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ParthKohli

  • one year ago

Interesting calc problem - please check my work.

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  1. ParthKohli
    • one year ago
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    What is the average distance randomly selected in a unit equilateral triangle to its center? Here's what I think...

  2. ParthKohli
    • one year ago
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    So if we instead think of discrete points and try to think of the average distance, we'd think of something like this:\[\sum \rm distance \cdot (fraction ~ of ~ points ~ that ~ are ~ that ~ distance ~ away)\]

  3. ParthKohli
    • one year ago
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    |dw:1443350101709:dw|

  4. ParthKohli
    • one year ago
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    Here, the average distance turns out to be \(2\cdot 2/3 + 4\cdot 1/3 = 8/3\) which makes sense.

  5. ParthKohli
    • one year ago
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    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.

  6. ganeshie8
    • one year ago
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    By unit equilateral triangle do you mean the side length is 1 ?

  7. ParthKohli
    • one year ago
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    Yes.

  8. ganeshie8
    • one year ago
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    Then simply setup a double integral for the distance expression

  9. ParthKohli
    • one year ago
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    How?!

  10. ParthKohli
    • one year ago
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    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.

  11. ganeshie8
    • one year ago
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    |dw:1443350378977:dw|

  12. ganeshie8
    • one year ago
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    \[\text{Average distance} = \dfrac{1}{A}\iint \sqrt{x^2+y^2}dA\]

  13. ParthKohli
    • one year ago
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    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.

  14. ganeshie8
    • one year ago
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    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

  15. ganeshie8
    • one year ago
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    Notice that the average velcity can be expressed using integral as : \[\text{Average velocity over (a,b)} = \dfrac{s(b)-s(a)}{b-a} = \dfrac{1}{b-a} \int\limits_{a}^b v(t)\,dt\]

  16. ganeshie8
    • one year ago
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    \(s\) is the displacement function

  17. ParthKohli
    • one year ago
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    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?

  18. ganeshie8
    • one year ago
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    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)

  19. ganeshie8
    • one year ago
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    Yes In single integral, \(dx\) refers to a small width element. In double integral, \(dA\) refers to a small area element.

  20. ParthKohli
    • one year ago
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    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.

  21. ganeshie8
    • one year ago
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    depends on the coordinate system, if you're using rectangular coordinates, then \(dA = dxdy\) if you're using polar coordinates, then \(dA = r*drd\theta\)

  22. ParthKohli
    • one year ago
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    So it's not a circular element. That makes sense because within the circle, the distance will kinda change.

  23. ganeshie8
    • one year ago
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    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..

  24. ParthKohli
    • one year ago
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    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 \(dA = dx dy\).

  25. anonymous
    • one year ago
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    yes it does

  26. ParthKohli
    • one year ago
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    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.

  27. ganeshie8
    • one year ago
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    Maybe, lets take this example problem. Find the center of mass of below 2 dimensional region : |dw:1443351868292:dw|

  28. ParthKohli
    • one year ago
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    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 \(-y/\sqrt{3} + 1/4 \).

  29. ParthKohli
    • one year ago
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    That's great - you understand my love for physics.\[x_{CM} = \frac{\displaystyle \int x dm}{ M}\]

  30. ParthKohli
    • one year ago
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    Area of the whole region: \(\displaystyle \int x^2dx/4 \) from 0 to ???

  31. ganeshie8
    • one year ago
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    Mass = density*Area lets just assume that the object shown in above diagram has uniform density of 1

  32. ParthKohli
    • one year ago
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    Wow I definitely know how to solve this problem - just tell me the upper limit.

  33. ganeshie8
    • one year ago
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    Yes you definitely knew it I took the easy one because I wanted to show you how to set it up using double integrals

  34. ganeshie8
    • one year ago
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    finding the coordinates of center of mass is not the goal setting up the integrals correctly is the goal here

  35. ParthKohli
    • one year ago
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    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).

  36. ParthKohli
    • one year ago
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    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.

