quadratic mean of all values of r of a circle

ray5450

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quadratic mean (xq) = /.. bb y2 dx
.......................................a

The hint I need is what expression to use for y, but I would think it has the radius in it and a to b would be r from 0 to R (of circle).

Thanks.
 
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Erm... it looks like you tried to "write" out your math using whitespace-based formatting, which, as you can see, the forum does not preserve. As such, I have no clue what you meant to type. Please reply with a clear statement of the problem at hand, noting that you can (and should) use the forum's built-in LaTeX parser to make it easier to read. For instance, if you type:

Code:
[tex]\displaystyle \sqrt{x} = \int_{a}^{b} \: f(x) \: dx[/tex]

It will render as:

\(\displaystyle \displaystyle \sqrt{x} = \int_{a}^{b} \: f(x) \: dx\)

Thank you.
 
In the Read Before Posting message, there is a link to a web page that shows how to type math expressions as text. That information is an easy alternative to learning the LaTex-coding language. :cool:
 
That's strange. If I view what I posted, even when I am not logged in, it looks perfect and clear. You must be seeing something different than I am. Maybe it depends on what browsers we are using.

I looked up a language reference for your LaTex and it is 140 pages. I guess I don't have time for that.

I think the "typed math expressions" are difficult to read, even if I write it, if I go back to it, it's hard to re-interpret.

How about plain English--quadratic mean is the square root of all:
--inverse of b--
--multiplied by--
--the integral from a to b of y squared, dx--
--end of square root.
 
No, it's not strange. What you see depends strongly on what browser you are using and even what version of that browser. And while the reference for Latex has 140 pages, most of what you need is on the first couple of pages.
 
If I view what I posted, even when I am not logged in, it looks perfect and clear.
The attached image (at the bottom of this post) is what your "drawing" looks like, in the latest version of Google Chrome.

\(\displaystyle \displaystyle \bar{x}_q = \sqrt{b^{-1} \cdot \int_{a}^{b} \: y^{2} \: dx}\)

Are you implying that:

x^2 + y^2 = r^2

So that:

y = sqrt(r^2 - x^2)

where r goes from 0 to R?

There is no circle, by the way, if r = 0.

(I hope that you find my texted math expressions not too difficult to read.) :cool:
 

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"No, it's not strange."/"...in the latest version of Google Chrome."--I think that's what I said...it is browser dependent. Thanks, for confirming.

I am not implying anything. y can be any function. In this case, I need one that would have some relation with all values of r in a circle.

Yes, there is no circle when r is 0, the same as for an area under a curve at the point of an inferior limit (at zero, for example)--there is no area as it is only a line.

I tried to use the relation you offered, but the only thing I can come up with is working backwards to manipulate the limits as expressions, but I do not think that is the right way.

Thanks.
 
I think that's what I said...
Please excuse me; I think I misunderstood. What is it that you've claimed is strange?

I am not implying anything. y can be any function.
Okay, but I thought we were talking about circles.

y = f(x), where function f can be any function. Is this correct?

It would help, if you were to define your variables. Thank you. :cool:
 
Thank-you, for your response.

Strange: I was talking about it being strange that when I view the post, my construction of the expression looks fine, but it appears sort of scrambled to other viewers. We seem to all be in agreement that it is because different browsers affect its formatting.

Talking about quadratic mean of a function, but in need of such function to have some relation with all values of r in a circle.

Okay, yes, definitions:
_
xq = quadratic mean

y = any function (of x, r, or whatever would work)

a = inferior limit of integral of y

b = superior limit of integral of y

R = radius of circle

r = radius of each concentric circle from 0 to R

r is a guess. It might be x as you say. I don't know. It depends what relation would be useful to find the quadratic mean of all values of r in a circle.

Thanks.
 
There must be some expression for y to find the quadratic mean of all values of r in a circle, but I do not know what it is.
 
I am confused as to what "all the values of r for a circle" means! I would expect "r" to be the radius of a circle but then there is only one "value of r".
 
The attached image (at the bottom of this post) is what your "drawing" looks like, in the latest version of Google Chrome.

\(\displaystyle \displaystyle \bar{x}_q = \sqrt{b^{-1} \cdot \int_{a}^{b} \: y^{2} \: dx}\)

Are you implying that:

x^2 + y^2 = r^2

So that:

y = sqrt(x^2 - r^2)
No, y would be sqrt(r^2- x^2). The integral above would be \(\displaystyle \sqrt{\frac{1}{b}\int_a^b r^2- x^2 dx}\)

where r goes from 0 to R?
"r" here is not a variable. If believe you mean that x goes from a= -r to b= r.

There is no circle, by the way, if r = 0.

(I hope that you find my texted math expressions not too difficult to read.) :cool:
 
"all values of r"--I should have said "all values of R" to be consistent, but I'm not sure what that means, either, but I am assuming it means: r = radius of each infinitesimal concentric circle from 0 to R, inside of circle of radius R.


Working backwards from the solution (R square roots of 2), y has to be: R multiplied by square root of 6. 6 is about 2π, so it is likely R multiplied by square root of 2π. The question is, how to deduce "R multiplied by square root of 2π" from "all values of R".
 
What relationship could "R multiplied by square root of 2π" be deduced from "all values of R"?
 
Still doesn't make much sense. Why would this value you seek be different from the same consideration on a straight line?
 
Okay. How would I deduce R multiplied by square root of 6, or R multiplied by square root of 2π, from "all values of R" by way of a straight line? Thanks.
 
Okay. How would I deduce R multiplied by square root of 6, or R multiplied by square root of 2π, from "all values of R" by way of a straight line? Thanks.
??

Please reply with the full and exact text of the original exercise, the complete instructions, and a clear listing of your thoughts and efforts so far, so we can perhaps make greater headway in figuring out what you're talking about. Thank you! ;)
 
Circle 2 is twice the area of circle 1. Show that the quadratic mean of all the values of r of circle 1, is equal to the radius of circle 2.

I have found that the relation between the radius of a circle that is twice the area of another is that the radius of the larger circle is equal to the radius of the smaller circle multiplied by square root of 2. Therefore, the result of the quadratic mean calculation must be radius of circle 1 multiplied by square root of 2. Using this information to work backwards, y must be r1 multiplied by square root of 6 (or r1 multiplied by square root of 2π). However, I do not believe the exercise is to work it backwards, but to find the correct relationship to deduce y.
 
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