let x be the horizontal distance to the dock.
x^2=R^2-100 where R is rope length to the pulley.
2x dx/dt=2RdR/dT
solve for dx/dt when R is 125? (x= sqrt(125^2-100). dR/dt= 20ft/min
A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the bow of the boat at a point 10 ft below the pulley. If the rope is pulled through the pulley at a rate of 20 ft/min, at what rate will the boat be approaching the dock when 125 ft of rope is out?
3 answers
2.oo6
yum