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1

The general form is $( |x|^p + |y|^p + |z|^p )^\frac{1}{p} = 1$, which is just the surface with $p$-norm equal to $1$. The $1$-norm gives an octahedron, the $2$-norm gives a sphere, the $\infty$-norm (defined as the limit of the $p$-norm as $p \to \infty$ gives a cube, and in-between looks like in-between shapes. For yours, it is the $\frac{2}{3}$-norm which ...

1

I always use mathematica for plotting things like these. ContourPlot3D[ x^(2/3) + y^(2/3) + z^(2/3) == 1, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]

1

If $(u, v, w)$ lies on the unit sphere in $\mathbf{R}^{3}$, i.e., if $u$, $v$, and $w$ are real numbers such that $u^{2} + v^{2} + w^{2} = 1$, then $$(x, y, z) = (u^{3}, v^{3}, w^{3}) = F(u, v, w)$$ lies on your surface. In fact, the mapping $F$ defines a "homeomorphism" between the unit sphere and your surface, i.e., a continuous bijection with continuous ...

1

Your surface $S$ initially has a representation of the form $$F:\quad {\bf w}=(u,v)\mapsto {\bf x}=\bigl(u,v,f(u,v)\bigr)\ .$$ You then rotate $S$ in space by means of some rotation map $R$, resp. some orthogonal matrix $[R]$: $${\bf x}\mapsto {\bf x}'=R{\bf x}\ .$$ The rotated surface $S':=R(S)$ then has a representation $$G:=R\circ F:\quad {\bf w}\mapsto ... 2 Your map F(x,y) = R\left(\matrix{x\\ y\\ f(x, y)}\right) is the function composition R \circ G where R is a linear map (the rotation) and G: (x,y) \mapsto (x,y,f(x,y)). Then you probably know what the derivative of a function composition is? If not, the derivative of F is F^\prime=R^\prime \circ G^\prime. As R is linear, it is equal to its ... 0 Consider \Gamma as 2 distinct curves, \gamma_1 and \gamma_2 between 2 distinct points on \Gamma. Parametrize the curves over [0,1]. The set$$ \{a \gamma_1(t) + (1-a) \gamma_2(t) | \ a \in [0,1] \ \& \ t \in [0,1]\} $$forms a surface with coordinates (a,t) and boundary \Gamma. The coordinate system defines an orientation. 1 Parametric Breakdown Start by dividing the function from polar form into a set of two parametric functions. That is, the horizontal and vertical component of the radial coordinate r:$$ x =r \cos{\theta}\qquad\quad y=r\sin{\theta}\\ \implies \dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta} \qquad\quad \dfrac{dy}{d\theta} = ...

0

Such a strong result as Proposition 3.4 is not necessary to do this problem. Perhaps you could get by with something much weaker. Here's a few suggestions. For instance, suppose you could prove a very weak version of Proposition 3.4 which says the following: for each $b$, if there exists $a_i$ such that $i(a_i,b) \ne 0$, then $i(M_a(b),b) \ne 0$. Here I use ...

2

$\newcommand{\Cpx}{\mathbf{C}}$Here's a fairly detailed sketch of the underlying framework, together with hints for applying the machinery to your situation. Generalities: Let $X$ be a connected holomorphic manifold, $\phi:X \to X$ a biholomorphism, and $A$ the cyclic group generated by $\phi$. To get a manifold quotient, we'll assume $A$ acts properly ...

1

Try $(x,y,z)\mapsto (x|x|,y|y|,z|z|)$. Note that this is invertible by dividing each term by positive square root of absolute value of each coordinate when the coordinate is not 0, and sending 0 to 0 otherwise. Since $\sqrt{|x|x||}=\sqrt{|x^2|}=|x|$, and dividing $x|x|$ by this gives $x$. Since both this function and its inverse are continuous, it is a ...

3

You want a more linear map. Note for that, that on every ray through the origin, there is precisely one point of each of those spheres. That defines you already a map. Another way of saying this, is that both spheres are unit norm balls. You can get a map from the cornered one to the normal one by $z \mapsto \frac z {||z||}$ where $||-||$ is the standard ...

