smooth curve that is tangent to a 1-form kernel in every point
$begingroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$.
Prove that $forall p,q in mathbb{R}^3, exists gamma: [0,1] rightarrow mathbb{R}^3$ smooth, such that $γ(0)=p, γ(1) =q$ and $gamma$ is tangent to $ker alpha$ for all $tin [0,1]$.
If we take $gamma(t) = p(1-t) + qt$ we math the conditions $γ(0)=p, γ(1) =q$, but how to match the last condition : $alpha (gamma'(t)) = 0$?
Thank you for any insights.
differential-geometry tangent-spaces
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
$endgroup$
add a comment |
$begingroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$.
Prove that $forall p,q in mathbb{R}^3, exists gamma: [0,1] rightarrow mathbb{R}^3$ smooth, such that $γ(0)=p, γ(1) =q$ and $gamma$ is tangent to $ker alpha$ for all $tin [0,1]$.
If we take $gamma(t) = p(1-t) + qt$ we math the conditions $γ(0)=p, γ(1) =q$, but how to match the last condition : $alpha (gamma'(t)) = 0$?
Thank you for any insights.
differential-geometry tangent-spaces
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
$endgroup$
add a comment |
$begingroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$.
Prove that $forall p,q in mathbb{R}^3, exists gamma: [0,1] rightarrow mathbb{R}^3$ smooth, such that $γ(0)=p, γ(1) =q$ and $gamma$ is tangent to $ker alpha$ for all $tin [0,1]$.
If we take $gamma(t) = p(1-t) + qt$ we math the conditions $γ(0)=p, γ(1) =q$, but how to match the last condition : $alpha (gamma'(t)) = 0$?
Thank you for any insights.
differential-geometry tangent-spaces
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
$endgroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$.
Prove that $forall p,q in mathbb{R}^3, exists gamma: [0,1] rightarrow mathbb{R}^3$ smooth, such that $γ(0)=p, γ(1) =q$ and $gamma$ is tangent to $ker alpha$ for all $tin [0,1]$.
If we take $gamma(t) = p(1-t) + qt$ we math the conditions $γ(0)=p, γ(1) =q$, but how to match the last condition : $alpha (gamma'(t)) = 0$?
Thank you for any insights.
differential-geometry tangent-spaces
differential-geometry tangent-spaces
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
asked Dec 10 '18 at 0:32
PerelManPerelMan
544311
544311
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The third condition says that for each $t$, we must have
$$
alpha(gamma(t)) [gamma'(t)] = 0. tag{*}
$$
If we write
$$
gamma(t) = (x(t), y(t), z(t))
$$
then
$$
gamma'(t) = (x'(t), y'(t), z'(t))
$$
and equation (*) becomes
$$
z'(t) - y(t) x'(t) = 0,
$$
which you can, perhaps, solve (or can at least prove has a solution).
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
$endgroup$
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The third condition says that for each $t$, we must have
$$
alpha(gamma(t)) [gamma'(t)] = 0. tag{*}
$$
If we write
$$
gamma(t) = (x(t), y(t), z(t))
$$
then
$$
gamma'(t) = (x'(t), y'(t), z'(t))
$$
and equation (*) becomes
$$
z'(t) - y(t) x'(t) = 0,
$$
which you can, perhaps, solve (or can at least prove has a solution).
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
$endgroup$
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
add a comment |
$begingroup$
The third condition says that for each $t$, we must have
$$
alpha(gamma(t)) [gamma'(t)] = 0. tag{*}
$$
If we write
$$
gamma(t) = (x(t), y(t), z(t))
$$
then
$$
gamma'(t) = (x'(t), y'(t), z'(t))
$$
and equation (*) becomes
$$
z'(t) - y(t) x'(t) = 0,
$$
which you can, perhaps, solve (or can at least prove has a solution).
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
$endgroup$
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
add a comment |
$begingroup$
The third condition says that for each $t$, we must have
$$
alpha(gamma(t)) [gamma'(t)] = 0. tag{*}
$$
If we write
$$
gamma(t) = (x(t), y(t), z(t))
$$
then
$$
gamma'(t) = (x'(t), y'(t), z'(t))
$$
and equation (*) becomes
$$
z'(t) - y(t) x'(t) = 0,
$$
which you can, perhaps, solve (or can at least prove has a solution).
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
$endgroup$
The third condition says that for each $t$, we must have
$$
alpha(gamma(t)) [gamma'(t)] = 0. tag{*}
$$
If we write
$$
gamma(t) = (x(t), y(t), z(t))
$$
then
$$
gamma'(t) = (x'(t), y'(t), z'(t))
$$
and equation (*) becomes
$$
z'(t) - y(t) x'(t) = 0,
$$
which you can, perhaps, solve (or can at least prove has a solution).
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
answered Dec 10 '18 at 0:49
John HughesJohn Hughes
64k24191
64k24191
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
add a comment |
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
$begingroup$
Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q
$endgroup$
– PerelMan
Dec 10 '18 at 1:14
add a comment |
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