Answer:
The boat is approaching the dock at a rate of <u>2.5 ft/s</u>.
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, 
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:

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

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

Now, substituting all the given values in equation (2) and solve for
. This gives,

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