Related Rates Calculus

Question

A man is running over a bridge at a rate of 20 ft/sec while a boat passes under the bridge and immediately below him at a rate of 10ft/sec/ The boat's course is at right angles to the man's and is 20 ft below it. How fast are the man and the boat seperating one second later?

Answer 1

D = sqrt(20^2 + (20t)^2 + (10t)^2) = sqrt(400+500t^2)

dD/dt = 500t/sqrt(400+500t^2)

t=1, so dD/dt = 500/sqrt(900) = 500/30