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

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    (Original post by kateyl)
    I just got 3m, not one over
    Yeah, dy/dx = k/(pir^2) -> r=3
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    (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?
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    (Original post by MoHoosen9167)
    We already had 1/2 when t=π/6
    yer i thought it was wrong but was the quadratic equation 8 sin^2 x + 2sinx -3?
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    I checked just before the exam, its definitely k over so 3m
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    (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
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    (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.
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    (Original post by Ual12)
    yer i thought it was wrong but was the quadratic equation 8 sin^2 x + 2sinx -3?
    Yup!
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    (Original post by Ual12)
    yer i thought it was wrong but was the quadratic equation 8 sin^2 x + 2sinx -3?
    Yes
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    (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
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    (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.
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    (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
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    Did anyone get the +c in the differential equation of 8a to be -4??


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    (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.
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    (Original post by Cattie2312)
    I got 1/2 too!!
    (1/2) was already given in part a
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    That was piss easy!
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    (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
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    Unofficial mark scheme anyone? :-)
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    (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 =
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    Best. Exam. Ever.
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    (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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