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Advanced Math/rotation trig. problem


ferris wheel
ferris wheel  
Hello Paul -- I seem to be stuck again on a problem that involves using trig. on a ferris wheel problem. COuld you take a look at it and maybe explain to me how you came to resolve your answer? Thanks very much again.

If the radius is 60 m, and the rotation time is 30 minutes, then you want a picture (which you should have drawn to start things off) that looks like this:

1. Draw a nice circle.  Put the center C at  (0,60), and radius = 60.

2. The bottom of the circle is at the origin, which we call O.

3. Mark the angle theta (I will write 'T' for that.) whose vertex is at the center and which is bounded by CO and CP.  (It is angle OCP)

4. Let P be any point on the circle.  Let's find the coordinates of P.

On the diagram, CP = 60,  CM = 60 cos T,  PM = 60 sin T

OM = CO - CM = 60 - 60 cos T, and that is the y-coordinate of P.

PM is the x-coordinate of P.

So the coordinates are P(60 sin T, 60(1- cos T) )

What about T as a function of time t?

Since you start with T = 0 (radians) at t = 0 (minutes)

and have T = 2pi at t = 30 minutes,

T = (pi/15) t = t pi/15

Finally, that gives you  P(60 sin(t pi/15), 60(1 - cos(t pi/15) )

That should be enough to get you started.  

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Paul Klarreich


I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.


I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.


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