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Distance between 2 planes
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shib
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#1
Distance between 2 planes
Well I'm in pre-calc but our teacher gave us a calculus problem that we shouldn't know how to do and is offering bonus points to anyone who can figure it out. I'll just type it up verbatum.

One dark rainy night Kim and Adam arrived at their job as air traffic controllers at Dayton international airport. There were only two airplanes flying in the direction of Dayton at the time. One airplane was a 747 jet approaching from 400 miles to the east at 325 MPH. Its altitude was 10000 ft and descending at 70ft/min. The other plane was a 10 passenger prop plane approaching from a point 260 miles to the south at 211.25MPH. The second plane's altiture was 3000 feet and climbing at 24.8 ft/min.
Suddenly the computers all lose power. Kim and Adam are equipped with only their TI's. An accident is considered likely if the two aircraft pass within 200 ft of eachother.
Find the minimum distance between the airplanes are when this will occur.

So right now I have
f(x)=sqrt(((10000-4200x)-(3000+1488x))^2+(400-325x)^2+(260-211.25x)^2)

the test is x=3 y=685.797124204

I don't want somone to flat out tell me the answer but if you could help tell me what I'm doing wrong.

**I realized that I posted this elsewhere, but I think maybe it belongs here more.

PostPosted: Thu Feb 10, 2005 7:41 pm  Back to top 
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Kent Merryfield
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#2
You need to answer some basic questions.

What is this variable you call x? What units of measurement does it carry?

What units of measurement to any of your expressions carry?

What does f(x) represent?

I ask these things in part becuase it looks like you may have an expression of the form (a\,\text{miles})^2+(b\,\text{feet})^2, which doesn't make sense.

PostPosted: Thu Feb 10, 2005 7:53 pm  Back to top 
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Soroban
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#3
Re: Distance between 2 planes
Hello, shib!

You said "distance between 2 planes" . . . and I thought "No problem! I know a formula for that!"
Then I saw that you meant airplanes . . . Ha!

I haven't worked out the answers yet, but I can give you my set-up for the problem.

Quote:
One dark rainy night Kim and Adam arrived at their job as air traffic controllers at Dayton international airport.
There were only two airplanes flying in the direction of Dayton at the time.
One airplane was a 747 jet approaching from 400 miles to the east at 325 MPH. Its altitude was 10000 ft and descending at 70ft/min.
The other plane was a 10 passenger prop plane approaching from a point 260 miles to the south at 211.25MPH.
The second plane's altitude was 3000 feet and climbing at 24.8 ft/min.
Suddenly the computers all lose power. Kim and Adam are equipped with only their TI's.
An accident is considered likely if the two aircraft pass within 200 ft of each other.
Find the minimum distance between the airplanes are when this will occur.

One huge problem is that we have miles-per-hour and feet-per-minute.

I placed the problem on an x-y-z coordinate system (measured in feet).

The 747 is at (0, 2,112,000, 10000)
The y-distance is changing at -28,600 ft/min.
The z-distance (altitude) is chaning at -70 ft/min.
The parametric equations for the 747 are: x = 0, y = 2,112,000 - 28,600t, z = 10,000 - 70t

The prop plane is at (1,372,800, 0, 3000)
The x-distance is changing at -18,590 ft/min.
The z-distance is changing at +28.8 ft/min.
The parametric equations for the prop plane are: x = 1,372,800 - 18,590t, y = 0, z = 3000 + 28.8t

We can use the Distance Formula to express D (the distance between the two planes) as a function of t.
We get something like: D = square root of [(1,372,800 - 18,590t)^2 + (2,112,000 - 28,600t)^2 + (7,000 - 98.8t)^2 ]

The bad news: we may overflow our calculators, trying to crank out those numbers.
The good news: that thing under the square root is only a quadratic expression
. . . it will be an upward-opening parabola, so we can find its minimum point (the vertex).

I'll try to work on it myself . . . but later.

PostPosted: Fri Feb 11, 2005 8:25 am  Back to top 
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Jimmy
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#4
Can you please clarify what the problem is asking?

PostPosted: Fri Feb 11, 2005 10:53 am  Back to top 
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