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hey, you keep posting problems i can't catch up, which do you want to work on?
haha sorry! i'm flustered. whatever you're heart desires to work on is good enough for me
with these types of problems i always start by drawing the picture, did you do that?
let me know if youre still here
yeah i drew one
sorry, you still there?
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?
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
triangle sounds good, what kind of triangle
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
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
say a is 180x and b is 225y then d(b-a)/dt=(180)x+(225)y just a guess
hmm, now a and b is what you are using for the sides of the triangle?
i was using x and y for that
so if you want you can take everything i said and substitute a for x and b for y, it's the same thing
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
what's a formula for d in terms of a and b
do you know the pythagorean theorem?
a^2 + b^2 = c^2
so what is c here in our problem
we don't know c?
wait 225^2 + 180^2 =83025 then take the square of that?
what does c stand for though
yes that is what c is when a and b are those numbers
the distance between planes
so what's the equation for the distance between the planes in terms of a and b
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
dt/dt = y' ?
192mi/hr. i never would have gotten that
lol....(dy/dt) = y' :)
was I right is what I want to know :)
haha i have to find that out....