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Chapter 3: Problem 10

Assume that the osculating plane of a line of curvature \(C \subset S\), which is nowhere tangent to an asymptotic direction, makes a constant angle with the tangent plane of \(S\) along \(C\). Prove that \(C\) is a plane curve.

Short Answer

Expert verified
The line of curvature is a plane curve because the osculating plane's constant angle with the tangent plane implies the normal's projection on the tangent plane is constant.

Step by step solution

01

- Understanding the Problem

We need to prove that a line of curvature, under given conditions, is a plane curve. Given that the osculating plane of the line of curvature makes a constant angle with the tangent plane of the surface, and the line of curvature is not tangent to an asymptotic direction.
02

- Define the Line of Curvature

A line of curvature on a surface is a curve where the normal curvature in the direction of the curve is an extremum. This implies that the principal normal of the curve lies in the direction of a principal curvature of the surface.
03

- Tangent and Normal Planes

The tangent plane to the surface at any point can be described by the surface normal vector at that point. The osculating plane of the curve contains the tangent vector of the curve and the principal normal vector of the curve.
04

- Constant Angle Condition

If the osculating plane makes a constant angle with the tangent plane, this means the angle between the osculating plane and the surface's tangent plane does not change as we move along the curve. This implies the dihedral angle between the planes is constant.
05

- Implication of the Constant Angle

Since the osculating plane's angle with the tangent plane is constant, the normal vectors to these planes maintain a constant angle. Let \(\theta\) be this constant angle.
06

- Using the Frenet-Serret Formulas

The Frenet-Serret formulas provide a relationship between the tangent vector \(\textbf{T}\), the normal vector \(\textbf{N}\), and the binormal vector \(\textbf{B}\) of a curve. With the principal normal \(\textbf{N}\) being normal to the curve, and knowing that \(\textbf{T}\) and \(\textbf{N}\) lie within the osculating plane, we derive that \(\textbf{N}\) has a constant projection onto the tangent plane.
07

- Projection of the Normal Vector

The fact that the projection of the normal vector \(\textbf{N}\) is constant on the tangent plane suggests that the curve must lie in a plane parallel to this projection. If the normal vector's projection remains unchanged, the curve does not deviate from this plane.
08

- Conclusion

Since the projection of the principal normal vector remains constant, the curve itself must lie in a single plane. Thus, we have shown that the line of curvature under the given conditions is necessarily a plane curve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line of Curvature
In differential geometry, a line of curvature is a special type of curve on a surface. It is a curve where the normal curvature, which measures the bending of the surface in a particular direction, is at an extremum. This means the principal normal of the curve aligns with one of the principal directions or principal curvatures of the surface. Understanding lines of curvature is crucial because they provide important insights into the geometric properties of the surface. They can help us identify directions of maximum and minimum bending, which can be essential in practical applications like engineering and computer graphics.
Osculating Plane
The osculating plane at a given point on a curve refers to the plane that best approximates the curve at that point. It contains the tangent vector and the principal normal vector of the curve. To visualize this, imagine a plane that 'kisses' the curve as closely as possible, following its path. In mathematical terms, the osculating plane is crucial because it captures the immediate direction and bend of the curve. For our problem, the constant angle between the osculating plane and the tangent plane of the surface suggests a specific geometric relationship that must be maintained as one moves along the curve.
Tangent Plane
The tangent plane is a flat surface that just touches a curved surface at a single point, and it is perpendicular to the surface's normal vector at that point. In the context of our problem, the tangent plane plays a central role because it helps in understanding the geometric positioning of the line of curvature relative to the surface. If the osculating plane of the curve makes a constant angle with this tangent plane, it imposes a constraint that significantly simplifies the spatial behavior of the line of curvature, leading to the conclusion that the curve lies in a single plane.
Principal Normal
The principal normal vector is an important concept in analyzing curves on surfaces. It is the normal vector to the curve that points in the direction of the principal curvature. This means it is a vector perpendicular to the tangent vector at any point on the curve. In our specific exercise, the principal normal's projection remains constant on the tangent plane, suggesting that the curve does not deviate from a plane defined by this projection. Thus, the principal normal aids in showing that the line of curvature must lie entirely in a plane, making it a plane curve.
Frenet-Serret Formulas
The Frenet-Serret formulas are a set of differential equations in space that describe the kinematic properties of a particle moving along a continuous, differentiable curve. These formulas involve three fundamental vectors: the tangent vector (\textbf{T}), the normal vector (\textbf{N}), and the binormal vector (\textbf{B}). They describe how these vectors change along the curve. The formulas can be written as follows:

\[ \frac{d\textbf{T}}{ds} = \kappa \textbf{N} \ \frac{d\textbf{N}}{ds} = -\kappa \textbf{T} + \tau \textbf{B} \ \frac{d\textbf{B}}{ds} = -\tau \textbf{N} \]\
where \( \kappa \) is the curvature and \( \tau \) is the torsion of the curve. In our problem, these formulas help demonstrate that the osculating plane maintaining a constant angle with the tangent plane implies the curve must stay within a single plane, as the normal vector \textbf{N} has a constant projection onto the tangent plane.

