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Thread: Unit tangent vector

  1. #1
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    Unit tangent vector

    I am trying to find the unit tangent vector T(t)
    Heres the problem

    $\displaystyle r(t) = ti + \frac{1}{t}j, t = 1$


    $\displaystyle v(t) = i - \frac {1}{t^2} j$

    then

    $\displaystyle a(t) = \frac {2}{t^3} j$

    I think the magnitude is:

    $\displaystyle ||v(t)|| = \sqrt \frac{2} {t^4} j$

    then the solutions manual has this:


    $\displaystyle T(t) = \frac {v(t)}{ || v(t) ||} = \frac{t^2} {\sqrt{ t^4 + 1}} (i - \frac{ 1}{t^2}j )$

    I am not sure how they are getting this.
    Can someone help me with the steps? I would appreciate this!!
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    MHF Contributor Reckoner's Avatar
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    Re: Unit tangent vector

    Quote Originally Posted by icelated View Post
    I think the magnitude is:

    $\displaystyle ||v(t)|| = \sqrt \frac{2} {t^4} j$
    This is way off. You do know that the magnitude of a vector is a scalar, right? What is that $\displaystyle \mathbf j$ doing there? And where did the 2 come from?
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    Re: Unit tangent vector

    so the magnitude of v(t)

    $\displaystyle v(t) = i - \frac {1}{t^2} j$

    would =

    $\displaystyle v(t) = \sqrt { ( 1- \frac { 1 }{t^2} )^2} $

    then,


    $\displaystyle v(t) = \sqrt { ( \frac {1 + 1 }{t^4} )} $

    I think i see a mistake? should be

    $\displaystyle ||v(t)|| = \sqrt { ( \frac {t^4 + 1 }{t^4} )} $

    How would i take the magnitude of v(t) then?
    Last edited by icelated; Jun 15th 2012 at 04:34 PM.
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    MHF Contributor Reckoner's Avatar
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    Re: Unit tangent vector

    Quote Originally Posted by icelated View Post
    would =

    $\displaystyle v(t) = \sqrt { ( 1- \frac { 1 }{t^2} )^2} j$
    No, we square the components individually.

    Quote Originally Posted by icelated View Post
    $\displaystyle v(t) = \sqrt { ( \frac {1 + 1 }{t^4} )} j$
    No, that's not how we add fractions.

    $\displaystyle \mathbf v(t) = \mathbf i - \frac1{t^2}\mathbf j$

    $\displaystyle \left\|\mathbf v(t)\right\| = \sqrt{1^2 + \left(-\frac1{t^2}\right)^2}$

    $\displaystyle = \sqrt{1 + \frac1{t^4}}$

    $\displaystyle = \sqrt{\frac{t^4+1}{t^4}}$

    $\displaystyle = \frac{\sqrt{t^4+1}}{t^2}$
    Thanks from icelated
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    Re: Unit tangent vector

    how are they getting the t squared on top?
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    Re: Unit tangent vector

    Quote Originally Posted by icelated View Post
    how are they getting the t squared on top?
    Division. What happens when you divide

    $\displaystyle \frac1{\sqrt{t^4+1}/t^2}?$
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