

EDIT: Actually not sure you can do that. I think you might need to use the A/(px+q) + Bx+C/(px+q)^2

(Original post by Amy-Rose)
By the way, a couple people mentioned integrating sin^3(x) as a possible question. Just found it in the MEI book and it says 'enrichment' which is beyond the criteria for the unit but included for completeness.

So that would make me think we don't need to know how to do it..?
It's not enrichment, even powers of cos(x)/sin(x) are enrichment but not odd powers. At least according to my book on page 267. So, you are required to know it. Then again, volumes of revolution about the y axis are marked as enrichment but they are definitely on the syllabus and in exams.
Last edited by JumpingFrog; 13-06-2012 at 14:00.
7. Re: OCR MEI C4 [June 14th] Discussion
(Original post by JumpingFrog)
It's not enrichment, even powers of cos(x)/sin(x) are enrichment but not odd powers. At least according to my book on page 267. So, you are required to know it. Then again, volumes of revolution about the y axis are marked as enrichment but they are definitely on the syllabus and in exams.
Okay thanks
8. Re: OCR MEI C4 [June 14th] Discussion
Just nailing Jan 2011 paper. It's actually quite nice
9. Re: OCR MEI C4 [June 14th] Discussion
(Original post by shiinkii)
For anyone else freaking out about the comprehension paper, this is what you need to know : http://www.mei.org.uk/files/pdf/09co...9HandoutB1.pdf

'Candidates are allowed to take a standard English dictionary into the exam' - what the buck?! I thought this was a maths exam..
10. Re: OCR MEI C4 [June 14th] Discussion
(Original post by JumpingFrog)
Yes, you could do that. Sometimes you lose marks for not simplifying enough, they seem to want the simplest form possible unless they give you the form they want in the question.

EDIT: Actually not sure you can do that. I think you might need to use the A/(px+q) + Bx+C/(px+q)^2
Hmmm, I'm going to do the exercises in the text book and I'll get back to you on that one!
11. Re: OCR MEI C4 [June 14th] Discussion
I would be really grateful to anyone who can help me with the Jan 12 C4 paper, Q8. I just don't know where to begin
12. Re: OCR MEI C4 [June 14th] Discussion
(Original post by EckoGecko)
I would be really grateful to anyone who can help me with the Jan 12 C4 paper, Q8. I just don't know where to begin
Hi which part? From the beginning?
13. Re: OCR MEI C4 [June 14th] Discussion
(Original post by EckoGecko)
I would be really grateful to anyone who can help me with the Jan 12 C4 paper, Q8. I just don't know where to begin
We did this for our mock and it really stumped me, too:

8i) you need to use the chain rule and find dy/dx from dx/dt and dy/dt

8ii) calculate the gradient of QP using (Py-Qy)/(Px-Qx). Note this is tan(phi).
now substitute tan(theta) in the tan(2theta) identity with (1/t), as found in 8i.
you should end up with tan(2theta) = tan(phi)
the angles are just due to angles on a straight line and complementary angles
graphical solution available here

8iii) y^2 = 16t^2, and x = 2t^2, so t^2 = x/2
hence y^2 = 16(x/2) = 8x as required
then use solids of revolution as usual.

Hope that helps!
Last edited by SaulRennison; 13-06-2012 at 14:41.
14. Re: OCR MEI C4 [June 14th] Discussion
(Original post by EckoGecko)
I would be really grateful to anyone who can help me with the Jan 12 C4 paper, Q8. I just don't know where to begin
It's a tough question, I'm still trying to get my head around it. I can do bits of the maths but I don't see how the MS simplifies on ii).

The volume of revolution is quite easy though, eliminate the parameter t, work out what x is when t = root 2 and then do the integration
15. Re: OCR MEI C4 [June 14th] Discussion
(Original post by Yash13)
Hi which part? From the beginning?

