Calculus Rates & speed question about airplanes

- anonymous

Calculus Rates & speed question about airplanes

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- anonymous

I know the answer of a; how do I find the equation for the distance between the planes at all times for part b?

- anonymous

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- amistre64

its easier to read when its not in the chinese mode ...

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- anonymous

yup

- anonymous

Is it solvable?

- amistre64

its solveble, buti have to read thru it first, and these old eyes get tired at times

- amistre64

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

- amistre64

how how do we read airplane headings? from the north line right?

- anonymous

yes

- amistre64

doesnt really matter, the orientation is the same regardless of our 'proper' placements

- amistre64

lets draw a picture, both planes, and Frada

- anonymous

Ohh ok I just assumed it was a right triangle and so drew a right triangle with 81 9 and 90

- amistre64

171-81 = 90, so your assumption was valid

- anonymous

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

- amistre64

|dw:1433270115412:dw|

- anonymous

the rate we got is 472.8 knots

- anonymous

yup we have that already

- amistre64

d^2 = x^2 + y^2
rate of change is derivatives, so what is our derivative?

- amistre64

jut making sure we are on the same page here

- amistre64

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

- amistre64

what is our derivative? lets make sure we are working it thru and have not developed any errors

- anonymous

hmm..how did you draw the triangle?

- perl

|dw:1433453806233:dw|

- anonymous

yup we have that

- perl

|dw:1433453888308:dw|

- anonymous

we got part a already, we got ds/dt = 494.7

- anonymous

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!

- perl

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

- anonymous

that makes sense! we got t = .0879

- perl

@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

- perl

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

- perl

|dw:1433456029343:dw|

- perl

do you agree with my work, how i got ds/dt = -614.27

- anonymous

ohh!! we screwed it up thank you!!

- anonymous

44 * (-465) + 32 * (-405) I think you had an error here...

- anonymous

I think the speed of American 1003 is 405 not 465

- anonymous

oops again!!!!!!

- anonymous

you're right

- anonymous

hmm okay

- perl

we can think of this problem as two dots approaching each other. The path of the dots can be represented by parametric equations

- perl

woops i need a negative sign there

- anonymous

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?"

- perl

t=0 is 1:30 PM

- anonymous

mhm

- perl

|dw:1433457025732:dw|

- perl

+465 since it is moving to the right

- perl

|dw:1433457096742:dw|

- perl

now if you plug in t= .08 into both equations you can see exactly where the points are

- perl

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

- perl

if you change the altitude that could change the minimum 5 nautical miles rule.
but we would have to go into three dimensions

- perl

|dw:1433457782758:dw|

- perl

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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