Posted by **sophie** on Sunday, May 22, 2011 at 12:16am.

illuminated by the rays of the setting Sun, Kelly rides along on a merry-go-round casting a shadow on a wall. The merry-go-round is turning 36 deg per sec. Kelly is 25 ft from its center, and the Sun's rays are perpendicular to the wall.

At what rate is the shadow moving when it is 15 ft from the point on the wall that is closest to the merry-go-round?

- math -
**drwls**, Sunday, May 22, 2011 at 9:37am
Draw a figure with a circle of radius 25 and a flat sceen behind it. Draw a line through the center of the circle that extends back to the screen, which will be perpendicular to the line. Let the angle of Kelly's location from the sun direction line be A.

A = sin^-1 15/25 = 36.8 degrees

The location of the shadow relative to the closest position (where the line meets the wall) is

X = R sin A

The rate the shadow moves is

dX/dt = R cos A* dA/dt

where dA/dt is in radians/s

dA/dt = (2 pi/10 rad)/1 s

= 0.6283 rad/s

dX/dt = 25 ft*0.8*0.6283 rad/s= 12.6 ft/s

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