Prove that if $alpha$ lies inside an osculating plane at any point, then it has zero torsion.Osculating plane and unit speed curveunit speed curves and frenet serretFrenet-Serret to show something lies in the planeShowing a characterisation of curves with constant slope.plane curves and osculating planeHow to prove this curve lies in the plane through $alpha(0)$ orthogonal to $u$?What is happening to a curve in space at a point where it has small curvature, but high torsionThe knowledge of $n=n(s)$ can be used to determine the curvature $k(s)$ and the torsion $tau (s)$Clarification regarding the notion of Ribbons given in WikipediaA point in every osculating plane of a curve
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Prove that if $alpha$ lies inside an osculating plane at any point, then it has zero torsion.
Osculating plane and unit speed curveunit speed curves and frenet serretFrenet-Serret to show something lies in the planeShowing a characterisation of curves with constant slope.plane curves and osculating planeHow to prove this curve lies in the plane through $alpha(0)$ orthogonal to $u$?What is happening to a curve in space at a point where it has small curvature, but high torsionThe knowledge of $n=n(s)$ can be used to determine the curvature $k(s)$ and the torsion $tau (s)$Clarification regarding the notion of Ribbons given in WikipediaA point in every osculating plane of a curve
$begingroup$
I know in words that torsion is the failure of a curve to remain within a plane so if a point on the curve is within the plane then torsion must be 0.
I also know the osculating plane is $x: langle x-alpha(t), B(t)rangle=0 $
I do not know how to write this proof mathematically. I tried using the Frenet-Serret theorem but was unable to approach a valid solution.
calculus differential-geometry
edited Mar 23 at 17:45
Ernie060
2,940719
asked Mar 21 at 14:17
AC06AC06
405
$endgroup$
add a comment |
$begingroup$
I know in words that torsion is the failure of a curve to remain within a plane so if a point on the curve is within the plane then torsion must be 0.
I also know the osculating plane is $x: langle x-alpha(t), B(t)rangle=0 $
I do not know how to write this proof mathematically. I tried using the Frenet-Serret theorem but was unable to approach a valid solution.
calculus differential-geometry
edited Mar 23 at 17:45
Ernie060
2,940719
asked Mar 21 at 14:17
AC06AC06
405
$endgroup$
$begingroup$
This statement isn't quite right: For all $t$, $alpha(t)$ itself lies in the osculating plane to the curve at $alpha(t)$, but not all curves in $Bbb R^3$ have zero torsion.
$endgroup$
– Travis
Mar 23 at 18:02
add a comment |
$begingroup$
I know in words that torsion is the failure of a curve to remain within a plane so if a point on the curve is within the plane then torsion must be 0.
I also know the osculating plane is $x: langle x-alpha(t), B(t)rangle=0 $
I do not know how to write this proof mathematically. I tried using the Frenet-Serret theorem but was unable to approach a valid solution.
calculus differential-geometry
edited Mar 23 at 17:45
Ernie060
2,940719
asked Mar 21 at 14:17
AC06AC06
405
$endgroup$
I know in words that torsion is the failure of a curve to remain within a plane so if a point on the curve is within the plane then torsion must be 0.
I also know the osculating plane is $x: langle x-alpha(t), B(t)rangle=0 $
I do not know how to write this proof mathematically. I tried using the Frenet-Serret theorem but was unable to approach a valid solution.
calculus differential-geometry
calculus differential-geometry
edited Mar 23 at 17:45
Ernie060
2,940719
asked Mar 21 at 14:17
AC06AC06
405
edited Mar 23 at 17:45
Ernie060
2,940719
asked Mar 21 at 14:17
AC06AC06
405
edited Mar 23 at 17:45
Ernie060
2,940719
edited Mar 23 at 17:45
Ernie060
2,940719
edited Mar 23 at 17:45
Ernie060
2,940719
2,940719
asked Mar 21 at 14:17
AC06AC06
405
asked Mar 21 at 14:17
AC06AC06
405
asked Mar 21 at 14:17
AC06AC06
405
405
$begingroup$
This statement isn't quite right: For all $t$, $alpha(t)$ itself lies in the osculating plane to the curve at $alpha(t)$, but not all curves in $Bbb R^3$ have zero torsion.
$endgroup$
– Travis
Mar 23 at 18:02
add a comment |
$begingroup$
This statement isn't quite right: For all $t$, $alpha(t)$ itself lies in the osculating plane to the curve at $alpha(t)$, but not all curves in $Bbb R^3$ have zero torsion.
$endgroup$
– Travis
Mar 23 at 18:02
$begingroup$
This statement isn't quite right: For all $t$, $alpha(t)$ itself lies in the osculating plane to the curve at $alpha(t)$, but not all curves in $Bbb R^3$ have zero torsion.
$endgroup$
– Travis
Mar 23 at 18:02
$begingroup$
This statement isn't quite right: For all $t$, $alpha(t)$ itself lies in the osculating plane to the curve at $alpha(t)$, but not all curves in $Bbb R^3$ have zero torsion.
$endgroup$
– Travis
Mar 23 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As was pointed out in the comments, the statement is not quite true. I guess you wanted to ask the following statement: If the curve lies in the osculating plane at some fixed point, then the torsion is zero.
Consider a unit speed curve $alphacolon Isubset mathbbRtomathbbR^3$ with positive curvature $kappa$ and a point $alpha(t_0)$, $t_0in I$, on the curve. The osculating plane at $alpha(t_0)$ consists of the points $x in mathbbR^3$ with $langle x- alpha(t_0), B(t_0)rangle=0$.
