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Thread: amusement park ride

  1. #1
    Eater of Worlds
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    amusement park ride

    Hello All and Happidy New Year. I am setting here sipping on some Tangueray Rangpur and thought I would pose a rather fun calc applications problem.

    "An amusement park ride has cars which move along a track (I hope I can get it across in my haphazard diagram). The cars are attached to arms that are 20 feet long and are connected to a common point on the axis of rotation(z-axis). The arms swing up and down along the track whose elevation is given by $\displaystyle z=4sin^{2}({\theta})$. Assuming all distances are measured in feet, find the length of the track".

    I tried drawing the picture, but a I think I got it a little askew. Not too bad, though. It's height is 0 where it touches the x-axis and at the points over the y-axis it is 4 feet in elevation. The circumference of a circle with radius 20 is 125.66, so we know it'll be a wee bit more than that.

    If anyone wants to try it, see what you come up with. I attempted it in both polar and spherical and arrived at the same answer using the arc length formulae. So, I am pretty sure I am correct.

    I used @ for theta because that's easier in Paint.
    Last edited by galactus; Nov 24th 2008 at 05:38 AM.
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  2. #2
    Eater of Worlds
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    It would appear no one seemed interested in tackling this problem. I hope my hackneyed diagram wasn't to blame
    Anyway, if anyone would like to see what I came up with, let me know.
    I thought it was a cool problem.
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  3. #3
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    Partial response

    Well, I think cylindrical coordinates would be the best to work with, parametrizing by the angle $\displaystyle \theta$. Then
    $\displaystyle z=4\sin^2\theta$, and (using that our ride is at the end of an arm of length 20)
    $\displaystyle r=\sqrt{20^2-z^2}=4\sqrt{25-\sin^4\theta}$. For cylindrical coordinates, $\displaystyle ds^2=dr^2+r^2d\theta^2+dz^2$, and so, in terms of $\displaystyle \theta$, $\displaystyle ds=\sqrt{r^2+(\frac{dr}{d\theta})^2+(\frac{dz}{d\t heta})^2}d\theta$. Taking these derivatives and combining, I get that the length is:
    $\displaystyle S=4\int_{0}^{2\pi}\frac{\sqrt{\sin^8\theta-150\sin^4\theta+100\sin^2\theta+625}}{\sqrt{25-\sin^4\theta}}\,d\theta$. I haven't tried to integrate it yet.

    --Kevin C.
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  4. #4
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    Integral

    In fact, I'm not sure how to do that integral (if it can be done analytically; it looks like it might reduce to an elliptic integral).

    --Kevin C.
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  5. #5
    Eater of Worlds
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    Very good, Twisted. That's the answer I got as well. I also tried it in spherical and got the same so that must be it. You're not sure how to do the integral?. I know exactly how to get it.......... Run it through technology.
    I used my TI-92. It gave me 125.985. Since you done the polar in fine fashion, here's the spherical.

    The spherical arc length formula is $\displaystyle \int_{0}^{2\pi}\sqrt{(\frac{d{\rho}}{d{\theta}})^{ 2}+{\rho}^{2}sin{\phi}+{\rho}^{2}(\frac{d{\phi}}{d {\theta}})^{2}}$

    Since $\displaystyle z=4sin^{2}{\theta}, \;\ z=20cos{\phi}$

    $\displaystyle 4sin^{2}{\theta}=20cos{\phi}$

    $\displaystyle {\phi}=cos^{-1}(\frac{sin^{2}{\theta}}{5})$

    We can find the derivatives and take the subs:

    $\displaystyle 20\int_{0}^{2\pi}\sqrt{1-\frac{sin^{4}{\theta}}{25}+\frac{4sin^{2}{\theta}c os^{2}{\theta}}{(sin^{2}{\theta}+5)(cos^{2}{\theta }+4)}}d{\theta}$

    This gives the same exact result as the polar method.

    Kind of a tough problem to set up. Don't even think about doing these integrals by hand. That's what computers are for.
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