Length of an ellipse
Clash Royale CLAN TAG#URR8PPP
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I am asked to give the length of the curve:
$gamma: [0,2pi]tomathbbR^2$, $gamma(t)=(acos(t), bsin(t))$ where $a,bin(0,infty)$
We get to solving the following:
$int_0^2pi sqrta^2sin(t)^2+b^2cos(t)^2, dt$
Which should not be possible to integrate probably.
For $a=b$ I can solve it, but thats it. We would get:
$2picdot a$
Is there any way, which does not involve using the form above and its solution with "second kind elliptic integrals" which wolframalpha refers to?
Thanks in advance.
integration arc-length
asked Aug 14 at 5:10
Cornman
2,61721128
add a comment |Â
up vote
-1
down vote
favorite
I am asked to give the length of the curve:
$gamma: [0,2pi]tomathbbR^2$, $gamma(t)=(acos(t), bsin(t))$ where $a,bin(0,infty)$
We get to solving the following:
$int_0^2pi sqrta^2sin(t)^2+b^2cos(t)^2, dt$
Which should not be possible to integrate probably.
For $a=b$ I can solve it, but thats it. We would get:
$2picdot a$
Is there any way, which does not involve using the form above and its solution with "second kind elliptic integrals" which wolframalpha refers to?
Thanks in advance.
integration arc-length
asked Aug 14 at 5:10
Cornman
2,61721128
1
You cannot avoid elliptic integrals but there are very good approximations.
– Claude Leibovici
Aug 14 at 5:30
@ClaudeLeibovici Thanks. I wonder because in the lecture notes I study, there are no such approximations or even anything about elliptic curves. I think that this homework was a "mistake". Or not well choosen.
– Cornman
Aug 14 at 5:34
About the perimeter of the ellipse, have a look at pages 138+ of my notes. I still have to mention Ramanujan's approximation.
– Jack D'Aurizio♦
Aug 14 at 8:44
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I am asked to give the length of the curve:
$gamma: [0,2pi]tomathbbR^2$, $gamma(t)=(acos(t), bsin(t))$ where $a,bin(0,infty)$
We get to solving the following:
$int_0^2pi sqrta^2sin(t)^2+b^2cos(t)^2, dt$
Which should not be possible to integrate probably.
For $a=b$ I can solve it, but thats it. We would get:
$2picdot a$
Is there any way, which does not involve using the form above and its solution with "second kind elliptic integrals" which wolframalpha refers to?
Thanks in advance.
integration arc-length
asked Aug 14 at 5:10
Cornman
2,61721128
I am asked to give the length of the curve:
$gamma: [0,2pi]tomathbbR^2$, $gamma(t)=(acos(t), bsin(t))$ where $a,bin(0,infty)$
We get to solving the following:
$int_0^2pi sqrta^2sin(t)^2+b^2cos(t)^2, dt$
Which should not be possible to integrate probably.
For $a=b$ I can solve it, but thats it. We would get:
$2picdot a$
Is there any way, which does not involve using the form above and its solution with "second kind elliptic integrals" which wolframalpha refers to?
Thanks in advance.
integration arc-length
asked Aug 14 at 5:10
Cornman
2,61721128
asked Aug 14 at 5:10
Cornman
2,61721128
asked Aug 14 at 5:10
Cornman
2,61721128
asked Aug 14 at 5:10
Cornman
2,61721128
2,61721128
1
You cannot avoid elliptic integrals but there are very good approximations.
– Claude Leibovici
Aug 14 at 5:30
@ClaudeLeibovici Thanks. I wonder because in the lecture notes I study, there are no such approximations or even anything about elliptic curves. I think that this homework was a "mistake". Or not well choosen.
– Cornman
Aug 14 at 5:34
About the perimeter of the ellipse, have a look at pages 138+ of my notes. I still have to mention Ramanujan's approximation.
– Jack D'Aurizio♦
Aug 14 at 8:44
add a comment |Â
1
You cannot avoid elliptic integrals but there are very good approximations.
– Claude Leibovici
Aug 14 at 5:30
@ClaudeLeibovici Thanks. I wonder because in the lecture notes I study, there are no such approximations or even anything about elliptic curves. I think that this homework was a "mistake". Or not well choosen.
– Cornman
Aug 14 at 5:34
About the perimeter of the ellipse, have a look at pages 138+ of my notes. I still have to mention Ramanujan's approximation.
– Jack D'Aurizio♦
Aug 14 at 8:44
1
1
You cannot avoid elliptic integrals but there are very good approximations.
– Claude Leibovici
Aug 14 at 5:30
You cannot avoid elliptic integrals but there are very good approximations.
– Claude Leibovici
Aug 14 at 5:30
@ClaudeLeibovici Thanks. I wonder because in the lecture notes I study, there are no such approximations or even anything about elliptic curves. I think that this homework was a "mistake". Or not well choosen.
