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Vector Funtion

  1. Jul 9, 2007 #1
    1. The problem statement, all variables and given/known data
    The position of a spaceship is:
    [tex] r(t)=(3+t)i +(2+ln(t))j+(7-\frac{4}{t^2+1})k[/tex]

    and the coordinated of the space station are (6,4,9). The captian wants the spaceship to coast into the the space station. When should the engines be turned off?

    2. Relevant equations
    [tex] r(t)=(3+t)i +(2+ln(t))j+(7-\frac{4}{t^2+1})k[/tex]

    3. The attempt at a solution
    Ok the ship coasts(uniform velocity) into the space ship.
    So max/min problem right? Find [tex] \frac{d^2r}{dx^2}[/tex] set equall to zero and solve for t right?
     
  2. jcsd
  3. Jul 9, 2007 #2

    Dick

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    Nooo. You want r'(t) to be parallel to r(t)-(6,4,9) and pointing in the right direction. Your turn. Why????
     
  4. Jul 9, 2007 #3
    So r'(t)=<6,4,9> because it heads in the spacestations direction
     
    Last edited: Jul 9, 2007
  5. Jul 9, 2007 #4

    Dick

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    The sense of your answer is correct. But the direction of the station is <6,4,9>-r(t) from the position of the ship, right? Difference of two positions is the direction.
     
  6. Jul 9, 2007 #5
    Ok, so the position vector of the space station is r(t)=<6,4,9>, and r'(t) has to be parallel being r'(t)=<6,4,9> or some scalar multiple.
    This means we must solve for t in r'(t) when r'(t)=<6,4,9>?
    [tex]r'(t)=<1,\frac{1}{t},\frac{4t}{(t^2+1)^2}>[/tex]
     
  7. Jul 10, 2007 #6
  8. Jul 10, 2007 #7

    HallsofIvy

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    You've already been told, twice, that this is wrong. The vector from the ship to the space station is <6, 4, 9>- r(t). You must have r' equal to that.
     
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