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7

First, look at some pictures of hyperboloids, to get a feeling for their shape and symmetry. There are two ways to think of your hyperboloid. Firstly, it's a surface of revolution. You can form it by drawing the hyperbola $x^2 - z^2 = 1$ in the plane $y=0$, and then rotating this around the $z$-axis. Another way to get your hyperbola is as a "ruled" ...


3

HINT: Note that our surface is a surface of revolution, putting us a general context, let $S$ be a surface of revolution with parametrization $X\left(u,v\right)=\left(f\left(u\right)\cos \left(v\right),f\left(u\right)\sin \left(v\right),g\left(u\right)\right)$. Let $\gamma$ a curve in $S$, this is, $\gamma ...


3

In general, an orientation is not exactly a ``smooth choice of unit normal''. This definition is only really helpful for 2D surfaces embedded in 3D. I'm guessing that you need to consider more general manifolds. The best way to generalise is as follows. The normal vector $\mathbf{n}$ to a 2D surface in 3D is really a machine that maps pairs of tangent ...


2

Using your first definition of orientable, let $$\{(U_i,\phi_i)\}_{i \in I} $$ be an atlas for $S$ satisfying the orientation requirement. To be clear on the notation, $U_i \subset S$ is open and $\phi_i : U_i \to \mathbb{R}^2$ is a homeomorphism, and the orientation requirement says that each overlap map $$\phi_j \circ \phi_i^{-1} : \phi_i(U_i \cap U_j) ...


2

Hint For two geodesics, consider the planes $\Pi$ of symmetry of the hyperboloid $H$ through $(1,0,0)$, and use symmetry and the uniqueness of geodesics to argue that the curves $\Pi \cap H$ must be geodesic. For the other two, one can use that the hyperboloid of one sheet is doubly ruled.


2

There are infinitely many geodesics on it in each direction. The meridian, the circumference at neck ( minimum radius), two ruled straight line asymptotes are the 4 principal geodesics you refer to. Their normal curvatures follow Euler's law $$ k_n = k_1 \cos^2 \alpha + k_2 \sin ^2 \alpha \tag{1} $$ respectively for 180 degree rotation the four $k_n's ...


2

I'm not sure I understand this but it looks like you found a typo. In the last line,assuming $x_{u t}=x_{t u}$ then from the previous line, we have $x_{u t}=x_{t u}=w'$. From a previous line,$x_t\wedge x_u=\beta' \wedge w+u w'\wedge w$, and since $u w'\wedge w$ is orthogonal to $w'$ we should have $(x_t,x_u,x_{u t})=(x_t\wedge x_u,w')=(\beta' \wedge w+u ...


2

Let $R$ be the region of $S$ between the intersection curves $c_0$ and $c_1$ ($c_i$ denotes the intersection curve of the surface of revolution $S$ with the plane $z=i,\ i=0,1$) Thus, if both were geodesics then by the Gauss-Bonnet theorem we would have $$\int_R K=2\pi\chi(R) , \ \ \ (\dagger)$$ where $K$ is the Gaussian curvature and $\chi(R)$ is the ...


2

Let's call the three families the constant-$u$ family ($C_u$), the constant-$v$ family ($C_v$), and the constant-$w$ family ($C_w$). A surface in the family $C_u$ is of the form $$ G(v,w) = \Sigma(u_0,v,w) $$ for some fixed $u=u_0$. Similarly, a surface in the family $C_v$ is of the form $$ H(u,w) = \Sigma(u,v_0,w) $$ for some fixed $v=v_0$. The ...


2

The surface is contained inside the sphere, but the tangent plane is outside the sphere, except at $p$. Therefore as you leave $p$ in any direction, the surface will curve away from the tangent plane at least as fast as the sphere does.


2

Take a point on $S$, $p = (x,y,z)$. We know that $x^2 + y^2 - z^2 - 1 = 0$. Thus $(x+z)(x-z) = (1+y)(1-y)$. Suppose $(x-z) \not = 0$ and $(1+y) \not = 0$ (you can do these two cases by hands). Then of course $\frac{x+z}{1+y} = \frac{1-y}{x-z} = a$ for some $a \in \mathbb{R}$. Since $ctg(x)$ is onto, there is $\theta$ such that $a = ctg(\theta) = ...


1

From $z^2=x^2+y^2-1$ the surface K can be recover by the images of these paramétrizations : $f(x,y)=(x,y,\sqrt{x^2+y^2-1})$ and $g=(x,y,-\sqrt{x^2+y^2-1})$. The two functions are defined on the region ${(x,y) \in \mathbb {R}^2} / x^2+y^2-1>=0$ .


1

First, let $S$ be a regular surface of $\mathbb{R}^3$with Gauss map $N$. Let $c\colon I:=(-\varepsilon,\varepsilon)\subset \mathbb{R}\longrightarrow S$ be a curve parametrized by arc length with $c(0)=p\in S$ and $c'(0)=v\in T_pS$. Then we have the following: $$II_p(v)=-\langle dN_p(v),v\rangle=-\langle dN_{c(0)}(c'(0)),c'(0)\rangle=-\langle (N\circ ...


