Expert: Alon Mandes - 6/4/2009

Question
1. At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 15 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.)

i tried working this problem out but i don't think i set it up right. i did:
after doing the derivatives and solving for dz/dt i got:

dz
-- = 2x (dx/dt) + 2y (dy/dt)
dt   -----------------------
            2z
then did:

= (30)(15)+(32)(16)
 ----------------
 (sqrt(30^2+32^2))

= 21.9317

how do I do this?

2. A plane flying with a constant speed of 24 km/min passes over a ground radar station at an altitude of 11 km and climbs at an angle of 20 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 2 minutes later?  

Answer
Hello Brook,
1. I assume you took z=√(x²+y²) . Thus,

dz    2x (dx/dt) + 2y (dy/dt)
-- =  -----------------------
dt           2√(x²+y²)
           
dz    2(40)(15)*(40) + 2(16)(15)*(16)
-- =  -------------------------------
dt           2√[(40*15)²+(16*15)²]

As for 2 , I have solved that kind of question before, you may
browse through my Answered questions queue, & you will find it.

Alon.