(Original post by SaulRennison)
We did this for our mock and it really stumped me, too:

8i) you need to use the chain rule and find dy/dx from dx/dt and dy/dt
8ii) calculate the gradient of QP using (Py-Qy)/(Px-Qx). Note this is tan(phi).
now substitute tan(theta) in the tan(2theta) identity with (1/t), as found in 8i.
you should end up with tan(2theta) = tan(phi)
the angles are just due to angles on a straight line and complementary angles

8iii) y^2 = 16t^2, and x = 2t^2, so t^2 = x/2
hence y^2 = 16(x/2) = 8x as required
then use solids of revolution as usual.

Hope that helps!
Thanks!
For part i does anyone understand why tan(theta) = 1/t ?
As in where did tan(theta) come from?
16. Re: OCR MEI C4 [June 14th] Discussion
(Original post by ILM16)
It's a tough question, I'm still trying to get my head around it. I can do bits of the maths but I don't see how the MS simplifies on ii).

The volume of revolution is quite easy though, eliminate the parameter t, work out what x is when t = root 2 and then do the integration
I love the volume of revolution questions and I agree about the rest of this question, I've tried to work out things but it's just such a strange question...
17. Re: OCR MEI C4 [June 14th] Discussion
(Original post by EckoGecko)
Thanks!
For part i does anyone understand why tan(theta) = 1/t ?
As in where did tan(theta) come from?

Where opposite is the vertical (i.e. change in y) and the adjacent is the horizontal (i.e. change in x)

So tan(theta) = dy / dx, which is the same as your gradient
18. Re: OCR MEI C4 [June 14th] Discussion
(Original post by SaulRennison)
We did this for our mock and it really stumped me, too:

8i) you need to use the chain rule and find dy/dx from dx/dt and dy/dt
8ii) calculate the gradient of QP using (Py-Qy)/(Px-Qx). Note this is tan(phi).
now substitute tan(theta) in the tan(2theta) identity with (1/t), as found in 8i.
you should end up with tan(2theta) = tan(phi)
the angles are just due to angles on a straight line and complementary angles

8iii) y^2 = 16t^2, and x = 2t^2, so t^2 = x/2
hence y^2 = 16(x/2) = 8x as required
then use solids of revolution as usual.

Hope that helps!
This is all correct.

First with 8i)

You have V right. Differentiate with respect to x. Thus DV/DX.

V= pi(10x^2-1/3x^3) dv/dx = pi(20x - x^2)

Yes?

Now using chain rule we know :

dv/dx x dx/dt = dv/dt

DV/DT = k(20-x)

Rewrite dv/dx to pi.x(20-x). chain rule:

pi.x(20-x) x dx/dt = k(20-x)

divide by (20-x)

pi.x.dx/dt = k
Last edited by Yash13; 13-06-2012 at 14:20.
19. Re: OCR MEI C4 [June 14th] Discussion
(Original post by SaulRennison)

Where opposite is the vertical (i.e. change in y) and the adjacent is the horizontal (i.e. change in x)

So tan(theta) = dy / dx, which is the same as your gradient
Damn, why didn't I realise that.
20. Re: OCR MEI C4 [June 14th] Discussion
(Original post by SaulRennison)
We did this for our mock and it really stumped me, too:

8i) you need to use the chain rule and find dy/dx from dx/dt and dy/dt

8ii) calculate the gradient of QP using (Py-Qy)/(Px-Qx). Note this is tan(phi).
now substitute tan(theta) in the tan(2theta) identity with (1/t), as found in 8i.
you should end up with tan(2theta) = tan(phi)
the angles are just due to angles on a straight line and complementary angles

8iii) y^2 = 16t^2, and x = 2t^2, so t^2 = x/2
hence y^2 = 16(x/2) = 8x as required
then use solids of revolution as usual.

Hope that helps!
How do you know the gradient of TS = tan(theta)?

Also once you substitute on ii), how do you tidy that up to simply tan(2 theta).

Thanks

Edit: Just seen you've answered the first part, thanks.
Last edited by ILM16; 13-06-2012 at 14:24.

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