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anonymous

  • one year ago

finding the curve with minimum length between two points

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  1. anonymous
    • one year ago
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    and when i tried \[ax ^{4}+bx ^{3}+cx ^{2}+\]

  2. anonymous
    • one year ago
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    so my original arc length is 27.3712 from ax^2+bx+c and i want it shorter length so u sugested use ax^4+bx^3.... and when i did i am getting 28.07

  3. ganeshie8
    • one year ago
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    @Astrophysics looks it requires calculus of variations try this

  4. ganeshie8
    • one year ago
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    @zepdrix @freckles

  5. Astrophysics
    • one year ago
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    So we have to find function g(x) ye? from (10,10) to (30,0) I guess we will have to put a constraint, lets say length l, with g(10) = 10, and g(30) = 0, the length l comes from a differential length along your curve dl

  6. anonymous
    • one year ago
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    Yes where arc length is shorter than 27

  7. anonymous
    • one year ago
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    27.3712

  8. Astrophysics
    • one year ago
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    No I don't think that makes sense, so what did you have originally as g(x) then?

  9. Astrophysics
    • one year ago
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    Because your arc length would be \[\int\limits_{10}^{30} \sqrt{1+[g'(x)]^2}dx\]

  10. anonymous
    • one year ago
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    let me post the question whole solution

  11. anonymous
    • one year ago
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    this is the solution for whole question and other part of this question is to get shorter length than 27.3712

  12. anonymous
    • one year ago
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    have to come up with some function that satisfies all the requirement that ax^2+bx+c satisfies and gives shorter arc length

  13. Astrophysics
    • one year ago
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    Oooh ok I see, hmm interesting

  14. ParthKohli
    • one year ago
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    Hey you made this thread yesterday as well :|

  15. ParthKohli
    • one year ago
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    Oh great, looks like you found the fourth-degree polynomial.

  16. anonymous
    • one year ago
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    yes i did @ParthKohli but ended up getting larger value , still scratching my head over how to get a function that gives me smaller arc length also satisfies all the requirement

  17. anonymous
    • one year ago
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    i tried even fraction function but gets too ugly

  18. ParthKohli
    • one year ago
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    I don't know - you must have done it wrong somewhere. It's not a 4x4 equation as I told you yesterday. What we first do is use four equations in five variables. The solution to that is NOT a bunch of constants, but you should get all variables in terms of one variable. Then you put the value of all variables in terms of that one variable in the arc-length function, and then minimise that value you finally get. I'm going to repeat what I posted yesterday: \[g(x) = ax^4 + bx^3 + cx^2 + dx + e\]\[10^4 a + 10^3 b + 10^2 c + 10 d + e = 10\]\[30^4 a + 30^3 b + 30^2 c + 30 d + e=0\]\[\int_{10}^{30} g(x)dx=200\]\[4a(10)^3 + 3b(10)^2 + 2c(10) + d(10) =1\]Now the last one (I was referring to the arc-length) is tricky.

  19. ParthKohli
    • one year ago
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    @ganeshie8 if you're still there, could you plug that into Wolfram? It apparently needs pro features to get going.

  20. anonymous
    • one year ago
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    i did

  21. ParthKohli
    • one year ago
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    Yeah! What's the one that you equated with 30?

  22. ganeshie8
    • one year ago
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    thats the area under curve, integral..

  23. ParthKohli
    • one year ago
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    Oh, of course it should be. Did you simplify the equation? The right-hand-side in the original one was 200. So I'm going to take it that you did simplify the equation. Hmm.

  24. ganeshie8
    • one year ago
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    Yes, very slightly.. to reduce the pain of writing out fractions..

  25. ParthKohli
    • one year ago
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    Perfect! So Wolfram did return all variables in terms of \(a\). Our function is\[g(x) =ax^4 + bx^3 + cx^2 + dx + e\]So if you plug in the values of each of the coefficients in terms of \(a\) and then tell Wolfram to calculate its arc-length, it should return another huge expression in terms of \(a\) which you then minimise, and there - you have the value of \(a\) and subsequently all other variables that are expressed in terms of \(a\). Thanks Ganeshie.

