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March 26

Ratios

If there are five numbers, A,B,C,D and E, aranged in order from least to greatest and the ratio A:E is equal to 1:2 and the ratio between any of the consecutive numbers is equal, how can I calculate the ratio A:B (which is equal to B:C, C:D, etc.)? Thegreenj 01:10, 26 March 2007 (UTC)[reply]

Hint: Let

n={\frac {A}{B}}={\frac {B}{C}}={\frac {C}{D}}={\frac {D}{E}}

, do some rearranging and find an equation where

{\frac {A}{E}}

equals some unknown which equals to a half. x42bn6 Talk 01:16, 26 March 2007 (UTC)[reply]

Thank you. I am now at

n={\frac {A}{E}}={\frac {B^{2}}{CE}}={\frac {CB}{DE}}={\frac {DB}{E^{2}}}={\frac {1}{2}}

, but I'm still uncertain of what to do now. Thegreenj 02:12, 26 March 2007 (UTC)[reply]

I think the most important thing you need to remember is that the ratio between any consecutive numbers is equal. Perhaps using the old standarized test trick of plugging numbers for A and E might help you along? --Ybbor ^T 02:15, 26 March 2007 (UTC)[reply]

Another hint: what if you multiplied the ratios together? --Spoon! 03:06, 26 March 2007 (UTC)[reply]

More hinting: In your second message, you're defining n differently than in x42bn6's hint. The way x42bn6 defines n is really more useful, because in his case, n is the answer to the problem. It's usually a good idea in a math problem to define things such that the answer is one variable. They way you're trying to do things, the answer would be a ratio of two variables, which would be more complicated. The advantage of having the answer being one variable is that if you can figure out an equation that only involves one or more instances of that one variable and some constants, then it's usually easy to solve for the one variable. Can you do that in this case? MrRedact 03:49, 26 March 2007 (UTC)[reply]

I disagree with the methods these folks are proposing. Think mean proportionals; a mean proportional Y between X and Z is the number such that X:Y = Y:Z. Neon Merlin 03:51, 26 March 2007 (UTC)[reply]

Isn't this somewhat begging the question? --Lambiam ^Talk 06:36, 26 March 2007 (UTC)[reply]

The way I see it, there are 5 equations and 6 unknowns, using matrices (or elimination), you can solve for n, and then you can confirm it by setting any value for A and following the equations through. I had a go at it and it seems to work. Might not be the easiest way though... - Akamad 06:01, 26 March 2007 (UTC)[reply]

This requires somewhat advanced mathematics for a problem that can be solved by much more elementary means. There is nothing wrong with x42bn6's and MrRedact's hints. Yet another approach:

Let n be the unknown ratio A:B, so A/B = n. Then we can write B = .... Also C/B = n, so C = .... Eliminate B from this expression (so you now have an expression giving C in terms of A and n). Continue likewise for D and E. You now have E as an expression in A and n. Eliminate A and E together by using A/E = 1/2. Solve the final equation obtained this way for n. --Lambiam ^Talk 06:36, 26 March 2007 (UTC)[reply]

I don't suppose yet another view will really help, but I'll offer one anyway. First, I would not use n for a ratio, because we typically use that to denote an integer, which is inappropriate here. Second, I would work from the time-tested all-purpose plan of George Pólya in How to Solve It. (Our summary is not as helpful as the one here.)

What is the unknown? — The ratio A:B.
What is the data? — The ratio A:E is 1:2. We have equal ratios A:B = B:C = C:D = D:E.
What is the connection between the data and the unknown?

Let us stop here, because I believe this is the key step where everyone is offering hints. We see it so clearly, and perhaps so quickly, that we find it challenging to hint, not answer! I believe the essential step for most of us concerns notation. We want to change from colons to fractions, simultaneously changing from thinking of integers to thinking of real numbers.

Turn that inappropriate n upside down, and denote the ratio B:A (note the reversal!) by the real number u. Likewise, the ratio E:A is simply 2. What connects u and 2?

The first part of the answer is plain: The connection is a chain of equal ratios. The second part requires insight: How can we convert that chain into a mathematical expression we can solve?

Well, suppose A is a definite number, say π = 3.14…; then we know that, in symbols, B = uA. For example, if u is 1.25 then B = 3.9…. So if we know u, then we can find B from A, then C from B, and so on until we find E.

