Posted by Michael on Thursday, November 11, 2010 at 9:21pm.
Two people start from the same point. One walks east at 1 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changing after 15 minutes?

Related rates  Reiny, Thursday, November 11, 2010 at 9:43pm
At a time of t hours
let the distance covered by the eastbound person be 1t km
let the distance covered by the northeastbound person be 2t km
let the distance between them be d km
by the Cosine Law
d^2 = t^2 + 4t2  2(t)2t)cos45°
= 5t^2  4t^2(√2/2)
= 5t^2  2√2 t^2
when t = 15 min or .25 hours
d^2 = 5(.25)^2  2√2(.25)^2 = .489277
d = .6995
2d dd/dt = 10t 4√2t
dd/dt = (5t  2√2t)/d
= (5(.25)  2√2(.25))/.6995 = .776
at that moment they are separating at .776 km/h
check my arithmetic

Related rates  Michael, Thursday, November 11, 2010 at 9:55pm
Incorrect

Related rates  Reiny, Thursday, November 11, 2010 at 10:08pm
If you had checked my arithmetic like I suggested you would have found my error to be in
d^2 = 5(.25)^2  2√2(.25)^2 = .489277
It should have been
d^2 = 5(.25)^2  2√2(.25)^2 = .135723
then
d = .3684
I will let you make the rest of the corrections.
BTW, I also used km/h instead of mi/h, but that has no effect on the calculation
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