  26. ganeshie8
    • one year ago
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    it seems we have missed the constraint that g(x) has to be positive... the curve might go below the x axis

  27. ParthKohli
    • one year ago
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    hmm, isn't that constraint only for the region \(x=10\) to \(x=30\)? otherwise this means that if \(30\) is a root, it's gotta have a multiplicity of two, if I'm not wrong.

  28. ganeshie8
    • one year ago
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    yes

  29. ParthKohli
    • one year ago
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    wait, is that function the final answer? it shouldn't be - it should probably resemble a straight line

  30. ParthKohli
    • one year ago
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    ok, so does the question say that it's positive for the entirety of its existence?

  31. ParthKohli
    • one year ago
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    if that's so it makes our work easier\[g(x) = (x-30)^2( ax^2 + bx +c)\]Additionally the quadratic can only have no real roots or only one.

  32. ganeshie8
    • one year ago
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    g(x) is defined only in (10, 30)

  33. ParthKohli
    • one year ago
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    Err, so then we don't even need that positive constraint...

  34. ganeshie8
    • one year ago
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    somebody in MSE says circle gives the minimum length, try https://www.desmos.com/calculator/6wcuykttko

  35. ganeshie8
    • one year ago
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    But I do not see how that circle can give a minimum length https://www.desmos.com/calculator/nbunq0u1m1

  36. ParthKohli
    • one year ago
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    yeah! it's got to be something like a straight line ugh!!!

  37. ganeshie8
    • one year ago
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    also it doesn't meet the area requirement

  38. ganeshie8
    • one year ago
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    because there are only 3 variables to play with : h, k, r we have 4 requirements, so one requirement is always sacrificed if we model the curve with a circle

  39. ParthKohli
    • one year ago
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    Where is that MSE thread?

  40. ParthKohli
    • one year ago
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    You come up with pretty... weird usernames...

  41. ganeshie8
    • one year ago
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    diversifying my usernames haha

  42. ParthKohli
    • one year ago
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    But you do agree with the idea that if we increase the degree of the polynomial, we get closer to the true answer right? And what's a polynomial with an infinite degree? It's now a series. Maybe the ultimate answer is a series of some function that's not elementary. How does the closed form of the answer even matter though? We know what the graph looks like.

  43. ParthKohli
    • one year ago
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    |dw:1443606482275:dw|

  44. ParthKohli
    • one year ago
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    Ohhh, I always seem to forget that constraint about area = 200.

  45. ganeshie8
    • one year ago
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    yeah we also need to maintain the area at 200

  46. anonymous
    • one year ago
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    |dw:1443606802829:dw|

  47. anonymous
    • one year ago
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    i know i might sound stupid but visually is there nay function that has curve on top but afterwards straight line

  48. ParthKohli
    • one year ago
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    Yes there is. Whether it is simple to find its closed form - that's a concern.

  49. ParthKohli
    • one year ago
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    I mean we can define functions in a piecewise manner... there really is nothing wrong with that.

  50. anonymous
    • one year ago
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    can we split the eq like (10,20) then (20,30) straight line

  51. anonymous
    • one year ago
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    and we can find the slope at each end point curve for both straight ,so we know if it is smooth

  52. anonymous
    • one year ago
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    this is almost surely meant to be solved using calculus of variations like in the MSE answer

  53. anonymous
    • one year ago
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    the Lagrangian for arclength action with the constrained area is just (using Lagrange multipliers) $$L(x,g,g')=\sqrt{1+(g')^2}+\lambda g\\\implies \frac{\partial L}{\partial g}=\lambda,\quad\frac{\partial L}{\partial g'}=\frac{g'}{\sqrt{1+(g')^2}}$$ so we get the Euler-Lagrange equation as follows: $$\frac{\partial L}{\partial g}-\frac{d}{dx}\left[\frac{\partial L}{\partial g'}\right]=0\\\implies \lambda x -\frac{g'}{\sqrt{1+(g')^2}}=C\\\implies \frac{g'}{\sqrt{1+(g')^2}}=\lambda x+C\\\implies \frac{(g')^2}{1+(g')^2}=(\lambda x+C)^2\\\implies (1-(\lambda x+C)^2)(g')^2=(\lambda x+C)^2\\\implies (g')^2=\frac{(\lambda x+C)^2}{1-(\lambda x+C)^2}\\\implies\frac{dg}{dx}=\pm\frac{\lambda x+C}{\sqrt{1-(\lambda x+C)^2}}$$ now let \(u=\lambda x+C\) so \(\frac{dg}{dx}=\frac{dg}{du}\cdot\frac{du}{dx}=\lambda\frac{dg}{du}\) giving $$\frac{dg}{du}=\pm\frac{u}{\sqrt{1-u^2}}\implies g(u)=\pm\sqrt{1-u^2}+C$$

