Mixing
Arc length of a point on ellipse from the vertex
$begingroup$
How is the arc length of an ellipse (measured from the vertex) defined by $x = a cos (theta)$, $y = b sin(theta)$ given by $s(psi) = a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right) $. Please see my attempt below
begin{align*}
s(psi) &~=~int_{0}^{psi} sqrt{left(frac{dx}{dtheta}right)^2+ left(frac{dy}{dtheta}right)^2} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 sin^2(theta)+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 (1-cos^2(theta))+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2+(b^2-a^2)cos^2(theta)} dtheta
\
&~=~ a int_{0}^{psi} sqrt{ 1-left(1-frac{a^2}{b^2}right)cos^2(theta)} dtheta,
end{align*}
which is not equal to
begin{align*}
s(psi) &~=~ a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right)
\
&~=~a int_{0}^{psi} sqrt{ 1-left(1-frac{b^2}{a^2}right)sin^2(theta)}.
end{align*}
calculus integration trigonometry conic-sections arc-length
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
$endgroup$
add a comment |
$begingroup$
How is the arc length of an ellipse (measured from the vertex) defined by $x = a cos (theta)$, $y = b sin(theta)$ given by $s(psi) = a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right) $. Please see my attempt below
begin{align*}
s(psi) &~=~int_{0}^{psi} sqrt{left(frac{dx}{dtheta}right)^2+ left(frac{dy}{dtheta}right)^2} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 sin^2(theta)+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 (1-cos^2(theta))+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2+(b^2-a^2)cos^2(theta)} dtheta
\
&~=~ a int_{0}^{psi} sqrt{ 1-left(1-frac{a^2}{b^2}right)cos^2(theta)} dtheta,
end{align*}
which is not equal to
begin{align*}
s(psi) &~=~ a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right)
\
&~=~a int_{0}^{psi} sqrt{ 1-left(1-frac{b^2}{a^2}right)sin^2(theta)}.
end{align*}
calculus integration trigonometry conic-sections arc-length
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
$endgroup$
$begingroup$
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind).
$endgroup$
– Jack D'Aurizio
Sep 18 '18 at 15:10
$begingroup$
@JackD'Aurizio Please see the last three lines added to the question. Thanks.
$endgroup$
– Pheobey
Sep 18 '18 at 15:23
add a comment |
$begingroup$
How is the arc length of an ellipse (measured from the vertex) defined by $x = a cos (theta)$, $y = b sin(theta)$ given by $s(psi) = a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right) $. Please see my attempt below
begin{align*}
s(psi) &~=~int_{0}^{psi} sqrt{left(frac{dx}{dtheta}right)^2+ left(frac{dy}{dtheta}right)^2} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 sin^2(theta)+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 (1-cos^2(theta))+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2+(b^2-a^2)cos^2(theta)} dtheta
\
&~=~ a int_{0}^{psi} sqrt{ 1-left(1-frac{a^2}{b^2}right)cos^2(theta)} dtheta,
end{align*}
which is not equal to
begin{align*}
s(psi) &~=~ a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right)
\
&~=~a int_{0}^{psi} sqrt{ 1-left(1-frac{b^2}{a^2}right)sin^2(theta)}.
end{align*}
calculus integration trigonometry conic-sections arc-length
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
$endgroup$
How is the arc length of an ellipse (measured from the vertex) defined by $x = a cos (theta)$, $y = b sin(theta)$ given by $s(psi) = a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right) $. Please see my attempt below
begin{align*}
s(psi) &~=~int_{0}^{psi} sqrt{left(frac{dx}{dtheta}right)^2+ left(frac{dy}{dtheta}right)^2} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 sin^2(theta)+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2 (1-cos^2(theta))+b^2cos^2(theta)} dtheta
\
&~=~ int_{0}^{psi} sqrt{ a^2+(b^2-a^2)cos^2(theta)} dtheta
\
&~=~ a int_{0}^{psi} sqrt{ 1-left(1-frac{a^2}{b^2}right)cos^2(theta)} dtheta,
end{align*}
which is not equal to
begin{align*}
s(psi) &~=~ a Ellipticleft(psi,sqrt{1-frac{b^2}{a^2}}right)
\
&~=~a int_{0}^{psi} sqrt{ 1-left(1-frac{b^2}{a^2}right)sin^2(theta)}.
end{align*}
calculus integration trigonometry conic-sections arc-length
calculus integration trigonometry conic-sections arc-length
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
edited Sep 18 '18 at 15:21
Pheobey
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
asked Sep 18 '18 at 15:07
PheobeyPheobey
32
32
$begingroup$
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind).
