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Thread: Unit Tangent Vector

  1. #1
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    Unit Tangent Vector

    Given the vector function r(t)=<\sqrt{2}t,e^t,e^{-t}> I need to find the tangent unit vector T(t)=\frac{r'(t)}{\sqrt{x'^2+y'^2+z'^2}}. The denominator is the absolute value of the vector derivative r'(t)=<\sqrt{2},e^t,-e^{-t}>. I write an expression for T(t):


    T(t)=\frac{1}{\sqrt{2+e^{2t}+e^{-2t}}}<\sqrt{2},e^t,-e^{-t}>

    =\frac{1}{\sqrt{2(1+\frac{e^{2t}+e^{-2t}}{2}})}<\sqrt{2},e^t,-e^{-t}>

    =\frac{1}{\sqrt{(1+cosh2t)}}(i+\frac{e^t}{\sqrt{2}  }j-\frac{e^{-t}}{\sqrt{2}}k)


    I don't think this could be wrong since it's pretty straight forward. It requires zero intelligence, you just apply the definition directly. However, I can't seem to obtain the same expression given in the book:

    T(t)=\frac{1}{e^{2t}+1}<\sqrt{2}e^t,e^{2t},-1>

    I can see that e^tr'(t)=<\sqrt{2}e^t,e^{2t},-1>, which suggests that they just multiplied the numerator and denominator of T(t) by e^t. However I can't get the demonator in the form e^{2t}+1. I'm confused by the algebra, I can't show that e^t\sqrt{2+e^{2t}+e^{-2t}}=e^{2t}+1.

    ????
    Last edited by adkinsjr; Jul 30th 2009 at 11:48 PM.
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  2. #2
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    One of the addition formulae for hyperbolic functions (very similar to the addition formulae for trig functions) is \cosh2x = 2\cosh^2x-1. As a consequence, \sqrt{1+\cosh2x} = \sqrt2\,\cosh x = \frac{e^{2x}+1}{\sqrt2\,e^x}.
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