calculus
posted by Anonymous .
a bridge is 10 m. above a canal. A motorboat going 3 m/sec passes under the center of the bridge at the same instant that a woman walking 2 m/sec reaches that point. How rapidly are they separating 3 sec. later?

At a time of t seconds
let the distance covered by the boat be 3t m
let the distance covered by the woman be 2t m
If D is the distance between them,
D^2 = (3t)^2 + 10^2 + (2t)^2
D^2 = 13t^2 + 100
2D dD/dt = 26t
dD/dt = 13t/D
at 3 seconds D^2 = 81 + 100 + 16 = 197
D = √197
so at 3 seconds , dD/t = 13(3)/√197 = 2.7786
= appr 2.8 m/s
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