# Math/Geometry Question

Discussion in 'Community' started by 4409723, Mar 4, 2005.

1. ### 4409723 Suspended

Joined:
Jun 22, 2001
#1
Hello,

I'm on the last question on a math assignment here is the premise.

1. Take a circle of flat paper, cut out a sector with measure θ.
2. Roll this into a cone.
3. Find the optimum value of θ to get the maximum volume. I got this value as shown below.

The final question:
When the circle is cut there are two sectors so it is possible to make two cones.
Find the value of values of x for which the sum of the volumes of the two cones is a maximum.

Now I'm wondering how I can factor the fact that second sectors measure is 2π minus the first into my substitution of:

Note: I'm assuming a slant height of 1 as the slant height does not matter when calculating the optimum angle.

One other small question: I've done quite a bit of this working with mathematica but I'm following it and verifying it but I can't for the life of me see why there isn't another solution to the derivative as shown in the graph but not in the solution of the derivative.

2. ### edesignuk Moderator emeritus

Joined:
Mar 25, 2002
Location:
London, England
#2
Wes, mate, you've just ruined my easy going evening

Hope someone can help you, I sure as hell can't...I seem to remember there are a lot of over educated people here though...now where are they?

3. ### Rend It macrumors 6502

Joined:
Oct 27, 2003
Location:
United States
#3
To get volume of both cones:

Volume = f(x) + f(y), where f(x) is as you have defined it above. Your 2nd sector angle is 2pi-theta. So, whereas x=theta/2*pi,

y = (2pi-theta)/2pi = 1 - x

So, making the substitution,

Volume = f(x) + (1/3)*pi*{(1-x)^2}*sqrt{1-(1-x)^2}
= f(x) + (1/3)*pi*{(1-x)^2}*sqrt{x*(2-x)}

This all assuming, of course, that I understood your question.

Joined:
Aug 10, 2004
5. ### Rend It macrumors 6502

Joined:
Oct 27, 2003
Location:
United States
#5
Assuming that your graph is the derivative of the volume, then I think it's simply wrong. Recall that your original equation for the Volume is cubic. So, you should end up getting 3 locations for extrema. In this case, you get 2 where the volume is maximized (but you throw out the negative one because it's redundant or physically unmeaningful), and one where the volume is minimized, namely theta = 0. Draw a graph of the volume, and then overlay the derivative on top of that to visualize.

Remember that sqrt(2/3) = 0.816497..., not 0.78... as indicated in the graph.

6. ### Mr. Durden macrumors 6502a

Joined:
Jan 13, 2005
Location:
#6
Hmmm. Either you or I forgot to move the decimal. I came up with .2

7. ### 4409723 thread starter Suspended

Joined:
Jun 22, 2001
#7
Thanks for your help and I've now fixed that graph .

Here is the graph of that result:

Now when I try to solve the derivative of THAT I get this?!?

The real parts of these answers correspond to the maximums but the tiny imaginary part has me puzzled.

8. ### Blackheart macrumors 6502a

Joined:
Mar 13, 2004
Location:
Seattle
#8
I hate calculus from the deep, DEEP depths of my heart. I think I hate linear algebra more though.

Joined:
Dec 6, 2004

Joined:
Jan 12, 2003
#10
Ditto

11. ### applemacdude macrumors 68040

Joined:
Mar 26, 2001
Location:
Over The Rainbow
12. ### Rend It macrumors 6502

Joined:
Oct 27, 2003
Location:
United States
#12
Wes, never mind the imaginary parts. They're unphysical (in this problem) and are simply an artifact of the algorithm Mathematica uses to find the roots of transcendental equations. If nothing else, a 16 order-of-magnitude difference tells you that they are meaningless, or can be set = 0.

After some algebra (by hand), the characteristic equation looks like this:

0 = sqrt(2x-x^2)*3x*(2/3-x^2) + sqrt(1-x^2)*(1-x-5x^2-3x^3)

There are very complicated ways of dealing with analytical solutions to fractional polynomials. See any handbook of mathematics, like Abramowitz and Stegun. But usually, when you're trying to find roots of a transcendental equation, you solve it graphically. The finite square well problem in quantum physics is solved exactly that way, where you have something like,

k*tan(k/2) = -sqrt(a-k),

and it needs to be solved graphically. The solutions tell you the energy levels for such a problem. Happy calculus!

13. ### 4409723 thread starter Suspended

Joined:
Jun 22, 2001
#13

Thanks for the help, it's a bit late to be doing this at 00:02; I'll pick it up in the morning and wrap this assignment up.

Cheers once again.

I'm always surprised with the knowledgable crowd here at MR.

14. ### Blackheart macrumors 6502a

Joined:
Mar 13, 2004
Location:
Seattle
#14
Are/were you a math major?

15. ### Mr. Anderson Moderator emeritus

Joined:
Nov 1, 2001
Location:
VA
#15
Its been a while since I was in school and had to do such things, but I never remember having done them with Mathmatica or any other tool - it was all by hand (and my last Math class was only 12 years ago or so).

What class is this for and what level?

D

16. ### patrick0brien macrumors 68040

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Oct 24, 2002
Location:
The West Loop
17. ### benbondu macrumors regular

Joined:
Jul 2, 2004
#17
Look at that flat top!

At the very least, that graph of the nasty looking equation tells you you can get very close to the maximum volume without much precision at all. Anything between 2/3*Pi and 4/3*Pi looks pretty good. Even the obvious choice of cutting the circle in half gives a pretty high volume. Looks to be within 1-2% of the true maximum. (Of course I guess 1-2% could be more significant if the paper was several meters wide, as rediculous as that sounds) Very interesting result nonetheless.

18. ### 4409723 thread starter Suspended

Joined:
Jun 22, 2001
#18
This is for IB higher math. Which equates to about AP Calculus in high school. I'm a senior.