  37. ganeshie8
    • one year ago
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    |dw:1443352602540:dw|

  38. ganeshie8
    • one year ago
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    I'm not looking at the original problem yet

  39. ParthKohli
    • one year ago
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    \[\int ^{+\sqrt{3}/4}_{-\sqrt{3}/4}\int^{-y/\sqrt{3}+1/4 }_{y/\sqrt{3}-1/4 } \sqrt{x^2 + y^2}dx dy\]

  40. ganeshie8
    • one year ago
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    Not so fast, do you see why the double integral gives the averge value ?

  41. ParthKohli
    • one year ago
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    Yeah, we've covered a lot of center of mass and moment of inertia in the class. :)

  42. ganeshie8
    • one year ago
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    but you said earlier that you don't know about double integrals

  43. ganeshie8
    • one year ago
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    So I'm confused how you're able to setup that double integral in two minutes hmm

  44. IrishBoy123
    • one year ago
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    .

  45. ParthKohli
    • one year ago
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    I realised that the inner integral is really just a function of the outer integral.

  46. ganeshie8
    • one year ago
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    Hmm that doesn't make sense to me but its okay, if you're good, im good.

  47. ganeshie8
    • one year ago
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    just evaluate the integral and we're done! there isn't much to it

  48. ParthKohli
    • one year ago
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    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?

  49. ParthKohli
    • one year ago
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    |dw:1443353241031:dw|

  50. ParthKohli
    • one year ago
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    oops

  51. ParthKohli
    • one year ago
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    the limits on the outer integral are not functions of x... just meant to ask about those on the inner integral

  52. ganeshie8
    • one year ago
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    Yes, but again, I don't really know what exactly you're asking... I feel you're rushing

  53. ParthKohli
    • one year ago
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    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.

  54. ganeshie8
    • one year ago
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    the bounds in a double integral are the "inequalities" that represent the "region" over which you integrate the function, f(x,y)

  55. ganeshie8
    • one year ago
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    maybe find the center of mass of that parabolic region you will see whats going on with the bounds..

  56. ParthKohli
    • one year ago
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    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.

  57. ParthKohli
    • one year ago
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    Could you input that double integral to Wolfram? I dunno how to. Also divide it by \(\sqrt{3}/4\) which is the area.

  58. ganeshie8
    • one year ago
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    wolfram understands most latex

  59. ParthKohli
    • one year ago
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    Interesting, that worked. Can I have the closed form though?

  60. ganeshie8
    • one year ago
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    try evaluating it manually \(\sqrt{a^2+x^2}\) has a closed form

  61. ParthKohli
    • one year ago
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    nooooooooooooooooooooooooooooooooooooo

  62. ParthKohli
    • one year ago
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    really ugly form, that.

  63. ganeshie8
    • one year ago
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    http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%7Bx%5E2+%2B+y%5E2%7D+dx+

  64. ganeshie8
    • one year ago
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    you may try changing it to polar but then the bounds will get ugly

  65. ParthKohli
    • one year ago
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    Maybe try evaluating just the inner thing?

  66. ganeshie8
    • one year ago
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    notice that the integrand is simply \(r\) in polar form

  67. ParthKohli
    • one year ago
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    is there an integrator that returns the closed form? can software like Mathematica do what WolframAlpha wasn't able to?

  68. ganeshie8
    • one year ago
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    what if it returns the answers in terms of logs and inverse hyperbolic tangnets ?

  69. ParthKohli
    • one year ago
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    The answer is supposed to be in a really simple form.

  70. ganeshie8
    • one year ago
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    i think double integral may be a bad idea if you want closed form there must be some geometric method..

  71. ParthKohli
    • one year ago
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    :(

  72. ParthKohli
    • one year ago
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    but it's a calculus problem

  73. ParthKohli
    • one year ago
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    yeah i mean there are geometrical methods in calc too but

  74. ganeshie8
    • one year ago
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    wolfram is not cooperating, so can't do much with that double integral

  75. ganeshie8
    • one year ago
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    need to think of somethign else..

  76. ParthKohli
    • one year ago
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    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

  77. ParthKohli
    • one year ago
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    never mind, it really was dumb

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