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ParthKohli
 one year ago
Interesting calc problem  please check my work.
ParthKohli
 one year ago
Interesting calc problem  please check my work.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0What is the average distance randomly selected in a unit equilateral triangle to its center? Here's what I think...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0So if we instead think of discrete points and try to think of the average distance, we'd think of something like this:\[\sum \rm distance \cdot (fraction ~ of ~ points ~ that ~ are ~ that ~ distance ~ away)\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443350101709:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Here, the average distance turns out to be \(2\cdot 2/3 + 4\cdot 1/3 = 8/3\) which makes sense.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Now I tried to apply the exact same fact to a continuous system of particles like the interior of an equilateral triangle. But nope. Didn't work because we're gonna have to take radial elements and taking triangular elements wouldn't work.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0By unit equilateral triangle do you mean the side length is 1 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Then simply setup a double integral for the distance expression

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I was wondering if we could use things like radius of gyration, but the radius of gyration is more weighted towards distance, if you know what I mean.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443350378977:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{Average distance} = \dfrac{1}{A}\iint \sqrt{x^2+y^2}dA\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That seems nice. So a way to look at the double integral is that we fix the outer thing and vary the inner one throughout that outer thing, then we also vary the outer thing.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Right, double integral over a region does this : 1) evaluate the function "sqrt(x^2+y^2)" at a particular point (x, y) in the region 2) multiply that by the area element in the xy plane : dA

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Notice that the average velcity can be expressed using integral as : \[\text{Average velocity over (a,b)} = \dfrac{s(b)s(a)}{ba} = \dfrac{1}{ba} \int\limits_{a}^b v(t)\,dt\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(s\) is the displacement function

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I can understand singlevariable integrals. Just trying to understand how this thing works. \(dA\) is basically a really small circular element around \((x,y)\) right? Or what is it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0similarly, in our case, average distance within an equilateral triangle can be obtained by taking the integral of the distance function over the triangle, and then dividing by the number of data points (Area)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes In single integral, \(dx\) refers to a small width element. In double integral, \(dA\) refers to a small area element.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Can we say that \(dA = dxdy\) or \(dA = \pi (x^2 + y^2) dxdy\) or something like that? Because I see the \(dxdy\) always occurring in double integrals.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0depends on the coordinate system, if you're using rectangular coordinates, then \(dA = dxdy\) if you're using polar coordinates, then \(dA = r*drd\theta\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0So it's not a circular element. That makes sense because within the circle, the distance will kinda change.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0don't worry about the conversion factors, they are not important as of now.. you need to study jacobians and other stuff which comes later..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Help me through this: it means that if I go \(dx\) units around the point in the xdirection and \(dy\) units around the point in the ydirection, it doesn't change the \(\sqrt{x^2+y^2 }\) term  which is why the area element is \(dA = dx dy\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Great. Also the limits of \(x\) depend on the limits of \(y\), so it basically means that the inner integral is a function of the outer one.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Maybe, lets take this example problem. Find the center of mass of below 2 dimensional region : dw:1443351868292:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0First, the height of the triangle is \(\sqrt{3}/2\) so \(y\) varies from \(\sqrt{3}/4\) to \(+\sqrt{3}/4\). When \(y = \sqrt{3}/4\), \(x\) varies from \(0\) to \(+0\). When \(y = \sqrt{3}/4\), \(x\) varies from \(1/2\) to \(+1/2\). In general, for a given \(y\), \(x\) varies from \(y/\sqrt{3}  1/4\) to \(y/\sqrt{3} + 1/4 \).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That's great  you understand my love for physics.\[x_{CM} = \frac{\displaystyle \int x dm}{ M}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Area of the whole region: \(\displaystyle \int x^2dx/4 \) from 0 to ???

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Mass = density*Area lets just assume that the object shown in above diagram has uniform density of 1

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Wow I definitely know how to solve this problem  just tell me the upper limit.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes you definitely knew it I took the easy one because I wanted to show you how to set it up using double integrals

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0finding the coordinates of center of mass is not the goal setting up the integrals correctly is the goal here

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0OK. If I just let the upper limit be \(X\), then the area is \(A = \dfrac{X^3}{12}\) Now if I go \(x\) units along the curve in the xdirection, and a small width \(dx\), then the area of this thing is \(\frac{x^2}{4}dx \) and the mass is \(dm = x^2 dx/4\) because dm = dA (density = 1).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0So yes, life's good so far. Coming back to the original problem... Did I write the limits there correctly though? The \(x\) and \(y\) ones.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443352602540:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I'm not looking at the original problem yet

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\int ^{+\sqrt{3}/4}_{\sqrt{3}/4}\int^{y/\sqrt{3}+1/4 }_{y/\sqrt{3}1/4 } \sqrt{x^2 + y^2}dx dy\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Not so fast, do you see why the double integral gives the averge value ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, we've covered a lot of center of mass and moment of inertia in the class. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0but you said earlier that you don't know about double integrals

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0So I'm confused how you're able to setup that double integral in two minutes hmm

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I realised that the inner integral is really just a function of the outer integral.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Hmm that doesn't make sense to me but its okay, if you're good, im good.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0just evaluate the integral and we're done! there isn't much to it

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Just one question to understand if I'm thinking correctly: the limits of the inner integral should be in terms of the variable involved in the outer integral right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443353241031:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0the limits on the outer integral are not functions of x... just meant to ask about those on the inner integral

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes, but again, I don't really know what exactly you're asking... I feel you're rushing

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0So we first evaluate the inner integral, and since its limits are in terms of \(y\), we recover a function that is in terms of \(y\), which is then evaluated as a singleintegral.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0the bounds in a double integral are the "inequalities" that represent the "region" over which you integrate the function, f(x,y)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0maybe find the center of mass of that parabolic region you will see whats going on with the bounds..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yes... here is how I'm thinking  if we go at a height \(y\), the bounds of \(x\) depend on the height \(y\) as this is not a simple figure like a circle.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Could you input that double integral to Wolfram? I dunno how to. Also divide it by \(\sqrt{3}/4\) which is the area.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0wolfram understands most latex

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Interesting, that worked. Can I have the closed form though?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0try evaluating it manually \(\sqrt{a^2+x^2}\) has a closed form

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0nooooooooooooooooooooooooooooooooooooo

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0really ugly form, that.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%7Bx%5E2+%2B+y%5E2%7D+dx+

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you may try changing it to polar but then the bounds will get ugly

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Maybe try evaluating just the inner thing?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0notice that the integrand is simply \(r\) in polar form

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0haha look at this http://www.wolframalpha.com/input/?i=%5Cint%5E%7By%2F%5Csqrt%7B3%7D%2B1%2F4+%7D_%7By%2F%5Csqrt%7B3%7D1%2F4+%7D+%5Csqrt%7Bx%5E2+%2B+y%5E2%7Ddx+

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0is there an integrator that returns the closed form? can software like Mathematica do what WolframAlpha wasn't able to?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what if it returns the answers in terms of logs and inverse hyperbolic tangnets ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0The answer is supposed to be in a really simple form.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i think double integral may be a bad idea if you want closed form there must be some geometric method..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0but it's a calculus problem

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0yeah i mean there are geometrical methods in calc too but

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0wolfram is not cooperating, so can't do much with that double integral

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0need to think of somethign else..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0here is how I think of it: let's go back to the sum of distance*fraction there. is there any reason whatsoever that the fraction \(r\) distance away in a circle and the fraction \(r\) distance away in a triangle should be different? may be a dumb question with a dumb answer but we need to think of this in order to go somewhere

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0never mind, it really was dumb
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