Question

Two people leave from the same spot and walk at 4ft/sec going north and 5ft/sec going northwest. At 30 seconds, how fast is the distance between them changing?

Answer 1

let the distance between them be x ft
after a time of t seconds,
distance of the northbound person = 4t ft
distance of the other person = 5t ft
They walk at an angle of 45° to each other.

By the cosine law:

x^2 = (4t)^2 + (5t)^2 - 2(4t)(5t)cos45
= 41t^2 - 40t^2 (√2/2)
= 41t^2 - 20√2 t^2
2x dx/dt = 82t - 40√2 t
dx/dt = (41t - 20√2t)/x
when t = 30 s
x^2 = 41(900) - 20√2(900)
x = appr 106.97736...

dx/dt = (41(30) - 20√2(30))/106.97736..
= appr 3.566 ft/s