Related rates - help!

Question

A man starts walking north at  4 ft/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?

Answer 1

Distance between walkers t minutes after woman starts: d(t) = {[5t + 4(t + 300)]² + 500²}^1/2 = (81t² + 21600t + 1690000)^1/2; d'(t) = (81t + 10800)/((81t² + 21600t + 1690000)^1/2.

d'(900) = (81·900 + 10800)/(81·900² + 21600·900 + 1690000)^1/2 = 8.987 ft/s

Answer 2

they're moving about 2 2/3 feet per minute apart

the man moves north from P at 4 ft/s 5 minutes before woman moves south.

she starts

5 minutes later a woman walks south from a point 500 east of P

at what rate are they moving apart 15 minutes after the woman starts walking? that's 20 minutes after the man started moving. he moved 4x20 = 80 feet. She moved 5x15 = 75 feet. horizontally they're 80+75= 155 feet apart. directly they're d = square root of (155^2 + 500^2) = about 523.47 feet apart

let d = the distance between them

d^2 = h^2 + b^2

where b = 500 feet

and h = 20(4) + 15(5) = 80+75=155

d = square root of (155^2+ 500^2) = 523.47

d^2 = h^2 + b^2

take the derivative and set = zero, plug in the values of d, h, b, h' and b to solve for d'

h=155, b = 500, h'=4+5 =9, b'=0, d= about 523.47

2dd' = 2hh' + 2bb'

divide by 2

dd' = hh' + bb' (b'=0, so the last term drops out)

leaving

dd' = hh'

d' = hh'/d = 155(4+5)/sqr(155^2 + 500^2) = 155(9)/523.47 = 2.665 ft per minute

d' = 2.665 feet per minute change in distance between the two people

the longer the time, the more d' approaches 9

the shorter the time, the closer d' is to 0