Calculus 1
Francis starts walking north at 4 ft/s from an intersection I. At the same time Aram starts walking south at a
rate of 5 ft/s from a intersection 1500 ft due east of I. At what rate are they moving apart 15 minutes later?
(Rounding to 1 decimal place is fine.)
asked by
Kristine

the separation distance d at time t is given by
d^2 = (4t+5t)^2 + 1500^2
d^2 = 81t^2 + 2250000
2d dd/dt = 162t
d^2 at t=15min = 900 sec = 81*900^2 + 1500^2
= 100^2 * (81*81 + 15*15)
d = 100 * sqrt(6786)
d = 8237.719
dd/dt = 162(900)/16475.438 = 8.85
That looks reasonable. It's less than 9, but pretty close, since at 8237 ft apart, the initial 1500ft separation is relatively small.posted by Steve
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