Rotating Beacon Light
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Please help me with this derivative problem geniuses!!?
A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that's closest to the beacon. As the beacon rotates at 10 revolutions per minute, the beam sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles downshore from point A?
1 revolution = 2(pi) radians
10 rev/min = 20(pi) rad/min
Location of beacon: Point B
Point on the shore closest to Point B: Point A
Point located downshore from Point A: Point C
AB = 2
AC = x
m < ABC = n radians
tan n = x / 2
tan n = (1/2)x
[(sec n)^2](dn/dt) = (1/2)(dx/dt)
x = 2
n = arctan (2/2) = arctan 1 = pi/4
dn/dt = 20(pi)
{[sec (pi/4)]^2}(20)(pi) = (1/2)(dx/dt)
{[sqrt(2)]^2}(20)(pi) = (1/2)(dx/dt)
40(pi) = (1/2)(dx/dt)
40(pi) * 2 = (1/2)(dx/dt) * 2
80(pi) = dx/dt
The point is moving at 80(pi) mi/min.
Best answer since I did all that hard math, please? 
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