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Expert: - 10/21/2009


Question
At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 25 knots. How fast (in knots) is the distance between the ships changin at 4 pm?

Answer
Take where ship B is as the origin, which puts A at –10.
Ship A’s position is given by P(A) = 10 + 15t, where t is in hours and P is in the -x direction.
Ship B’s position is given by Q(B) = 25t, where t is in hours and Q is in the y direction.

Since these form the length of two legs of a right triangle, the length of the hypotenuse is
√[(15t+10)²+(25t)²] = √(225t² + 300t + 100 + 625t²) = √(850t² + 300t + 100).

Thus, we had D(t) = √(850t² + 300t + 100), where D is the distance between them.

Divide everything by 25 and take 25 outside as √25 = 5.
This gives the distance as D(t) = 5√(34t² + 12t + 4).

The derivative of the inside is 68t + 12,
so the derivative of the whole thing is

D’(t) = 5(68t+12)/√(34t² + 12t + 4).  This is the speed.

Now 4 pm is 4 hours away, so t = 4.  Put t=4 into D’(t) and the answer is
5(272+12)/√(544 + 48 + 4) = 5(284)/√596 = 5(2)(2)71/√(4(149)) = 710/√149.

Now most of the time, square-roots and the denominator don’t get along, so this is
710√149/149.

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