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#37
Feb1211, 11:29 PM

HW Helper
P: 3,515

[tex]\sin\left(2\cos^{1}\left(x\right)\right)=2x\sqrt{1x^2}[/tex] It may look complex, but its purpose is simple. When you plug [itex]\theta[/itex] into the area equation [tex]A=\frac{r^2}{2}\left(\theta\sin\theta\right)[/tex] you're going to be left with edit: [tex]\sin\left(2\cos^{1}\left(\frac{rh}{r}\right)\right)[/tex] which is where you can simplify this with the equality I gave above. You already have the answer, but it's just if you wanted to simplify things a bit more. Markers wouldn't give full marks if you left an answer as [tex]\sin\left(\sin^{1}\left(x\right)\right)[/tex] so I doubt they would give full marks if you left it as [tex]\sin\left(\cos^{1}\left(x\right)\right)[/tex] either. 


#38
Feb1211, 11:38 PM

P: 709

Ok thanks alot for all your help



#39
Feb1211, 11:39 PM

PF Gold
P: 1,420

The cylinder on it's side problem reduces to y=x  sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem. There is a reason they put this problem in computer science text books and never in mathematics text books. 


#40
Feb1211, 11:45 PM

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P: 3,515




#41
Feb1211, 11:52 PM

P: 709

This question is for yr 11 math. My solution is I believe [tex] V=L*\sin\left(\cos^{1}\left(2\cdot\frac{rh}{r}\right)\right) [/tex].



#42
Feb1211, 11:58 PM

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P: 3,515




#43
Feb1311, 12:14 AM

P: 709

whoops I forgot the first theta, so V=L*(r^2/2(2arccos(rh/r)sin(2arccos(rh/r))) pretty messy.



#44
Feb1311, 12:43 AM

PF Gold
P: 1,420

All I know is that if you have a 10 gallon tank, it is not possible to place integer gallon marks on the dipstick. Unless of course you are clever enough to solve for x. And perhaps I should give some background regarding this particular problem before someone yells at me for playing mind games. 


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