Finding the inverse of the arc length equation of a parabola I need help in reversing the equation of the arc length of a parabola. 
$$y = \frac{1}{2} x \sqrt{1+4\cdot x^2} + \frac{1}{4} \cdot \ln(2\cdot x + \sqrt{1+4\cdot x^2})$$
Thank you.
 A: This problem seem to somewhat misunderstood. Let me express it this way: Find the function $y(x)$ whose arc length is given by
$$s=\int_0^x\sqrt{1+\left(\frac{dy}{dx} \right)^2}~dx=\frac{1}{2} x \sqrt{1+4\cdot x^2} + \frac{1}{4} \cdot \ln(2\cdot x + \sqrt{1+4\cdot x^2})$$
Differentiating both sides we get
$$\sqrt{1+\left(\frac{dy}{dx} \right)^2}=\sqrt{4x^2+1}\\
1+\left(\frac{dy}{dx} \right)^2=4x^2+1\\
\frac{dy}{dx}=2x\\
\fbox{$y=\pm x^2+C$}
$$
I have verified this solution numerically.
A: This problem does not have an analytical solution. You can invert this equation numerically with numerical solver. 
Alternatively, you can transform this problem into a differential equation problem, that can be solved numerically too:
We have by definition of the arc length 
$$y = \int \sqrt{ 1 + f'(x)^2} dx. $$
Here $$f(x) = x^2$$  So 
$$dy =\sqrt{ 1 + f'(x)^2} dx = \sqrt{ 1 + (2x)^2} dx $$
or $$ {dx\over dy}  = {1\over \sqrt{ 1 + 4x^2}}.$$
This is an explicit differential equation in x (as a function of y), whose 
solution is precisely the inverse function demanded by the OP. It can be solved numerically.