3

Hint: Try $f: \text{cornered sphere} \to \mathbb{S}^2(1)$ given by $$f(x,y,z) = (x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}}(x,y,z).$$ This map projects the cornered sphere into the unit sphere, radially.

0

Let $\Phi$ be a parameterisation for the surface $\Sigma\subset{\Bbb R}^3$. This is a map $\left(\begin{array}{c} v\\ w \end{array}\right) \stackrel{\Phi}\longmapsto \left(\begin{array}{c} x\\ y\\ z \end{array}\right).$ Suppose that the point $a=a^1e_1+a^2e_2$ of ${\Bbb R}^2$ in the domain of $\Phi$ is mapped to $p$ in $\Sigma$. Then ...

0

Let $f(x,y,z)=f(\vec r)=C$, where $C$ is a constant, define a surface. Then, for any curve $\vec r=\vec r(s)$ on the surface, we have $f'(s)=0$. By the chain rule we have $$f'(s)=\nabla f(\vec r)\cdot \vec r'(s)=0$$ Thus, since $\vec r'(s)$ is tangent to the surface, the gradient at any point on the surface is normal to the surface there. Thus, the ...

1

Hint: I think you want the vertices of the convex hull of the given set of points ( the vertices of a polyhedron). I just discovered that you can do this reasonably fast in Mathematica. Here is how to find the vertices of the convex hull for a random set of 50 points in a cube: pts = RandomReal[ {-1, 1}, {50, 3}]; polyh = ConvexHullMesh[pts]; ...

1

Answer to 1: Use the right hand rule, with respect to an orientation on the surface. The local model for using the right hand rule is that, when applied in the upper half plane of $\mathbb{R}^2$, it gives you the positive orientation on the $x$-axis. Answer to 2: The external angles are not dependent on orientation. In your example, each of the four ...

1

Note that the cross product $e_{1} \times e_{2}$ depends on whether you're interpreting these vectors as elements of the plane (in which case the product is customarily viewed as the scalar $1$) or as elements of space (in which case $e_{1} \times e_{2} = e_{3}$). The first line of your question suggests $X$ is defined (only) on $U$, in which case $dX(e_{1} ... 1 The plane is the graph of the function$g(x, y) = 5x + 3y$. The gradient of this function is$(5, 3)$. The gradient of$f(x, y)$is$(3x^2 + 2y, 2x + 1)$. Now set one equal to the other, and solve. 0 Another way to see this is from the following description of the Ricci tensor evaluated on a unit length vector$X \in T_pM$: For a manifold$(M,g)$of dimension$n$, it is given by $$\text{Ric}(X,X) = \sum_{j=2}^{n}k(\pi_{X,e_j})$$ where$(X,e_2, \ldots, e_n)$is an orthonormal basis of$T_pM$and where$k(\pi_{X,e_j})$is the sectional curvature of the ... 3 For a surface$(M, g)$, there is only independent component of the curvature tensor$R$, namely, $$R_{1212} = - R_{1221} = -R_{2112} = R_{2121},$$ and this quantity (which depends on the choice of coordinates) is related to the Gaussian curvature$K$, by $$R_{1212} = -\det(g) K$$ (which is independent of coordinates). (Beware that this sign is a matter of ... 3 The answer is basically no for continuous functions: the key lies in the phrase "of two sheets". On the other hand, through some mild discontinuous devilry one can construct such a parametrisation: it suffices to look for a parametrisation$(x(t),y(t))$of the hyperbola $$x^2-y^2=a^2,$$ and then one can rotate this about the axis of$x$to obtain the ... 0 Chappers pointed out a problem with edges; those could be handled by imposing an inequality constraint. That is, it is conceivable to have an implicit equation for the Möbius strip of the form $$M=\{(x,y,z) : F(x,y,z) = 0, \ \text{some inequality with x,y,z}\}$$ Another issue is the non-orientability. If the gradient of$F$did not vanish anywhere on$M\$, ...

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