  54. anonymous
    • one year ago
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    so using the fact that \(g(10)=10,g(30)=0,g\ge 0\) we can see $$g(x)=\sqrt{1-(\lambda x+C)^2}\\g(10)=10\implies 100=1-(10\lambda+C)^2\\\implies g(30)=0\implies 30\lambda+C=\pm1$$ and now to find \(\lambda\) to ensure we get the right area: $$\int_{10}^{30}\sqrt{1-(\lambda x+C)^2}\, dx=200$$

  55. anonymous
    • one year ago
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    oops, I guess I should've written: $$\frac{dg}{du}=\pm\frac1{\lambda}\frac{u}{\sqrt{1-u^2}}\\\implies g(x)=\frac1{\lambda}\sqrt{1-(\lambda x+C)^2}$$

  56. anonymous
    • one year ago
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    so $$100\lambda^2=1-(10\lambda +C)^2\\30\lambda+C=\pm 1$$ and then we integrate $$\int_{10}^{30}\sqrt{1-(\lambda x+C)^2}\, dx=200$$ so we let \(\lambda x+C=\sin\theta\) so \(\cos d\theta=\lambda dx\) giving $$\frac1{\lambda}\int_\alpha^\beta\cos^2\theta\, d\theta=\frac1{\lambda}\left[\frac12\theta+\frac12\sin(\theta)\cos(\theta)\right]_\alpha^\beta$$ with bounds: $$\sin \alpha= 10\lambda+C\\\sin\beta=30\lambda+C=1\\\implies \cos\beta=0,\quad \beta=\frac\pi2$$ so let's try to simplify $$\frac1{2\lambda}\left[\frac\pi2-\arcsin(10\lambda+C)-(10\lambda+C)\sqrt{1-(10\lambda+C)^2}\right]=200\\\implies \frac\pi2-\arcsin(10\lambda+C)-10\lambda(10\lambda+C)=400\lambda$$ hmm i don't see how this is solvable using elementary tricks

  57. anonymous
    • one year ago
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    basically we plug everything back into the constraints to figure out what \(\lambda,C\) must be but it is correct that the solution is a circular arc

  58. ParthKohli
    • one year ago
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    Nineteen? Aww, come on. That's only three older than me. What does one have to do to, uh, be you?

  59. anonymous
    • one year ago
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    well, to learn how to solve this type of problem just look up anything involving calculus of variations and integral constraints, e.g. http://liberzon.csl.illinois.edu/teaching/cvoc/node38.html I just read a lot of ebooks and course notes I find online because I enjoy figuring out the math, but I haven't learned that much over the past couple years because of school work

  60. ParthKohli
    • one year ago
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    So most of your current knowledge was there back when you were... seventeen... That kills self-confidence.

  61. anonymous
    • one year ago
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    I don't study math in school, I'm an undergraduate in a biomedical and computer engineering program so I just read about math independently for fun

  62. anonymous
    • one year ago
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    I don't know that much math -- you should check out ##math on freenode and math.SE/mathoverflow if you want your self-confidence to die :p I doubt I know very much if any more math than you do, and I imagine I knew less than you when I was 16

  63. Astrophysics
    • one year ago
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    I had something similar on paper but ended up with a cycloid X_X

  64. Astrophysics
    • one year ago
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    The fact you know how to do this, amazes me @ParthKohli

  65. ParthKohli
    • one year ago
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    I don't - I was just solving for polynomials because that's what the OP wanted to do. Arcs of circles never really hit me.

