Another AP Physics question?

Question

A lifeguard standing on a tower throws a buoy to a swimmer 18 m from the tower . The lifeguard, positioned 3 m above the water, pulls in the rope at a speed of 1 m/s.

How fast is the swimmer coming when he is 3 m from the water's edge?

Answer 1

Let:

x be the distance of the swimmer from the base of the tower,

y be the length of the rope.

y^2 = x^2 + 3^2 ...(1)

Differentiating with respect to time t:

2y dy / dt = 2x dx / dt

dx / dt = y (dy / dt) / x ...(2)

When dy / dt = - 1 m/sec, x = 3m,  (1) gives:

y = sqrt(18) = 3 sqrt(2) m.

Substituting these values in (2):

dx / dt = 3 sqrt(2)(- 1) / 3

= - sqrt(2) m / sec.

The swimmer is coming in at sqrt(2) m/sec.