Wikipedia:Reference desk/Archives/Mathematics/2007 November 9

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November 9

Convert Area to Circumference directly and vice versa

I need someone with maths skills and wiki familiarity to prove and post the following. I couldn't find it and 6th graders everywhere will appreciate your efforts. It works.

C^2 = 4 * Pi * A or C^2 = 12.566 A

I couldn't find it when my daughter needed it so we contacted a math whiz at Dr. Math online to help us come up with a simple elegant equation that allows you to go directly from Area to Circumference and back.

Please post it on wiki.

Channelchannel 03:57, 9 November 2007 (UTC)Channel Sanderson

Circumference of a circle $C=2\pi r$ (where r is the radius). Area of a circle $A=\pi r^{2}$. From those it follows pretty directly with basic algebra. --Spoon! 04:02, 9 November 2007 (UTC)

It might not be as obvious as you thing spoon, I want to point out that the key here is that both the Circuference Formula and the Area formula have the radius r in them, by solving one of them for r and substituting it into the other you can write Area in terms of Circumference or vis versa. A math-wiki 10:58, 10 November 2007 (UTC)

That's one way to do it, but the easy way is to simply say that $C^{2}=(2\pi r)^{2}=4\pi ^{2}r^{2}$ and $4\pi A=4\pi (\pi r^{2})=4\pi ^{2}r^{2}$, hence $C^{2}=4\pi A$. Spoon was correct in the description, "follows pretty directly with basic algebra" - it was never implied that basic algebra comes easily to everyone, and it was assumed that the OP would ask if he required further assistance. -- Meni Rosenfeld (talk) 12:57, 10 November 2007 (UTC)

To tease more mathematics from this exercise, we may point out a few interesting facts. Archimedes showed millennia ago that the area of a disk is the same as that of a right triangle which has one side with length equal to the circumference and a perpendicular side with length equal to the radius. For any circle, no matter what the size, the ratio of circumference (perimeter) to diameter (twice the radius) is the constant π. Given a radius R, the diameter is then 2R, the circumference is 2πR, and the area is ¹⁄₂⋅2πR⋅R — which is πR². The proof by Archimedes employs simple geometric arguments that might be of interest to some "6th graders". As well, he put the dual role of π to good use, calculating areas of regular polygons and using those areas to narrow the bounds on an approximate value for π.

Note, however, the distinction between the use of π, which is the same in both area and circumference, and the use of the radius R. In particular, for the circumference we use R, and for the area we use R². This has implications for pizzas and wounds. If one pizza has twice the diameter (hence twice the radius) of another, its area will be four times as great. This we like. Wounded skin heals from the edge, so the rate of healing depends on the circumference; a circular wound with twice the circumference (or diameter, or radius) will have four times the area, thus can take much longer to heal. This we don't like!

I can't recall ever having a practical need for a squared circumference, so this seems like an artificial problem. As my two examples show, however, the relationship between area and unsquared circumference is of some interest. --KSmrq^T 01:35, 11 November 2007 (UTC)

Sets

Please could someone help me out with the following questions related to mathematical sets, i can't work them out.

• 5P3
• 5C3

according to the article on sets, P involves denoting the set of all primes, while C involves denoting the set of all complex numbers, but i don't understand how that works. Thanks--Jac16888 12:42, 9 November 2007 (UTC)

Where do these questions come from? The correct articles to look at are Permutation and Combination, which belong to the subject of combinatorics. These would be denoted there $P_{3}^{5}$ and $C_{3}^{5}$. They are easy to compute directly, and most calculators have a special button to calculate them. -- Meni Rosenfeld (talk) 12:50, 9 November 2007 (UTC)

IMHO original symbols were ₅P₃ and ₅C₃ --CiaPan 13:32, 9 November 2007 (UTC)

That's still a representation for permutation and combination. There are several ways to represent them. Gscshoyru 13:37, 9 November 2007 (UTC)

That's quite obvious.... :) CiaPan —Preceding comment was added at 13:43, 9 November 2007 (UTC)

Yes, it seems most likely the questions ask for Permutations of five distinct items taken three at a time, and Combinations of five distinct items taken three at a time.

Consider related questions: How many permutation are there for four items taken three at a time? Combinations?

Permutations drawing from ABCD

ABC	BCA	CAB	ACB	CBA	BAC
ABD	BDA	DAB	ADB	DBA	BAD
ACD	CDA	DAC	ADC	DCA	CAD
BCD	CDB	DBC	BDC	DCB	CBD

Each entry contributes one permutation (there are 24); each row contributes one combination (there are 4). --KSmrq^T 19:00, 9 November 2007 (UTC)

Sorry, i forgot to reply, but yes, combination and permutation were the right way to go thanks for your help--Jac16888 19:06, 9 November 2007 (UTC)

Prime Factor

Please help with the following problem: Any positive integer of the form 3k+2 has a prime factor of the same form; similarly for 4k+3 & 6k+5. Thank you. --Shahab 16:36, 9 November 2007 (UTC)

Not too tough if you work with modular arithmetic. For the first one, we have that n = 3k+2, so n mod 3 = 2. Since (a mod 3)*(b mod 3) = ab mod 3, we can tell that if the prime factors of n are all equal to 1 mod 3, then n is equal to 1 mod 3. Since this is not the case, there must be at least one factor equal to 2 mod 3. The others work similarly, but may require a bit more thought. Crabula 17:25, 9 November 2007 (UTC)

Unfortunately modular arithmetic is in the next chapter and the problem is intended to be solved without it. If I understand your solution, then:

1. Clearly 3 or any of its powers is not a prime factor as that would contradict the uniqueness of the division algorithm.

2.If all the prime factors give remainder 1 (when divided by 3) then their product also will give remainder 1 as (3a+1)(3b+1)=3(a+3ab+b)+1

3.So some prime factor would give remainder 2.

This looks correct to me. Thanks. But how do we proceed for the 4k+3 case. Specifically if there are prime factors of the form 4k+1 and 4k+2 then their possible products are of the form 4k, 4k+1, 4k+2. What now?--Shahab 18:10, 9 November 2007 (UTC)

OK I have got it now.--Shahab 18:15, 9 November 2007 (UTC)

A minor remark: powers of 3 (except of course 3¹) are not prime numbers anyway, and therefore not prime factors of any number.  --Lambiam 22:03, 9 November 2007 (UTC)

AA Battery Hack from 6volt Lantern Battery??

I just opened an e-mail attchment that showed a 6 volt lantern battery (the bulky square one that fits in my flashlight) being prised open to reveal 32 standard AA Batteries - the message being that anyone who uses a lot of AA batteries can save themselves shed loads of dosh by following the simple instructions above. Am I being naive in wanting to believe that story? Or is it true? 81.145.240.17 17:31, 9 November 2007 (UTC)

Higher voltage batteries often do contain smaller dry cells inside, but these aren't likely in the right shape to be useable as AA batteries. There is a page about this on snopes. - Rainwarrior 17:41, 9 November 2007 (UTC)

There is the 9V battery hack, tested by xkcd here. risk 21:41, 9 November 2007 (UTC)

If you're using lots of AA batteries, buy rechargables (there are now rechargable batteries with really 1.5 V voltage, not the traditional 1.2 V, but there's less need for that because most new machines are designed to work with 1.2 V as well). If you don't want to do that, buy the cheap yellow batteries from IKEA. – b_jonas 20:41, 10 November 2007 (UTC)