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Maximizing arc length

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MarkFL

Administrator
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Feb 24, 2012
13,775
Yesterday on another forum, someone asked for the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is the only force on the projectile after the launch.

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.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Yesterday on another forum, someone asked for the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is the only force on the projectile after the launch.

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:

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

My question is, what did I do wrong when trying to apply the FTOC? I suspect I am not correctly applying the chain rule.
You need to account for the fact that the integrand is also dependent on \(\theta\).

If I have done this right the FTOC you need is:

If:

\[s(\theta)=\int_a^{f(\theta)} h(x,\theta) dx\]

Then:

\[s'(\theta)=f'(\theta)h(f(\theta),\theta)+\int_a^{f(\theta)}\frac{\partial h(x,\theta)}{\partial \theta} dx \]

CB
 
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MarkFL

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Feb 24, 2012
13,775
Thank you! It seems the FTOC wouldn't really help me in this case.