$endgroup$
– Jack D'Aurizio
Sep 18 '18 at 15:10
$begingroup$
@JackD'Aurizio Please see the last three lines added to the question. Thanks.
$endgroup$
– Pheobey
Sep 18 '18 at 15:23
add a comment |
$begingroup$
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind).
$endgroup$
– Jack D'Aurizio
Sep 18 '18 at 15:10
$begingroup$
@JackD'Aurizio Please see the last three lines added to the question. Thanks.
$endgroup$
– Pheobey
Sep 18 '18 at 15:23
$begingroup$
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind).
$endgroup$
– Jack D'Aurizio
Sep 18 '18 at 15:10
$begingroup$
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind).
$endgroup$
– Jack D'Aurizio
Sep 18 '18 at 15:10
$begingroup$
@JackD'Aurizio Please see the last three lines added to the question. Thanks.
$endgroup$
– Pheobey
Sep 18 '18 at 15:23
$begingroup$
@JackD'Aurizio Please see the last three lines added to the question. Thanks.
$endgroup$
– Pheobey
Sep 18 '18 at 15:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For clockwise sense:
begin{align}
(x,y) &= (asin theta,bcos theta) \
s(theta) &= aE(theta, varepsilon) \
varepsilon &= sqrt{1-frac{b^2}{a^2}}
end{align}
For anti-clockwise sense:
begin{align}
(x,y) &= (acos theta,bsin theta) \
s(theta) &= bEleft( theta, frac{iavarepsilon}{b} right) \
&= aEleft( frac{pi}{2}-theta, varepsilon right) \
frac{iavarepsilon}{b} &= sqrt{1-frac{a^2}{b^2}}
end{align}
Parametrized with Jacobi elliptic functions:
begin{align}
(x,y) &= (aoperatorname{sn} (u,varepsilon),
boperatorname{cn} (u,varepsilon)) \
s(u) &= aE( operatorname{am} ( u,varepsilon ), varepsilon ) \
end{align}
See also the arclength formula in polar coordinates here.
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
add a comment |
$begingroup$
You just have to exchange $a$ and $b$, and use $cos^2=1-sin^2$. Then the two expressions agree. So it's just a question of how you define your ellipse.
Alternatively, you can look on the elliptic integral as measuring the arc length anticlockwise from the point $(0,b)$.
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
$endgroup$
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For clockwise sense:
begin{align}
(x,y) &= (asin theta,bcos theta) \
s(theta) &= aE(theta, varepsilon) \
varepsilon &= sqrt{1-frac{b^2}{a^2}}
end{align}
For anti-clockwise sense:
begin{align}
(x,y) &= (acos theta,bsin theta) \
s(theta) &= bEleft( theta, frac{iavarepsilon}{b} right) \
&= aEleft( frac{pi}{2}-theta, varepsilon right) \
frac{iavarepsilon}{b} &= sqrt{1-frac{a^2}{b^2}}
end{align}
Parametrized with Jacobi elliptic functions:
begin{align}
(x,y) &= (aoperatorname{sn} (u,varepsilon),
boperatorname{cn} (u,varepsilon)) \
s(u) &= aE( operatorname{am} ( u,varepsilon ), varepsilon ) \
end{align}
See also the arclength formula in polar coordinates here.
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
add a comment |
$begingroup$
For clockwise sense:
begin{align}
(x,y) &= (asin theta,bcos theta) \
s(theta) &= aE(theta, varepsilon) \
varepsilon &= sqrt{1-frac{b^2}{a^2}}
end{align}
For anti-clockwise sense:
begin{align}
(x,y) &= (acos theta,bsin theta) \
s(theta) &= bEleft( theta, frac{iavarepsilon}{b} right) \
&= aEleft( frac{pi}{2}-theta, varepsilon right) \
frac{iavarepsilon}{b} &= sqrt{1-frac{a^2}{b^2}}
end{align}
Parametrized with Jacobi elliptic functions:
begin{align}
(x,y) &= (aoperatorname{sn} (u,varepsilon),
boperatorname{cn} (u,varepsilon)) \
s(u) &= aE( operatorname{am} ( u,varepsilon ), varepsilon ) \
end{align}
See also the arclength formula in polar coordinates here.
