## anonymous one year ago finding the curve with minimum length between two points

1. anonymous

and when i tried $ax ^{4}+bx ^{3}+cx ^{2}+$

2. anonymous

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

@Astrophysics looks it requires calculus of variations try this

4. ganeshie8

@zepdrix @freckles

5. Astrophysics

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

Yes where arc length is shorter than 27

7. anonymous

27.3712

8. Astrophysics

No I don't think that makes sense, so what did you have originally as g(x) then?

9. Astrophysics

Because your arc length would be $\int\limits_{10}^{30} \sqrt{1+[g'(x)]^2}dx$

10. anonymous

let me post the question whole solution

11. anonymous

this is the solution for whole question and other part of this question is to get shorter length than 27.3712

12. anonymous

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

Oooh ok I see, hmm interesting

14. ganeshie8
15. ParthKohli

16. ParthKohli

Oh great, looks like you found the fourth-degree polynomial.

17. anonymous

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

18. anonymous

i tried even fraction function but gets too ugly

19. ParthKohli

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.

20. ParthKohli

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

21. ganeshie8
22. anonymous

i did

23. ParthKohli

Yeah! What's the one that you equated with 30?

24. ganeshie8

thats the area under curve, integral..

25. ParthKohli

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.

26. ganeshie8

Yes, very slightly.. to reduce the pain of writing out fractions..

27. ParthKohli

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.

28. ParthKohli
29. ganeshie8

it seems we have missed the constraint that g(x) has to be positive... the curve might go below the x axis

30. ganeshie8
31. ParthKohli

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.

32. ganeshie8

yes

33. ParthKohli

wait, is that function the final answer? it shouldn't be - it should probably resemble a straight line

34. ParthKohli

ok, so does the question say that it's positive for the entirety of its existence?

35. ParthKohli

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.

36. ganeshie8

g(x) is defined only in (10, 30)

37. ParthKohli

Err, so then we don't even need that positive constraint...

38. ganeshie8

somebody in MSE says circle gives the minimum length, try https://www.desmos.com/calculator/6wcuykttko

39. ganeshie8

But I do not see how that circle can give a minimum length https://www.desmos.com/calculator/nbunq0u1m1

40. ParthKohli

yeah! it's got to be something like a straight line ugh!!!

41. ganeshie8

also it doesn't meet the area requirement

42. ganeshie8

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

43. ParthKohli

44. ganeshie8
45. ParthKohli

You come up with pretty... weird usernames...

46. ganeshie8

47. ParthKohli

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.

48. ParthKohli

|dw:1443606482275:dw|

49. ParthKohli

Ohhh, I always seem to forget that constraint about area = 200.

50. ganeshie8

yeah we also need to maintain the area at 200

51. anonymous

|dw:1443606802829:dw|

52. anonymous

i know i might sound stupid but visually is there nay function that has curve on top but afterwards straight line

53. ParthKohli

Yes there is. Whether it is simple to find its closed form - that's a concern.

54. ParthKohli

I mean we can define functions in a piecewise manner... there really is nothing wrong with that.

55. anonymous

can we split the eq like (10,20) then (20,30) straight line

56. anonymous

and we can find the slope at each end point curve for both straight ,so we know if it is smooth

57. anonymous

this is almost surely meant to be solved using calculus of variations like in the MSE answer

58. anonymous

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 $$\frac{dg}{du}=\pm\frac{u}{\sqrt{1-u^2}}\implies g(u)=\pm\sqrt{1-u^2}+C$$

59. anonymous

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: $$\int_{10}^{30}\sqrt{1-(\lambda x+C)^2}\, dx=200$$

60. anonymous

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

61. anonymous

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

62. anonymous

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

63. ParthKohli

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

64. anonymous

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

65. ParthKohli

So most of your current knowledge was there back when you were... seventeen... That kills self-confidence.

66. anonymous

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

67. anonymous

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

68. Astrophysics

I had something similar on paper but ended up with a cycloid X_X

69. Astrophysics

The fact you know how to do this, amazes me @ParthKohli

70. ParthKohli

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.

71. anonymous

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

72. anonymous

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

73. ParthKohli

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)

74. ParthKohli

by "specific" i really mean high-level ones.

75. anonymous

@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

76. anonymous

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

77. anonymous
78. Astrophysics

@oldrin.bataku Yeah I figured as much, I see where I went wrong, thanks!

79. ParthKohli

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.

80. ParthKohli

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

81. anonymous

@Hero

82. anonymous
83. anonymous
84. anonymous

My question is similar to arc length contest thing

85. anonymous

@freckles u got any idea

86. anonymous

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

87. anonymous

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

88. anonymous

@oldrin.bataku tank u so much but can't be a circle

89. anonymous

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)

90. ganeshie8

@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