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Tangential Component of Acceleration

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    I was following through the notes I took in my Calc class, and either I wrote down the formulas incorrectly, my professor did this problem incorrectly, or I wrote the answer down incorrectly. I'm just going to redo it here, so could someone give me a thumbs up or thumbs down if I did it correctly? Thanks.

    Problem:

    Find the tangential component of the acceleration vector of the given partial motion vector:

    [tex]r(t)=\langle e^{-t},\sqrt{2}t,e^t\rangle[/tex]


    2. Relevant equations

    [tex]a_T=\frac{d}{dt}|v|[/tex]

    3. The attempt at a solution

    [tex]v(t)=\frac{dr(t)}{dt}=\langle -e^{-t},\sqrt{2},e^t \rangle[/tex]

    [tex]|v|=\sqrt{e^{-2t}+2+e^{2t}}[/tex]
    [tex]|v|=\sqrt{(e^{-t}+e^t)^2}[/tex]
    [tex]|v|=e^{-t}+e^t[/tex]


    [tex]a_T=\frac{d}{dt}|v|=-e^{-t}+e^t[/tex]
     
  2. jcsd
  3. Jun 19, 2011 #2

    LCKurtz

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    No, that is not how you calculate it. You need to calculate the unit tangent vector T which you get by dividing the velocity vector by its length. Then you get the acceleration vector A by differentiating your velocity vector. Then you get the tangential component of the acceleration by calculating

    [tex] A_T = \vec A\cdot T[/tex]
     
  4. Jun 19, 2011 #3
    Okay then I'll do it that way:


    (from before)
    [tex]v(t)=\langle -e^{-t},\sqrt{2},e^t \rangle[/tex]
    [tex]|v|=e^{-t}+e^t[/tex]

    (new)
    [tex]T(t)=\frac{v(t)}{|v|}=\langle \frac{-e^{-t}}{e^{-t}+e^t}, \frac{\sqrt{2}}{e^{-t}+e^t}, \frac{e^t}{e^{-t}+e^t}\rangle[/tex]

    [tex]a(t)=\frac{d}{dt}v(t)=\langle e^{-t},0,e^t \rangle[/tex]

    [tex]a_T=a(t) \centerdot T(t) = e^{-t}\frac{-e^{-t}}{e^{-t}+e^t} + 0 + e^t\frac{e^t}{e^{-t}+e^t}=\frac{e^{-2t}-e^{2t}}{e^{-t}+e^t}=\frac{(e^{t}-e^{-t})(e^{t}+e^{-t})}{e^{-t}+e^t}=-e^{-t}+e^t[/tex]

    So apparently I CAN do it the original way...and it was a bit easier too. So am I safe to assume I have done it correctly?
     
  5. Jun 19, 2011 #4

    LCKurtz

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    Yes, I made a mistake in my calculation and had a different answer, sorry for the misstep.
     
  6. Jun 19, 2011 #5
    No worries, even my professor with a PHD flubbed it up! Thank the lord for technology :p
     
  7. Jun 19, 2011 #6

    Char. Limit

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    One note: e^(t) - e^(-t) = sinh(t), if you want to make your answer look more elegant.
     
  8. Jun 19, 2011 #7

    LCKurtz

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    You mean = 2 sinh(t).
     
  9. Jun 19, 2011 #8

    Char. Limit

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    Ooops :redface: Of course. 2 sinh(t).
     
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