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July 25

Cylinders and Spheres

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Could someone please check my answer to this question and tell me if I'm right or wrong? Thanks.

A cylindrical biscuit tin has volume V and surface area S (including the ends). Show that the minimum possible surface area for a given value of V is $S=3(2\pi V^{2})^{\frac {1}{3}}$ . For this value of S, find the volume of the smallest sphere into which the tin fits.

I get the volume of the sphere as ${\frac {5^{\frac {3}{2}}V}{12}}$ . Is this right? Thanks 92.3.252.234 (talk) 11:17, 25 July 2008 (UTC)[reply]

Unlikely, as

{\frac {5^{\frac {3}{2}}}{12}}<1

. AndrewWTaylor (talk) 12:51, 25 July 2008 (UTC)[reply]

Was it sphere into tin, or tin into sphere?

Shouldn't S=3(2pi V²)^1/3. 87.102.86.73 (talk) 12:57, 25 July 2008 (UTC)[reply]

Yes, I think we should have

S=3(2\pi V^{2})^{\frac {1}{3}}

. To the original poster - when surface area is minimised, what value do you get for the ratio of the cylinder's height to its radius ? Gandalf61 (talk) 13:14, 25 July 2008 (UTC)[reply]

I'll third that factor-of-three correction. :-) Hopefully that was just a typo and the original homework question wasn't wrong! --_{tiny plastic} Grey Knight ⊖ 13:44, 25 July 2008 (UTC)[reply]

Your answer is wrong.. Did you get an intermediate result (eg I think r_{minimal surface area}=cuberoot(V/2pi) ) ?

ie

r_{\text{minimal surface area}}={\sqrt[{3}]{\tfrac {V}{2\pi }}}

I think you need to use differentiation to get the minima, one way or another - if you're stuck post your method so far and I'll have a look at it.87.102.86.73 (talk) 14:00, 25 July 2008 (UTC)[reply]

Sorry. I missed out the three in the surface area, I've added it in. I managed to do that correctly and got the value of the radius right but it looks like my algebraic skills let me down on the last bit. I'll have another go now. 92.3.252.234 (talk) 19:22, 25 July 2008 (UTC)[reply]

Tried again and discovered one (and hopefully the only) error in my calculation; I made the noobish mistake of using the radius of the tin to determine the sphere's radius, when I should have been using the diameter. Sickening I know. I now have the volume as

{\frac {4{\sqrt {2}}}{3}}V

. For my own peace of mind if nothing else, I hope this is correct. 92.3.252.234 (talk) 19:49, 25 July 2008 (UTC)[reply]

That's the same as what I got, so I hope it's right too. ;-)

Incidentally, I don't quite get your comment about radius/diameter, as the method I used to determine it employed the radius. Obviously it's only a factor of 2 different, so you might have just formulated it differently, but I was just idly wondering if you'd done something different altogether. What was your approach, if you don't mind me asking?

--Grey Knight ⊖ 20:17, 25 July 2008 (UTC)[reply]

(edit conflict) Of course I don't. If you look at the cylinder 'side-on' - ie ignoring the top and bottom - then you see it as a rectangle width equal to the tin's diameter and of length equal to the tin's height. You can then determine the diagonal of this rectangle using Pythagoras and if you halve the length of the diagonal, then you have the radius of the smallest sphere into which the tin will fit. I hope I explained that clearly. How did you do it? 92.3.252.234 (talk) 20:23, 25 July 2008 (UTC)[reply]

Basically the same but I split up the projected rectangle into four quadrants and solved straight to the radius. Probably no faster. --Grey Knight ⊖ 22:39, 25 July 2008 (UTC)[reply]

Is it the sphere fitting inside the tin or the tin fitting inside the sphere?

If it's the sphere fitting inside the tin the I get V=4/3 pi r³ = 4/3 pi (V/2pi) = 2V/3 ? 87.102.86.73 (talk) 20:22, 25 July 2008 (UTC)[reply]

Note that the tins height=tins diameter - so the tin touches the sphere both places. (I may have made a mistake?)87.102.86.73 (talk) 20:30, 25 July 2008 (UTC)[reply]

The tin has to just fit in the sphere. Sorry for not making it clear. 92.3.252.234 (talk) 20:24, 25 July 2008 (UTC)[reply]

Yep I get the same.87.102.86.73 (talk) 20:30, 25 July 2008 (UTC)[reply]

Music and Math

Is there any way that the Pythagorean Theorum can be related to and used in Music? And also, how is sine related to music? 207.171.214.26 (talk) 21:17, 25 July 2008 (UTC)[reply]

Pythagorus is attributed numerous musical theories/methods including Pythagorean interval (and links therein) also see Pythagorus#Musical theories and investigations - but those don't relate to Pythagorus's Theorum.. There probably has been (experimental) music based on it, (as there has many other mathematical relationships) but you would be better placed to ask about that on the humanities desk.

As an example you could make 'music' based on Pythagorean triples

Here is an example http://www.oocities.com/Vienna/9349/numbers.html#pythagorean I've no idea what it is like. Here is the music file (midi?) http://www.oocities.com/Vienna/9349/triples.mid It has a certain maudlin charm..

You could construct a 'note' using a synthesizer consisting of harmonics that are Pythagorean triples - again I've no idea if they would share common musical character , or even have a pleasing sound..87.102.86.73 (talk) 21:45, 25 July 2008 (UTC)[reply]

A sine wave with the amplitude varying with time is a musical note. Such a note is called a pure tone.87.102.86.73 (talk) 21:36, 25 July 2008 (UTC)[reply]

Just FYI (which in this case stands for for y'all's info, since this is addressed to more than one person), it's Pythagoras with an a, and theorem with an e. --Trovatore (talk) 08:54, 27 July 2008 (UTC)[reply]

Please: the correct spellings are theorem (there's no "u"!) and Pythagoras (also no "u"!). Michael Hardy (talk) 18:12, 28 July 2008 (UTC)[reply]