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Confusion on the relationship between asymptotic directions and $dN_p$ for a point on a surface

parabolic or planar pointsQuestions on surfaces touching along a curve (why is it a curvature line?)Any Straight Line Contained in a Surface is Asymptotic and Hyperbolic “Squares”question from do carmo diff. geometrySome confusion about normal vector, curvature and normal curvature in Do Carmo's textbook.Gauss map and the curvature of a regular curve on a surfaceFinding Parabolic and Planar PointsCurvature of a curve on a surface$K=0$ at each point of a line on a surface that locally lies on one side of the tangent planeQuestion on Do Carmo Diff Geo about Gaussian Maps

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I am using Do Carmo's Differential Geometry of Curves and Surfaces textbook and I am a little confused on section 3.2. An asymptotic direction of S at p is defined to be a direction for which the normal curvature is zero. Exercise 1 of 3-2 implies that hyperbolic points have asymptotic directions. Exercise 2 asks to show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar, and the solution makes use of the fact that $dN_p(v) = 0$ implies $det[dN_p]=0$. (Thus, point is parabolic or planar)

The part where I am confused about is I thought that the normal curvature equalling zero along some direction meant that the normal vector would stay constant along that direction, and thus $dN_p(v)$ would be 0. But, since hyperbolic points have points where the normal curvature is 0, this would invalidate the solution for exercise 2. So, is my understanding of normal curvature and differential of the Gauss map flawed? On a surface $S$ at a point $p$ with a direction $v$, how does the normal curvature at $p$ along $v$ relate to $dN_p(v)$ if at all?

Solution for exercise 2 found here:
http://mathhelpforum.com/differential-geometry/57838-parabolic-planar-points.html

Sorry for being quite wordy, as this is my first question asked here and I wasn't quite sure how to phrase it.

differential-geometry

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asked Mar 14 at 16:37

Alan HuangAlan Huang

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$begingroup$

I am using Do Carmo's Differential Geometry of Curves and Surfaces textbook and I am a little confused on section 3.2. An asymptotic direction of S at p is defined to be a direction for which the normal curvature is zero. Exercise 1 of 3-2 implies that hyperbolic points have asymptotic directions. Exercise 2 asks to show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar, and the solution makes use of the fact that $dN_p(v) = 0$ implies $det[dN_p]=0$. (Thus, point is parabolic or planar)

The part where I am confused about is I thought that the normal curvature equalling zero along some direction meant that the normal vector would stay constant along that direction, and thus $dN_p(v)$ would be 0. But, since hyperbolic points have points where the normal curvature is 0, this would invalidate the solution for exercise 2. So, is my understanding of normal curvature and differential of the Gauss map flawed? On a surface $S$ at a point $p$ with a direction $v$, how does the normal curvature at $p$ along $v$ relate to $dN_p(v)$ if at all?

Solution for exercise 2 found here:
http://mathhelpforum.com/differential-geometry/57838-parabolic-planar-points.html

Sorry for being quite wordy, as this is my first question asked here and I wasn't quite sure how to phrase it.

differential-geometry

share|cite|improve this question

asked Mar 14 at 16:37

Alan HuangAlan Huang

1

$endgroup$

add a comment | 

$begingroup$

I am using Do Carmo's Differential Geometry of Curves and Surfaces textbook and I am a little confused on section 3.2. An asymptotic direction of S at p is defined to be a direction for which the normal curvature is zero. Exercise 1 of 3-2 implies that hyperbolic points have asymptotic directions. Exercise 2 asks to show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar, and the solution makes use of the fact that $dN_p(v) = 0$ implies $det[dN_p]=0$. (Thus, point is parabolic or planar)

The part where I am confused about is I thought that the normal curvature equalling zero along some direction meant that the normal vector would stay constant along that direction, and thus $dN_p(v)$ would be 0. But, since hyperbolic points have points where the normal curvature is 0, this would invalidate the solution for exercise 2. So, is my understanding of normal curvature and differential of the Gauss map flawed? On a surface $S$ at a point $p$ with a direction $v$, how does the normal curvature at $p$ along $v$ relate to $dN_p(v)$ if at all?

Solution for exercise 2 found here:
http://mathhelpforum.com/differential-geometry/57838-parabolic-planar-points.html

Sorry for being quite wordy, as this is my first question asked here and I wasn't quite sure how to phrase it.

differential-geometry

share|cite|improve this question

asked Mar 14 at 16:37

Alan HuangAlan Huang

1

$endgroup$

share|cite|improve this question

asked Mar 14 at 16:37

Alan HuangAlan Huang

1

share|cite|improve this question

asked Mar 14 at 16:37

Alan HuangAlan Huang

1

asked Mar 14 at 16:37

Alan HuangAlan Huang

1

asked Mar 14 at 16:37

Alan HuangAlan Huang

1

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Welcome to MSE. Normal curvature in the direction of a unit vector $v$ is $pm langle dN_p(v),vrangle$. So normal curvature $0$ merely signifies that $dN_p(v)$ is orthogonal to $v$, not that it is $0$ itself. A good example is to consider a helicoid along one of the rulings: The tangent plane twists as you move along the ruling.
(You might enjoy reading my differential geometry text as an additional resource.)

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answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692

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$begingroup$

Welcome to MSE. Normal curvature in the direction of a unit vector $v$ is $pm langle dN_p(v),vrangle$. So normal curvature $0$ merely signifies that $dN_p(v)$ is orthogonal to $v$, not that it is $0$ itself. A good example is to consider a helicoid along one of the rulings: The tangent plane twists as you move along the ruling.
(You might enjoy reading my differential geometry text as an additional resource.)

share|cite|improve this answer

answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

add a comment | 

0

$begingroup$

Welcome to MSE. Normal curvature in the direction of a unit vector $v$ is $pm langle dN_p(v),vrangle$. So normal curvature $0$ merely signifies that $dN_p(v)$ is orthogonal to $v$, not that it is $0$ itself. A good example is to consider a helicoid along one of the rulings: The tangent plane twists as you move along the ruling.
(You might enjoy reading my differential geometry text as an additional resource.)

share|cite|improve this answer

answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

add a comment | 

$begingroup$

Welcome to MSE. Normal curvature in the direction of a unit vector $v$ is $pm langle dN_p(v),vrangle$. So normal curvature $0$ merely signifies that $dN_p(v)$ is orthogonal to $v$, not that it is $0$ itself. A good example is to consider a helicoid along one of the rulings: The tangent plane twists as you move along the ruling.
(You might enjoy reading my differential geometry text as an additional resource.)

share|cite|improve this answer

answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692

$endgroup$

share|cite|improve this answer

answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692

answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692

answered Mar 14 at 17:29

Ted ShifrinTed Shifrin

64.5k44692