# Thread: x-intercept of a capture zone curve

1. ## 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.

2. 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.