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How can I tell if an arc length integral represents a length or a functional?
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I have been trying to understand functionals in the sense of variational calculus and there was an example of finding the shortest path between two points.
The arc length integral which is easy to derive measures the distance between those stationary points let us say x1 and x2. The functional is a function where the input is another function.
The solution to this functional is the Euler Lagrange equation which I assume is the differential equation that solves for the arc length integrals that are stationary . ( a min or a max I assume )
So how do we know the arc length integral is the functional that the Euler Lagrange equation is solving or just another instance of one particular length ?
P.S. Maybe I am not understanding the definition of a functional ....it seems to me it's like a composite function but the second function just happens to be an integral that you feed the first function into so they found a new name, functional. Where as in composite function no integral is involved but it's still the same business, a function gets fed into anther.
multivariable-calculus classical-mechanics
asked Jul 20 at 23:33
Sedumjoy
617314
add a comment |Â
up vote
1
down vote
favorite
I have been trying to understand functionals in the sense of variational calculus and there was an example of finding the shortest path between two points.
The arc length integral which is easy to derive measures the distance between those stationary points let us say x1 and x2. The functional is a function where the input is another function.
The solution to this functional is the Euler Lagrange equation which I assume is the differential equation that solves for the arc length integrals that are stationary . ( a min or a max I assume )
So how do we know the arc length integral is the functional that the Euler Lagrange equation is solving or just another instance of one particular length ?
P.S. Maybe I am not understanding the definition of a functional ....it seems to me it's like a composite function but the second function just happens to be an integral that you feed the first function into so they found a new name, functional. Where as in composite function no integral is involved but it's still the same business, a function gets fed into anther.
multivariable-calculus classical-mechanics
asked Jul 20 at 23:33
Sedumjoy
617314
1
I'm not totally sure I understand what you are asking, but the arc length integral $S(f) = int_a^b sqrt1 + f'^2(x)rm dx$ is a functional that takes in a function $f(x)$ (with some fixed boundary conditions $f(a) = c$ and $f(b) = d$) and outputs a number that is the arc-length of this function. The solution to the Euler-Lagrange equation for this functional gives the function that minimizes $S$, i.e. it gives the function that has the shortest length between two points (which is the straight line $f(x) = f(a) + [f(b)-f(a)](x-a)/(b-a)$)
– Winther
Jul 20 at 23:40
oh ...you mean the arc length integrals we study in multivariable calculus are functionals ? I am a little confused ...why don't they call them that in my multivariable calculus book? if what you are saying is true that would solve my confusion
– Sedumjoy
Jul 20 at 23:46
1
It depends on what you are looking at. If you are to compute the arc-length of a given function you don't have to consider the concept of a functional: you simply compute the arc-length integral for this function. The concept of a functional is of interest when you consider problems like "what is the function that has the shortest arc length between two points" (i.e. when the unknown is a function not just a number). Then you would start by putting up the arc-length functional, compute the Euler-Lagrange equations and solve it in order to determine this function.
– Winther
Jul 20 at 23:52
gotcha ...this clears it up for me. I was going in circles ....thanx
– Sedumjoy
Jul 21 at 0:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been trying to understand functionals in the sense of variational calculus and there was an example of finding the shortest path between two points.
The arc length integral which is easy to derive measures the distance between those stationary points let us say x1 and x2. The functional is a function where the input is another function.
The solution to this functional is the Euler Lagrange equation which I assume is the differential equation that solves for the arc length integrals that are stationary . ( a min or a max I assume )
So how do we know the arc length integral is the functional that the Euler Lagrange equation is solving or just another instance of one particular length ?
P.S. Maybe I am not understanding the definition of a functional ....it seems to me it's like a composite function but the second function just happens to be an integral that you feed the first function into so they found a new name, functional. Where as in composite function no integral is involved but it's still the same business, a function gets fed into anther.
multivariable-calculus classical-mechanics
asked Jul 20 at 23:33
Sedumjoy
617314
I have been trying to understand functionals in the sense of variational calculus and there was an example of finding the shortest path between two points.
