[hhh123]

(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.

They had better not.

[Benniboi1]

(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!

[Benniboi1]

(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.

[hhh123]

(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.

[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

Get your freak on!

[JumpingFrog]

(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.

[Amy-Rose]

(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

[hhh123]

Just nailing Jan 2011 paper. It's actually quite nice

[Benniboi1]

(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

Get your freak on!

'Candidates are allowed to take a standard English dictionary into the exam' - what the buck?! I thought this was a maths exam..

[Benniboi1]

(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!

[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

[Yash13]

(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?

[SaulRennison]

(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!

[ILM16]

(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

[EckoGecko]

(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?

[EckoGecko]

(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...

[SaulRennison]

(Original post by EckoGecko)
Thanks!
For part i does anyone understand why tan(theta) = 1/t ?
As in where did tan(theta) come from?

tan(theta) = opposite / adjacent

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

[Yash13]

(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

[EckoGecko]

(Original post by SaulRennison)
tan(theta) = opposite / adjacent

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.

[ILM16]

(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.