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
add a comment |
$begingroup$
For clockwise sense:
begin{align}
(x,y) &= (asin theta,bcos theta) \
s(theta) &= aE(theta, varepsilon) \
varepsilon &= sqrt{1-frac{b^2}{a^2}}
end{align}
For anti-clockwise sense:
begin{align}
(x,y) &= (acos theta,bsin theta) \
s(theta) &= bEleft( theta, frac{iavarepsilon}{b} right) \
&= aEleft( frac{pi}{2}-theta, varepsilon right) \
frac{iavarepsilon}{b} &= sqrt{1-frac{a^2}{b^2}}
end{align}
Parametrized with Jacobi elliptic functions:
begin{align}
(x,y) &= (aoperatorname{sn} (u,varepsilon),
boperatorname{cn} (u,varepsilon)) \
s(u) &= aE( operatorname{am} ( u,varepsilon ), varepsilon ) \
end{align}
See also the arclength formula in polar coordinates here.
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
For clockwise sense:
begin{align}
(x,y) &= (asin theta,bcos theta) \
s(theta) &= aE(theta, varepsilon) \
varepsilon &= sqrt{1-frac{b^2}{a^2}}
end{align}
For anti-clockwise sense:
begin{align}
(x,y) &= (acos theta,bsin theta) \
s(theta) &= bEleft( theta, frac{iavarepsilon}{b} right) \
&= aEleft( frac{pi}{2}-theta, varepsilon right) \
frac{iavarepsilon}{b} &= sqrt{1-frac{a^2}{b^2}}
end{align}
Parametrized with Jacobi elliptic functions:
begin{align}
(x,y) &= (aoperatorname{sn} (u,varepsilon),
boperatorname{cn} (u,varepsilon)) \
s(u) &= aE( operatorname{am} ( u,varepsilon ), varepsilon ) \
end{align}
See also the arclength formula in polar coordinates here.
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
edited Jan 2 at 10:00
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
answered Sep 19 '18 at 2:45
Ng Chung TakNg Chung Tak
14.3k31334
14.3k31334
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
add a comment |
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
I think for clockwise sense $(x,y)=left(acos(t),-bsin(t)right)$. How do you define the clockwise sense?
$endgroup$
– Pheobey
Sep 19 '18 at 5:52
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
$begingroup$
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{circ}$.
$endgroup$
– Ng Chung Tak
Sep 19 '18 at 5:58
add a comment |
$begingroup$
You just have to exchange $a$ and $b$, and use $cos^2=1-sin^2$. Then the two expressions agree. So it's just a question of how you define your ellipse.
Alternatively, you can look on the elliptic integral as measuring the arc length anticlockwise from the point $(0,b)$.
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
$endgroup$
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
add a comment |
$begingroup$
You just have to exchange $a$ and $b$, and use $cos^2=1-sin^2$. Then the two expressions agree. So it's just a question of how you define your ellipse.
Alternatively, you can look on the elliptic integral as measuring the arc length anticlockwise from the point $(0,b)$.
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
$endgroup$
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
add a comment |
$begingroup$
You just have to exchange $a$ and $b$, and use $cos^2=1-sin^2$. Then the two expressions agree. So it's just a question of how you define your ellipse.
Alternatively, you can look on the elliptic integral as measuring the arc length anticlockwise from the point $(0,b)$.
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
$endgroup$
You just have to exchange $a$ and $b$, and use $cos^2=1-sin^2$. Then the two expressions agree. So it's just a question of how you define your ellipse.
Alternatively, you can look on the elliptic integral as measuring the arc length anticlockwise from the point $(0,b)$.
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
answered Sep 18 '18 at 16:10
TonyKTonyK
42.1k355134
42.1k355134
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
add a comment |
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
$begingroup$
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree.
$endgroup$
– Pheobey
Sep 18 '18 at 17:29
add a comment |
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$begingroup$
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind).
$endgroup$
– Jack D'Aurizio
Sep 18 '18 at 15:10
$begingroup$
@JackD'Aurizio Please see the last three lines added to the question. Thanks.
$endgroup$
– Pheobey
Sep 18 '18 at 15:23