anonymous
  • anonymous
finding the curve with minimum length between two points
Mathematics
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anonymous
  • anonymous
finding the curve with minimum length between two points
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
and when i tried \[ax ^{4}+bx ^{3}+cx ^{2}+\]
anonymous
  • 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
ganeshie8
  • ganeshie8
@Astrophysics looks it requires calculus of variations try this

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ganeshie8
  • ganeshie8
Astrophysics
  • 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
anonymous
  • anonymous
Yes where arc length is shorter than 27
anonymous
  • anonymous
27.3712
Astrophysics
  • Astrophysics
No I don't think that makes sense, so what did you have originally as g(x) then?
Astrophysics
  • Astrophysics
Because your arc length would be \[\int\limits_{10}^{30} \sqrt{1+[g'(x)]^2}dx\]
anonymous
  • anonymous
let me post the question whole solution
anonymous
  • anonymous
this is the solution for whole question and other part of this question is to get shorter length than 27.3712
anonymous
  • anonymous
have to come up with some function that satisfies all the requirement that ax^2+bx+c satisfies and gives shorter arc length
Astrophysics
  • Astrophysics
Oooh ok I see, hmm interesting
ganeshie8
  • ganeshie8
http://math.stackexchange.com/questions/1457780/finding-the-smooth-curve-of-minimum-length-between-two-points-with-some-constrai
ParthKohli
  • ParthKohli
Hey you made this thread yesterday as well :|
ParthKohli
  • ParthKohli
Oh great, looks like you found the fourth-degree polynomial.
anonymous
  • 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
anonymous
  • anonymous
i tried even fraction function but gets too ugly
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
@ganeshie8 if you're still there, could you plug that into Wolfram? It apparently needs pro features to get going.
ganeshie8
  • ganeshie8
like this ? http://www.wolframalpha.com/input/?i=solve+a*10%5E4%2Bb*10%5E3%2Bc*10%5E2%2Bd*10%2Be%3D10%2C+a*30%5E4%2Bb*30%5E3%2Bc*30%5E2%2Bd*30%2Be%3D0%2C++a*4*10%5E3%2Bb*3*10%5E2%2Bc*2*10%2Bd%3D1%2C726000*a%2B30000*b%2B1300*c%2B60*d%2B3*e%3D30
anonymous
  • anonymous
i did
ParthKohli
  • ParthKohli
Yeah! What's the one that you equated with 30?
ganeshie8
  • ganeshie8
thats the area under curve, integral..
ParthKohli
  • 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.
ganeshie8
  • ganeshie8
Yes, very slightly.. to reduce the pain of writing out fractions..
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
uh oh http://www.wolframalpha.com/input/?i=arc+length+x%3D10+to+x%3D30+of+ax%5E4%2B%283%2F200-88a%29x%5E3%2B%282760a-39%2F40%29x%5E2%2B%2819-36000a%29x%2B%28162000a-195%2F2%29
ganeshie8
  • ganeshie8
it seems we have missed the constraint that g(x) has to be positive... the curve might go below the x axis
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=arc+length+g%28x%29%3D-x%5E4%2B72x%5E3-72003%2F40x%5E2%2B36805%2F2x-132015%2F2%2C+10..30
ParthKohli
  • 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.
ganeshie8
  • ganeshie8
yes
ParthKohli
  • ParthKohli
wait, is that function the final answer? it shouldn't be - it should probably resemble a straight line
ParthKohli
  • ParthKohli
ok, so does the question say that it's positive for the entirety of its existence?
