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

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    (Original post by HennersPD)
    Thank you i really appreciate your help are there any questions you'd like help with as I'd like to return the favour
    Hopefully this helps

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    And yeah, I'm struggling with 7 b) i) on Jan 13! But I think I'll be able to sort it when I look at the mark scheme.
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    (Original post by 2014_GCSE)
    Hopefully this helps

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    And yeah, I'm struggling with 7 b) i) on Jan 13! But I think I'll be able to sort it when I look at the mark scheme.
    Thank you this really helped
    Ill send you my solution in a minute for that
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    guys what shapes could they ask?~~~[vectors]
    Rhombus
    isosceles triangle
    trapeziums
    kites
    rectangles
    squares
    perpendicular a.b=0
    parallel=a.b=-1
    cut put circles in there
    maybe use cosine rule/sine rule also given that we find the magnitudes out???
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    (Original post by Chung224)
    guys what shapes could they ask?~~~[vectors]
    Rhombus
    isosceles triangle
    trapeziums
    kites
    rectangles
    squares
    perpendicular a.b=0
    parallel=a.b=-1
    cut put circles in there
    maybe use cosine rule/sine rule also given that we find the magnitudes out???
    Parallelograms in general; not just a rhombus.
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    June 2013 Question 7 anyone?

    3 marks but I just do not get it
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    (Original post by amyrah)
    June 2013 Question 7 anyone?3 marks but I just do not get it
    It tells you the largest value that the range of change can be is 1.3

    The largest number cos(ANYTHING) can be is 1.
    Therefore to get 1.3, A must equal 1.3.

    Then you have to find a point where cos() will equal 1 at every 12 hour occurrence. (i.e. when t=0, 12, 24, etc...)

    If you draw a cos graph, you will see that this is at Pi/6.
    Therefore k = Pi/6

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    (Original post by 2014_GCSE)
    It tells you the largest value that the range of change can be is 1.3

    The largest number cos(ANYTHING) can be is 1.
    Therefore to get 1.3, A must equal 1.3.

    Then you have to find a point where cos() will equal 1 at every 12 hour occurrence. (i.e. when t=0, 12, 24, etc...)

    If you draw a cos graph, you will see that this is at Pi/6.
    Therefore k = Pi/6


    Hi, thanks for your reply! I get up to the 1.3 part now but not much past that.. So do I make cos(kt)=1/12?
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    Does anyone know if there have been any recent spec changes at all, because I have been using past papers as my only revision source so if something comes up that isn't on there than I shall not be prepared!


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    (Original post by amyrah)
    June 2013 Question 7 anyone?

    3 marks but I just do not get it
    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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    (Original post by jjsnyder)
    Does anyone know if there have been any recent spec changes at all, because I have been using past papers as my only revision source so if something comes up that isn't on there than I shall not be prepared!


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    Don't worry, any changes to the spec will not affect this year's exam - only next year's and onwards.
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    (Original post by datpr0)
    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).
    Thank you so much!!!! It was too tricky for me
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    June 2014, Q4ci - http://prntscr.com/7e7il0
    they started in the mark scheme with the opening line

     5000p^{T-10} = 2500q^T

    would this:

     5000p^{T} = 2500q^{T+10}

    still give the right answer, and if so can someone post a solution using this to start with?
    Thanks
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    Anybody here who's offer is higher than A*AA?
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    (Original post by edd1234)
    June 2014, Q4ci - http://prntscr.com/7e7il0
    they started in the mark scheme with the opening line

     5000p^{T-10} = 2500q^T

    would this:

     5000p^{T} = 2500q^{T+10}

    still give the right answer, and if so can someone post a solution using this to start with?
    Thanks
    I haven't seen the question but it wouldn't because p and q are independent variables. If p = q then it would.
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    (Original post by edd1234)
    June 2014, Q4ci - http://prntscr.com/7e7il0
    they started in the mark scheme with the opening line

     5000p^{T-10} = 2500q^T

    would this:

     5000p^{T} = 2500q^{T+10}

    still give the right answer, and if so can someone post a solution using this to start with?
    Thanks
    It's really difficult to solve it in the form they want that way.
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    (Original post by amyrah)
    Thank you so much!!!! It was too tricky for me
    No problem .
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    Can someone write out a step by step on June 2014, Q4)C)I) whilst it's been brought up.
    Can't do it at all.
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    (Original post by 2014_GCSE)
    Can someone write out a step by step on June 2014, Q4)C)I) whilst it's been brought up.
    Can't do it at all.
    Sure if you give me about 5 minutes . I'll edit this post with working.
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    (Original post by datpr0)
    Sure if you give me about 5 minutes . I'll edit this post with working.
    Brilliant, thank you!
    • Thread Starter
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    (Original post by edd1234)
    June 2014, Q4ci - http://prntscr.com/7e7il0
    they started in the mark scheme with the opening line

     5000p^{T-10} = 2500q^T

    would this:

     5000p^{T} = 2500q^{T+10}

    still give the right answer, and if so can someone post a solution using this to start with?
    Thanks
    The whole point is that the years start from 1991.

    As painting V starts counting 10 years after painting W, its value of t since 1991 is T-10 when W's value of t is T.


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