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A boat is pulled in to 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 6 ft higher than the front of the boat. The rope is being pulled through the ring at the rate of 0.4 ft/sec. How fast is the boat approaching the dock when 10 ft of rope is out?

Answer :

davidlcastiblanco

Answer:

The speed of the boat when is approaching the dock is

v=0.5 [tex]\frac{ft}{s}[/tex]

Explanation:

When the boat is 10ft os the rope using trigonometry can find the length to know the speed  the boat is

[tex]h^{2} =x^{2} +y^{2} \\h=10ft\\y=6ft\\10^{2}=x^{2}+6^{2}\\x^{2}=10^{2}-6^{2} \\ x=\sqrt{100-36} =\sqrt{64}\\  x=8 ft[/tex]

Now the velocity final is just at x axis so iis derivate the function can find the velocity

[tex]\frac{dh}{dt}*h=\frac{dx}{dt}*x+\frac{dy}{dt}*y\\\frac{dy}{dt}=0\\0.4\frac{ft}{s}*10=\frac{dx}{dt}*8+0\frac{ft}{s}*6\\\frac{dx}{dt}*8ft=0.4\frac{ft}{s}*10ft\\\frac{dx}{dt}=0.5\frac{ft}{s}[/tex]

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