At time t = 0 one plane is directly above a second plane at an altitude of 1 mil

Question

At time t = 0 one plane is directly above a second plane at an altitude of 1 mile. The second plane is at an altitude of 1/2 mile. The first plane is flying along a somewhat elliptical path described by
r(t) = < 4 - 4cos(t) , 3sin(t) , 1 + t/(2pi) >
and the second plane is flying along a hyperbolic path described by
r(t) = < (8/31/2)tan(t/3) , 0 , -1/2 + sec(t/3) >.
These do represent the actual position functions for the airplanes with distance in miles and time in minutes. In how many minutes after t = 0 will the planes collide? Approximate the distance between the planes and the rate of change of the distance between the planes 30 seconds before they collide. Approximate the rate of change of the distance between the planes at the instant just before they collide.

Explanation / Answer

equating both r<t>

we get t = sec = 180 sec

displacement vector between r<t> 30 sec before

d<t= -30> = ( 9.8756, 2.598, 0.61125)

distance = 10.229 units

v<t=> = (-2.1584, -3, 0.1591)

speed = 3.699 units

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