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Thread: Logarithmic spiral

  1. #1
    jpi
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    Question Logarithmic spiral

    Hi
    I have some questions regarding a puzzle.



    This is a logarithmic spiral, and i know that from "Startpunkt" to "A" the direct distance is 86,23. From "Startpunkt" to "B" the direct distance is 75,41.
    If i start at "Startpunkt" and measure the spiral where will the point be when the distance is 3000 ?

    I have found the definition r = ae^(b*theta), but i don't know how to use it. How do i find the two constants a and b ?

    And how do i find the length ?

    Regards
    Jacob
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  2. #2
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    Quote Originally Posted by jpi View Post
    Hi
    I have some questions regarding a puzzle.



    This is a logarithmic spiral, and i know that from "Startpunkt" to "A" the direct distance is 86,23. From "Startpunkt" to "B" the direct distance is 75,41.
    If i start at "Startpunkt" and measure the spiral where will the point be when the distance is 3000 ?

    I have found the definition r = ae^(b*theta), but i don't know how to use it. How do i find the two constants a and b ?
    For $\displaystyle \theta= 0$,at startpunkt, r= a and when $\displaystyle \theta= \pi$, at A, $\displaystyle r= ae^(b\pi)$. And we are told that the distance between the points, which is the difference between those two, is 86,23: $\displaystyle a- ae^{b\pi}= 86,23$. When $\displaystyle \theta= 2\pi$, point B, [tex]r= ae^{2b\pi}. The distance between startpunkt and B, we are told, is 7541: ae^{2b\pi}- a= 7541. Solve those two equations for a and b.

    And how do i find the length ?

    Regards
    Jacob
    As for the arclength, the general formula, given that x(t) and y(t) are functions of the parameter t, is
    $\displaystyle \int \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt$

    Here, since this is given in terms of r and $\displaystyle \theta$ so you need $\displaystyle x= rcos(\theta)$ and $\displaystyle y= rsin(\theta)$ as well as the fact that $\displaystyle r= ae^{b\theta}$. That is, $\displaystyle x= ae^{b\theta}cos(\theta)$ and $\displaystyle y= ae^{b\theta}sin(\theta)$.

    The derivatives messy but the formula reduces nicely (since $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$) to
    $\displaystyle \int a\sqrt{b^2+ 1}e^{b\theta} d\theta$
    which is easy to integrate:
    $\displaystyle \frac{a}{b}\sqrt{b^2+1}e^{b\theta}+ C$
    to find the arclength between $\displaystyle \theta_1$ and the startpunkt, evaluate that at the two points and subtract:
    $\displaystyle \frac{a}{b}\sqrt{b^2+1}\left(e^{b\theta}- 1\right)$
    set that equal to 3000 and solve for $\displaystyle \theta$.
    Last edited by HallsofIvy; Jan 12th 2009 at 07:49 AM.
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  3. #3
    jpi
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    Quote Originally Posted by HallsofIvy View Post
    For $\displaystyle \theta= 0$,at startpunkt, r= a and when $\displaystyle \theta= \pi$, at A, $\displaystyle r= ae^(b\pi)$. And we are told that the distance between the points, which is the difference between those two, is 86,23: $\displaystyle a- ae^{b\pi}= 86,23$.
    Shouldn't it be $\displaystyle a+ ae^{b\pi}= 86,23$ ?
    r = a is the piece from (0,0) to startpunkt and $\displaystyle r = ae^{b\pi}$ is the part from (0,0) to A. Or am i wrong here ?
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  4. #4
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    Quote Originally Posted by jpi View Post
    Shouldn't it be $\displaystyle a+ ae^{b\pi}= 86,23$ ?
    r = a is the piece from (0,0) to startpunkt and $\displaystyle r = ae^{b\pi}$ is the part from (0,0) to A. Or am i wrong here ?
    Yes, you are right: one point is left of (0,0), the other to the right. I though I had corrected that by editting!
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  5. #5
    jpi
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    I have now solved the puzzle, and got the answer correct !!!

    Thanks a lot. It is higly appreciated

    /Jacob
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