ParthKohli
  • 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.
ganeshie8
  • ganeshie8
g(x) is defined only in (10, 30)
ParthKohli
  • ParthKohli
Err, so then we don't even need that positive constraint...
ganeshie8
  • ganeshie8
somebody in MSE says circle gives the minimum length, try https://www.desmos.com/calculator/6wcuykttko
ganeshie8
  • ganeshie8
But I do not see how that circle can give a minimum length https://www.desmos.com/calculator/nbunq0u1m1
ParthKohli
  • ParthKohli
yeah! it's got to be something like a straight line ugh!!!
ganeshie8
  • ganeshie8
also it doesn't meet the area requirement
ganeshie8
  • 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
ParthKohli
  • ParthKohli
Where is that MSE thread?
ganeshie8
  • ganeshie8
http://math.stackexchange.com/questions/1457780/finding-the-smooth-curve-of-minimum-length-between-two-points-with-some-constrai
ParthKohli
  • ParthKohli
You come up with pretty... weird usernames...
ganeshie8
  • ganeshie8
diversifying my usernames haha
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
|dw:1443606482275:dw|
ParthKohli
  • ParthKohli
Ohhh, I always seem to forget that constraint about area = 200.
ganeshie8
  • ganeshie8
yeah we also need to maintain the area at 200
anonymous
  • anonymous
|dw:1443606802829:dw|
anonymous
  • anonymous
i know i might sound stupid but visually is there nay function that has curve on top but afterwards straight line
ParthKohli
  • ParthKohli
Yes there is. Whether it is simple to find its closed form - that's a concern.
ParthKohli
  • ParthKohli
I mean we can define functions in a piecewise manner... there really is nothing wrong with that.
anonymous
  • anonymous
can we split the eq like (10,20) then (20,30) straight line
anonymous
  • anonymous
and we can find the slope at each end point curve for both straight ,so we know if it is smooth
anonymous
  • anonymous
this is almost surely meant to be solved using calculus of variations like in the MSE answer
anonymous
  • 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 $$\frac{dg}{du}=\pm\frac{u}{\sqrt{1-u^2}}\implies g(u)=\pm\sqrt{1-u^2}+C$$
anonymous
  • 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: $$\int_{10}^{30}\sqrt{1-(\lambda x+C)^2}\, dx=200$$
anonymous
  • anonymous
oops, I guess I should've written: $$\frac{dg}{du}=\pm\frac1{\lambda}\frac{u}{\sqrt{1-u^2}}\\\implies g(x)=\frac1{\lambda}\sqrt{1-(\lambda x+C)^2}$$
anonymous
  • 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
anonymous
  • 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
ParthKohli
  • ParthKohli
Nineteen? Aww, come on. That's only three older than me. What does one have to do to, uh, be you?
anonymous
  • 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
ParthKohli
  • ParthKohli
So most of your current knowledge was there back when you were... seventeen... That kills self-confidence.
anonymous
  • 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
anonymous
  • 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
Astrophysics
  • Astrophysics
I had something similar on paper but ended up with a cycloid X_X
Astrophysics
  • Astrophysics
The fact you know how to do this, amazes me @ParthKohli
ParthKohli
  • 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.
anonymous
  • 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
anonymous
  • anonymous
here's another good set of course notes @ParthKohli https://www.math.ucdavis.edu/~hunter/m280_09/title.pdf
ParthKohli
  • 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)
ParthKohli
  • ParthKohli
by "specific" i really mean high-level ones.
anonymous
  • 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
anonymous
  • 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
anonymous
  • anonymous
http://www.physics.miami.edu/~nearing/mathmethods/variational.pdf
Astrophysics
  • Astrophysics
@oldrin.bataku Yeah I figured as much, I see where I went wrong, thanks!
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
oh wow i'm reading that article and i do understand much of the content. the link with physics makes things interesting. thanks!
anonymous
  • anonymous
anonymous
  • anonymous
http://math.stackexchange.com/questions/796409/shortest-distance-between-two-points-calculus-of-variation?rq=1
anonymous
  • anonymous
http://math.stackexchange.com/questions/1122929/arc-length-contest-minimize-the-arc-length-of-fx-when-given-three-condition
anonymous
  • anonymous
My question is similar to arc length contest thing
anonymous
  • anonymous
@freckles u got any idea
anonymous
  • 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
anonymous
  • 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
anonymous
  • anonymous
@oldrin.bataku tank u so much but can't be a circle
anonymous
  • 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)
ganeshie8
  • 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

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