Calculus
posted by Arvin .
A ladder 10 ft leans against a vertical wall. The upper end slips down the wall at 5 ft/sec. How fast is the ladder turning when it takes an angle of 20 degrees with the ground.

let the top of the ladder by y ft up on the wall
let the angle made at the foot be Ø
sinØ = y/10
10 sinØ = y
10 cosØ dØ/dt = dy/dt
given: when Ø = 20° , dy/dt = 5 ( y is decreasing)
10 cos20° dØ/dt = 5
dØ/dt = 1/(2cos20°) = .532 radians/sec
or angle is decreasing at .532 radians / sec
which is appr. 30.5° / sec
(notice the angle is changing very rapidly at that moment, since the ladder has almost fallen)
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