Question

The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approaching the dock at the moment when there is 10 feet of rope still left to pull in

Answer 1

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, $\frac{dl}{dt}=2\ ft/s$

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:

$l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)$

Now, differentiating the above equation with time 't', we get:

$2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)$

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

$b^2=10^2-36\\\\b=\sqrt{64}=8\ ft$

Now, substituting all the given values in equation (2) and solve for $\frac{db}{dt}$. This gives,

$\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s$

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.