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Mechanics: Displacement Vector

  1. May 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the vector r(t)= (1-t)i + (3+2t)j + (t-4)k which depends on scalar parameter t. For which t0 is the length of r(t0) minimal?

    2. Relevant equations

    Equations of motion

    3. The attempt at a solution

    So I approached this question by saying the length would be minimal when the magnitude is minimal ie. |r|=0.

    |r|=√[(1-t)2+(3+2t)2+(t-4)2]

    Simplified down I got 3t2+t+13=0

    I tried using the quadratic formula to solve this then but couldn't as the value under the square root was negative.

    Am I approaching this question wrong then? Any help would be much appreciated.
     
  2. jcsd
  3. May 18, 2014 #2

    Ray Vickson

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    Why would you assume that the smallest value of |r| is zero? That would be the case only if the line passes through the origin. Draw a sketch of a similar situation in two dimensions, and look carefully at the picture.
     
  4. May 18, 2014 #3
    So if t is the x-axis and r(t) the y-axis when t=0 r(t) is at a minimum?
     
  5. May 18, 2014 #4
    Actually is it when the x-compenent is zero. So in this case 1-t=0 so t=1?
     
  6. May 18, 2014 #5

    vela

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    ##\vec{r}(t)## describes a curve in three dimensions, and ##t## is a parameter telling you where you are on that curve. To sketch the curve, you need to do so in three dimensions. Ray is suggesting you try a simpler example where the curve lies in two dimensions so you can get an idea of how to analyze the problem correctly. For instance, you could try a curve like ##\vec{r}(t) = (t-1)\hat{i} + (t+2)\hat{j}##.

    Back to the original problem… You found the length ##\| \vec{r}(t) \| = \sqrt{2(3t^2+t+13)}## as a function of ##t##. You want to find where this function attains a minimum. How do you do that?
     
  7. May 18, 2014 #6
    Differentiate and let it equal to zero. Then sub t values back into r(t)?
     
  8. May 18, 2014 #7

    vela

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    Right. The problem is only asking for the value of ##t##, so you don't need to plug the value back into ##\vec{r}(t)##, though you might try it to see if the answer is reasonable.
     
  9. May 18, 2014 #8
    Ok vela thanks for the help. I got t=1/3 in the end. There was chain rule involved in differentiating so it would take a while to type it out. Thanks for the help again.
     
  10. May 18, 2014 #9

    vela

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    You're off by a factor of 2.
     
  11. May 18, 2014 #10

    Ray Vickson

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    Even easier: ##|r|## is a minimum if and only if ##r^2## is a minimum (why?), so you might as well just minimize ##r^2 = (1-t)^2 + (3+2t)^2 + (t-4)^2 = 6 t^2 + 2t + 26.##
     
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