Calculus
posted by bryan .
A lighthouse is located on a small island 3 km away from the nearest point P on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P?

A is angle in radians between perpendicular to shore at P (length 3 ) and light at shore (x which is 1 at start).
tan A = x/3
dA/dt = w = rad/sec = 4* 2 pi/60 = .419 rad/s
dx/dt = 3 d tan A /dt
dx/dt = (3/sec^2 A) dA/dt
dx/dt = 1.26 /sec^2 A = 1.26 cos^2 A
at x = 1, tan A = 1/3
so A = 18,43 deg
so cos A = .949
1.26 cos^2 A = 1.13 km/s 
Consider a right angled triangle,
tan(A)=x/3
cos(A)=3/sqrt(x^2+9)
sec^2(A)=(x^2+9)/3^2
dA/dt=4 rev/min = 2*pi/15 rad/s
Also,
x=3 tan(A)
dx/dt=3 sec^2(A) dA/dt
dx/dt=(x^2+9)/3 * 0.419
At x=1, dx/dt=1.396 km/s or 5027 km/hr.
The above attempt confused sec^2(A) with 1/sec^2(A), and 1/cos^2(A) with cos^2(A).