ParthKohli
  • ParthKohli
Interesting calc problem - please check my work.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
What is the average distance randomly selected in a unit equilateral triangle to its center? Here's what I think...
ParthKohli
  • ParthKohli
So 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
  • ParthKohli
|dw:1443350101709:dw|

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ParthKohli
  • ParthKohli
Here, the average distance turns out to be \(2\cdot 2/3 + 4\cdot 1/3 = 8/3\) which makes sense.
ParthKohli
  • ParthKohli
Now 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
  • ganeshie8
By unit equilateral triangle do you mean the side length is 1 ?
ParthKohli
  • ParthKohli
Yes.
ganeshie8
  • ganeshie8
Then simply setup a double integral for the distance expression
ParthKohli
  • ParthKohli
How?!
ParthKohli
  • ParthKohli
I 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
  • ganeshie8
|dw:1443350378977:dw|
ganeshie8
  • ganeshie8
\[\text{Average distance} = \dfrac{1}{A}\iint \sqrt{x^2+y^2}dA\]
ParthKohli
  • ParthKohli
That 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
  • ganeshie8
Right, 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
  • ganeshie8
Notice that the average velcity can be expressed using integral as : \[\text{Average velocity over (a,b)} = \dfrac{s(b)-s(a)}{b-a} = \dfrac{1}{b-a} \int\limits_{a}^b v(t)\,dt\]
ganeshie8
  • ganeshie8
\(s\) is the displacement function
ParthKohli
  • ParthKohli
Yeah, I can understand single-variable 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
  • ganeshie8
similarly, 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
  • ganeshie8
Yes In single integral, \(dx\) refers to a small width element. In double integral, \(dA\) refers to a small area element.
ParthKohli
  • ParthKohli
Can 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
  • ganeshie8
depends 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
  • ParthKohli
So it's not a circular element. That makes sense because within the circle, the distance will kinda change.
ganeshie8
  • ganeshie8
don'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
  • ParthKohli
Help me through this: it means that if I go \(dx\) units around the point in the x-direction and \(dy\) units around the point in the y-direction, it doesn't change the \(\sqrt{x^2+y^2 }\) term - which is why the area element is \(dA = dx dy\).
anonymous
  • anonymous
yes it does
ParthKohli
  • ParthKohli
Great. 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
  • ganeshie8
Maybe, lets take this example problem. Find the center of mass of below 2 dimensional region : |dw:1443351868292:dw|
ParthKohli
  • ParthKohli
First, 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
  • ParthKohli
That's great - you understand my love for physics.\[x_{CM} = \frac{\displaystyle \int x dm}{ M}\]
ParthKohli
  • ParthKohli
Area of the whole region: \(\displaystyle \int x^2dx/4 \) from 0 to ???
ganeshie8
  • ganeshie8
Mass = density*Area lets just assume that the object shown in above diagram has uniform density of 1
ParthKohli
  • ParthKohli
Wow I definitely know how to solve this problem - just tell me the upper limit.
ganeshie8
  • ganeshie8
Yes you definitely knew it I took the easy one because I wanted to show you how to set it up using double integrals
ganeshie8
  • ganeshie8
finding the coordinates of center of mass is not the goal setting up the integrals correctly is the goal here
ParthKohli
  • ParthKohli
OK. 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 x-direction, 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
  • ParthKohli
So 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
  • ganeshie8
|dw:1443352602540:dw|
ganeshie8
  • ganeshie8
I'm not looking at the original problem yet
ParthKohli
  • ParthKohli
\[\int ^{+\sqrt{3}/4}_{-\sqrt{3}/4}\int^{-y/\sqrt{3}+1/4 }_{y/\sqrt{3}-1/4 } \sqrt{x^2 + y^2}dx dy\]
ganeshie8
  • ganeshie8
Not so fast, do you see why the double integral gives the averge value ?
