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Thread: x-intercept of a capture zone curve

  1. #1
    Oct 2008

    x-intercept of a capture zone curve

    Hi all, not sure if this is the correct forum for this, but here goes. I'm trying to prove that the stagnation point (x-intercept) of a capture zone curve is \frac{-Q}{\pi Bv}

    Here are the equations I've found in the book.
    y=\frac{Q}{2Bv} -\frac{Q}{2\pi Bv} *arctan\frac{y}{x}

    tan\phi = \frac{y}{x}

    y=\frac{Q}{2Bv} (1-\frac{\phi}{\pi}

    Here is what the problem reads:

    A stagnation point in a capture-zone type curve is a spot where the groundwater would have no movement. For the case of a single extraction well, the stagnation point is located where the capturze-zone curve crosses the x-axis. Use the fact that for small angles tan\theta \approx \theta to show that the x-axis intercept of the capture-zone curve for a single well is x= -\frac{Q}{ 2Bv\pi }

    Q, B, and v are all constants

    All I've done so far is gotten \phi \approx \frac{y}{x} , but any attempts to solve for x with what I have isn't working, x ends up getting cancelled and I get 0= a constant. I've also tried getting the derivative to have slope = \infty , but I haven't had any success with that either. Thanks in advance for the help.

    I've attached one image from my book of how the capture zone curve looks.
    Last edited by Crysolice; April 25th 2011 at 01:41 PM. Reason: Edit: apparently can't get the math formatting to work...
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  2. #2
    Oct 2008
    Here is an attachment with the formulas I'm given and what I'm trying to isolate since I can't get the forum math to work.
    Attached Files Attached Files
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