calculus
posted by Anonymous .
At noon, ship A is 150 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

If we assume A is at (0,0) at noon, the the distance z is
z^2 = (15025t)^2 + (20t)^2
now get z when t=4
and find dz/dt from
2z dz/dt = 50(15025t) + 40(20t)
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