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Llamas
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#1
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I've just come out of P5 and thought it was a horrible paper (compared to last year's anyway). The first six questions were fine, but I got into a real mess on the last two. How did everyone else find it?

edit: You can now find a copy of the paper on page 3!
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kriztinae
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(Original post by Llamas)
I've just come out of P5 and thought it was a horrible paper (compared to last year's anyway). The first six questions were fine, but I got into a real mess on the last two. How did everyone else find it?
hey i did it
the first 4 questions were fine, the reduction formulae was a bit of a mess, intrinsic coordinates?what are they! didnt even bother attempting that one!
number 7 was ok, i think i got the right answer and if not my method was all corect.
number 8a) was piss easy the rest was whaT?
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beauford
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I thought it was all fairly easy to be honest, some bits were a bit technical but nothing too hard, i think i got them all right...
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kriztinae
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(Original post by beauford)
I thought it was all fairly easy to be honest, some bits were a bit technical but nothing too hard, i think i got them all right...
it wasnt too hard no, but parts of it got messy
some parts where just... hmm how do u put it? well they looked pretty, a nice white clean page!
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beauford
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(Original post by kriztinae)
it wasnt too hard no, but parts of it got messy
some parts where just... hmm how do u put it? well they looked pretty, a nice white clean page!
It was just a little fiddley at times i think...
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kriztinae
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(Original post by beauford)
It was just a little fiddley at times i think...
yeah, question 7, that got really mess, but i think i got the right answer
did u find the coordinats of Q? for question 8
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beauford
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(Original post by kriztinae)
yeah, question 7, that got really mess, but i think i got the right answer
did u find the coordinats of Q? for question 8
I made q. 7 to be 6a and 12a^2pi/5 and for q. 8, the coordinates were (-c/t^3,-ct^3) i think.
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kriztinae
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(Original post by beauford)
I made q. 7 to be 6a and 12a^2pi/5 and for q. 8, the coordinates were (-c/t^3,-ct^3) i think.
quetions 7 looks good
Q 8 i attempted that part but didnt get it so i guess urs is right
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chetan
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evry1 i knw seems 2 fink it was ***** hard n evry1 got diff answers for evrythin :mad:
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Llamas
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on 7 I ended up with INT(root(Sin^6[x] - Sin^4[x] + Cos^6[x] - Cos^4[x]))dx

I figured I was on the right track but time wasn't looking so good I moved on to 8 because I was usually good at that kind of question. The first part of 8 was easy, then I made a stupid error on the second part and that was the mess which consumed my remaining time. Overall I think the problem with that paper wasn't so much it's difficulty (although I think it was definitely harder than the June 2003 paper) but that the questions were fairly long. Maybe an extra 30 mins and I would've finished it all ^_^
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janetjanet
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hi,I got the radian of curveture like 25/9 root 3 ,something,did anyway get that???
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beauford
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(Original post by janetjanet)
hi,I got the radian of curveture like 25/9 root 3 ,something,did anyway get that???
as long as you mean root 2 i got that (i put it as -25/9 root(2) actually, don't think it matters)
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janetjanet
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(Original post by beauford)
as long as you mean root 2 i got that (i put it as -25/9 root(2) actually, don't think it matters)
yes,thanks.it was 2 ,I think....
haha////so happy....
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janetjanet
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i did not get the last part of q 8,it was only 2 marks though..
what do people think the boundry would be???
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Llamas
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(Original post by janetjanet)
hi,I got the radian of curveture like 25/9 root 3 ,something,did anyway get that???
I think that question went something like this:

t= ln3

x = Cosh(t) - t

dx = Sinh(t) - 1
d2x = Cosh(t)

y = Cosh(t) + t

dy = Sinh(t) + 1
d2y = Cosh(t)

dx = 1/3
d2x = 5/3

dy = 7/3
d2y = 5/3

Radius of Curvature = (dx^2 + dy^2)^3/2 / (dx*d2y - dy*d2x)

= (1/9 + 49/9)^(3/2) / (5/9 - 35/9)
= (50/9)^(3/2) / (-30/9)
= (50^(3/2) / 27) / (-30/9)

= -(50^(3/2) / 90)

I think that's right, anyway.
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username9303
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cor i found that so hard its unreal. intrinsic co-ordinates?!?! help! lol. 8(a) was easy but the rest was a joke. i got 6a as well for the length but i just didnt have time to do the surface area of revolution. i filled up 18 sides of working, cant be a good sign.
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kriztinae
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(Original post by olliemccowan)
cor i found that so hard its unreal. intrinsic co-ordinates?!?! help! lol. 8(a) was easy but the rest was a joke. i got 6a as well for the length but i just didnt have time to do the surface area of revolution. i filled up 18 sides of working, cant be a good sign.
intrinsic was a joke! i didnt get any of it in the end
8a) was easy the rest a joke? yes!
i like you!
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infekt
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(Original post by olliemccowan)
cor i found that so hard its unreal. intrinsic co-ordinates?!?! help! lol. 8(a) was easy but the rest was a joke. i got 6a as well for the length but i just didnt have time to do the surface area of revolution. i filled up 18 sides of working, cant be a good sign.
ok ...igot 6a(sin^2a) .... ... and 24pi(a^2)(u^5 / 5)

u was either cos or sine ... cantremember ...

im totally****ed ... i got eveything rite till q 6 ... then itwas a massacre
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username9303
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yer ur rite - everything was going so well until i hit the last 3 questions and then my life got shattered lol. the woman in the exam was lookin at me very funny when after finishin the 16 page answer booklet i asked for "lots more sheets of paper please".

:mad: so annoyed wiv that paper. Anyone who says that it was easy shud be hung, drawn and quartered for lying.
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qqqwwweee
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(Original post by beauford)
I made q. 7 to be 6a and 12a^2pi/5 and for q. 8, the coordinates were (-c/t^3,-ct^3) i think.
I got the same too!
So i think you are right
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