  66. anonymous
    • one year ago
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    well, the way this problem is phrased makes it clear they want a solution via calculus of variations, and you just have to figure out how to incorporate the integral constraint into our Lagrangian -- luckily, Lagrange already figured that out with his method of multipliers

  67. anonymous
    • one year ago
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    here's another good set of course notes @ParthKohli https://www.math.ucdavis.edu/~hunter/m280_09/title.pdf

  68. ParthKohli
    • one year ago
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    those are very specific ones. do you remember any that you used when you were around my age? I could probably use them to brush my basics up. also I'd suppose that you're equally good at physics and chemistry (at least at my level)

  69. ParthKohli
    • one year ago
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    by "specific" i really mean high-level ones.

  70. anonymous
    • one year ago
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    @Astrophysics the idea behind Lagrangian multipliers here is that we want to incorporate our constraint into our Lagrangian so that deviating from the constraint has an intrinsic 'penalty' that we optimize against: $$\text{minimize }\int_{10}^{30}\sqrt{1+(g')^2}\, dx\text{ subject to }\int_{10}^{30} g\, dx=200\\\implies \text{minimize }\int_{10}^{30}\sqrt{1+(g')^2}\, dx+\lambda\left(\int_{10}^{30} g\, dx-200\right)$$and that is equivalent to minimizing $$\int_{10}^{30}\left(\sqrt{1+(g')^2}+\lambda g\right)\, dx-200\lambda$$now the constant is irrelevant, so throw it away and we get the Lagrangian that I used. so you can see here that if we violate our constraint, we get penalized; \(\lambda\) controls how badly this penalizes these candidate solutions, which is why in economics it's called a 'shadow price' -- the cost of breaking our constraint rules to some 'unit' degree

  71. anonymous
    • one year ago
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    they might be specific, maybe try a softer introduction first and then try reading through those? I often find material out of my reach but that just gives you a goal to work towards, idk. there are usually good notes/explanations accessible on places like math.SE, physics.SE, and in public course notes, which is what I usually look up but I'll link you to a brief exposition of calculus of variations taught in a university-level math-for-physicists class

  72. anonymous
    • one year ago
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    http://www.physics.miami.edu/~nearing/mathmethods/variational.pdf

  73. Astrophysics
    • one year ago
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    @oldrin.bataku Yeah I figured as much, I see where I went wrong, thanks!

  74. ParthKohli
    • one year ago
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    noooo, i'm not even clear with lots of math which comes before that. think high-school level stuff, but more towards the olympiad spectrum. i know you're very good at that.

  75. ParthKohli
    • one year ago
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    oh wow i'm reading that article and i do understand much of the content. the link with physics makes things interesting. thanks!

  76. anonymous
    • one year ago
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    @Hero

  77. anonymous
    • one year ago
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    My question is similar to arc length contest thing

  78. anonymous
    • one year ago
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    @freckles u got any idea

  79. anonymous
    • one year ago
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    well the solution looks like $$g(x)=\frac1A\sqrt{1-(Ax+B)^2}+C$$for some constants \(A,B,C\) we have to figure out so that it meets these constraints: $$g(10)=10\\\implies A^2(10-C)^2=1-(10A+B)^2\\g(30)=0\\\implies (30A+B)^2=1\\\int_{10}^{30}\left(\frac1A\sqrt{1-(Ax+B)^2}+C\right)\, dx=200\\\implies \dots$$ the system of equations you get is very nonlinear, though, so we might have to just solve numerically

  80. anonymous
    • one year ago
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    i dropped the C prematurely before, but the point is that the curve \(g\) is constrained to be an arc of some circle passing through the points

  81. anonymous
    • one year ago
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    @oldrin.bataku tank u so much but can't be a circle

  82. anonymous
    • one year ago
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    it is an arc of a circle, and to figure out precisely the expression you need to solve the system I gave for \(A,B,C\), but I suspect it cannot be done analytically (i.e. you need approximate numerical methods)

  83. ganeshie8
    • one year ago
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    @oldrin.bataku I think, it cannot be arc of any circle because there is only one circle that passes through (10, 10) and (30, 0) such that the slope of tangent line at (10,10) is 1. And this circle doesn't meet the area requirement. https://www.desmos.com/calculator/nbunq0u1m1

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