Posted by soffy on Monday, June 11, 2012 at 10:00am.
If x is the distance from the closest point on the wall, then
x = 10 tanθ
dx/dt = 10 sec^2θ dθ/dt
when x=50, tanθ = 5, so secθ = √26
dx/dt = 10(26) (75)(2π) = 39000π ft/sec
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