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
[tex]\frac{dx}{dt}=-0.6ft/sec[/tex]
x=13 feet
Using Pythagoras theorem
[tex]H^2=P^2+B^2[/tex]
[tex]x^2=z^2+y^2[/tex]
[tex]x^2=5^2+y^2=25+y^2[/tex]
[tex](13)^2=25+y^2[/tex]
[tex]169-25=y^2[/tex]
[tex]144=y^2[/tex]
[tex]y=12 feet[/tex]
Differentiate w.r.t t
[tex]2x\frac{dx}{dt}=2y\frac{dy}{dt}[/tex]
[tex]x\frac{dx}{dt}=y\frac{dy}{dt}[/tex]
[tex]13(-0.6)=12\frac{dy}{dt}[/tex]
[tex]\frac{7.8}{12}=\frac{dy}{dt}[/tex]
[tex]\frac{dy}{dt}=-0.65feet/sec[/tex]
Hence,the boat is approaching the dock at 0.65 ft/sec when 13 feet of rope are out.