Condense the chain of calculation of E from A into one formula. Done properly, the formula will involve A, E, and u, but B, C, and D will not appear.

I suspect that if you get this far you can finish the solution on your own. If not, post the formula and explain your remaining difficulty. --KSmrq^T 14:01, 26 March 2007 (UTC)[reply]

I think we are all making this difficult. :O You might want to see geometric series as well, if it helps. x42bn6 Talk 17:15, 26 March 2007 (UTC)[reply]

The difficult part is to explain the thinking that makes solution easy — without handing over an explicit solution. It would take less than a minute to find and post a solution, say approximately ³⁷⁄₄₄, but that teaches nothing.

I have a hunch that if the question involved differences rather than ratios it would pose little challenge for the poster. For us, they are equally easy; why? --KSmrq^T 18:22, 26 March 2007 (UTC)[reply]

Perhaps because we are very familiar with the isomorphism between the additive group of real numbers and the multiplicative group of positive reals. --Lambiam ^Talk 23:05, 26 March 2007 (UTC)[reply]

When I work with this type of problem but involving differences, I would hit upon the idea that if you keep subtracting, this equals some sort of division or multiplication operation. When the operation is division or multiplication, however, where do you go from there? Powers, simply - but this is not an easy concept to draw upon without practice. I only hit upon the geometric series part a few hours ago, partially because of the way the question is worded. Also, ratios can be hard to understand at first, because, quite simply, why aren't they fractions?!. x42bn6 Talk 19:04, 26 March 2007 (UTC)[reply]

Okay... I used KSmrq's idea and got

{\sqrt[{4}]{.}}5

as the ratio A:B or A/B for A=1. I checked it and it works! Thanks for all the input. Thegreenj 20:36, 26 March 2007 (UTC)[reply]

Weighted mean and median where weights equal values

Suppose a university's statistics department has three first-year classes: Statistics 101 with 100 students, Statistics 102 with 200 students, and Statistics 103 with 300 students. We'd say the average class size is 200, whether we use a mean or median. But there are 300 students in the class of 300 and only 100 in the class of 100, so the mean student is in a class of 233 ande the median student is in a class of 250. Therefore, we need a weighted mean and median where the weights equal the values. (The former is the sum of squares divided by the sum of values.) Do terms for these quantities exist? Neon Merlin 02:05, 26 March 2007 (UTC)[reply]

That's a rather specialized problem, so I doubt if general terms exist for it, but I could be wrong. StuRat 02:32, 26 March 2007 (UTC)[reply]

But there are 300 students in the class of 300 and only 100 in the class of 100, so the mean student is in a class of 233 ande the median student is in a class of 250.

Eh? What the hell is a mean student? 202.168.50.40 05:04, 26 March 2007 (UTC)[reply]

I believe the median class size for the students is 250, in that exactly half the students are in classes larger than 250, and exactly half in classes smaller than 250. (In the case of an even number of items, it is customary for the median to be the average of the number immediately higher than the halfway point, 300, and the number immediately lower, 200.) StuRat 16:02, 26 March 2007 (UTC)[reply]

The mean student is a really mean dude. --Lambiam ^Talk 06:40, 26 March 2007 (UTC)[reply]

Women can be more cruel than men! (I'm assuming, of course, that at least a few women will study statistics.) --KSmrq^T 14:08, 26 March 2007 (UTC)[reply]

I think you are calculating two slightly different statistics here. The mean class size per class is 200 - here you have three observations, 100, 200 and 300. But the mean class size per student is 233 1/3 - now you have 600 observations with 100 at a value of 100, 200 at a value of 200 and 300 at a value of 300. Gandalf61 08:43, 26 March 2007 (UTC)[reply]

I concur there. There is a difference between mean class size, and mean perceived class size (same with median). This is related to the observation that the transportaion people tell us that traffic isn't that bad, yet most people seem to think that they are usually in a traffic jam. Think about it. Baccyak4H (Yak!) 16:21, 26 March 2007 (UTC)[reply]

Arc Length with Integrals

How can I find the length of the curve defined by (y+1)²=4x³ between the points (0,-1) and (1,1)?

