
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(zaxis). The arms swing up and down along the track whose elevation is given by . 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 xaxis and at the points over the yaxis 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.

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.

Partial response
Well, I think cylindrical coordinates would be the best to work with, parametrizing by the angle . Then
, and (using that our ride is at the end of an arm of length 20)
. For cylindrical coordinates, , and so, in terms of , . Taking these derivatives and combining, I get that the length is:
. I haven't tried to integrate it yet.
Kevin C.

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.

Very good, Twisted(Bigsmile). 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 TI92. It gave me 125.985. Since you done the polar in fine fashion, here's the spherical.
The spherical arc length formula is
Since
We can find the derivatives and take the subs:
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.