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Math Help - Modelling and Problem Solving Calculus

  1. #1
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    Modelling and Problem Solving Calculus

    ***ANSWERED***
    Last edited by MathsThief; August 15th 2006 at 12:17 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by MathsThief
    Hello. This is a crazy modelling i found in my maths work. i've been making assumptions and they are crap >_< plz help:
    A function f(x) is considered to be “smooth” in the interval [a,b] if f'(x) is continuous throughout the interval.

    The arc length of a smooth function in the interval (that is, the length of the curve) is given by:



    Totally Twak (TT) Industries manufactures exercise equipment. The company wants its new model of exercise bike to feature a “hill profile” that simulates riding the bike over two mountains (one small and one large). The approximate shape of the mountains involved is shown below:
    When the profile is chosen, the bike alters the resistance to simulate riding uphill (greater resistance) and downhill (less resistance). Although the starting and finishing points are to be only 15 kilometres apart horizontally, the distance that is to be “ridden” is 20 kilometres because of the mountains.

    TT hires you to design the profile (but not to calculate the sizes of the resistances). Your task is to find intervals of a minimum of 2 functions that can be placed together to closely match the shape of the hills. Both x and y for the functions are to be measured in kilometres. The following conditions must be met:

    • the total arc length must be as close to 20 kilometres as possible (the range 19.5<=L<=20.5 is acceptable).

    • at the point(s) where the intervals join up, the derivatives of the functions must not differ significantly so that a “smooth” ride is guaranteed.

    Checklist

    (i) The equations of the functions and the intervals that have been chosen.

    (ii) A graph that clearly shows the profile.

    (iii) Verification that the derivatives at the “join” point(s) do not differ greatly.

    (iv) Verification that the total arc length lies in the required range. (All definite integrals may be performed on your graphics calculator.)

    (v) A detailed explanation of how you approached the exercise (how you chose your functions, etcetera).
    Choose two cubics, one of the form a x^3+b x^2+c x +d, with a>0, and the other
    \alpha x^3 + \beta x^2 + \gamma x +\delta, with \alpha<0.

    Arrange for the first to have a local minimum at x=5, and the
    second to have a local maximum at the same point, and then arrange that the
    two curves are equal at the point. Then if this point is the join the merged curve will be smooth.

    Now arrange for the total length of the merged curve to be 20km

    RonL
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  3. #3
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    Thanks! Please delete thread this i'm done now. ^_^
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by MathsThief
    Thanks! Please delete thread this i'm done now. ^_^
    No, threads are generaly left up to inform others and stimulated discussion

    RonL
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    Do not delete your question, ever!
    That violates the principles by which this site was founded.
    -=USER WARNED=-
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    Choose two cubics, one of the form a x^3+b x^2+c x +d, with a>0, and the other
    \alpha x^3 + \beta x^2 + \gamma x +\delta, with \alpha<0.

    Arrange for the first to have a local minimum at x=5, and the
    second to have a local maximum at the same point, and then arrange that the
    two curves are equal at the point. Then if this point is the join the merged curve will be smooth.

    Now arrange for the total length of the merged curve to be 20km

    RonL
    I'm attempting a similar problem -- would you be able to expand upon that? I'm having problems manipulating my cubic function(s) to fit accordingly (namely, setting local max/mins and getting them to lie between two points).

    Also, those two cubic function would only cover the first 'part' of the 'profile', and not the final slope downwards -- how would you go about that?

    Many thanks,

    Faux
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