ParthKohli
  • ParthKohli
Yeah, we've covered a lot of center of mass and moment of inertia in the class. :)
ganeshie8
  • ganeshie8
but you said earlier that you don't know about double integrals
ganeshie8
  • ganeshie8
So I'm confused how you're able to setup that double integral in two minutes hmm
IrishBoy123
  • IrishBoy123
.
ParthKohli
  • ParthKohli
I realised that the inner integral is really just a function of the outer integral.
ganeshie8
  • ganeshie8
Hmm that doesn't make sense to me but its okay, if you're good, im good.
ganeshie8
  • ganeshie8
just evaluate the integral and we're done! there isn't much to it
ParthKohli
  • ParthKohli
Just 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
  • ParthKohli
|dw:1443353241031:dw|
ParthKohli
  • ParthKohli
oops
ParthKohli
  • ParthKohli
the limits on the outer integral are not functions of x... just meant to ask about those on the inner integral
ganeshie8
  • ganeshie8
Yes, but again, I don't really know what exactly you're asking... I feel you're rushing
ParthKohli
  • ParthKohli
So 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 single-integral.
ganeshie8
  • ganeshie8
the bounds in a double integral are the "inequalities" that represent the "region" over which you integrate the function, f(x,y)
ganeshie8
  • ganeshie8
maybe find the center of mass of that parabolic region you will see whats going on with the bounds..
ParthKohli
  • ParthKohli
Yes... 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
  • ParthKohli
Could you input that double integral to Wolfram? I dunno how to. Also divide it by \(\sqrt{3}/4\) which is the area.
ganeshie8
  • ganeshie8
wolfram understands most latex
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=%5Cint%5E%7B%2B%5Csqrt%7B3%7D%2F4%7D_%7B-%5Csqrt%7B3%7D%2F4%7D%5Cint%5E%7B-y%2F%5Csqrt%7B3%7D%2B1%2F4+%7D_%7By%2F%5Csqrt%7B3%7D-1%2F4+%7D+%5Csqrt%7Bx%5E2+%2B+y%5E2%7D+dx+dy
ParthKohli
  • ParthKohli
Interesting, that worked. Can I have the closed form though?
ganeshie8
  • ganeshie8
try evaluating it manually \(\sqrt{a^2+x^2}\) has a closed form
ParthKohli
  • ParthKohli
nooooooooooooooooooooooooooooooooooooo
ParthKohli
  • ParthKohli
really ugly form, that.
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%7Bx%5E2+%2B+y%5E2%7D+dx+
ganeshie8
  • ganeshie8
you may try changing it to polar but then the bounds will get ugly
ParthKohli
  • ParthKohli
Maybe try evaluating just the inner thing?
ganeshie8
  • ganeshie8
notice that the integrand is simply \(r\) in polar form
ParthKohli
  • ParthKohli
haha look at this http://www.wolframalpha.com/input/?i=%5Cint%5E%7B-y%2F%5Csqrt%7B3%7D%2B1%2F4+%7D_%7By%2F%5Csqrt%7B3%7D-1%2F4+%7D+%5Csqrt%7Bx%5E2+%2B+y%5E2%7Ddx+
ParthKohli
  • ParthKohli
is there an integrator that returns the closed form? can software like Mathematica do what WolframAlpha wasn't able to?
ganeshie8
  • ganeshie8
what if it returns the answers in terms of logs and inverse hyperbolic tangnets ?
ParthKohli
  • ParthKohli
The answer is supposed to be in a really simple form.
ganeshie8
  • ganeshie8
i think double integral may be a bad idea if you want closed form there must be some geometric method..
ParthKohli
  • ParthKohli
:(
ParthKohli
  • ParthKohli
but it's a calculus problem
ParthKohli
  • ParthKohli
yeah i mean there are geometrical methods in calc too but
ganeshie8
  • ganeshie8
wolfram is not cooperating, so can't do much with that double integral
ganeshie8
  • ganeshie8
need to think of somethign else..
ParthKohli
  • ParthKohli
here 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
  • ParthKohli
never mind, it really was dumb

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