anonymous
  • anonymous
ALKJFSALJF. Two small planes approach an airport, one flying due west at 120mi/hr and the other flying due north at 150mi/hr. Assuming they fly at the same constant elevation, how fast is the distance between the planes changing when the westbound plane is 180mi from the airport and the northbound plane is 225mi from the air port?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
hey, you keep posting problems i can't catch up, which do you want to work on?
anonymous
  • anonymous
haha sorry! i'm flustered. whatever you're heart desires to work on is good enough for me
anonymous
  • anonymous
your*

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anonymous
  • anonymous
with these types of problems i always start by drawing the picture, did you do that?
anonymous
  • anonymous
let me know if youre still here
anonymous
  • anonymous
yeah i drew one
anonymous
  • anonymous
sorry, you still there?
anonymous
  • anonymous
still here!
anonymous
  • anonymous
ok so what does your picture look like? the question is asking for the rate of change of the distance. so we need an equation for the distance and take the derivative. how can we get an equation for the distance?
anonymous
  • anonymous
i kind of made a triangle between the two planes. i'm use to doing rate of change of distance with a point and equation of a line given
anonymous
  • anonymous
triangle sounds good, what kind of triangle
anonymous
  • anonymous
right
anonymous
  • anonymous
yes, so we are given rate of change of the sides of the triangle, so we want an equation for the distance as a function of the sides of the triangle
anonymous
  • anonymous
basically we want d = (something with x's and y's), because we know x,y, dx/dt, dy/dt, so if we take the derivative we can find dd/dt
anonymous
  • anonymous
say a is 180x and b is 225y then d(b-a)/dt=(180)x+(225)y just a guess
anonymous
  • anonymous
hmm, now a and b is what you are using for the sides of the triangle?
anonymous
  • anonymous
i was using x and y for that
anonymous
  • anonymous
oh whoops!
anonymous
  • anonymous
so if you want you can take everything i said and substitute a for x and b for y, it's the same thing
anonymous
  • anonymous
now don't confuse 180,225 and the variables. it's 180 at a certain time, and x (or a), when you don't know what it is
anonymous
  • anonymous
what's a formula for d in terms of a and b
anonymous
  • anonymous
do you know the pythagorean theorem?
anonymous
  • anonymous
a^2 + b^2 = c^2
anonymous
  • anonymous
so what is c here in our problem
anonymous
  • anonymous
we don't know c?
anonymous
  • anonymous
wait 225^2 + 180^2 =83025 then take the square of that?
anonymous
  • anonymous
what does c stand for though
anonymous
  • anonymous
yes that is what c is when a and b are those numbers
anonymous
  • anonymous
the distance between planes
anonymous
  • anonymous
right
anonymous
  • anonymous
so what's the equation for the distance between the planes in terms of a and b
anonymous
  • anonymous
?
amistre64
  • amistre64
the distance formula is a^2 + b^2 = c^2 and we want to derive with respect to time: (d/dt)(x^2) + (d/dt)(y^2) = (d/dt)(c^2) (dx/dt)(2x) + (dy/dt)(2y) = (dc/dt)(2c) To clean this up: (dx/dt) = x'; (dt/dt) = y'; and (dc/dt) = c'. x'2x + y'2y = c'2c solve for c': x'2x + y'2y ----------- = c' 2c the twos cancel to give us: x'(x) + y'(y) --------- = c' c y' = 150; x'=120; and c = whatever you got with the pythag. theorum. x and y values are their current values of x=180; y=225 So lets plug those in: 120(180) + 150(225) ------------------- = c' 288.14
anonymous
  • anonymous
dt/dt = y' ?
anonymous
  • anonymous
192mi/hr. i never would have gotten that
amistre64
  • amistre64
lol....(dy/dt) = y' :)
amistre64
  • amistre64
was I right is what I want to know :)
anonymous
  • anonymous
haha i have to find that out....

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