Thanks! —The preceding unsigned comment was added by 75.144.0.89 (talk) 14:27, 26 March 2007 (UTC).[reply]

I believe the first step is to put it into an equation in the "y =" form. StuRat 15:45, 26 March 2007 (UTC)[reply]

I don't think so. Not for something as implicit as that. Take the derivative, and solve for dy/dx.

$(y+1)^{2}=4x^{3}\!$
$2(y+1){\frac {dy}{dx}}=12x^{2}\!$

And then use the formula for arc length:

$s=\int _{a}^{b}{\sqrt {1+\left[{\frac {dy}{dx}}\right]^{2}}}\,dx.$

--Ķĩřβȳ ♥♥♥Ťįɱé Ø 16:14, 26 March 2007 (UTC)[reply]

And in this case, I believe it may be simpler to solve for x, so actually it would be better to differentiate with respect to y:

$(y+1)^{2}=4x^{3}\!$
$x={\sqrt[{3}]{\frac {(y+1)^{2}}{4}}}$
${\frac {dx}{dy}}=$ ...

And then using the arc length in terms of y:

$s=\int _{a}^{b}{\sqrt {1+\left[{\frac {dx}{dy}}\right]^{2}}}\,dy.$ and of course a and b would be -1 and 1, respectively.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 16:24, 26 March 2007 (UTC)[reply]

Hello, mathematicians! Why don't you shift the curve first by 1, so it gets an equation z²=4x³ (where z is 'shifted' y)?
CiaPan 16:38, 26 March 2007 (UTC)[reply]

Good thinking! For this problem, that makes solution much easier.

Otherwise, we have two questions. First, do we have a clearly defined unique piece of curve between the two points? (Suppose the implicit equation was a circle.) We'll assume the problem is well-posed. Second, how do we compute the arclength of a curve segment? This has a two-part answer.

The typical computation integrates differential steps along the curve, compensating for parameterization as necessary.
We need to find a parameterization and a derivative.

But we know that implicit curves of degree three and above are not so easy to parameterize (except "locally"). --KSmrq^T 17:03, 26 March 2007 (UTC)[reply]

Oblique Curvature?

If north-south curvature on an ellipsoid is considered "meridional curvature" and east-west curvature, perpendicular to meridional curvature, is regarded as "normal curvature", would any randomly angled curvature——i.e., plane curve arc——properly be termed "oblique curvature"? Likewise its radius——"radius of arc" or "arcradius" equals "oblique radius of curvature"? ~Kaimbridge~15:07, 26 March 2007 (UTC)[reply]

Some math

Transferred here from Science desk -- atropos235 ✄ (blah blah, my past) 23:22, 26 March 2007 (UTC)[reply]

You have the following set: {a, b, c}. How many different three-letter combinations can you form using the letters of this set (you may repeat any letter)? I know the question is silly. It's even possible to do all the possibilities by hand, but I would really appreciate it if you should the exact mathematical concepts you use (combinations, permutations,...). I don't know if you'll believe me or not but this is not a homework question. I'd be very grateful for an answer. —LestatdeLioncour t 18:29, 26 March 2007 (UTC)[reply]

3! = 6. For the first letter in the sequence, there are 3 choices, for the second you then have 2 choices, and the third is selected: 1 choice. 3 times 2 times 1 = 6

abc acb bca bac cba cab That is all. WilyD 18:36, 26 March 2007 (UTC)[reply]

Allow me to say this again: you may repeat any letter. For example, bba is a valid combination. Although I just remembered that I have asked this question in a wrong way (the answer is 3³, if order doesn't matter). The matter I'm having trouble with actually deals with the probabilities of each combination, and the letters have weights. So, never mind. I'll find a better way to rephrase this question later. —LestatdeLioncour t 19:02, 26 March 2007 (UTC)[reply]

It's actually 3³ with order, fewer without. I'm not sure what you mean by weight. Are you talking about the probability of each letter being used, or the value of each resulting combination? Clarityfiend 19:24, 26 March 2007 (UTC)[reply]

Since you can repeat letters, permutations and combinations aren't involved. The general answer is mⁿ for a string of length n that draws from a set of m letters. Do a Google search for "Generalized Basic Principle of Counting" to find a description of the basic principle involved in getting this answer. MrRedact 19:30, 26 March 2007 (UTC)[reply]

Agreed. The 3³ or 3x3x3 or 27 possibilities in your example are:

AAA, AAB, AAC, 
ABA, ABB, ABC,
ACA, ACB, ACC,

BAA, BAB, BAC, 
BBA, BBB, BBC,
BCA, BCB, BCC,

CAA, CAB, CAC, 
CBA, CBB, CBC,
CCA, CCB, CCC.

