Spherical Triangle: Law of Sines with Clairaut's theorem
$begingroup$
The spherical law of sines states that
On the sphere $mathbb{S}^2$ we consider a triangle, i.e. three points connected by geodesics. We denote the side lengths with $a, b, c < pi$ and the angles opposite to these sides with $alpha, beta, gamma in [0, 2pi]$.
Then:
$$ frac{sin alpha}{sin a} = frac{sin beta}{sin b} = frac{sin gamma}{sin c}.$$
As part of my studies, I have been given the task of proving the above theorem.
However, I need to somehow apply Clairaut's theorem, that is:
Let $f$ be a rotation surface, i.e. $f(t, varphi) := left( r(t)cosvarphi, r(t) sin varphi, h(t)right)$ and let $gamma(s) := left( t(s), varphi(s)right)$ be a regular curve and $c := f circ gamma$. Futher, we denote by $theta(s)$ the angle between the curve $c(s)$ and the line of latitude through $c(s)$.
That means:
$$ cos theta(s) = frac{langle c'(s), partial_2f(gamma(s)) rangle}{lvert c'(s) rvert lvert partial_2f(gamma(s)) rvert}.$$
Then:
$$ s mapsto rleft(t(s)right) cos theta(s) ; text{is constant}.$$
What I did: I proved Clairaut's theorem.
Remark: I know that many answers are already given here – however I think I need a little help to apply Clairaut's theorem.
IDEA: Maybe it could be benificial to put one corner of the triangle onto the north pole.
differential-geometry spherical-geometry geodesic
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
$endgroup$
add a comment |
$begingroup$
The spherical law of sines states that
On the sphere $mathbb{S}^2$ we consider a triangle, i.e. three points connected by geodesics. We denote the side lengths with $a, b, c < pi$ and the angles opposite to these sides with $alpha, beta, gamma in [0, 2pi]$.
Then:
$$ frac{sin alpha}{sin a} = frac{sin beta}{sin b} = frac{sin gamma}{sin c}.$$
As part of my studies, I have been given the task of proving the above theorem.
However, I need to somehow apply Clairaut's theorem, that is:
Let $f$ be a rotation surface, i.e. $f(t, varphi) := left( r(t)cosvarphi, r(t) sin varphi, h(t)right)$ and let $gamma(s) := left( t(s), varphi(s)right)$ be a regular curve and $c := f circ gamma$. Futher, we denote by $theta(s)$ the angle between the curve $c(s)$ and the line of latitude through $c(s)$.
That means:
$$ cos theta(s) = frac{langle c'(s), partial_2f(gamma(s)) rangle}{lvert c'(s) rvert lvert partial_2f(gamma(s)) rvert}.$$
Then:
$$ s mapsto rleft(t(s)right) cos theta(s) ; text{is constant}.$$
What I did: I proved Clairaut's theorem.
Remark: I know that many answers are already given here – however I think I need a little help to apply Clairaut's theorem.
IDEA: Maybe it could be benificial to put one corner of the triangle onto the north pole.
differential-geometry spherical-geometry geodesic
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
$endgroup$
add a comment |
$begingroup$
The spherical law of sines states that
On the sphere $mathbb{S}^2$ we consider a triangle, i.e. three points connected by geodesics. We denote the side lengths with $a, b, c < pi$ and the angles opposite to these sides with $alpha, beta, gamma in [0, 2pi]$.
Then:
$$ frac{sin alpha}{sin a} = frac{sin beta}{sin b} = frac{sin gamma}{sin c}.$$
As part of my studies, I have been given the task of proving the above theorem.
However, I need to somehow apply Clairaut's theorem, that is:
Let $f$ be a rotation surface, i.e. $f(t, varphi) := left( r(t)cosvarphi, r(t) sin varphi, h(t)right)$ and let $gamma(s) := left( t(s), varphi(s)right)$ be a regular curve and $c := f circ gamma$. Futher, we denote by $theta(s)$ the angle between the curve $c(s)$ and the line of latitude through $c(s)$.
That means:
$$ cos theta(s) = frac{langle c'(s), partial_2f(gamma(s)) rangle}{lvert c'(s) rvert lvert partial_2f(gamma(s)) rvert}.$$
Then:
$$ s mapsto rleft(t(s)right) cos theta(s) ; text{is constant}.$$
What I did: I proved Clairaut's theorem.
Remark: I know that many answers are already given here – however I think I need a little help to apply Clairaut's theorem.
IDEA: Maybe it could be benificial to put one corner of the triangle onto the north pole.
differential-geometry spherical-geometry geodesic
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
$endgroup$
The spherical law of sines states that
On the sphere $mathbb{S}^2$ we consider a triangle, i.e. three points connected by geodesics. We denote the side lengths with $a, b, c < pi$ and the angles opposite to these sides with $alpha, beta, gamma in [0, 2pi]$.
Then:
$$ frac{sin alpha}{sin a} = frac{sin beta}{sin b} = frac{sin gamma}{sin c}.$$
As part of my studies, I have been given the task of proving the above theorem.
However, I need to somehow apply Clairaut's theorem, that is:
Let $f$ be a rotation surface, i.e. $f(t, varphi) := left( r(t)cosvarphi, r(t) sin varphi, h(t)right)$ and let $gamma(s) := left( t(s), varphi(s)right)$ be a regular curve and $c := f circ gamma$. Futher, we denote by $theta(s)$ the angle between the curve $c(s)$ and the line of latitude through $c(s)$.
That means:
$$ cos theta(s) = frac{langle c'(s), partial_2f(gamma(s)) rangle}{lvert c'(s) rvert lvert partial_2f(gamma(s)) rvert}.$$
Then:
$$ s mapsto rleft(t(s)right) cos theta(s) ; text{is constant}.$$
What I did: I proved Clairaut's theorem.
Remark: I know that many answers are already given here – however I think I need a little help to apply Clairaut's theorem.
IDEA: Maybe it could be benificial to put one corner of the triangle onto the north pole.
differential-geometry spherical-geometry geodesic
differential-geometry spherical-geometry geodesic
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
edited Dec 12 '18 at 8:19
fpmoo
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
asked Dec 11 '18 at 17:59
fpmoofpmoo
382113
382113
add a comment |
add a comment |
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