Question

finding the curve with minimum length between two points

Answer 1

and when i tried \[ax ^{4}+bx ^{3}+cx ^{2}+\]

Answer 2

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

Answer 3

@Astrophysics looks it requires calculus of variations
try this

Answer 4

@zepdrix @freckles

Answer 5

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

Answer 6

Yes where arc length is shorter than 27

Answer 7

27.3712

Answer 8

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

Answer 9

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

Answer 10

let me post the question whole solution

Answer 11

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

Answer 12

have to come up with some function that satisfies all the requirement that ax^2+bx+c satisfies and gives shorter arc length

Answer 13

Oooh ok I see, hmm interesting

Answer 14

http://math.stackexchange.com/questions/1457780/finding-the-smooth-curve-of-minimum-length-between-two-points-with-some-constrai

Answer 15

Hey you made this thread yesterday as well :|

Answer 16

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

Answer 17

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

Answer 18

i tried even fraction function but gets too ugly

Answer 19

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.

Answer 20

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

Answer 21

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

Answer 22

i did

Answer 23

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

Answer 24

thats the area under curve, integral..

Answer 25

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.

Answer 26

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

Answer 27

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.

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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.

Answer 32

yes

Answer 33

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

Answer 34

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

Answer 35

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.

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

also it doesn't meet the area requirement

Answer 42

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

Answer 43

Where is that MSE thread?

Answer 44

http://math.stackexchange.com/questions/1457780/finding-the-smooth-curve-of-minimum-length-between-two-points-with-some-constrai

Answer 45

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

Answer 46

diversifying my usernames haha

Answer 47

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.

Answer 48

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

Answer 49

yeah we also need to maintain the area at 200

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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.

Answer 69

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

Answer 70

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

Answer 71

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)

Answer 72

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

Answer 73

@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

Answer 74

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