New answers tagged

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

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 < ...


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

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 ...


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

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

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

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

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 ...


0

Slice the surface of the sphere with two horizontal planes just distant by $dy$. They enclose a portion of the cap that is a circular strip of length $2\pi x$ ($x$ being the radius). The height of the strip is $ds=\sqrt{dx^2+dy^2}$ by Pythagoras, because it is oblique. $ds$ is known as the element of arc, i.e. the length of an infinitesimal curve drawn on ...


0

The idea they are trying to put across is that a surface can be generated by an arc. In the sketch, they are highlighting an arc of the circle and indicating that when the arc revolves around the y-axis, the arc sweeps out an area. Then entire circle would generate the sphere, but the arc only generates the spherical cap. Until now we have been talking ...


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

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

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 ...


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...


0

Assuming that we are calculating at a normal coordinate $(x_1, x_2)$ at $x$. Then $$\begin{split}\nabla_i N &= \langle \nabla_i N, e_1\rangle e_1 + \langle \nabla_i N, e_1\rangle e_1 \\ &= -h_{1i} e_1 -h_{2i} e_2 \end{split}$$ Differentiate again, $$\nabla_i \nabla _i N = -(\nabla_i h_{1i}) e_1 - (\nabla_i h_{2i}) e_2 - (h_{1i})^2N - (h_{2i})^2 ...


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 ...


0

$\Phi$ is conformal is equivalent to saying that for every $x$ the Jacobian matrix is a similitude which is equivalent to saying that: $Jac(\Phi_x)=\pmatrix{a_x &b_x\cr -b_x & a_x}$ or $Jac(\Phi_x)=\pmatrix{a_x &b_x\cr b_x & -a_x}$ since $Jac(\Phi_x)=\pmatrix{\partial f_x &\partial f_y\cr \partial g_x & \partial g_y}$ you have 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 ...


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 ...


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 ...


0

HINT: Two parabolic cylinders cut by a transverse plane. The projections on $x,y,z $ planes are: $$\begin{cases} x^2-2 z+1=0 \\ x^2 - 2 y -1 = 0 \\ y-z +1 =0 \end{cases} $$ The constant has to be set to zero in the last equation in order that it become the generator of a cone passing through origin: $$ y=z $$


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 ...


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) ...


1

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


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 ...


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" ...


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.


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 ...


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 ...


0

Nevermind, please ignore my question, I figured it out. I had to make sure the X was greater than 0, and I included the dip in the calculator in the guess. Edit: After answering my own question and reloading the page, I realized @MartinR said r had to be greater than 0, which was the issue. So, good job on figuring it out, Martin.


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 ...


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) = ...


2

Let me give an explicit answer: Write $M$ for your surface. First, the $2-$nd fundamental form is typically defined as $II(u,v) := - I(d \nu (u) , v)$ where $\nu : M \rightarrow S^2$ is the Gauss map, i.e. the map that assigns to each point $p$ the unit normal vector to the tangent plane of $p$. (Note that I had a different sign in my comment. I was used ...


1

You know $$\sigma_u\sigma_v = 0 $$ which implies $$\sigma_{uu}\sigma_{v}+\sigma_{u}\sigma_{uv}=0\\ \sigma_{vu}\sigma_{v}+\sigma_{u}\sigma_{vv}=0$$ You also know $$\sigma_u^2=\sigma_v^2$$ which implies $$2\sigma_{uv}\sigma_u=2\sigma_{vv}\sigma_{v}\\ 2\sigma_{uu}\sigma_u=2\sigma_{uv}\sigma_{v} $$ Insert the first two implications into the second to arrive ...



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