If $alpha$ lies in the osculating plane, then $langle alpha(t)-alpha(t_0),B(t_0)rangle=0$ for all $tin I$. Deriving this equation twice and using the Frenet formulas gives
$$
langle T, B(t_0)rangle = 0 quad text and quad langle kappa N , B(t_0)rangle =0.
$$
Since we assume $kappa > 0$, the binormal vector field $B$ is always parallel with $B(t_0)$. So the direction and length of $B(t)$ is constant. Hence $B'=0$, so $-tau N= 0$ at every point by the Frenet-Serret formulas. This implies that $tau =0$.
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
$endgroup$
add a comment |
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1 Answer
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$begingroup$
As was pointed out in the comments, the statement is not quite true. I guess you wanted to ask the following statement: If the curve lies in the osculating plane at some fixed point, then the torsion is zero.
Consider a unit speed curve $alphacolon Isubset mathbbRtomathbbR^3$ with positive curvature $kappa$ and a point $alpha(t_0)$, $t_0in I$, on the curve. The osculating plane at $alpha(t_0)$ consists of the points $x in mathbbR^3$ with $langle x- alpha(t_0), B(t_0)rangle=0$.
If $alpha$ lies in the osculating plane, then $langle alpha(t)-alpha(t_0),B(t_0)rangle=0$ for all $tin I$. Deriving this equation twice and using the Frenet formulas gives
$$
langle T, B(t_0)rangle = 0 quad text and quad langle kappa N , B(t_0)rangle =0.
$$
Since we assume $kappa > 0$, the binormal vector field $B$ is always parallel with $B(t_0)$. So the direction and length of $B(t)$ is constant. Hence $B'=0$, so $-tau N= 0$ at every point by the Frenet-Serret formulas. This implies that $tau =0$.
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
$endgroup$
add a comment |
$begingroup$
As was pointed out in the comments, the statement is not quite true. I guess you wanted to ask the following statement: If the curve lies in the osculating plane at some fixed point, then the torsion is zero.
Consider a unit speed curve $alphacolon Isubset mathbbRtomathbbR^3$ with positive curvature $kappa$ and a point $alpha(t_0)$, $t_0in I$, on the curve. The osculating plane at $alpha(t_0)$ consists of the points $x in mathbbR^3$ with $langle x- alpha(t_0), B(t_0)rangle=0$.
If $alpha$ lies in the osculating plane, then $langle alpha(t)-alpha(t_0),B(t_0)rangle=0$ for all $tin I$. Deriving this equation twice and using the Frenet formulas gives
$$
langle T, B(t_0)rangle = 0 quad text and quad langle kappa N , B(t_0)rangle =0.
$$
Since we assume $kappa > 0$, the binormal vector field $B$ is always parallel with $B(t_0)$. So the direction and length of $B(t)$ is constant. Hence $B'=0$, so $-tau N= 0$ at every point by the Frenet-Serret formulas. This implies that $tau =0$.
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
$endgroup$
add a comment |
$begingroup$
As was pointed out in the comments, the statement is not quite true. I guess you wanted to ask the following statement: If the curve lies in the osculating plane at some fixed point, then the torsion is zero.
Consider a unit speed curve $alphacolon Isubset mathbbRtomathbbR^3$ with positive curvature $kappa$ and a point $alpha(t_0)$, $t_0in I$, on the curve. The osculating plane at $alpha(t_0)$ consists of the points $x in mathbbR^3$ with $langle x- alpha(t_0), B(t_0)rangle=0$.
If $alpha$ lies in the osculating plane, then $langle alpha(t)-alpha(t_0),B(t_0)rangle=0$ for all $tin I$. Deriving this equation twice and using the Frenet formulas gives
$$
langle T, B(t_0)rangle = 0 quad text and quad langle kappa N , B(t_0)rangle =0.
$$
Since we assume $kappa > 0$, the binormal vector field $B$ is always parallel with $B(t_0)$. So the direction and length of $B(t)$ is constant. Hence $B'=0$, so $-tau N= 0$ at every point by the Frenet-Serret formulas. This implies that $tau =0$.
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
$endgroup$
As was pointed out in the comments, the statement is not quite true. I guess you wanted to ask the following statement: If the curve lies in the osculating plane at some fixed point, then the torsion is zero.
Consider a unit speed curve $alphacolon Isubset mathbbRtomathbbR^3$ with positive curvature $kappa$ and a point $alpha(t_0)$, $t_0in I$, on the curve. The osculating plane at $alpha(t_0)$ consists of the points $x in mathbbR^3$ with $langle x- alpha(t_0), B(t_0)rangle=0$.
If $alpha$ lies in the osculating plane, then $langle alpha(t)-alpha(t_0),B(t_0)rangle=0$ for all $tin I$. Deriving this equation twice and using the Frenet formulas gives
$$
langle T, B(t_0)rangle = 0 quad text and quad langle kappa N , B(t_0)rangle =0.
$$
Since we assume $kappa > 0$, the binormal vector field $B$ is always parallel with $B(t_0)$. So the direction and length of $B(t)$ is constant. Hence $B'=0$, so $-tau N= 0$ at every point by the Frenet-Serret formulas. This implies that $tau =0$.
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
edited Mar 24 at 22:28
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
answered Mar 23 at 18:14
Ernie060Ernie060
2,940719
2,940719
add a comment |
add a comment |
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$begingroup$
This statement isn't quite right: For all $t$, $alpha(t)$ itself lies in the osculating plane to the curve at $alpha(t)$, but not all curves in $Bbb R^3$ have zero torsion.
$endgroup$
– Travis
Mar 23 at 18:02