Question 1160156: Andrew and David both leave their houses at 7 a.m. for their Sunday run. Andrew’s house is 22km
south of David’s house. Andrew runs north at 9km/h, while David runs west at 7km/h. Determine the
rate of change of the distance between the two runners at 1h. Include diagram in your solution.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
What does it mean, " . . . at 1h." --- Does it mean " at 1 pm" ?
Do they run 6 hours ?
Do they need an emergency medical help ?
Hurry with your answer --- their life may depend on it . . .
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Choose a point in the figure to be the origin. To make things easy (nice numbers), let the origin be Andrew's house. Then Andrew's house is (0,0) and David's house is (0,22).
Andrew is starting at his house and running north at 9km/h; his location after t hours is (0,9t).
David is starting at his house and running west at 7km/h; his location after t hours is (-7t,22).
The distance between Andrew and David after t hours is  . We want to find the rate at which that distance is changing at t=1 -- that is. we want to evaluate the derivative of the distance function at t=1.
It is often easier to work with the square of the distance function when taking a derivative, to avoid differentiating a square root function. So
At t=1, their positions are (-7,22) and (0,9); the distance d between them is 
And, finally, the rate at which the distance between them is changing at t=1 is, in km/h,
That is about -4.6 km/hr.
That indicates that they are actually getting closer together 1 hour after starting. That makes sense, because Andrew is running towards David's house at 9km/h while David is running away from his house at 7km/h and in a direction perpendicular to the direction Andrew is running.
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