1. ## related rates

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 took the derivitive of both sides dx/dt = 21 and dy/dt =18
and at 4pm, x= 20 +21*4, y=18*4
21+21+21+21+20 =104
18*4= 72
D = sqrt(x^2+y^2)
D' = 1/2(x^2+Y^2)^-1/2 *(2xx' +2xY')
Z= sqrt((104+72) + 20^2) = sqrt(31376)

104+72/sqrt(31376)*(21+18)
would you let me know if I am on the right path.
Thankx
keith

2. 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).