SOLUTION: 5. The position of a lost hiker, H, is given by the parametric equations {{{x= 4 - 3t}}} {{{y= 3 - 2t}}} , where position is measured in km, and time in hours. The position

Algebra ->  Vectors -> SOLUTION: 5. The position of a lost hiker, H, is given by the parametric equations {{{x= 4 - 3t}}} {{{y= 3 - 2t}}} , where position is measured in km, and time in hours. The position      Log On


   

Question 389493: 5. The position of a lost hiker, H, is given by the parametric equations
x=+4+-+3t
y=+3+-+2t
, where position is measured in km, and time in hours. The position of his rescue party, R, is given by
x=+3+-+t
y=+4t+-+5
.
a) Find the initial position of the hiker and his rescue party.
b) Find the velocity vector of each.
c) When is the rescue party closest to the hiker and how close are they at this time?

C is giving me the most trouble,

any help is appreciated ^.^
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
Set t+=+0
(4,3) and (3,-5)
(b)
For velocity vectors:
hiker:
sqrt%283%5E2+%2B+2%5E2%29 at 180 degrees + arc tan(2/3)
sqrt%2813%29 at 33.69 degrees
rescue party:
sqrt%281%5E2+%2B+4%5E2%29 at 180 degrees - arc tan(4)
sqrt%285%29 at 104.04 degrees
(c)
I need a general formula for the distance between them
It would be square root of(x2 - x1)^2 + (y2 - y1)^2)
sqrt%28%283+-+t+-+4+%2B+3t%29%5E2+%2B+%28-5+%2B+4t+-+3+%2B+2t%29%5E2%29
sqrt%28%28-1+%2B+2t%29%5E2+%2B+%28-8+%2B+6t%29%5E2%29
The 1st term is zero when t+=+1%2F2, then the distance
would be sqrt%28%28-5%29%5E2%29+=+5
The 2nd term is zero when t+=+8%2F6
Then distance = 5%2F3
I think they come the closest in 1 hr 20 min and
they are 1.67 km apart.
This is kind of a guess, I would get another opinion, too