<p>Cindy and George are running toward and intersection. Cindy's course is perpendicular to George's. If Cindy is 300 ft away from the intersection and running at 14 feet per second and George is 160 feet away and runing at 12 feet per second, how fast is the distance between them decreasing?</p>
<p>Ooh, related rates! You would use the Pythagorean Theorem to relate their distances from the intersection (legs of the right triangle) to their distance from each other (hypotenuse). Correct me if I'm wrong, but I think I was pretty good at that related rates stuff. Maybe someone else will get excited and actually do the problem.</p>
<p>c=300
dc/dt=-14
(The distance to the intersection is decreasing, because they are running towards it, so the derivative is negative.)
g=160
dg/dt=-12
give the hyp the letter h, then c^2 + g^2 = h^2
implicit differentiation yeilds 2c<em>dc/dt + 2g</em>dg/dt = 2h<em>dh/dt
so dh/dt = (c</em>dc/dt + g*dg/dt)/h
To solve for dh/dt, we need h, which we don't have - so we use the origional pythagorean theorum. h = sqrt(c^2 + g^2)
h=sqrt( 300^2 + 160^2) = 340 <--- (15, 8, 17 tripple)</p>
<p>Then we just plug and chug:
dh/dt = (300 * -14 + 160 * -12) / 340
dh/dt = -18</p>
<p>So the they are getting closer together at a rate of 18 ft/s (at this instant in time)</p>