A boy stands on a shore and pulls a rope connected to a boat. The velocity of the rope v_{r} is constant. Prove that the closer the boat to the shore, the faster it moves.

###### Solution.

We can start with the Pythagorean formula.

h^{2} + s^{2} = l^{2}

The length of the rope l and the distance between the boat and shore s are functions of time. The height from the sea level to the boy’s hand is constant.

We differentiate the Pythagorean expression with respect to time and we get

d(h^{2})/dt + d(s^{2})/dt = d(l^{2})/dt

or

2s*ds/dt = 2l*dl/dt

Now, we can use the fact that s/l = cos(α) , where α depends on time.

Also, it has to be clear that

ds/dt = v_{b}, where v_{b} is the velocity of the boat and

dl/dt = v_{r}.

Finally, we got the following expression

v_{b} = v_{r}/cos(α).

From the last expression, we can see that when the boat approaches the shore it moves faster, because cos(α) decreases as α increases, so v_{b} increases as the velocity of the rope v_{r} remains constant.