Posted by asdf on Tuesday, October 23, 2012 at 8:11pm.
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, Tuesday, October 23, 2012 at 8:40pm
if the man starts at t=0, his position at time t is (0,4t)
The woman is at (500+5(t-5),0)
the distance d is
d^2 = (500+5(t-5))^2 + (4t)^2 = 41t^2 + 4750t + 225625
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, Tuesday, October 23, 2012 at 8:47pm
Rats. I thought she was walking east.
she is at (500,-5(t-5))
d^2 = 500^2 + (4t+5(t-5))^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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