Calculus, Related Rates

A search light in a prison rotates counterclockwise at a rate of 3 revolutions per second. The light shines on a long straight wall that is 40 feet from the search light. How fast is the light beam moving across the wall when the beam is hitting the wall at a spot which is 80 feet from the light?

sahsjing made error i think:this is a basic 1:2 triangle so
tan(6pi*t) = 1/√3 and not 2 value he gives?

x'/40 = sec^2(6pi*t) (6pi) = (1+tan^2(6pi*t)) (6pi) = (1+2^2)(6pi)
becomes
= (1+1/√3^2)(6pi) = (4/3)(6pi)= 1005 ft/sec
hope sahsjing will clarify.

Nevermind, I understand where the 6pi came from!!! Now it makes sense.

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This is not right. The 80 is the distance from the light to the wall, the hypotenuse.
The tangent would be sqr(3), and the secant is 2.

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P(light)
|..\
|
|(40)
___x__

Tan P = x/40

X= 40tan P
Dx/dt = 40* sec^2(P) *dP/dt
DP/dt = 3rev= 6pi rads/sec

Dx/dt= 40sec^2(P) * 6pi

When the spot is 80 feet from the light, secP = 80/40=2

Dx/dt = 40(2^2)(6pi) = 960pi ft/sec

Hoping this helps!

x/40 = tan(6pi*t), where x = 80 at the time, and t is the time in sec.
Differentiate with respect to t,
x'/40 = sec^2(6pi*t) (6pi) = (1+tan^2(6pi*t)) (6pi) = (1+2^2)(6pi)
x' = 1200 pi ft/sec