1

Let $X\colon U\subset \mathbb{R}^2\longrightarrow S$ be a parametrization of a regular surface $S$ and let $p=X(u,v)$. If $f\colon S\rightarrow \mathrm{R}$ is a differentiable function then $\mathrm{grad} f(p)\in T_pS$. Thus, $$\mathrm{grad} f (p)=\alpha X_u+\beta X_v,\ \ \ (\dagger)$$ where $\alpha,\beta $ are functions defined on $U$. From $(\dagger)$ you ...


1

The intersection of two surfaces is a curve $\Sigma(u_0,v_0,w)$ resp. ($\Sigma(u_0,v,w_0)$, $\Sigma(u,v_0,w_0)$) which has two of three parameters fixed. These curves are coordinate lines on a surfaces patches defined by $\Sigma$. From the second part of the exercise you know that matrices of both first and second fundamental form are diagonal and ...


1

I think you are right that the example needs some work. A regular plane curve could have figure $8$ intersections with itself. In such cases the exterior and interior of the curve can be confusing. After rotation it is not hard to imagine the resulting surface, which tends to have a cusp-type singularity. It is very unlikely De Carmo meant this when writing ...


1

I am not sure if I understood you properly, ... whether self intersection is for the 3D curves on surface of revolution or for the meridian itself.By generating curve do you mean the meridian or the non-planar 3D space curve written on it? (f(v),g(v)) is a parameterization determining a single unique point on a meridian through which any number of curves ...


1

Though you've got an excellent answer, here's a lower-tech argument for posterity: If the equation of the surface in cylindrical coordinates is $r = f(z)$ for some positive, $C^{2}$ function $f$, it's well-known that: A latitude $z = c$ is a geodesic if and only if $f'(c) = 0$. The Gaussian curvature and $f''$ have opposite sign. Particularly, if $K < ...


1

LONG SOLUTION. The equations defining the parabola can be rewritten as parametric equations as follows: $$ \cases{ x=t \cr \displaystyle y={1\over2}t^2-{1\over2}\cr \displaystyle z={1\over2}t^2+{1\over2}\cr } $$ The parametric equation of the line passing through the origin and a point $P(t)=(t,\ t^2/2-{1/2},\ t^2/2+{1/2})$ of the parabola is just ...


1

Your isosurface is a sphere, having center at the centroid of the $p_i$ and whose radius $r$ is given by $$ r^2={v\over n}-{1\over n^2}\sum_{i\ne j}(\vec p_i-\vec p_j)^2. $$ That follows from the identity $$ \sum_{i=1}^n(\vec x-\vec p_i)^2= n\left(\vec x -\sum_{i=1}^n{\vec p_i\over n}\right)^2+ {1\over n}\sum_{i\ne j}{(\vec p_i-\vec p_j)^2}. $$ Once $r$ is ...


1

Since $p$ is a hyperbolic point there exists a neighborhood $U_p\subset S$ of $p$ such that $$K(q)<0,\ \forall q\in U_p.$$ Consider now a parametrization of $p$ in $U_p$, i.e. $$X\colon V\subset \mathbb{R}^2\longrightarrow U_p\subset S, \ \ (u,v)\mapsto X(u,v),\ \ \&\ p\in X(V),$$ where $V$ is an open subset of $\mathbb{R}^2$. Since every point of ...


1

If $T$ is a two dimensional vector space, an euclidian structure is given by a symmetric bilinear form, positive, non degenerate. In a given coordinate system $(x,y)$ it is given by $ Ex^2+2Fxy+Gy^2$. $E,F,G$ depends on the coordinate system, but the euclidian structure do not. This is what happen in your case, just you have a two parameter family of ...


1

First, if $S$ is a regular surface and $c\colon I\subset \mathbb{R}\longrightarrow S$ is a geodesic then $$N (c(s))\parallel \vec{n}(s),\ \ \forall s\in I,\ \ \ \ \ \ \ \ (\dagger)$$ where $N$ is the unit normal of $S$ and $\vec{n}$ the first normal of the curve $c$. Thus, in your problem you are interested in finding the parallels of your surface of ...


1

A surface is not orientable if and only if it contains a Möbius band embedded. This property is invariant by diffeomorphims or homeomorphism.


1

You were indeed correct to take a look at the Euler Characteristic. Recall that for a surface, $\chi(X) = V-E+F$. Now let n be the number of hexagons and m be the number of heptagons. The number of faces is then $F= n+m$. Now each hexagon has 6 edges and each heptagon has 7 edges, and furthermore each edge is shared by two of the figures. Therefore, the ...


1

Take any curve $z=f(r)$. Then $z=f(\sqrt{x^2+y^2})$ will be a surface of revolution. The other question is unclear...


1

For the first part: The surface of revolution is parameterized as : $(x, y, z) = ( u \cos(v), u \sin(v), z= f(u) )$ u can be taken as any other smooth function g(u) also. As you know for a cone $$ u = z \tan (\alpha) ;\, z= u \cot (\alpha), $$ and sphere $$ u = \sqrt{ 1-z^2} ;\, z = \sqrt{ 1-u^2}, $$ it can be used for sweeping any meridian like a ...



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