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Derivation of vector r(t)

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  1. Oct 26, 2014 #1
    Hello, I am new here, glad I found this Forum :0)
    I am german and english is not my mother-tongue, so sorry in advance for any language-related mistakes

    1. The problem statement, all variables and given/known data

    i don't know how to make a vector arrow on a letter so I declare a v for vector as the following:
    [PLAIN]http://upload.wikimedia.org/math/8/8/b/88b537be1c5870ce272a4dc9515bada9.png=vr(t) [Broken]


    A particle moves on a space-curve, described by following position-vector:

    [PLAIN]http://upload.wikimedia.org/math/8/8/b/88b537be1c5870ce272a4dc9515bada9.png= [Broken] (t²+t)vex + (3t-2)vey + (2t³-4t²)vez


    2. Relevant equations


    3. The attempt at a solution

    I think I might differentiate each part, but I am not sure,

    I would do it like this

    vr'(t)= (2t)vex + (3)vey + (6t² - 8t)vez
    so this would be the velocity


    and vr''(t) = (2)vex + (0)vey + (12t - 8)vez
    would be the acceleration
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 26, 2014 #2

    Orodruin

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    Your problem statement does not include a question. Are you simply interested in finding the velocity and acceleration?

    The basis vectors (in Cartesian coordinates) are constant so you can just differentiate the prefactors as you have done.

    Edit: You can write mathematical notation on Physics Forums using LaTeX commands. Here is a very brief introduction. You can also press Reply on my post and you can see how I have done the following:
    $$
    \vec v(t) = \vec r'(t) = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z
    $$
     
  4. Oct 26, 2014 #3
    I am sorry, yes correct, my task is to find the velocity and the acceleration.
    does your post mean that the two equations I've done are correct? so i could simply put in a t, for example t=2 and calculate it
     
  5. Oct 26, 2014 #4

    Orodruin

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    The principle is correct, but one of your derivatives is not. I suggest you double check your computations (alternatively that you typed it in correctly, it may be a typo as well).
     
  6. Oct 26, 2014 #5
    ahhh it is

    $$
    \vec r'(t) = (2t+1)\vec e_x + (3)\vec e_y + (6t²-8t)\vec e_z
    $$
     
  7. Oct 26, 2014 #6

    Orodruin

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    Correct.
     
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