Two particles move according to x1 = A + Bt^2 and x2 = Dt^3, t ≥ 0 , where A, B, and D are positive constants. At t = 0 the particles are a distance A apart, and as t increases they move apart, but there is a time t 0 at which the particles are as close as they will be ever again; in other words, |x1 − x2| is as small as it will get. What is this smallest value of |x1 − x2| for t 0?

Question

Two particles move according to x1 = A + Bt^2 and x2 = Dt^3, t ≥ 0 , where A, B, and D are positive constants. At t = 0 the particles are a distance A apart, and as t increases they move apart, but there is a time t > 0 at which the particles are as close as they will be ever again; in other words, |x1 − x2| is as small as it will get. What is this smallest value of |x1 − x2| for t > 0?

anonymous · Answer

Seems we should define s = x1 -x2
then set ds/dt = 0 to find max or min
and even check (d/dt)(ds/dt) to see whether we have a max or min.

ds/dt = 2 B t - 3 D t^2 =0 which can occur at t=0 and when
2 B = 3 D t
t= (2B/3D)

(d/dt)(ds/dt) = 2B - 6 D t = 2B - 6 D (2B/3D) = 2B-4B = -2B <0

unfortunately, that indicates we have a maximum, which is hard to believe, as t^3 will dominate as t gets very  large. Puzzling.

anonymous · Answer

1. A again
2. A +4/27 (B^3/D^2)

3. A +1/2 (B^3/D^2)

4. A + 5/42 (B^3/D^2)

5. A + 1/6 (B^3/D^2)

6. A +(B^3/D^2)
7. (A)( B)/D
8. A + 2/27 (B^3/D^2)

9. Zero

anonymous · Answer

i thought it ws A I still dont understand

anonymous · Answer

I guess that the two curves, x1 and x2, cross at some t>0, which would mean that the minimum separation is 0.