The arc length integral which is easy to derive measures the distance between those stationary points let us say x1 and x2. The functional is a function where the input is another function.
The solution to this functional is the Euler Lagrange equation which I assume is the differential equation that solves for the arc length integrals that are stationary . ( a min or a max I assume )
So how do we know the arc length integral is the functional that the Euler Lagrange equation is solving or just another instance of one particular length ?
P.S. Maybe I am not understanding the definition of a functional ....it seems to me it's like a composite function but the second function just happens to be an integral that you feed the first function into so they found a new name, functional. Where as in composite function no integral is involved but it's still the same business, a function gets fed into anther.
multivariable-calculus classical-mechanics
asked Jul 20 at 23:33
Sedumjoy
617314
asked Jul 20 at 23:33
Sedumjoy
617314
asked Jul 20 at 23:33
Sedumjoy
617314
asked Jul 20 at 23:33
Sedumjoy
617314
617314
1
I'm not totally sure I understand what you are asking, but the arc length integral $S(f) = int_a^b sqrt1 + f'^2(x)rm dx$ is a functional that takes in a function $f(x)$ (with some fixed boundary conditions $f(a) = c$ and $f(b) = d$) and outputs a number that is the arc-length of this function. The solution to the Euler-Lagrange equation for this functional gives the function that minimizes $S$, i.e. it gives the function that has the shortest length between two points (which is the straight line $f(x) = f(a) + [f(b)-f(a)](x-a)/(b-a)$)
– Winther
Jul 20 at 23:40
oh ...you mean the arc length integrals we study in multivariable calculus are functionals ? I am a little confused ...why don't they call them that in my multivariable calculus book? if what you are saying is true that would solve my confusion
– Sedumjoy
Jul 20 at 23:46
1
It depends on what you are looking at. If you are to compute the arc-length of a given function you don't have to consider the concept of a functional: you simply compute the arc-length integral for this function. The concept of a functional is of interest when you consider problems like "what is the function that has the shortest arc length between two points" (i.e. when the unknown is a function not just a number). Then you would start by putting up the arc-length functional, compute the Euler-Lagrange equations and solve it in order to determine this function.
– Winther
Jul 20 at 23:52
gotcha ...this clears it up for me. I was going in circles ....thanx
– Sedumjoy
Jul 21 at 0:00
add a comment |Â
1
I'm not totally sure I understand what you are asking, but the arc length integral $S(f) = int_a^b sqrt1 + f'^2(x)rm dx$ is a functional that takes in a function $f(x)$ (with some fixed boundary conditions $f(a) = c$ and $f(b) = d$) and outputs a number that is the arc-length of this function. The solution to the Euler-Lagrange equation for this functional gives the function that minimizes $S$, i.e. it gives the function that has the shortest length between two points (which is the straight line $f(x) = f(a) + [f(b)-f(a)](x-a)/(b-a)$)
– Winther
Jul 20 at 23:40
oh ...you mean the arc length integrals we study in multivariable calculus are functionals ? I am a little confused ...why don't they call them that in my multivariable calculus book? if what you are saying is true that would solve my confusion
– Sedumjoy
Jul 20 at 23:46
1
It depends on what you are looking at. If you are to compute the arc-length of a given function you don't have to consider the concept of a functional: you simply compute the arc-length integral for this function. The concept of a functional is of interest when you consider problems like "what is the function that has the shortest arc length between two points" (i.e. when the unknown is a function not just a number). Then you would start by putting up the arc-length functional, compute the Euler-Lagrange equations and solve it in order to determine this function.
