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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

User Discordian
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1 Answer

9 votes

Answer:

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

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


(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(dx)/(dt)=2y(dy)/(dt)


x(dx)/(dt)=y(dy)/(dt)


13(-0.6)=12(dy)/(dt)


(7.8)/(12)=(dy)/(dt)


(dy)/(dt)=-0.65feet/sec

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

User LebRon
by
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