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.
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.