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?

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**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

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

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**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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