Posted by **Anonymous** on Saturday, April 3, 2010 at 8:21am.

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 15 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

- Calculus -
**bobpursley**, Saturday, April 3, 2010 at 9:16am
draw the figure, at noon, then at 4PM.

At 4 pm, I have a right triangle of A,B of sides (base 10+19*4; altitude 15*4)

d^2=(10+va*t)^2 + (vb*t)^2

2d dd/dt=2(10*va*t)va+2(vb*t)vb

so when t=4, solve for dd/dt

- Calculus -
**Anonymous**, Tuesday, October 5, 2010 at 9:11pm
9898

- Calculus -
**Anonymous**, Sunday, October 12, 2014 at 6:34pm
khj

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