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Math Help - Determining the length of a logarithmic spiral.

  1. #1
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    Determining the length of a logarithmic spiral.

    Unfortunately I couldnt convert the pdf into a picture so that I could upload it but i will try to explain my problem as briefly i can
    here is a pdf link btw:
    <a target="_blank" title="mathhelplength (pdf)" href="http://pdfcast.org/pdf/mathhelplengt...hhelplength</a>
    Right so I am faced with the following problem
    "Determine the length of the lograrithmic spiral": x = e^-t * cos t y = e^-t * sin t
    0<= t <= 2pi
    I recal a formula
    l(length symbol) = $(integral from a to b) SQRT [x'(t)]^2 + [y'(t)}^ dt
    I don't know how to evaluate this, could someone show me?


    EDIT: New link mathhelplength - PDFCast.org
    Last edited by Riazy; February 13th 2011 at 08:37 AM. Reason: See new link at the bottom
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  2. #2
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    Quote Originally Posted by Riazy View Post
    "Determine the length of the lograrithmic spiral":
    x = e^-t * cos t
    y = e^-t * sin t 0<= t <= 2pi
    x' = -e^{-t}(\sin{t}+\cos{t})

    (x')^2 = e^{-2t}(\sin{t}+\cos{t})^2

    y' = e^{-t}(\cos{t}-\sin{t})

    (y')^2 = e^{-2t}(\cos{t}-\sin{t})^2


    (x')^2 + (y')^2 = 2e^{-2t}

    \displaystyle L = \int_0^{2\pi} \sqrt{2e^{-2t}} \, dt

    finish it.
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  3. #3
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    Sorry, but the my problem is really how you added together (x')^2+(y')^2 = 2e^-2t

    I would like to see that step in detail, because thats what I dont understand, i.e how they are added together to 2e^-2t

    THanks alot
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  4. #4
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    Hello, Riazy!

    This requires some very careful work . . . are you up for it?


    \text{Determine the length of the lograrithmic spiral:}

    . . \begin{Bmatrix}x &=& e^{\text{-}t}\cos t \\ y &=& e^{\text{-}t}\sin t \end{Bmatrix}\;\;0\,\le\, t\,\le\,2\pi

    \displaystyle \text{Formula: }\;L \;=\; \int^b_a \sqrt{[x'(t)]^2 + [y'(t)]^2}\, dt

    \text{Derivatives: }\;\begin{Bmatrix}x'(t) &=& \text{-}e^{\text{-}t}\sin t - e^{\text{-}t}\cos t &=& \text{-}e^{\text{-}t}(\sin t + \cos t) \\ \\[-3mm] y'(t) &=& e^{\text{-}t}\cos t - e^{\text{-}t}\sin t &=&  e^{\text{-}t}(\cos t - \sin t ) \end{Bmatrix}


    \text{Squares: }\;\begin{Bmatrix}[x'(t)]^2 &=& e^{\text{-}2t}(\sin^2\!t + 2\sin t\cos t + \cos^2\!t) \\ \\[-3mm] [y'(t)]^2 &=& e^{\text{-}2t}(\cos^2\!t - 2\sin t\cos t + \sin^2\!t) \end{Bmatrix}


    \text{Add: }\;[x'(t)]^2 + [y'(t)]^2 \;=\;e^{\text{-}2t}\bigg[(\underbrace{\sin^2\!t + \cos^2\!t)}_{\text{This is 1}} + \underbrace{(\cos^2\!t + \sin^2\!t)}_{\text{This is 1}}\bigg]
    . . . . . . . . . . . . . . . . . =\;2e^{\text{-}2t}


    \text{Hence: }\;\sqrt{[x'(t)]^2 + [y'(t)]^2} \;=\;\sqrt{2e^{\text{-}2t}} \;=\;\sqrt{2}\,e^{\text{-}t}


    \displaystyle \text{Therefore: }\;L \;=\;\sqrt{2}\!\int^{2\pi}_0 \!\!e^{\text{-}t}\,dt


    Got it?

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  5. #5
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    (x')^2 = e^{-2t}(\sin{t}+\cos{t})^2

    (x')^2 = e^{-2t}(\sin^2{t}+2\sin{t}\cos{t}+\cos^2{t})

    (x')^2 = e^{-2t}(1+2\sin{t}\cos{t})



    (y')^2 = e^{-2t}(\cos{t}-\sin{t})^2

    (y')^2 = e^{-2t}(\cos^2{t}-2\cos{t}\sin{t} + \sin2{t})

    (y')^2 = e^{-2t}(1-2\cos{t}\sin{t})



    (x')^2 + (y')^2 = e^{-2t}(1+2\sin{t}\cos{t}) + e^{-2t}(1-2\cos{t}\sin{t}) = e^{-2t}[(1+2\sin{t}\cos{t}) + (1-2\sin{t}\cos{t})]

    can you finish it now?
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