Answer:
The boat is approaching the dock at a rate of 2.5 ft/s.
Explanation:
Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.
Given:
The height of dock above water (h) = 6 feet
Rate of pull of rope or rate of change of rope is, [tex]\frac{dl}{dt}=2\ ft/s[/tex]
As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.
Now, the above situation represents a right angled triangle as shown below.
Using Pythagoras Theorem, we have:
[tex]l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)[/tex]
Now, differentiating the above equation with time 't', we get:
[tex]2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)[/tex]
Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,
[tex]b^2=10^2-36\\\\b=\sqrt{64}=8\ ft[/tex]
Now, substituting all the given values in equation (2) and solve for [tex]\frac{db}{dt}[/tex]. This gives,
[tex]\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s[/tex]
Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.
