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# AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] watch

1. (Original post by kateyl)
I just got 3m, not one over
Yeah, dy/dx = k/(pir^2) -> r=3
2. (Original post by kateyl)
I just got 3m, not one over
A lot of others at my school got 3 as well... but im pretty certain its supposed to be 1/k rather than 1/ pi R squared... if you do it the second way, you get 3... anyone else able to clarify?
3. (Original post by MoHoosen9167)
yer i thought it was wrong but was the quadratic equation 8 sin^2 x + 2sinx -3?
4. I checked just before the exam, its definitely k over so 3m
5. (Original post by akashpatel)
A lot of others at my school got 3 as well... but im pretty certain its supposed to be 1/k rather than 1/ pi R squared... if you do it the second way, you get 3... anyone else able to clarify?
From reading the question it seemed obvious to me to be k/(pir^2) as it was inversely proportional to the area. I am not saying I am definitely correct, but I think that this is the answer
6. (Original post by akashpatel)
A lot of others at my school got 3 as well... but im pretty certain its supposed to be 1/k rather than 1/ pi R squared... if you do it the second way, you get 3... anyone else able to clarify?
Yes because it's inversely proportional and k is the constant of proportionality, therefore k^-1 which is 1/k. I'm pretty sure you're correct.
7. (Original post by Ual12)
yer i thought it was wrong but was the quadratic equation 8 sin^2 x + 2sinx -3?
Yup!
8. (Original post by Ual12)
yer i thought it was wrong but was the quadratic equation 8 sin^2 x + 2sinx -3?
Yes
9. (Original post by jjsnyder)
From reading the question it seemed obvious to me to be k/(pir^2) as it was inversely proportional to the area. I am not saying I am definitely correct, but I think that this is the answer
This. dr/dt is proportional to 1/pir^2 and therefore = k * 1/pir^2... so the answer is 3
10. (Original post by laurenn1995)
Yes because it's inversely proportional and k is the constant of proportionality, therefore k^-1 which is 1/k. I'm pretty sure you're correct.
No this is incorrect, it is inversely proportional to the area not the constant.
11. (Original post by laurenn1995)
Yes because it's inversely proportional and k is the constant of proportionality, therefore k^-1 which is 1/k. I'm pretty sure you're correct.
No, if y is inversely proportional to x, then y=k/x
12. Did anyone get the +c in the differential equation of 8a to be -4??

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13. (Original post by akashpatel)
Did people get 1/3m for the last answer? I set up the equation as 1/k x pi R squared giving you pi R squared over k... pretty sure this is right...can anyone confirm? 1/k as it was inversely proportional to the area btw
The answer at the end was 3m, since dr/dt was inversely proportional to area, the equation was k/piR^2, you set it up inverted.

All things considered the exam wasn't that bad, wish I had more time, still needed 2 minutes for the last vector question.
14. (Original post by Cattie2312)
I got 1/2 too!!
(1/2) was already given in part a
15. That was piss easy!
16. (Original post by SomeGuy96)
The answer at the end was 3m, since dr/dt was inversely proportional to area, the equation was k/piR^2, you set it up inverted.

All things considered the exam wasn't that bad, wish I had more time, still needed 2 minutes for the last vector question.
Oh thank god for that

That last vector question was so not worth 4 marks
17. Unofficial mark scheme anyone? :-)
18. (Original post by laurenn1995)
Yes because it's inversely proportional and k is the constant of proportionality, therefore k^-1 which is 1/k. I'm pretty sure you're correct.
It can be written as dr/dt ∝ 1/pir^2 then you multiply by k so that proportional becomes =
19. Best. Exam. Ever.
20. (Original post by jjsnyder)
From reading the question it seemed obvious to me to be k/(pir^2) as it was inversely proportional to the area. I am not saying I am definitely correct, but I think that this is the answer
Ah man, I really dont know if im right now...

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