My Math Forum max length of trajectory

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October 18th, 2012, 12:33 PM   #1
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max length of trajectory

What is the angle needed to launch an object from ground level to achieve the maximum length of trajectory? The initial velocity is 96 feet per second.

Here's what i have: http://i.imgur.com/zsRxu.jpg
[attachment=0:1o0mkv9k]zsRxu.jpg[/attachment:1o0mkv9k]

When i use the graphing calculator to graph $s(\theta)$, the graph is undefined at 56.5 degrees. What am i doing wrong?

The solution manual has this equation (or something like it):
$\int_0^{6\sin{\theta}} \! [96^2 - (6144\sin{\theta})*t + 1024*t^2]_{dt}^{\frac{1}{2}} \$ which is probably $\int_0^{6\sin{\theta}} \! sqrt{96^2 - (6144\sin{\theta})*t + 1024*t^2} \mathrm{d} t$

Edit: I used the ti-89 to evaluate the integral.
Edit2: Oh wow, lol nevermind. i got it now. I had a negative for my -1024t^2 instead of a positive.
Attached Images
 zsRxu.jpg (180.7 KB, 352 views)

 October 18th, 2012, 11:22 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: max length of trajectory Eliminating the parameter $t$, we find the object's trajectory will be given by: $y=\tan$$\theta$$x-\frac{g\sec^2(\theta)}{2v_0^2}x^2$ $\frac{dy}{dx}=\tan(\theta)-\frac{g\sec^2(\theta)}{v_0^2}x$  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. So, here goes... Let: $u=\frac{g}{v_0^2\sin(2\theta)}x\:\therefore\:du=\f rac{g}{v_0^2\sin(2\theta)}\,dx$ and we have: $1+$$\tan(\theta)-\frac{g\sec^2(\theta)}{v_0^2}x$$^2=1+$$\tan(\theta )-2\tan(\theta)u$$^2=\frac{\cos^2(\theta)+\sin^2(\th eta)(1-2u)^2}{\cos^2(\theta)}=$ $\frac{\cos^2(\theta)+\sin^2(\theta)$$1-4u+4u^2$$}{\cos^2(\theta)}=\frac{\cos^2(\theta)+\s in^2(\theta)-4u\sin^2(\theta)+4u^2\sin^2(\theta)\)}{\cos^2(\the ta)}=$ $\frac{1-4u\sin^2(\theta)(1-u)}{\cos^2(\theta)}$ And now we may write:   If we observe that this integral is symmetric about the line $u=\frac{1}{2}$ we may write:  Let: $v=4u(1-u)\:\therefore\:dv=4(1-2u)\,du=4\sqrt{1-v}\,du$ and we have:  Let: $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:  Let: $w=\sin(\theta)\csc(\alpha)\:\therefore\:dw=-\sin(\theta)\csc(\alpha)\cot(\alpha)\,d\alpha$ and we have:  With a bit of algebra, we may write this in a form we may integrate directly: $\sec^3$$\alpha$$=\frac{1}{2}$$\sec^3\(\alpha$$+\se c^3$$\alpha$$\)=\frac{1}{2}$$\sec^3\(\alpha$$+\sec $$\alpha$$$$\tan^2\(\alpha$$+1\)\)=$ $\frac{1}{2}$$\sec^3\(\alpha$$+\sec$$\alpha$$\tan^2 $$\alpha$$+\sec$$\alpha$$\frac{\tan$$\alpha$$+\sec $$\alpha$$}{\tan$$\alpha$$+\sec$$\alpha$$}\)=$ $\frac{1}{2}$$\(\sec\(\alpha$$\sec^2$$\alpha$$+\sec $$\alpha$$\tan^2$$\alpha$$\)+$$\frac{\sec\(\alpha\ )\tan\(\alpha$$+\sec^2$$\alpha$$}{\sec$$\alpha$$+\ tan$$\alpha$$}\)\)=$ $\frac{1}{2}\frac{d}{d\theta}$$\sec\(\alpha$$\tan$$\alpha$$+\ln\left|\sec$$\alpha$$+\tan$$\alpha$$\ri ght|\)$ and so we have (given $0<\theta<\frac{\pi}{2}$):  $s(\theta)=\frac{v_0^2\cos^2(\theta)}{g}$$\sec\(\th eta$$\tan$$\theta$$+\ln$$\sec\(\theta$$+\tan$$\the ta$$\)\)$ $s(\theta)=\frac{v_0^2}{g}$$\sin\(\theta$$+\cos^2(\ theta)\ln$$\sec\(\theta$$+\tan$$\theta$$\)\)$ Now, differentiating and equating to zero, we find: $s'$$\theta$$=\frac{v_0^2}{g}$$\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$$\)\)=0$ $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: $f$$\theta$$=\sin$$\theta$$\ln$$\sec\(\theta$$+\tan $$\theta$$\)-1=0$ Using Newton's method, we will need: $f'$$\theta$$=\tan$$\theta$$+\cos$$\theta$$\ln\ (\sec$$\theta$$+\tan$$\theta$$\)$ And so Newton's method gives us the recursion: $\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 $\theta_0=\frac{\pi}{4}$ 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: $\theta\approx56.4658351275^{\circ}$
 October 20th, 2012, 12:58 AM #3 Member   Joined: Dec 2010 Posts: 51 Thanks: 0 Re: max length of trajectory WOW! That's some impressive mathematical calculations. I'll saved this and exam it more carefully once my workload from ODE and calc3 clears.

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