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

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    (Original post by Jimmy20002012)
    Wonder what the differential in context will be one this time it's always either ever on paintings or bacteria.


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    Or money in a bank or footballers wages.
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    Done four past papers today in preparation for tomorrow's exam. I'm just going to brush up on any lose info in my revision guide. Not much more I can do. All the best for tomorrow people!
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    (Original post by saeed97)
    New technological advancement has just been released which you could use to find it. Heres the link: google.co.uk
    uwotm9?
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    (Original post by saeed97)
    New technological advancement has just been released which you could use to find it. Heres the link: google.co.uk
    Yeah I could look at it individually but people in the past have made spreadsheets with just past C4 grade boundaries, also learn to hyperlink.
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    Super easy way to approach Vector questions are diagrams. Makes them so much easier.
    Can picture what lengths you are trying to find, where angles are, etc.
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    (Original post by Mudkipzz)
    Yeah I could look at it individually but people in the past have made spreadsheets with just past C4 grade boundaries, also learn to hyperlink.
    Sorry, im obviously not skilled in the art of forum hyperlinking.
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    (Original post by datpr0)
    uwotm9?
    u wunt sum? i'll giv it 2 u
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    (Original post by saeed97)
    Or money in a bank or footballers wages.
    there was one on a colony of rats
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    (Original post by RhyaCeri)
    there was one on a colony of rats
    Eh, vermin... bacteria... same difference.

    Kappa
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    Differential equation is the rate of forum posts on TSR is inversely proportional to time after the exam.


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    (Original post by saeed97)
    u wunt sum? i'll giv it 2 u
    gr8 b8 m8 I r8 8/8
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    Excuse me, but can one of you explain how to do question 7 on this paper?
    I know what the answer is, but I don't know how to get there.
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    Do we get any marks if we made one or two slips, like signs ?? Because its so easy to make silly mistakes in C4
    How many method marks can we get max???
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    (Original post by Plasmapause)
    Excuse me, but can one of you explain how to do question 7 on this paper?
    I know what the answer is, but I don't know how to get there.
    Excellent description from datpr0 earlier in the thread:

    Basically it's modelling the change in height of the tide over a period of time (i.e. rate of change of the height of the tide) as a cosine graph, whereby the maximum and minimum values for this are 1.3 and -1.3 (dh/dt = +/- 1.3).

    From the question you know: dh/dt = acos(kt).

    From a cosine graph you know that the maximum and minimum points occur at 1 and -1 respectively (i.e. cos = 1 or -1). Therefore a = 1.3 or -1.3. You can consider this as a stretch parallel to the y-axis of SF 1.3 if you like.

    k is adjacent to the t in the model and t is being measured on the x-axis. You know the period of a cosine graph is 2pi and the tides occur every 12 hours (i.e. t = 12). This means that k becomes 2pi/12 in order to cancel with the 12 introduced through t so that you end up with cos(2pi) not cos(12).
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    I know this may sound stupid but how do you integrate this quickly Name:  ImageUploadedByStudent Room1433791146.300579.jpg
Views: 65
Size:  47.7 KB
    Thanks


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    (Original post by 2014_GCSE)
    Excellent description from datpr0 earlier in the thread:

    Basically it's modelling the change in height of the tide over a period of time (i.e. rate of change of the height of the tide) as a cosine graph, whereby the maximum and minimum values for this are 1.3 and -1.3 (dh/dt = +/- 1.3).

    From the question you know: dh/dt = acos(kt).

    From a cosine graph you know that the maximum and minimum points occur at 1 and -1 respectively (i.e. cos = 1 or -1). Therefore a = 1.3 or -1.3. You can consider this as a stretch parallel to the y-axis of SF 1.3 if you like.

    k is adjacent to the t in the model and t is being measured on the x-axis. You know the period of a cosine graph is 2pi and the tides occur every 12 hours (i.e. t = 12). This means that k becomes 2pi/12 in order to cancel with the 12 introduced through t so that you end up with cos(2pi) not cos(12).
    Thanks!
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    (Original post by gloria0816)
    I know this may sound stupid but how do you integrate this quickly Name:  ImageUploadedByStudent Room1433791146.300579.jpg
Views: 65
Size:  47.7 KB
    Thanks


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    It's fine. That's equivalent to 1/2 x y^-1/2, so you just integrate from there (it becomes y^1/2).
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    (Original post by gloria0816)
    I know this may sound stupid but how do you integrate this quickly Name:  ImageUploadedByStudent Room1433791146.300579.jpg
Views: 65
Size:  47.7 KB
    Thanks


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    1/2rootY is equal to 1/2(y^-1/2)
    So integrates to y^1/2 or root Y
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    (Original post by datpr0)
    gr8 b8 m8 I r8 8/8
    thnx m(7+1)
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    (Original post by datpr0)
    It's fine. That's equivalent to 1/2 x y^-1/2, so you just integrate from there (it becomes 1/4 x y^1/2).
    You've done 1/2 x 1/2. It's 1/2 / 1/2

    so it's y^1/2

 
 
 
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