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anonymous

  • one year ago

Calculus Rates & speed question about airplanes

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  1. anonymous
    • one year ago
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    I know the answer of a; how do I find the equation for the distance between the planes at all times for part b?

  2. anonymous
    • one year ago
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  3. amistre64
    • one year ago
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    its easier to read when its not in the chinese mode ...

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  4. anonymous
    • one year ago
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    yup

  5. anonymous
    • one year ago
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    Is it solvable?

  6. amistre64
    • one year ago
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    its solveble, buti have to read thru it first, and these old eyes get tired at times

  7. amistre64
    • one year ago
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    both at 33000 ft they intersect directly over Frada hgts near misses shortly after 130pm --------------------------------------- at 130, Am2003 was 32 Nm from Frada, heading 171o at a rate of 405knots at 130 Un366 was 44 Nm from Frada, heading 81o at a rate of 465 knots

  8. amistre64
    • one year ago
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    how how do we read airplane headings? from the north line right?

  9. anonymous
    • one year ago
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    yes

  10. amistre64
    • one year ago
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    doesnt really matter, the orientation is the same regardless of our 'proper' placements

  11. amistre64
    • one year ago
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    lets draw a picture, both planes, and Frada

  12. anonymous
    • one year ago
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    Ohh ok I just assumed it was a right triangle and so drew a right triangle with 81 9 and 90

  13. amistre64
    • one year ago
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    171-81 = 90, so your assumption was valid

  14. anonymous
    • one year ago
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    we have that, we got the rate of the distance decreasing between them two, but we need to find the actual distance separation at the closest point. that is the part we are having trouble with

  15. amistre64
    • one year ago
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    |dw:1433270115412:dw|

  16. anonymous
    • one year ago
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    the rate we got is 472.8 knots

  17. anonymous
    • one year ago
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    yup we have that already

  18. amistre64
    • one year ago
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    d^2 = x^2 + y^2 rate of change is derivatives, so what is our derivative?

  19. amistre64
    • one year ago
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    jut making sure we are on the same page here

  20. amistre64
    • one year ago
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    the shortest distance between them is when one of them is at Frada ... that is when they are in line with each other.

  21. amistre64
    • one year ago
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    what is our derivative? lets make sure we are working it thru and have not developed any errors

  22. anonymous
    • one year ago
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    hmm..how did you draw the triangle?

  23. perl
    • one year ago
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    |dw:1433453806233:dw|

  24. anonymous
    • one year ago
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    yup we have that

  25. perl
    • one year ago
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    |dw:1433453888308:dw|

  26. anonymous
    • one year ago
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    we got part a already, we got ds/dt = 494.7

  27. anonymous
    • one year ago
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    yyup thanks we already have that part; we got ds/dt = 497.7 already. we are only stuck on part b - thank you so much!

  28. perl
    • one year ago
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    for part b) i set AA = 32-405t distance from Frada UA = 44 -465t distance from frada Distance between AA and UA D= sqrt((32-405*t)^2+(44-465*t)^2) minimize this by taking the derivative dD/dt = (1/2)*(-66840+760500*t)/sqrt((32-405*t)^2+(44-465*t)^2) set dD/dt = 0

  29. anonymous
    • one year ago
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    that makes sense! we got t = .0879

  30. perl
    • one year ago
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    @amistre64 please check for part a) i got a different ds/dt x^2 + y^2 = s^2 2x x' + 2y y' = 2s s' x* x' + y* y' = s * s ' 44 * (-465) + 32 * (-405) = sqrt( 44^2 + 32^2 ) * s ' s' = [44 * (-465) + 32 * (-405)]/ sqrt( 44^2 + 32^2 ) s' = -614.27 knots

  31. perl
    • one year ago
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    right, t = 1114/12675 = .0878 is the time when their distance is closest. and if you plug in you get D = 4.7677 miles .

  32. perl
    • one year ago
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    |dw:1433456029343:dw|

  33. perl
    • one year ago
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    do you agree with my work, how i got ds/dt = -614.27

  34. anonymous
    • one year ago
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    ohh!! we screwed it up thank you!!

  35. anonymous
    • one year ago
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    44 * (-465) + 32 * (-405) I think you had an error here...

  36. anonymous
    • one year ago
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    I think the speed of American 1003 is 405 not 465

  37. anonymous
    • one year ago
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    oops again!!!!!!

  38. anonymous
    • one year ago
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    you're right

  39. anonymous
    • one year ago
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    hmm okay

  40. perl
    • one year ago
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    we can think of this problem as two dots approaching each other. The path of the dots can be represented by parametric equations

  41. perl
    • one year ago
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    woops i need a negative sign there

  42. anonymous
    • one year ago
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    do you know "whether ATC would have time to take action? Might a slight altitude change for one of the lights help prevent any five mile rule violation?"

  43. perl
    • one year ago
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    t=0 is 1:30 PM

  44. anonymous
    • one year ago
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    mhm

  45. perl
    • one year ago
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    |dw:1433457025732:dw|

  46. perl
    • one year ago
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    +465 since it is moving to the right

  47. perl
    • one year ago
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    |dw:1433457096742:dw|

  48. perl
    • one year ago
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    now if you plug in t= .08 into both equations you can see exactly where the points are

  49. perl
    • one year ago
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    clearly AA gets to Frada before UA because 32-405t = 0 , t= 32/405 ~ .079 sec -44 + 465t = 0, t= 44/464 ~ .094 sec but at t = 1114/12675 they are closest together plug in this time into AA and UA AA is below F by 3.59 nautical miles UA is to the left of Frada by 3.13 nautical miles

  50. perl
    • one year ago
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    if you change the altitude that could change the minimum 5 nautical miles rule. but we would have to go into three dimensions

  51. perl
    • one year ago
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    |dw:1433457782758:dw|

  52. perl
    • one year ago
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    without loss of generality we can set one of the planes to 30,000 feet and the other one to some value A. But note that we must change units to nautical miles to be consistent with our earlier calculations.

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