– Winther
Jul 20 at 23:52
gotcha ...this clears it up for me. I was going in circles ....thanx
– Sedumjoy
Jul 21 at 0:00
1
1
I'm not totally sure I understand what you are asking, but the arc length integral $S(f) = int_a^b sqrt1 + f'^2(x)rm dx$ is a functional that takes in a function $f(x)$ (with some fixed boundary conditions $f(a) = c$ and $f(b) = d$) and outputs a number that is the arc-length of this function. The solution to the Euler-Lagrange equation for this functional gives the function that minimizes $S$, i.e. it gives the function that has the shortest length between two points (which is the straight line $f(x) = f(a) + [f(b)-f(a)](x-a)/(b-a)$)
– Winther
Jul 20 at 23:40
I'm not totally sure I understand what you are asking, but the arc length integral $S(f) = int_a^b sqrt1 + f'^2(x)rm dx$ is a functional that takes in a function $f(x)$ (with some fixed boundary conditions $f(a) = c$ and $f(b) = d$) and outputs a number that is the arc-length of this function. The solution to the Euler-Lagrange equation for this functional gives the function that minimizes $S$, i.e. it gives the function that has the shortest length between two points (which is the straight line $f(x) = f(a) + [f(b)-f(a)](x-a)/(b-a)$)
– Winther
Jul 20 at 23:40
oh ...you mean the arc length integrals we study in multivariable calculus are functionals ? I am a little confused ...why don't they call them that in my multivariable calculus book? if what you are saying is true that would solve my confusion
– Sedumjoy
Jul 20 at 23:46
oh ...you mean the arc length integrals we study in multivariable calculus are functionals ? I am a little confused ...why don't they call them that in my multivariable calculus book? if what you are saying is true that would solve my confusion
– Sedumjoy
Jul 20 at 23:46
1
1
It depends on what you are looking at. If you are to compute the arc-length of a given function you don't have to consider the concept of a functional: you simply compute the arc-length integral for this function. The concept of a functional is of interest when you consider problems like "what is the function that has the shortest arc length between two points" (i.e. when the unknown is a function not just a number). Then you would start by putting up the arc-length functional, compute the Euler-Lagrange equations and solve it in order to determine this function.
– Winther
Jul 20 at 23:52
It depends on what you are looking at. If you are to compute the arc-length of a given function you don't have to consider the concept of a functional: you simply compute the arc-length integral for this function. The concept of a functional is of interest when you consider problems like "what is the function that has the shortest arc length between two points" (i.e. when the unknown is a function not just a number). Then you would start by putting up the arc-length functional, compute the Euler-Lagrange equations and solve it in order to determine this function.
– Winther
Jul 20 at 23:52
gotcha ...this clears it up for me. I was going in circles ....thanx
– Sedumjoy
Jul 21 at 0:00
gotcha ...this clears it up for me. I was going in circles ....thanx
– Sedumjoy
Jul 21 at 0:00
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
To summarize the comments above: The arc-length integral defines a length. If you consider it as a functional then plugging in any function and computing the integral will give you the arc-length of this function.
If you only care about the arc-length of a given function then you don't need to consider the concept of a functional at all, you simply compute the arc-length integral for your function (and the Euler-Lagrange equations are not relevant).
The concept of a functional is useful when we consider problems like "what is the function that has the shortest arc-length between two points", i.e. when the solution we are after is a function. Then we would start by putting up a functional that describes the quantity we want to extremize (the arc-length), derive the Euler-Lagrange equations for this functional ($f''(x) = 0$) and solve it to get the desired function (a straight line connecting the two points).
answered Jul 21 at 1:03
Winther
20.1k33054
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
To summarize the comments above: The arc-length integral defines a length. If you consider it as a functional then plugging in any function and computing the integral will give you the arc-length of this function.
If you only care about the arc-length of a given function then you don't need to consider the concept of a functional at all, you simply compute the arc-length integral for your function (and the Euler-Lagrange equations are not relevant).
The concept of a functional is useful when we consider problems like "what is the function that has the shortest arc-length between two points", i.e. when the solution we are after is a function. Then we would start by putting up a functional that describes the quantity we want to extremize (the arc-length), derive the Euler-Lagrange equations for this functional ($f''(x) = 0$) and solve it to get the desired function (a straight line connecting the two points).
answered Jul 21 at 1:03
Winther
20.1k33054
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
add a comment |Â
up vote
2
down vote
accepted
To summarize the comments above: The arc-length integral defines a length. If you consider it as a functional then plugging in any function and computing the integral will give you the arc-length of this function.
