Related rates problem

[chocoholic]

This is the question.

"Consider two ships which are joined by a cable attached to each ship at the water line. Suppose the two
ships are 200 metres apart, with the cable stretched tight and attached to a pulley which is anchored
halfway between the ships at a depth of 45 metres. If one ship moves away from the other at 3 km/h,
how quickly is the other ship moving after one minute?"

I know that initially I can use the Pythagorean theorem to make an equation :

let the distance between anchor and ship B be s; the distance between ship A and B is 200m, so the midway through would be 100m.
s^2= 45^2+100^2
s= sqrt (45^2+100^2)



Then I'm stuck...Please help!!

 

[Unknown008]

From this, you get half the length of the cable, which you can use to get the full length of the cable. This will remain constant throughout the problem.

Let A be the ship which moves first (at 3 km/h) and B be the ship we need to find the speed, while C be the position of the anchor of the cable and D the point vertically above the anchor, between the two ships.

The relation between the distance AD and AC is:
\(\displaystyle AC^2 = 45^2 + AD^2\)

The rate of change of AD becomes:

\(\displaystyle \dfrac{d(AC)}{dt} = \dfrac{d(\sqrt{45^2 + AD^2})}{dt}\)

Since the cable has a constant length, the rate of change of the length BC becomes equal to the rate of change of AD, only negative.

\(\displaystyle \dfrac{d(AC)}{dt} = -\dfrac{d(BC)}{dt}\)

The rate of change of BC and BD is (again Pythagoras):

\(\displaystyle \dfrac{d(BC)}{dt} = \dfrac{d(\sqrt{45^2 + BD^2})}{dt}\)


This gives us:
\(\displaystyle \dfrac{d(AC)}{dt} = \dfrac{d(\sqrt{45^2 + AD^2})}{dt} = -\dfrac{d(BC)}{dt} = -\dfrac{d(\sqrt{45^2 + BD^2})}{dt}\)

\(\displaystyle 3 = -\dfrac{d(\sqrt{45^2 + BD^2})}{dt}\)

\(\displaystyle 3 = -\dfrac12(45^2 + BD^2)^{-0.5}\cdot 2BD\dfrac{d(BD)}{dt}\)

Remember, you need to get the rate of change of BD. You can get the value of BD by using the time elapsed. Can you take it from here?