– Cornman
Aug 14 at 5:34
@ClaudeLeibovici Thanks. I wonder because in the lecture notes I study, there are no such approximations or even anything about elliptic curves. I think that this homework was a "mistake". Or not well choosen.
– Cornman
Aug 14 at 5:34
About the perimeter of the ellipse, have a look at pages 138+ of my notes. I still have to mention Ramanujan's approximation.
– Jack D'Aurizio♦
Aug 14 at 8:44
About the perimeter of the ellipse, have a look at pages 138+ of my notes. I still have to mention Ramanujan's approximation.
– Jack D'Aurizio♦
Aug 14 at 8:44
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Quoting the Wikipedia page, the circumference $C$ of an ellipse is given by
$$C = 4 a, E(e)qquad textwith qquad e=sqrt1-fracb^2a^2$$
The exact infinite series is given by
$$C=2pi a left[1 - sum_n=1^infty left(frac(2n-1)!!(2n)!!right)^2 frace^2n2n-1right]$$ but a much faster convergen series is
$$C = pi (a+b) left[1 + sum_n=1^infty left(frac(2n-1)!!2^n(2n-1) n!right)^2 h^nright]qquad textwith qquad h= frac (a-b)^2 (a+b)^2$$
In the linked page, you will find a series of approximations wich can be more than sufficient for practical usage.
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
1
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
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
Quoting the Wikipedia page, the circumference $C$ of an ellipse is given by
$$C = 4 a, E(e)qquad textwith qquad e=sqrt1-fracb^2a^2$$
The exact infinite series is given by
$$C=2pi a left[1 - sum_n=1^infty left(frac(2n-1)!!(2n)!!right)^2 frace^2n2n-1right]$$ but a much faster convergen series is
$$C = pi (a+b) left[1 + sum_n=1^infty left(frac(2n-1)!!2^n(2n-1) n!right)^2 h^nright]qquad textwith qquad h= frac (a-b)^2 (a+b)^2$$
In the linked page, you will find a series of approximations wich can be more than sufficient for practical usage.
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
1
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
add a comment |Â
up vote
2
down vote
accepted
Quoting the Wikipedia page, the circumference $C$ of an ellipse is given by
$$C = 4 a, E(e)qquad textwith qquad e=sqrt1-fracb^2a^2$$
The exact infinite series is given by
$$C=2pi a left[1 - sum_n=1^infty left(frac(2n-1)!!(2n)!!right)^2 frace^2n2n-1right]$$ but a much faster convergen series is
$$C = pi (a+b) left[1 + sum_n=1^infty left(frac(2n-1)!!2^n(2n-1) n!right)^2 h^nright]qquad textwith qquad h= frac (a-b)^2 (a+b)^2$$
In the linked page, you will find a series of approximations wich can be more than sufficient for practical usage.
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
1
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Quoting the Wikipedia page, the circumference $C$ of an ellipse is given by
$$C = 4 a, E(e)qquad textwith qquad e=sqrt1-fracb^2a^2$$
The exact infinite series is given by
$$C=2pi a left[1 - sum_n=1^infty left(frac(2n-1)!!(2n)!!right)^2 frace^2n2n-1right]$$ but a much faster convergen series is
$$C = pi (a+b) left[1 + sum_n=1^infty left(frac(2n-1)!!2^n(2n-1) n!right)^2 h^nright]qquad textwith qquad h= frac (a-b)^2 (a+b)^2$$
In the linked page, you will find a series of approximations wich can be more than sufficient for practical usage.
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
Quoting the Wikipedia page, the circumference $C$ of an ellipse is given by
$$C = 4 a, E(e)qquad textwith qquad e=sqrt1-fracb^2a^2$$
The exact infinite series is given by
$$C=2pi a left[1 - sum_n=1^infty left(frac(2n-1)!!(2n)!!right)^2 frace^2n2n-1right]$$ but a much faster convergen series is
$$C = pi (a+b) left[1 + sum_n=1^infty left(frac(2n-1)!!2^n(2n-1) n!right)^2 h^nright]qquad textwith qquad h= frac (a-b)^2 (a+b)^2$$
In the linked page, you will find a series of approximations wich can be more than sufficient for practical usage.
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
answered Aug 14 at 5:54
Claude Leibovici
112k1055127
112k1055127
1
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
add a comment |Â
1
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
1
1
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
(+1) A practical approximations for a small eccentricity is given by $$ Capprox 2pileft(fraca^3/2+b^3/22right)^2/3. $$
– Jack D'Aurizio♦
Aug 14 at 8:46
add a comment |Â
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1
You cannot avoid elliptic integrals but there are very good approximations.
– Claude Leibovici
Aug 14 at 5:30
@ClaudeLeibovici Thanks. I wonder because in the lecture notes I study, there are no such approximations or even anything about elliptic curves. I think that this homework was a "mistake". Or not well choosen.
– Cornman
Aug 14 at 5:34
About the perimeter of the ellipse, have a look at pages 138+ of my notes. I still have to mention Ramanujan's approximation.
– Jack D'Aurizio♦
Aug 14 at 8:44