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    I know how to do every part, apart from part C) where it asks to find t
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    (Original post by chrlhyms)
    I know how to do every part, apart from part C) where it asks to find t
    Ah the ferris wheel question - a very common question in the maths forum

    You say you can't find t but does that mean you've already found the maximum value of H? Please post all your working / thoughts.
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    Try rewriting the H equastion in th Rcos ("") form and then think about ways you can maximise your H value
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    (Original post by chrlhyms)
    I know how to do every part, apart from part C) where it asks to find t
    what a classic, i remember i struggled on this , now if i remember.

    The max value of H means to make H as big as possible.

    So what can you do to the cos thing you now got from part a to make H as big as possible?
    Set it equal to 0? -1? or 1? what gives you the biggest value of H
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    (Original post by Notnek)
    Ah the ferris wheel question - a very common question in the maths forum

    You say you can't find t but does that mean you've already found the maximum value of H? Please post all your working / thoughts.
    So i've found H as 10 + root(85)
    and the equation of H as

    H=10-root85 cos(pit/5 +0.2187)
    but cant think how I would find the maximum value of t as surely this would be 1 but the MS says -1?
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    (Original post by chrlhyms)
    So i've found H as 10 + root(85)
    and the equation of H as

    H=10-root85 cos(pit/5 +0.2187)
    but cant think how I would find the maximum value of t as surely this would be 1 but the MS says -1?
    If cos(pit/5 +0.2187) = 1 then

    10-root85 cos(pit/5 +0.2187) = 10 - root85

    But you said the maximum was 10 + root85 so cos(pit/5 +0.2187) must be -1.
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    (Original post by Notnek)
    If cos(pit/5 +0.2187) = 1 then

    10-root85 cos(pit/5 +0.2187) = 10 - root85

    But you said the maximum was 10 + root85 so cos(pit/5 +0.2187) must be -1.
    Sorry, I still dont understand!
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    (Original post by chrlhyms)
    Sorry, I still dont understand!
    How did you find that the maximum value of H is 10 + root(85)? Please explain fully.
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    (Original post by Notnek)
    How did you find that the maximum value of H is 10 + root(85)? Please explain fully.
    well since the original equation was stretched by a factor of root 85, the new equation was +10 onto that, therefore the new maximum was 10 + root85 and minimum 10-root85
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    (Original post by Notnek)
    If cos(pit/5 +0.2187) = 1 then

    10-root85 cos(pit/5 +0.2187) = 10 - root85

    But you said the maximum was 10 + root85 so cos(pit/5 +0.2187) must be -1.
    if cos(pit/5) +0.2187)=-1
    then surely 10-root85 cos(pit/5) +0.2187)= -10+root85
    which isnt the max?
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    (Original post by chrlhyms)
    if cos(pit/5) +0.2187)=-1
    then surely 10-root85 cos(pit/5) +0.2187)= -10+root85
    which isnt the max?
    I'll write it using brackets to make it clearer:

    \displaystyle H = 10-\left[\sqrt{85} \cos\left(\frac{\pi t}{5} +0.2187\right)\right]

    So when

    \displaystyle \cos\left(\frac{\pi t}{5} +0.2187\right) = -1

    you get

    \displaystyle H = 10 - \left[\sqrt{85} \times -1\right] = 10 - [-\sqrt{85}] = 10 + \sqrt{85}
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    (Original post by Notnek)
    I'll write it using brackets to make it clearer:

    \displaystyle H = 10-\left[\sqrt{85} \cos\left(\frac{\pi t}{5} +0.2187\right)\right]

    So when

    \displaystyle \cos\left(\frac{\pi t}{5} +0.2187\right) = -1

    you get

    \displaystyle H = 10 - \left[\sqrt{85} \times -1\right] = 10 - [-\sqrt{85}] = 10 + \sqrt{85}
    perfect! I understand now thank you! that question was horrible
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    (Original post by chrlhyms)
    perfect! I understand now thank you! that question was horrible
    Yes it was a bit nasty. They've asked a ferris wheel type question a few times in C3 exams so it's worth practicing them
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    (Original post by Notnek)
    Yes it was a bit nasty. They've asked a ferris wheel type question a few times in C3 exams so it's worth practicing them
    at least now i will know how to solve it! thank you!
 
 
 
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