A man on the end of a tall pier is fishing. When he has caught a fish, it is reeled in at a rate of 0.6 feet per minute from a point 15 feet above the water. Find the rate at which the angle the string makes with the water is changing when there is 36 feet of line out. Answer (in radians per minute): 7.639*10^(-3)

A man on the end of a tall pier is fishing. When he has caught a fish, it is reeled in at a rate of 0.6 feet per minute from a point 15 feet above the water. Therefore:

sin θ = 15/x

If we differentiate the LHS and RHS of the equation above with respect to the variable 't', we have:

cos θ dθ/dt = -15/x^2 dx/dt equation (1.0)

With the rate of (-)0.6 ft/min and x =36 ft, we can complete the right-angle triangle formed: hypotenuse = 36, opposite = 15, and adjacent (y) is unknown. Thus:

y = sqrt(36^2 - 15^2) = sqrt(1296-225) = sqrt(1071) = 32.73 ft

cos θ = 32.73/36 = 0.9091

Using equation (1.0) above:

0.9091 dθ/dt = -15/36^2 (-0.6) = 0.00694

Thus:

dθ/dt = 0.00694/0.9091 = 7.639*10^(-3) radians per minute