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Thread: Path Integral

  1. #1
    Senior Member vincisonfire's Avatar
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    Path Integral

    I have to compute the arc length of $\displaystyle r = 1 + cos(\theta) $
    I found the formula $\displaystyle \int_0^{2\pi} \sqrt{r^2+(\frac{dr}{d\theta})^2} d\theta = \int_0^{2\pi} \sqrt{2+2cos(\theta)} d\theta $*
    I don't know how to do this integral.
    Can you please help me?
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  2. #2
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    Dear vincisonfire,

    one of tricks: cos(theta) = cos(2*(theta/2)) and
    cos(2x) = cos^2(x)-sin^2(x).
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  3. #3
    Senior Member vincisonfire's Avatar
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    I would have $\displaystyle *\int \sqrt{1+cos(\theta)} = \int \sqrt{1+cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2})} = \int \sqrt{2cos^2(\frac{\theta}{2})} $
    Back to my problem *
    $\displaystyle 2 \int_0^{2\pi} cos(\frac{\theta}{2}) d\theta= 4 [sin(\frac{\theta}{2})]_0^{2\pi} = 0 $
    I should have 8 no ?
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  4. #4
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    khm...khm...
    $\displaystyle \sqrt{x^2} \neq x$
    such in x = -3
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  5. #5
    Eater of Worlds
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    Integrate from 0 to Pi and multiply by 2.

    $\displaystyle 8sin(\frac{\theta}{2})|_{0}^{\pi}=8$
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  6. #6
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    Hello, vincisonfire!

    I have to compute the arc length of $\displaystyle r \:=\: 1 + cos(\theta) $

    I found the formula: .$\displaystyle \int_0^{2\pi} \sqrt{r^2+(\tfrac{dr}{d\theta})^2} d\theta \;\;= \int_0^{2\pi} \sqrt{2+2\cos\theta}\, d\theta $

    We're expected to know this identity: . $\displaystyle \cos^2\tfrac{\theta}{2} \:=\:\frac{1+\cos\theta}{2} \quad\Rightarrow\quad 1 + \cos\theta \:=\:2\cos^2\!\tfrac{\theta}{2}$

    We have: .$\displaystyle \sqrt{2(1 + \cos\theta)} \;=\;\sqrt{2\left(2\cos^2\!\tfrac{\theta}{2}\right )} \;=\;2\cos\tfrac{\theta}{2}$


    Therefore: .$\displaystyle L \;=\;2\int^{2\pi}_o\!\!\cos\tfrac{\theta}{2}\,d\th eta\quad\hdots $ . Got it?

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    It is not true!

    See above!
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  8. #8
    Math Engineering Student
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    Quote Originally Posted by Soroban View Post

    We have: .$\displaystyle \sqrt{2(1 + \cos\theta)} \;=\;\sqrt{2\left(2\cos^2\!\tfrac{\theta}{2}\right )} \;=\;2\cos\tfrac{\theta}{2}$


    Therefore: .$\displaystyle L \;=\;2\int^{2\pi}_o\!\!\cos\tfrac{\theta}{2}\,d\th eta\quad\hdots $ . Got it?
    Here's where the mistake is, that should be $\displaystyle \int_{0}^{2\pi }{\left| \cos \frac{\theta }{2} \right|\,d\theta },$ hence,

    $\displaystyle \begin{aligned}
    \int_{0}^{2\pi }{\left| \cos \frac{\theta }{2} \right|\,d\theta }&=2\int_{0}^{\pi }{\left| \cos \theta \right|\,d\theta } \\
    & =2\left( \int_{0}^{\pi /2}{\cos (\theta )\,d\theta }-\int_{\pi /2}^{\pi }{\cos (\theta )\,d\theta } \right) \\
    & =2(1+1)=4.
    \end{aligned}$
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