Posted by **CRYSTAL** on Monday, October 24, 2011 at 10:07am.

At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship B is sailing north at 19 knots. How fast (in knots) is the distance between the ships changing at 5 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

- CALCULUS -
**Steve**, Monday, October 24, 2011 at 11:40am
After t hours, distance d is

d^2 = (40+23t)^2 + (19t)^2

2d dd/dt = 2(40+23t)(23) + 2(19t)(19)

When t=5, d^2 = 155^2 + 95^2, d=181.8

2(181.8) dd/dt = 2(155)(23) + 2(95)(19)

363.6 dd/dt = 10740

dd/dt = 29.5 knots

check my math . . .

- CALCULUS -
**John Lee**, Thursday, March 8, 2012 at 4:36pm
Your math be wrong.

## Answer this Question

## Related Questions

- Calculus - At noon, ship A is 50 nautical miles due west of ship B. Ship A is ...
- calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
- calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
- calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
- calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...
- Calculus - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...
- Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
- Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
- calculus 1 - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...
- calculus 1 - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...