If you only care about the arc-length of a given function then you don't need to consider the concept of a functional at all, you simply compute the arc-length integral for your function (and the Euler-Lagrange equations are not relevant).
The concept of a functional is useful when we consider problems like "what is the function that has the shortest arc-length between two points", i.e. when the solution we are after is a function. Then we would start by putting up a functional that describes the quantity we want to extremize (the arc-length), derive the Euler-Lagrange equations for this functional ($f''(x) = 0$) and solve it to get the desired function (a straight line connecting the two points).
answered Jul 21 at 1:03
Winther
20.1k33054
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
To summarize the comments above: The arc-length integral defines a length. If you consider it as a functional then plugging in any function and computing the integral will give you the arc-length of this function.
If you only care about the arc-length of a given function then you don't need to consider the concept of a functional at all, you simply compute the arc-length integral for your function (and the Euler-Lagrange equations are not relevant).
The concept of a functional is useful when we consider problems like "what is the function that has the shortest arc-length between two points", i.e. when the solution we are after is a function. Then we would start by putting up a functional that describes the quantity we want to extremize (the arc-length), derive the Euler-Lagrange equations for this functional ($f''(x) = 0$) and solve it to get the desired function (a straight line connecting the two points).
answered Jul 21 at 1:03
Winther
20.1k33054
To summarize the comments above: The arc-length integral defines a length. If you consider it as a functional then plugging in any function and computing the integral will give you the arc-length of this function.
If you only care about the arc-length of a given function then you don't need to consider the concept of a functional at all, you simply compute the arc-length integral for your function (and the Euler-Lagrange equations are not relevant).
The concept of a functional is useful when we consider problems like "what is the function that has the shortest arc-length between two points", i.e. when the solution we are after is a function. Then we would start by putting up a functional that describes the quantity we want to extremize (the arc-length), derive the Euler-Lagrange equations for this functional ($f''(x) = 0$) and solve it to get the desired function (a straight line connecting the two points).
answered Jul 21 at 1:03
Winther
20.1k33054
answered Jul 21 at 1:03
Winther
20.1k33054
answered Jul 21 at 1:03
Winther
20.1k33054
answered Jul 21 at 1:03
Winther
20.1k33054
20.1k33054
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
add a comment |Â
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
i am continuing with the Euler-Lagrange derivation and it makes sense now...your explanation got me past my confusion . thank you
– Sedumjoy
Jul 21 at 14:16
add a comment |Â
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1
I'm not totally sure I understand what you are asking, but the arc length integral $S(f) = int_a^b sqrt1 + f'^2(x)rm dx$ is a functional that takes in a function $f(x)$ (with some fixed boundary conditions $f(a) = c$ and $f(b) = d$) and outputs a number that is the arc-length of this function. The solution to the Euler-Lagrange equation for this functional gives the function that minimizes $S$, i.e. it gives the function that has the shortest length between two points (which is the straight line $f(x) = f(a) + [f(b)-f(a)](x-a)/(b-a)$)
– Winther
Jul 20 at 23:40
oh ...you mean the arc length integrals we study in multivariable calculus are functionals ? I am a little confused ...why don't they call them that in my multivariable calculus book? if what you are saying is true that would solve my confusion
– Sedumjoy
Jul 20 at 23:46
1
It depends on what you are looking at. If you are to compute the arc-length of a given function you don't have to consider the concept of a functional: you simply compute the arc-length integral for this function. The concept of a functional is of interest when you consider problems like "what is the function that has the shortest arc length between two points" (i.e. when the unknown is a function not just a number). Then you would start by putting up the arc-length functional, compute the Euler-Lagrange equations and solve it in order to determine this function.
– Winther
Jul 20 at 23:52
gotcha ...this clears it up for me. I was going in circles ....thanx
– Sedumjoy
Jul 21 at 0:00