StuRat 21:08, 26 March 2007 (UTC)[reply]

Thank you guys for all your replies, but if you remember I said that the question was not phrased correctly. This isn't what I meant to ask. It somehow came off the wrong way. I will think of a better way to phrase it and post it here soon. —LestatdeLioncour t 12:21, 27 March 2007 (UTC)[reply]

Do you perhaps mean 3-element multisets (also known as "bags")? Then you have 10 possibilities: aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, ccc. In general, for a set of size n and multisets of size k, the number of different combinations you can form equals C(n+k−1, k) (and, if n = k = 0, 1). See Multiset coefficients. --Lambiam ^Talk 12:59, 27 March 2007 (UTC)[reply]

Thank you Lambiam! I think this is what I meant. I kind of have it at the back of my mind right now so I'm not giving it much thought. Can you recommend any books on this topic? (including probabilities and related topics). Something self-sufficient that doesn't require an instructor. —LestatdeLioncour t 17:13, 27 March 2007 (UTC)[reply]

It is ages ago since I last touched an elementary mathematics book, and I don't have access to a library, so it is difficult for me to recommend a specific book. If you have access to a university library, they probably have one or more books with titles like Introduction to Combinatorics and Introduction to Probability, and you could check out if you like any of them. There is also an online Introduction to Probability. People who have read Concrete Mathematics seem to like that book a lot. For a textbook it is relatively cheap, and if you can master it, it is definitely worth the money, if only for the joy. --Lambiam ^Talk 22:52, 27 March 2007 (UTC)[reply]

Volume

Hi! I was wondering if someone could tell me which is correct because i'm a little stumped with a volume question. The question asks for the volume between X²-Y²=1 , X=3 about X=-2. Some possible answers i got was...
(a) π ʃ 2 + [(Y²+1)^1/2]² - (5)² dy
(b) 2π ʃ [ (x²-1)^1/2-(3) ](x+2)dx
also i am having trouble finding the intergral's end points. Thanks for any feedback! --Agester 23:53, 26 March 2007 (UTC)[reply]

The region you describe lies in the plane (and is not even well defined there). What volume are you looking for exactly? Algebraist 00:24, 27 March 2007 (UTC)[reply]

btw, it will almost certainly help a lot to draw a picture of the curves involved (that's what I've just done) Algebraist 00:27, 27 March 2007 (UTC)[reply]

Do you mean "what is the area within X²-Y²=1 between X=2 and X=3 ?" ? If so, I estimate it to be somewhere around 5 square units. StuRat 00:32, 27 March 2007 (UTC)[reply]

Rotated about an axis?--Ķĩřβȳ ♥♥♥Ťįɱé Ø 01:11, 27 March 2007 (UTC)[reply]

I'm sorry if i wasn't clear. The axis of rotation is X = -2, that was a little unclear. And the region is between the line x = 3 and the curve X²-Y²=1. Sorry for being a bit vague. --Agester 01:23, 27 March 2007 (UTC)[reply]

I see, so we have a solid of revolution, a full 360 degree revolution, I assume. StuRat 03:30, 27 March 2007 (UTC)[reply]

It appears that you are trying to do in (a) disk integration and in (b) shell integration; but neither of the integrals you have written are right.

For disk integration, you want $\pi \int _{a}^{b}([R_{O}(y)]^{2}-[R_{I}(y)]^{2})dy$ . In this case the outer radius (R_O) should be $3-(-2)=5$ ; and the inner radius (R_I) should be ${\sqrt {y^{2}+1}}-(-2)=2+{\sqrt {y^{2}+1}}$ . It seems that you switched the inner (smaller) and outer (bigger) radii, and you also left the "2 +" outside of the square of the inner radius by accident. What are the bounds? They are the y values at the bottom and top ends of your shape.