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Most popular questions from this chapter

(The Hessian.) Let \(h: S \rightarrow R\) be a differentiable function on a surface \(S\), and let \(p \in S\) be a critical point of \(h\) (i.e., \(d h_{p}=0\) ). Let \(w \in T_{p}(S)\) and let $$ \alpha:(-\epsilon, \epsilon) \rightarrow S $$ be a parametrized curve with \(\alpha(0)=p, \alpha^{\prime}(0)=w\). Set $$ H_{p} h(w)=\left.\frac{d^{2}(h \circ \alpha)}{d t^{2}}\right|_{t=0} $$ a. Let \(\mathbf{x}: U \rightarrow S\) be a parametrization of \(S\) at \(p\), and show that (the fact that \(p\) is a critical point of \(h\) is essential here) $$ H_{p} h\left(u^{\prime} \mathbf{x}_{u}+v^{\prime} \mathbf{x}_{v}\right)=h_{u u}(p)\left(u^{\prime}\right)^{2}+2 h_{u v}(p) u^{\prime} v^{\prime}+h_{v v}(p)\left(v^{\prime}\right)^{2} . $$ Conclude that \(H_{p} h: T_{p}(S) \rightarrow R\) is a well-defined (i.e., it does not depend on the choice of \(\mathbf{x}\) ) quadratic form on \(T_{p}(S) . H_{p} h\) is called the Hessian of \(h\) at \(p\). b. Let \(h: S \rightarrow R\) be the height function of \(S\) relative to \(T_{p}(S)\); that is, \(h(q)=\langle q-p, N(p)\rangle, q \in S\). Verify that \(p\) is a critical point of \(h\) and thus that the Hessian \(H_{p} h\) is well defined. Show that if \(w \in T_{p}(S)\), \(|w|=1\), then \(H_{p} h(w)=\) normal curvature at \(p\) in the direction of \(w .\) Conclude that the Hessian at \(\mathrm{p}\) of the height function relative to \(\mathrm{T}_{\mathrm{p}}(\mathrm{S})\) is the second fundamental form of \(\mathrm{S}\) at \(\mathrm{p}\).

Determine the asymptotic curves and the lines of curvature of the helicoid \(x=v \cos u, y=v \sin u, z=c u\), and show that its mean curvature is zero.

(Morse Functions on Surfaces.) A critical point \(p \in S\) of a differentiable function \(h: S \rightarrow R\) is nondegenerate if the self-adjoint linear map \(A_{P} h\) associated to the quadratic form \(H_{p} h\) (cf. the appendix to Chap. 3) is nonsingular (here \(H_{p} h\) is the Hessian of \(h\) at \(p\); cf. Exercise 22). Otherwise, \(p\) is a degenerate critical point. A differentiable function on \(S\) is a Morse function if all its critical points are nondegenerate. Let \(h,: S \subset R^{3} \rightarrow R\) be the distance function from \(S\) to \(r ;\) i.e., $$ h_{r}(q)=\sqrt{\langle q-r, q-r\rangle}, \quad q \in S, \quad r \in R^{3}, \quad r \notin S . $$ a. Show that \(p \in S\) is a critical point of \(h\), if and only if the straight line \(p r\) is normal to \(S\) at \(p\). b. Let \(p\) be a critical point of \(h_{x}: S \rightarrow R\). Let \(w \in T_{p}(S),|w|=1\), and let \(\alpha:(-\epsilon, \epsilon) \rightarrow S\) be a curve parametrized by arc length with \(\alpha(0)=p, \alpha^{\prime}(0)=w\). Prove that $$ H_{p} h_{r}(w)=\frac{1}{h_{r}(p)}-k_{n}, $$ where \(k_{n}\) is the normal curvature at \(p\) along the direction of \(w\). Conclude that the orthonormal basis \(\left\\{e_{1}, e_{2}\right\\}\), where \(e_{1}\) and \(e_{2}\) are along the principal directions of \(T_{P}(S)\), diagonalizes the self-adjoint linear map \(A_{p} h_{r}\). Conclude further that \(p\) is a degenerate critical point of \(h_{r}\) if and only if either \(h_{r}(p)=1 / k_{1}\) or \(h_{r}(p)=1 / k_{2}\), where \(k_{1}\) and \(k_{2}\) are the principal curvatures at \(p\). c. Show that the set $$ B=\left\\{r \in R^{3} ; h, \text { is a Morse function }\right\\} $$ is a dense set in \(R^{3}\); here dense in \(R^{3}\) means that in each neighborhood of a given point of \(R^{3}\) there exists a point of \(B\) (this shows that on any regular surface there are "many" Morse functions).

Let \(S\) be a surface and \(\mathbf{x}: U \rightarrow S\) be a parametrization of \(S\). Then $$ a(u, v) u^{\prime}+b(u, v) v^{\prime}=0, $$ where \(a\) and \(b\) are differentiable functions, determines a field of directions \(r\) on \(\mathbf{x}(U)\), namely, the correspondence which assigns to each \(\mathbf{x}(u, v)\) the straight line containing the vector \(b \mathbf{x}_{u}-a \mathbf{x}_{v}\). Show that a necessary and sufficient condition for the existence of an orthogonal field \(r^{\prime}\) on \(\mathbf{x}(U)\) (cf. Example 3 ) is that both functions $$ E b-F a, \quad F b-G a $$ are nowhere simultaneously zero (here \(E, F\), and \(G\) are the coefficients of the first fundamental form in \(\mathbf{x}\) ) and that \(r^{\prime}\) is then determined by $$ (E b-F a) u^{\prime}+(F b-G a) v^{\prime}=0 . $$

Let \(\lambda_{1}, \ldots, \lambda_{m}\) be the normal curvatures at \(p \in S\) along directions making angles \(0,2 \pi / m, \ldots,(m-1) 2 \pi / m\) with a principal direction, \(m>2\). Prove that $$ \lambda_{1}+\cdots+\lambda_{m}=m H, $$ where \(H\) is the mean curvature at \(p\).
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