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Math Help - Length of Curve (Parametric)

  1. #1
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    Length of Curve (Parametric)

    Find the length of the following parametric curve

    y= a(cos t + ln tan \frac {t}2)
    y= a sint



    from the point (0,a) to the point (x,y).

    I need to use the formula
    L= \int \sqrt{( \frac {dx}{dt}^2)+ (\frac {dy}{dt}^2)}dt

    but I'm not sure what the limits of integration are so I can evaluate the result.

    As well, would \frac {dx}{dt} be a(-sin t+ \frac {1}{tan \frac {t}2})? Or am I differentiating the ln incorrectly?
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  2. #2
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    Quote Originally Posted by Hellreaver View Post

    Find the length of the following parametric curve \color{red}x = a(\cos t + \ln \tan(t/2)), \ \ y= a \sin t, from the point (0,a) to the point (x,y).

    I need to use the formula
    L= \int \sqrt{( \frac {dx}{dt}^2)+ (\frac {dy}{dt}^2)}dt

    but I'm not sure what the limits of integration are so I can evaluate the result.

    As well, would \frac {dx}{dt} be a(-sin t+ \frac {1}{tan \frac {t}2})? Or am I differentiating the ln incorrectly?
    the curve is defined over 0 < t < \pi. the points (0,a) and (x,y) correspond to t=\frac{\pi}{2} and, say t_0, where t_0 is any number between 0 and \pi. we also have \frac{dx}{dt}=a(-\sin t + \csc t) and \frac{dy}{dt}=a \cos t.

    so the length of the curve between two given points is: \ell = \int_{t_0}^{\frac{\pi}{2}} \sqrt{a^2(-\sin t + \csc t)^2 + a^2 \cos^2t} \ dt=-a \ln \sin t_0. (i'm assuming that a>0)
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  3. #3
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    So there is no definite answer then? Thats a pain in the butt...
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  4. #4
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    There is no "definite answer" because there is no definite question! You asked for the arclength from the specific point (0, a) to the variable point (x, y). Obviously the answer will depend upon (x,y). Since you are integrating with respect to t, NonCommAlg converted that to the variable t_0, the value of t corresponding to (x,y).
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