Suppose we are asked to compute the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is constant and is the only force on the projectile after the launch.
Eliminating the parameter t, we find the object's trajectory will be given by:
y=tan(θ)x−gsec2(θ)2v20x2
dydx=tan(θ)−gsec2(θ)v20x
s=∫v20gsin(2θ)0√1+(dydx)2dx
At this point I wanted to be slick and use the derivative form of the FTOC to differentiate directly, but I couldn't make it work. :lol:
So, here goes...
Let:
u=gv20sin(2θ)x∴
and we have:
\displaystyle 1+\left(\tan(\theta)-\frac{g\sec^2(\theta)}{v_0^2}x \right)^2=1+\left(\tan(\theta)-2\tan(\theta)u \right)^2=\frac{\cos^2(\theta)+\sin^2(\theta)(1-2u)^2}{\cos^2(\theta)}=
\displaystyle \frac{\cos^2(\theta)+\sin^2(\theta)(1-4u+4u^2)}{\cos^2(\theta)}=\frac{\cos^2(\theta)+\sin^2(\theta)-4u\sin^2(\theta)+4u^2\sin^2(\theta)}{\cos^2(\theta)}=
\displaystyle \frac{1-4u\sin^2(\theta)(1-u)}{\cos^2(\theta)}
And now we may write:
\displaystyle s=\int_0^1\sqrt{\frac{1-4u\sin^2(\theta)(1-u)}{\cos^2(\theta)}}\,\frac{v_0^2\sin(2\theta)}{g}\,du
\displaystyle s=\frac{2v_0^2\sin(\theta)}{g}\int_0^1\sqrt{1-4\sin^2(\theta)u(1-u)}\,du
If we observe that this integral is symmetric about the line \displaystyle u=\frac{1}{2}, we may write:
\displaystyle s=\frac{4v_0^2\sin(\theta)}{g}\int_0^{\frac{1}{2}}\sqrt{1-4\sin^2(\theta)u(1-u)}\,du
Let:
\displaystyle v=4u(1-u)\:\therefore\:dv=4(1-2u)\,du=4\sqrt{1-v}\,du and we have:
\displaystyle s=\frac{v_0^2\sin(\theta)}{g}\int_0^1\sqrt{\frac{1-\sin^2(\theta)v}{1-v}}\,dv
Let:
\displaystyle w=\sqrt{\frac{1-\sin^2(\theta)v}{1-v}}\:\therefore\:dw=\frac{(w^2-\sin^2(\theta))^2}{2\cos^2(\theta)w}\,dv
and we have:
\displaystyle s=\frac{2v_0^2\sin(\theta)\cos^2(\theta)}{g}\int_1^{\infty}\frac{w^2}{(w^2-\sin^2(\theta))^2}\,dw
Let:
\displaystyle w=\sin(\theta)\csc(\alpha)\:\therefore\:dw=-\sin(\theta)\csc(\alpha)\cot(\alpha)\,d\alpha
and we have:
\displaystyle s=\frac{2v_0^2\cos^2(\theta)}{g}\int_0^{\theta}\sec^3(\alpha)\,d\alpha
With a bit of algebra, we may write this in a form we may integrate directly:
\displaystyle \sec^3(\alpha)=\frac{1}{2}(\sec^3(\alpha)+\sec^3(\alpha))=\frac{1}{2}(\sec^3(\alpha)+\sec(\alpha)(\tan^2(\alpha)+1))=
\displaystyle \frac{1}{2}\left(\sec^3(\alpha)+\sec(\alpha)\tan^2(\alpha)+\sec(\alpha)\frac{\tan(\alpha)+\sec(\alpha)}{\tan(\alpha)+\sec(\alpha)} \right)=
\displaystyle \frac{1}{2}\left(\sec(\alpha)\sec^2(\alpha)+\sec(\alpha)\tan^2(\alpha)+\frac{\sec(\alpha)\tan(\alpha)+\sec^2(\alpha)}{\sec(\alpha)+\tan(\alpha)} \right)=
\displaystyle \frac{1}{2}\frac{d}{d\theta}\left(\sec(\alpha)\tan(\alpha)+\ln\left|\sec(\alpha)+\tan(\alpha) \right| \right)
and so we have (given \displaystyle 0<\theta<\frac{\pi}{2}):
\displaystyle s=\frac{v_0^2\cos^2(\theta)}{g}\int_0^{\theta}\,d\left(\sec(\alpha)\tan(\alpha)+\ln(\sec(\alpha)+\tan(\alpha)) \right)
\displaystyle s(\theta)=\frac{v_0^2\cos^2(\theta)}{g}\left(\sec(\theta)\tan(\theta)+\ln(\sec(\theta)+\tan(\theta)) \right)
\displaystyle s(\theta)=\frac{v_0^2}{g}\left(\sin(\theta)+\cos^2(\theta)\ln(\sec(\theta)+\tan(\theta)) \right)
Now, differentiating and equating to zero, we find:
\displaystyle s'(\theta)=\frac{v_0^2}{g}\left(\cos(\theta)+\cos^2(\theta)\frac{\sec(\theta)\tan(\theta)+\sec^2(\theta)}{\sec(\theta)+\tan(\theta)}-2\cos(\theta)\sin(\theta)\ln(\sec(\theta)+\tan(\theta)) \right)=0
\displaystyle s'(\theta)=\frac{2v_0^2\cos(\theta)}{g}(1-\sin(\theta)\ln(\sec(\theta)+\tan(\theta)))=0
And so we find we want to satisfy:
\displaystyle f(\theta)=\sin(\theta)\ln(\sec(\theta)+\tan(\theta))-1=0
Using Newton's method, we will need:
\displaystyle f'(\theta)=\tan(\theta)+\cos(\theta)\ln(\sec(\theta)+\tan(\theta))
And so Newton's method gives us the recursion:
\displaystyle \theta_{n+1}= \theta_n-\frac{\sin(\theta_n)\ln(\sec(\theta_n)+\tan(\theta_n))-1}{\tan(\theta_n)+\cos(\theta_n)\ln(\sec(\theta_n)+\tan( \theta_n))}
Using \displaystyle \theta_0=\frac{\pi}{4} as the first guess, we find:
\theta_1\approx1.01751306067
\theta_2\approx0.986290147513
\theta_3\approx0.985515191499
\theta_4\approx0.985514737863
\theta_5\approx0.985514737862
\theta_6\approx0.985514737862
Converting from radians to degrees, we find the launch angle which maximizes the arc length is:
\theta\approx56.4658351275^{\circ}
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