For shell integration, you want $2\pi \int _{a}^{b}p(x)h(x)\,dx$ . p(x) is the "radius" from the axis of rotation, which you correctly got as x - (-2) = (x+2). What is the height of the shell (h(x))? It should be the difference between the top and bottom edges of the hyperbola, and since the hyperbola is symmetric about the x-axis, this distance is twice the distance between the hyperbola and 0, i.e. $2{\sqrt {x^{2}-1}}$ . You lacked this factor of 2, and added an extra "- 3" in there for some reason. What are the bounds? They are the x values at the left and right ends of your shape.

These should give the same answer. --Spoon! 04:12, 27 March 2007 (UTC)[reply]

i see. I added the extra 3 in the shell method because I was having trouble as to where to add in the straight line (X = 3). I understand how to get the end points but i've practiced these problems extremely hard on how to get the end points but this one is just a bit difficult for me. Usually, when i have two curves i set them equal to each other to find their intersection points, as for this one i'm not sure what to set what equal to each other. Also, doesn't the factor of 2 i added infront of the intergral make up for the multiple of two i missed in the intergral that is suppose to be put infront of the square root? Also with shell method i can disregard constants such as that X = 3 if my answer isn't going to be in terms of that line? Thanks for the help! --Agester 11:49, 27 March 2007 (UTC)[reply]

(This will be an elaboration of what Spoon! has said.) As presented to us, the question must be interpreted cautiously. Focus attention on the xy plane. My reading of "between" would say we want the region whose left side is delimited by the right branch of the hyperbola, and whose right side is the line x = 3. The y extent of this region lies between the intersection of these two bounds (which we will not need, but could find by substituting the x value of the line into the hyperbola equation and solving for y). We ignore the left branch of the hyperbola, because it does not satisfy the "between" requirement.

It will be convenient to describe the hyperbola boundary as a function of x. With this approach, we need the x bounds. When y is 0, we find the minimum x (of interest) is 1. The bottom half of the region is the mirror image, thus the same area, as the top half. So we have winnowed the data (for the top half) down to y = (x²−1)^1/2, with 1 ≤ x ≤ 3.

When we spin this around the line x = −2, each x value gives us a cylindrical shell segment whose y extent is given by the formula we just derived. The radius of the cylinder for a given x value is, of course, x+2. Then we double the final volume to include the bottom half.

My purpose in deriving the same shell method a second time is to try to make the reasoning clearer. Good luck! --KSmrq^T 15:16, 27 March 2007 (UTC)[reply]

I believe the answer was:
π ʃ (5)² - [2 + (Y²+1)^1/2]² dy
and the endpoints of the intergral was from negative radical 8 to radical 8 i believe was what my professor told the class. Unless i remembered it incorrectly that is what the answer is. Also, is it possible to do shell method in this case since we do have a problem dealing with the straight line X = 3 because we have it in terms of y and x=3 can't be expressed in y in this case? --Agester 01:52, 28 March 2007 (UTC)[reply]

As has already been mentioned, the shell method is not the only approach. But it can be used, and I gave you everything but the final integral. Now is a good time to use <math> markup. The integral you recall from class,

\pi \int _{-{\sqrt {8}}}^{\sqrt {8}}\left(5^{2}-\left(2+{\sqrt {y^{2}+1}}\right)^{2}\right)\,dy,

steps through y. For each y value we have an annulus (a flat ring) whose area is the difference of two disks. The outer disk has constant radius, 5, and area π5². The inner disk has variable radius, found by solving x²−y² = 1 for x, then adding 2; its area is π times that squared. Substituting x = 3 gives y = ±√8, so we have our limits of integration. That's the "stacking rings" approach, and the integral you give is correct.

The "nesting shells" approach is a little simpler to write. Putting all the pieces together, the integral is

2\int _{1}^{3}2\pi (x+2){\sqrt {x^{2}-1}}\,dx.

Each shell has radius x+2, circumference 2π times that, and area given by circumference times height. The (half-)height is, as noted, (x²−1)^1/2. Now look again; see any problem with the line boundary? No, it's just the outermost shell, the upper bound of the integral.

The remaining challenge is to solve both of these integrals, and to verify that they give the same answer. (They do.) --KSmrq^T 03:34, 28 March 2007 (UTC)[reply]