Question
A barge, whose deck is 10 ft below the level of a Dock, is being drawn in by means of a cable attached to the deck and passing through a ring on the dock. When the barge is 24ft away and. Approaching the Dock at 3/4 per second. How fast is the cable being pulled?
Answers
R_scott
d = distance to barge , c = length of cable
d^2 + 10^2 = c^2
implicit differentiation ... 2 d dd/dt = 2 c dc/dt ... d dd/dt = c dc/dt
24 * 3/4 = √(24^2 + 10^2) * dc/dt
d^2 + 10^2 = c^2
implicit differentiation ... 2 d dd/dt = 2 c dc/dt ... d dd/dt = c dc/dt
24 * 3/4 = √(24^2 + 10^2) * dc/dt