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Math Help - Unit normal vector function

  1. #1
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    Unit normal vector function

    I know that the unit vector function is equal to T'(t) / |T'(t)|

    If T(t) is < \frac{2t}{2t^2} , \frac{2t^2}{2t^2}, \frac{1}{2t^2} >


    To find T'(t), would I simply take the derivative of each element using the quotient rule? I did this on paper, and was getting results that were different than the answers in the back of my book. Any ideas? Thanks


    Edit: what on earth is wrong with my latex
    Last edited by zellster87; June 2nd 2008 at 06:31 PM.
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  2. #2
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    Hello, zellster87

    Close LaTex with [/math] . . . not [\math]


    I know that the unit vector function is equal to: \frac{T\,'(t)}{|T\,'(t)|}

    If T(t) \;=\;\left\langle \frac{2t}{2t^2},\:\frac{2t^2}{2t^2},\:\frac{1}{2t^  2}\right\rangle . . . . Why aren't these reduced?

    We have: . T(t) \;=\;\left\langle\frac{1}{t},\:1,\:\frac{1}{2t^2}\  right\rangle \;=\;\left\langle t^{-1},\:1,\:\frac{1}{2}t^{-2}\right\rangle

    Then: . T\,'(t) \;=\;\left\langle -t^{-2},\:0,\:-t^{-3}\right\rangle \;=\;\left\langle -\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle

    And: . |T\,'(t)| \;=\;\sqrt{\left(-\frac{1}{t^2}\right)^2 + 0^2 + \left(-\frac{1}{t^3}\right)^2} \;=\;\sqrt{\frac{1}{t^4} + \frac{1}{t^6}} \;=\;\sqrt{\frac{t^2+1}{t^6}} \;=\;\frac{\sqrt{t^2+1}}{t^3}


    Therefore: . \frac{T\,'(t)}{|T\,'(t)|} \;=\;\frac{\left\langle-\dfrac{1}{t^2},\:0,\:-\dfrac{1}{t^3}\right\rangle}{\dfrac{\sqrt{t^2+1}}{  t^3}} \;=\;\frac{t^3}{\sqrt{t^2+1}}\left\langle-\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle

    . . . . . . = \;\left\langle\frac{-t}{\sqrt{t^2+1}},\:0,\:\frac{-1}{\sqrt{t^2+1}}\right\rangle

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, zellster87

    Close LaTex with [/math] . . . not [\math]



    We have: . T(t) \;=\;\left\langle\frac{1}{t},\:1,\:\frac{1}{2t^2}\  right\rangle \;=\;\left\langle t^{-1},\:1,\:\frac{1}{2}t^{-2}\right\rangle

    Then: . T\,'(t) \;=\;\left\langle -t^{-2},\:0,\:-t^{-3}\right\rangle \;=\;\left\langle -\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle

    And: . |T\,'(t)| \;=\;\sqrt{\left(-\frac{1}{t^2}\right)^2 + 0^2 + \left(-\frac{1}{t^3}\right)^2} \;=\;\sqrt{\frac{1}{t^4} + \frac{1}{t^6}} \;=\;\sqrt{\frac{t^2+1}{t^6}} \;=\;\frac{\sqrt{t^2+1}}{t^3}


    Therefore: . \frac{T\,'(t)}{|T\,'(t)|} \;=\;\frac{\left\langle-\dfrac{1}{t^2},\:0,\:-\dfrac{1}{t^3}\right\rangle}{\dfrac{\sqrt{t^2+1}}{  t^3}} \;=\;\frac{t^3}{\sqrt{t^2+1}}\left\langle-\frac{1}{t^2},\:0,\:-\frac{1}{t^3}\right\rangle

    . . . . . . = \;\left\langle\frac{-t}{\sqrt{t^2+1}},\:0,\:\frac{-1}{\sqrt{t^2+1}}\right\rangle


    Thanks a lot Soroban, I appreciate the help.
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