An airplane, flying east at 400mph...goes over a certain town at 11:30 AM and a second plane, flying northeast at 500mph, goes over the same town at noon. How fast are they separating at 1:00 PM?
Let a be the distance from the town to the plane flying due east, and let b be the distance to the plane flying northeast.
Then at 1:00 pm we have `a=600,(da)/(dt)=400,b=500,(db)/(dt)=500`
(`(da)/(dt)` is the rate at which the distance between the town and the plane going east is increasing; in this case the plane is flying at 400mph)
((a=600 since the plane passed over the town at 11:30am, so has been flying 1.5 hours at 400mph))
Now the distance between the planes can be found using the Law of Cosines. The angle between them is `45^@` ; `cos45^@=sqrt(2)/2` .
`d=sqrt(a^2+b^2-sqrt(2)ab)` or `d=(a^2+b^2-sqrt(2)ab)^(1/2)`
The rate at which the distance between the planes is changing is `(dd)/(dt)`
At 1pm, the distance between the planes is increasing at approximately 316.6mph.