Question

A boat is pulled into a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 5 feet higher than the front of the boat. The rope is being pulled through the ring at the rate of 0.6 ft/sec. How fast is the boat approaching the dock when 13 feet of rope are out

Answer 1

Answer:

The boat is approaching the dock  at 0.65 ft/sec when 13 feet of rope are out.

Step-by-step explanation:

Let x be the length of rope and y be the distance between boat and dock.

Height of point where a ring attached =z=5 feet

$\frac{dx}{dt}=-0.6ft/sec$

x=13 feet

Using Pythagoras theorem

$H^2=P^2+B^2$

$x^2=z^2+y^2$

$x^2=5^2+y^2=25+y^2$

$(13)^2=25+y^2$

$169-25=y^2$

$144=y^2$

$y=12 feet$

Differentiate w.r.t t

$2x\frac{dx}{dt}=2y\frac{dy}{dt}$

$x\frac{dx}{dt}=y\frac{dy}{dt}$

$13(-0.6)=12\frac{dy}{dt}$

$\frac{7.8}{12}=\frac{dy}{dt}$

$\frac{dy}{dt}=-0.65feet/sec$

Hence,the boat is approaching the dock  at 0.65 ft/sec when 13 feet of rope are out.