calculus
posted by asdf .
A man starts walking north at 4ft/s from a point P. Five minutes later a woman starts walking south at 5ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?
I'm confused about the diagram and what to do with time

calculus 
Steve
if the man starts at t=0, his position at time t is (0,4t)
The woman is at (500+5(t5),0)
the distance d is
d^2 = (500+5(t5))^2 + (4t)^2 = 41t^2 + 4750t + 225625
at t=15,
d^2 = 306100, so d=553.26
2d dd/dt = 82t + 4750
1106.52 dd/dt = 5980
dd/dt = 5.4 ft/s
makes sense, since the 500' lead makes most of the distance due to the woman's motion, at 5 ft/s 
calculus  oops 
Steve
Rats. I thought she was walking east.
she is at (500,5(t5))
d^2 = 500^2 + (4t+5(t5))^2 = 81t^2  450t + 250625
at t=15, d = 512
2d dd/dt = 162t  450
1024 dd/dt = 1980
dd/dt = 1.93 ft/s
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