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July 24, 2014

July 24, 2014

Posted by **Bailey** on Monday, October 7, 2013 at 4:11am.

- Trig -
**Steve**, Monday, October 7, 2013 at 4:23amwe know we want

y = a sin(bx+c) + d

diameter is 2, so radius=1, and we have

y = a sin(bx+c) + 1

Starting at t=0, the bead starts going downward, so

y = -sin(bx+c)+1

If we make it so the axle is at (0,1) at t=0, we need to shift the graph to the right by 1, since the bead is on the rim

y = -sin(b(x-1)) + 1

The period is 15, so b = 2π/15

y = -sin(2π/15 (x-1))

= -sin(2π/15x - 2π/15) + 1

Naturally, if we took the bead starting at (0,0), we'd have

y = -sin(2π/15 x)

but the axle-centered graph shifts things to the right and up.

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