Wave speed derivation 
The wave speed derivation approximates the wave as a circle. It uses that to know that $$a=\frac{v^2}{R}$$However, numerous functions can approximate the wave. A straight line, $x^2$, $x^3$, etc. If I used those I would get a different equation for a. So why is a circle the correct approximation choice?
 A: The wave can be any shape $f(x)$. But when you focus in on a sufficiently small element of the curve, you can do a Taylor expansion about the point $x_0$:
$$f(x) = f(x_0) + (x-x_0)f' + \frac{(x-x_0)^2}{2!}f'' + ...$$
As the distance $(x-x_0)$ gets smaller, higher order terms can be neglected. If you consider a point with horizontal slope, then $f'=0$ and the first significant term is the quadratic term.
The Taylor expansion of a circle of the right radius happens to match that exactly; and this gives certain "nice" mathematical properties that makes the rest of the calculation easier. But note that if the real function was of the form $x^3$, meaning that the curvature will change with position, it will still have a certain value of curvature at a particular point - and therefore there will still be a circle that "matches" the curve at that point.
A: A single line isn't very useful for approximating a curve. You could use small segments, but then you'll need several, and the calculation would be more complicated. As noted in the comments, nothing stops you from using any other second order curve, i. e., a plane curve whose rectangular Cartesian coordinates satisfy an algebraic equation of the second degree, (a line is first order, since its corresponding equation is first degree, $y=ax+b$), except that doing the derivation with a circle is easier. Also keep in mind that you're approximating a single symmetrical pulse, not the whole wave.
A: circle as it is easy for approximation . you could approximate curve to any other possible curve but the equation you get may be complicated and after solving you will again return to this formula only. we use circle as it is easy to calculate .
