Answer :
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
[tex]x^{2} + y^{2} = z^{2}[/tex]
Differentiating the above equation with respect to time yields,
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex]
[tex]x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}[/tex]
dx/dt = 0 since dock height doesn't change
[tex]y\frac{dy}{dt} = z\frac{dz}{dt}[/tex]
[tex]\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}[/tex]
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,
[tex]\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}[/tex]
[tex]\frac{dy}{dt} = \frac{13.93}{13}(2)[/tex]
[tex]\frac{dy}{dt} = 2.14[/tex] [tex]ft/s[/tex]
Therefore, the boat is approaching the dock at rate of 2.14 ft/s
So, the boat approached the dock with a speed of 2.1337 m/sec.
Pythagoras Theorem:
Pythagoras Theorem is an important topic in Maths, which explains the relation between the sides of a right-angled triangle.
So, the formula is,
[tex]a^2+b^2=c^2[/tex]
Differentiating the above equation,
[tex]2a\frac{da}{dt} +2b\frac{db}{dt} =2c\frac{dc}{dt} ...(1)[/tex]
It is given that,
[tex]a=5m\\\frac{da}{dt}=0\\ c=13\\\frac{dc}{dt}=2 m/s[/tex]
[tex]b=\sqrt{13^2-5^2} \\b=12 m/s[/tex]
Substituting the above values in equation (1) we get,
[tex]2\times5\times0+2\times12\frac{db}{dt} =2\times13\times2\\\frac{db}{dt} =\frac{26}{12}\\ \frac{db}{dt} =2.1337 m/s[/tex]
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