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ololo11 [35]
3 years ago
7

A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p

ost on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet

Mathematics
1 answer:
irina [24]3 years ago
8 0

Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

x^{2} + y^{2} = z^{2}

Differentiating the above equation with respect to time yields,

2x\frac{dx}{dt}  + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}

x\frac{dx}{dt}  + y\frac{dy}{dt} = z\frac{dz}{dt}

dx/dt = 0  since dock height doesn't change

y\frac{dy}{dt} = z\frac{dz}{dt}

\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}

\frac{dy}{dt} = \frac{13.93}{13}(2)

\frac{dy}{dt} = 2.14 ft/s

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

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