calculus related rates help?

Question

I understand how to draw the picture, however I don't know how to relate the distances in an equation:

A man walks north 4ft/s from Point R. 5 minutes later a woman 500 ft due east of of point P walks 5ft/s south. what rate are the people moving apart 15 minutes after the woman.

m = distance of man
w= distance of woman
b = distance of the two apart

I let dm/dt = rate of man [4ft/s]
I let dw/dt = rate of woman [5ft/s]
I must solve for db/dt

Answer 1

Both the man's walk and the woman's walk increase the north/south separation. It would be mathematically equivalent to say:

dy/dt = dm/dt + dw/dt = 4 + 5 = 9 ft/sec

The east/west separation is constant at 500 ft.

15 minutes after the woman starts walking (and 20 minutes after the man starts) the north/south separation is:

y = 20*60*4 + 15*60*5 = 4800 + 4500 = 9300

The total distance between them is:

b = √(500² + y²) = √(500² + 9300²) = √86,740,000
b = 100√8674 ≈ 9313.4312

b² = 500² + y²
2b(db/dy) = 2y
db/dy = 2y/(2b) = y/b

db/dt = (db/dy)(dy/dt) = (y/b)(9) = 9*9300/(100√8674)
db/dt = 837/√8674 ≈ 8.9870208 ft/sec

Answer 2

try it like that:
the man starts at point (0:0) and his position is y1=4t, x1=0.
the woman starts at point (0:500) and her position is y2=5(t-300), x2=500.
the distance between them is y1-y2=4t-5(t-300), x1-x2=-500