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# Ocr - P3 watch

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1. (Original post by CharlC)
I thought the exam today was quite a good one, at least I think I got roughly the same answers as yous. Got pi²/16 - 1/4 tho. Anyone else get that for the integration q??
Yeh I got pi²/16 - 1/4
2. Could well have been that, although doing the integral on my calculator now neither seem to be right
3. Can anyone tell me how they did 8 (ii) and (iii)? I've got the paper in front of me and still can't get it! Its starting to annoy me

(i) Express 1 / p(1 - p) as partial fractions.

(ii) Hence find the general solution of the differential equation (You do not need to express p explicitly in terms of t.)

(iii) It is given that a proportion 0.01 of the population is infected when t = 0 and that a proportion 0.05 is infected when t = 1. Find the value of t when a proportion 0.95 of the population is infected.

I don't normally bother looking through and checking answers, probably should be revising physics, but this one is bugging me!
4. Yeah same for the integration question.

t=4.56 for the last question.
I went for the acute angle between the two lines (you just modulus the number you are inverse cosing).. 63ish degrees perhaps? I don't really remember.
5. (going from part (ii))
I found that (1/p + 1/(1-p))dp=kdt

so lnp - ln (1-p)= kt + c
ln(p/(1-p)) = kt + c
p/(1-p)= Ae^kt

(iii) Sub in p=0.01 and t=0 to find that A=0.01/0.99
Sub in p=0.05 and t=1 to find k=ln(19/99) (i think... going on memory here and skipping several steps, if you want them then say)
then sub in p=0.95 and use ln to find t (I think it was something like ln1881/(ln(19/99))
t=4.56
6. Didn't like that, well I mean, I could do most of it, as In knew what I needed to do, but always go so messed up in the basic algebra/simpiifying and making basic arithemtic erros I screwed things up somewhat.

Can anyone post the parametric equation equations plz?
7. (Original post by naystar)
(going from part (ii))
I found that (1/p + 1/(1-p))dp=kdt

so lnp - ln (1-p)= kt + c
ln(p/(1-p)) = kt + c
p/(1-p)= Ae^kt

(iii) Sub in p=0.01 and t=0 to find that A=0.01/0.99
Sub in p=0.05 and t=1 to find k=ln(19/99) (i think... going on memory here and skipping several steps, if you want them then say)
then sub in p=0.95 and use ln to find t (I think it was something like ln1881/(ln(19/99))
t=4.56
I messed that up as I didnt know where to put the k.
8. ln(p/(1-p)) = kt + c
p/(1-p)= Ae^kt

All same as mine, I just don't get going from a '+ c', to an A. I'm guessing its a bit of maths I'm not familiar with. Other than that I got all the right values of k etc, shame bout the important bit - the answer!

Cheers
9. because e^(kt+c) is like saying e^kt * e^c, and since e^c is a constant, you can just relabel it A or some other letter.
10. Gotcha I was rushing at that point of the exam so forgot about taking expodentials, altho i should of still got the right answer not taking them, oh well!
11. (Original post by Fluent in Lies)
Didn't like that, well I mean, I could do most of it, as In knew what I needed to do, but always go so messed up in the basic algebra/simpiifying and making basic arithemtic erros I screwed things up somewhat.

Can anyone post the parametric equation equations plz?
the end result for that was a circle
12. I went through the whole paper thinking "this is easy"...then went back to do the questions I missed. Didn't realise how many of them there were...oops.

And I hate vectors. I got the first part...but know I've done the rest wrong.

Still, I think I only needed about 20 UMS, so hopefully I'll be ok
13. (Original post by absentmindedone)

Still, I think I only needed about 20 UMS, so hopefully I'll be ok
wow! how does that work out!
14. ARRRGH! I got the very last question wrong! You know how you had to determine the constants in the equation in (iii)? Well, I worked them out fine, but then when it said work out t at p=0.95, I FORGOT THE FIRST CONSTANT (i.e. 1/99). I got something like 1.78! It didn't seem right, so I kept checking it, but I failed to notice that I'd just overlooked that one small thing. I only realised when I read this thread! I really wanted to get 100% as well...
15. Yeah I only need 20ish UMS as well... anyone got a definitive answer on the integral?
Also, did anyone think 4 marks was a lot for the binomial expansion, especially since you only needed three terms. I'm thinking I might have missed something there.
16. Yeah I thought four marks was a lot just for a binomial expansion up to x squared, but that is what they asked for. I'm guessing you get a mark per term, plus one for having simplified the coefficients?
17. I thought that was weird too. I only used to lines for the 4 marks whereas on the 5 mark questions i used over a page! I also used loads of lines to work out the vector question that was worth one mark!
18. (Original post by *hedgehog*)
I thought that was weird too. I only used to lines for the 4 marks whereas on the 5 mark questions i used over a page! I also used loads of lines to work out the vector question that was worth one mark!
lol, yh same here! weird mark allocation going on! shorter paper though, only two sides. hope it went well.

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