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I eliminated the parameter

*t*to get the trajectory:

$\displaystyle y=\tan(\theta)x-\frac{g\sec^2(\theta)}{2v_0^2}x^2$

and so the arc-length is given by:

$\displaystyle s=\int_0^{\frac{v_0^2}{g}\sin(2\theta)} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx$

Now, at this point I wanted to differentiate this using the derivative form of the FTOC, and what I got was:

$\displaystyle \frac{ds}{d\theta}=\frac{2v_0^2}{g}\cos(2\theta) \sqrt{1+\left(\frac{dy}{dx} \right)^2}$

This implies that the maximum occurs for:

$\displaystyle \theta=45^{\circ}$

Which I knew was incorrect. The OP stated the solution was approximately:

$\displaystyle \theta=56.5^{\circ}$

I gave up this route and went ahead and integrated through a myriad of substitutions to finally find:

$\displaystyle s(\theta)=\frac{v_0^2}{g}(\sin(\theta)+\cos^2( \theta)\ln(\sec(\theta)+\tan(\theta)))$

Differentiating and equating to zero eventually led to:

$\displaystyle f(\theta)=\sin(\theta)\ln(\sec(\theta)+\tan(\theta))-1=0$

and using Newton's method, I found:

$\displaystyle \theta\approx56.4658351275^{\circ}$

My question is, what did I do wrong when trying to apply the FTOC? I suspect I am not correctly applying the chain rule.