Calculus: Early Transcendentals, Chapter 3, 3.9, Section 3.9, Problem 44

If you look at the image bellow you see that a represents distance traveled eastward and b represents distance traveled in northeast direction. Clearly the angle between is 45°. We can now apply law of cosine to our triangle to get distance d.
d^2=a^2+b^2-2ab cos45^circ
We can now write d as a function of time.
d(t)=sqrt(a^2+b^2-2ab sqrt2/2)=sqrt(a^2+b^2-sqrt2 ab)
If a man is walking at 3 mi/h, then after t hours he will have traveled 3t miles. Hence, a(t)=3t and b(t)=2t Now we plug that into equation for d.
d(t)=sqrt(9t^2+4t^2-sqrt2cdot3t cdot2t)=sqrt(13t^2-6sqrt2t^2)=t sqrt(13-6sqrt2)
To find rate of change we need to find derivation of d.
d'(t)=sqrt(13-6sqrt2)
As we can see the rate of change is constant at every point in time including t=0.25 (15 min is a quarter of an hour).
The distance between the people after 15 minutes is changing at the rate of sqrt(13-6sqrt2)m//s