Hello, Keith!

At noon, ship A is 20 nautical miles due west of ship B.

Ship A is sailing west at 21 knots and ship B is sailing north at 18 knots.

How fast (in knots) is the distance between the ships changing at 4 PM?

(Note: 1 knot is a speed of 1 nautical mile per hour.)

I have an approach that eliminates dealing with the two different speeds.

I'll refer to "miles" (nautical miles) and "miles per hour" (knots). Code:

* B
* |
* |
x * | 18t
* |
* |
* |
* - - - - - - - * - - - - - *
A 21t P 20 Q

Ship A starts at point P, 20 miles west of point Q and sails west at 21 mph.

. . In t hours, it has sailed 21t miles to point A.

Ship B starts at point Q and sails north at 18 mph.

. . In t hours, it has sailed 18t miles to point B.

Let x = distance AB.

. . . . . . . . . . _________________

Then: .x .= .√(21t + 20)² + (18t)² .= .(765t² + 840t + 400)^½

Differentiate with respect to time:

. . dx

. . --- . = .½(765t² + 840t + 400)^(-½)·(1530t + 840)

. . dt

At 4 PM (t = 4), we have:

. . dx . . . . . . . 1530·4 + 840 . . . . . . . . . . . 6960

. . --- . = . ---__---------------------------__- . = . ---__---------__-

. . dt . . . . 2√765·4² + 840·4 + 400 . . . . .2√16,000

Therefore: . dx/dt .≈ .27.5 mph (knots).