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Math Help - Ferris wheel!

  1. #1
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    Ferris wheel!

    A spider attached itself to the point P on the side of car C when the point P was at its lowest point at 1.00pm.

    The height, h meters, of the point P above ground level, at the time t hours after 1.00pm is given by;

    h(t) = 62 + 60sin( (5t-1)pi / 2)


    Two questions to the information;
    i. Find the time, after 1.00pm, when P first reaches a height of 92 meters above ground level.

    ii. Find the number of minutes during one rotation that P is at least 92 meters above ground level.


    All help is much appreciated.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mercy99 View Post
    A spider attached itself to the point P on the side of car C when the point P was at its lowest point at 1.00pm.

    The height, h meters, of the point P above ground level, at the time t hours after 1.00pm is given by;

    h(t) = 62 + 60sin( (5t-1)pi / 2)


    Two questions to the information;
    i. Find the time, after 1.00pm, when P first reaches a height of 92 meters above ground level.

    ii. Find the number of minutes during one rotation that P is at least 92 meters above ground level.


    All help is much appreciated.
    For part (i), set h(t)=92:

    92=62+60\sin\bigg(\frac{(5t-1)\pi}{2}\bigg)

    Now find t:

    30=60\sin\bigg(\frac{(5t-1)\pi}{2}\bigg)

    \implies \frac{1}{2}=\sin\bigg(\frac{(5t-1)\pi}{2}\bigg)

    \implies \sin^{-1}\bigg(\frac{1}{2}\bigg)=\frac{(5t-1)}{2} **

    \implies \frac{\pi}{6}=\frac{(5t-1)\pi}{2}

    \implies \frac{\pi}{3}=(5t-1)\pi

    \implies \frac{1}{3}=5t-1

    \implies \color{red}\boxed{t=\frac{4}{15} \ \text{hours} = 16 \ \text{minutes}}

    For part (ii), find the other time value when h(t)=92

    Let's go back to **

    \implies \sin^{-1}\bigg(\frac{1}{2}\bigg)=\frac{(5t-1)}{2}

    This time, we will let \sin^{-1}\bigg(\frac{1}{2}\bigg)=\frac{5\pi}{6}

    \implies \frac{5\pi}{6}=\frac{(5t-1)\pi}{2}

    \implies \frac{5\pi}{3}=(5t-1)\pi

    \implies \frac{5}{3}=5t-1

    \implies \color{red}\boxed{t=\frac{8}{15} \ \text{hours} = 32 \ \text{minutes}}

    Thus the total time that h(t) is at least 92 meters above the ground is \frac{8}{15}-\frac{4}{15}=\color{red}\boxed{\frac{4}{15} \ \text{hours}=16 \ \text{minutes}}.

    I hope you can follow this.

    If you need me to clarify things, feel free to ask.

    --Chris

    EDIT : Used wrong arcsine value...
    Last edited by Chris L T521; July 13th 2008 at 12:12 AM. Reason: screwed up...>_<. Its 5 pi/6....not 7 pi/6
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  3. #3
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    Very well done. I appreciate your help Chris, again
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