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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and when i tried \[ax ^{4}+bx ^{3}+cx ^{2}+\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.1@Astrophysics looks it requires calculus of variations try this

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes where arc length is shorter than 27

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0No I don't think that makes sense, so what did you have originally as g(x) then?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Because your arc length would be \[\int\limits_{10}^{30} \sqrt{1+[g'(x)]^2}dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me post the question whole solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is the solution for whole question and other part of this question is to get shorter length than 27.3712

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0have to come up with some function that satisfies all the requirement that ax^2+bx+c satisfies and gives shorter arc length

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Oooh ok I see, hmm interesting

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Hey you made this thread yesterday as well :

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Oh great, looks like you found the fourthdegree polynomial.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 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
 one year ago
Best ResponseYou've already chosen the best response.0i tried even fraction function but gets too ugly

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I 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 arclength 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 arclength) is tricky.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 if you're still there, could you plug that into Wolfram? It apparently needs pro features to get going.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yeah! What's the one that you equated with 30?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1thats the area under curve, integral..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Oh, of course it should be. Did you simplify the equation? The righthandside in the original one was 200. So I'm going to take it that you did simplify the equation. Hmm.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes, very slightly.. to reduce the pain of writing out fractions..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Perfect! 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 arclength, 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1it seems we have missed the constraint that g(x) has to be positive... the curve might go below the x axis

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1hmm, 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1wait, is that function the final answer? it shouldn't be  it should probably resemble a straight line

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1ok, so does the question say that it's positive for the entirety of its existence?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1if that's so it makes our work easier\[g(x) = (x30)^2( ax^2 + bx +c)\]Additionally the quadratic can only have no real roots or only one.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1g(x) is defined only in (10, 30)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Err, so then we don't even need that positive constraint...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1somebody in MSE says circle gives the minimum length, try https://www.desmos.com/calculator/6wcuykttko

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1But I do not see how that circle can give a minimum length https://www.desmos.com/calculator/nbunq0u1m1

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yeah! it's got to be something like a straight line ugh!!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1also it doesn't meet the area requirement

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1because 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
 one year ago
Best ResponseYou've already chosen the best response.1Where is that MSE thread?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1You come up with pretty... weird usernames...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1diversifying my usernames haha

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1But 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443606482275:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Ohhh, I always seem to forget that constraint about area = 200.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah we also need to maintain the area at 200

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443606802829:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know i might sound stupid but visually is there nay function that has curve on top but afterwards straight line

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes there is. Whether it is simple to find its closed form  that's a concern.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I mean we can define functions in a piecewise manner... there really is nothing wrong with that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can we split the eq like (10,20) then (20,30) straight line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and we can find the slope at each end point curve for both straight ,so we know if it is smooth

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is almost surely meant to be solved using calculus of variations like in the MSE answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 EulerLagrange 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 $$\frac{dg}{du}=\pm\frac{u}{\sqrt{1u^2}}\implies g(u)=\pm\sqrt{1u^2}+C$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0oops, I guess I should've written: $$\frac{dg}{du}=\pm\frac1{\lambda}\frac{u}{\sqrt{1u^2}}\\\implies g(x)=\frac1{\lambda}\sqrt{1(\lambda x+C)^2}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$100\lambda^2=1(10\lambda +C)^2\\30\lambda+C=\pm 1$$ 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
 one year ago
Best ResponseYou've already chosen the best response.0basically 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
 one year ago
Best ResponseYou've already chosen the best response.1Nineteen? Aww, come on. That's only three older than me. What does one have to do to, uh, be you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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
 one year ago
Best ResponseYou've already chosen the best response.1So most of your current knowledge was there back when you were... seventeen... That kills selfconfidence.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0I don't know that much math  you should check out ##math on freenode and math.SE/mathoverflow if you want your selfconfidence 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
 one year ago
Best ResponseYou've already chosen the best response.0I had something similar on paper but ended up with a cycloid X_X

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0The fact you know how to do this, amazes me @ParthKohli

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0well, 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
 one year ago
Best ResponseYou've already chosen the best response.0here's another good set of course notes @ParthKohli https://www.math.ucdavis.edu/~hunter/m280_09/title.pdf

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1those 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
 one year ago
Best ResponseYou've already chosen the best response.1by "specific" i really mean highlevel ones.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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: $$\text{minimize }\int_{10}^{30}\sqrt{1+(g')^2}\, dx\text{ subject to }\int_{10}^{30} g\, dx=200\\\implies \text{minimize }\int_{10}^{30}\sqrt{1+(g')^2}\, dx+\lambda\left(\int_{10}^{30} g\, dx200\right)$$and that is equivalent to minimizing $$\int_{10}^{30}\left(\sqrt{1+(g')^2}+\lambda g\right)\, dx200\lambda$$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
 one year ago
Best ResponseYou've already chosen the best response.0they 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 universitylevel mathforphysicists class

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.physics.miami.edu/~nearing/mathmethods/variational.pdf

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku Yeah I figured as much, I see where I went wrong, thanks!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1noooo, i'm not even clear with lots of math which comes before that. think highschool level stuff, but more towards the olympiad spectrum. i know you're very good at that.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1oh wow i'm reading that article and i do understand much of the content. the link with physics makes things interesting. thanks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My question is similar to arc length contest thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles u got any idea

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well 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: $$g(10)=10\\\implies A^2(10C)^2=1(10A+B)^2\\g(30)=0\\\implies (30A+B)^2=1\\\int_{10}^{30}\left(\frac1A\sqrt{1(Ax+B)^2}+C\right)\, dx=200\\\implies \dots$$ the system of equations you get is very nonlinear, though, so we might have to just solve numerically

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku tank u so much but can't be a circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it 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
 one year ago
Best ResponseYou've already chosen the best response.1@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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