I have the beginning points (0,1) and end points (180, 141.732) of a curve. The function I am currently using is f(x) = Ae^kx. However, when deriving the original function, I end up with 0 (from ln(1)) or an indefinite length (from ln(0)). The answer should be between 100 and 200. When I use the end points in the function, my answer ends up being 141.6841.
Current assumption is A = 1.
Perhaps I should change the exponential function, change the A assumption, or I'm messing up. Any suggestions would be helpful. Sorry for the poor formatting, my professor is making me turn off the laptop.
P.S. This is not for class. Just an afternoon personal project.
$$L = \int dL = \int \sqrt{dx^2 + dy^2}$$
$$dy = f'(x)dx \rightarrow dL = \sqrt{1 + (f')^2}dx$$
So consulting Wolfram Integrator, there is a way to do this analytically. Start with:
$$ \int \sqrt{1 + e^x}dx = 2\left(\sqrt{1 + e^x} - tanh^{-1}\left(\sqrt{1 + e^x}\right)\right) + C $$
Now find $f'(x) = kAe^{kx} \rightarrow (f')^2 = k^2A^2e^{2kx}$ and the integral becomes:
$$\int \sqrt{1 + k^2A^2e^{2kx}}dx$$
Try a u-substitution (using the result from Wolfram):
$$ e^u = k^2A^2e^{2kx} \\ e^u du = 2k^3A^2e^{2kx}dx = 2ke^udx \rightarrow dx = \frac{1}{2k}du $$
Now we find the following integral:
$$ \frac{1}{2k}\int \sqrt{1 + e^u}du = \frac{1}{k}\left(\sqrt{1 + k^2A^2e^{2kx}} - tanh^{-1}\left(\sqrt{1 + k^2A^2e^{2kx}}\right)\right) + C $$
It's nasty, but theoretically possible once you find $k$ and $A$ (which I will leave up to you).