# 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?

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### 1 Answer

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^2=a^2+b^2-2abcostheta`

`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)`

`(dd)/(dt)=1/2(a^2+b^2-sqrt(2)ab)^(-1/2)(2a(da)/(dt)+2b(db)/(dt)-sqrt(2)(a(db)/(dt)+b(da)/(dt)))`

So `(dd)/(dt)=(2(600)(400)+2(500)(500)-sqrt(2)((600)(500)+(500)(400)))/(2sqrt(600^2+500^2-sqrt(2)(600)(500)))`

`=(980000-sqrt(2)(500000))/(2sqrt(610000-300000sqrt(2)))`

`~~316.6`

**At 1pm, the distance between the planes is increasing at approximately 316.6mph.**

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