Question 1160156
<br>
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).<br>
Andrew is starting at his house and running north at 9km/h; his location after t hours is (0,9t).<br>
David is starting at his house and running west at 7km/h; his location after t hours is (-7t,22).<br>
The distance between Andrew and David after t hours is {{{d(t) = sqrt((0-(-7t))^2+(9t-22)^2)}}}.  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.<br>
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<br>
{{{d(t)^2 = (0-(-7t))^2+(9t-22)^2 = 49t^2+81t^2-396t+484 = 130t^2-396t+484}}}<br>
{{{2d*(dd/dt) = 260t-396}}}
{{{dd/dt = (260t-396)/2d = (130t-198)/d}}}
At t=1, their positions are (-7,22) and (0,9); the distance d between them is {{{sqrt(7^2+13^2) = sqrt(49+169) = sqrt(218)}}}<br>
And, finally, the rate at which the distance between them is changing at t=1 is, in km/h,<br>
{{{(130-198)/sqrt(218) = -68/sqrt(218)}}}<br>
That is about -4.6 km/hr.<br>
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.<br>