Two people start from the same point.  One walks east at 3 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changing after 15 minutes.

Textbook Question

Chapter 3, 3.9 - Problem 44 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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tiburtius | High School Teacher | (Level 2) Educator

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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`

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