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If ${\bf x} = A \bf X$, then ${\bf x}^T Q {\bf x} = {\bf X}^T A^T Q A \bf X$.
For completeness, I add some numerical details to the excellent answer by Noam D. Elkies. Let $C$ be the cone such that the lateral surface unrolls to a circular sector of angle $2\theta$. There are two sources of distortion: Between two points on lateral surface (worst case: same level, diametrally opposite). This gives $c\ge (\pi/\theta) ... 0 I suggest please look up Euler's formula: kn = k1 cos(si)^2 + k2 sin(si)^2 , draw a graph of kn as a function of si.It makes things clearer about extremality of k1 and k2. Also it is very instructive to draw a Mohr's circle with kn's on x-axis and geodesic torsion taug = (k1 - k2)*sin(si)*cos(si) on y-axis.si is the parameter, your reference direction. The ... 4 suppose the surface is described using$z=f(x,y)d\overrightarrow{S} = \hat{n}dS = \langle -f_x, -f_y, 1\rangle dxdy\hat{n} = \dfrac{\langle -f_x, -f_y, 1\rangle}{||\langle -f_x, -f_y, 1\rangle||}dS = ||\langle -f_x, -f_y, 1\rangle|| dxdy $In your particular problem :$d\overrightarrow{S} = \hat{n}dS = \langle 2x, 2y, 1\rangle dxdy$... 0 The " big rim" is a cuspidal edge or equator; it has infinity and zero principal curvatures. In between them are zero normal curvature asymptotics having constant geodesic torsion continuously from - infinity to + infinity, across the rim. 15 How about a sharp cone? Suppose the cone's lateral surface unrolls to a circular sector of angle$2\theta$for some small positive$\theta$. Then:$\bullet$the base is flat, so any$a,b$on the base are joined by a line of length$|a-b|$.$\bullet$if$a$is on the base and$b$on the side, then we can choose$\gamma$to go straight down from$b$to the ... 0 If$\mathbf{A}: [0,1] \to \mathbb{R}^3$and$\mathbf{B}: [0,1] \to \mathbb{R}^3$are any two parametric curves whatsoever, and$\mathbf{A}(0) = \mathbf{B}(0) = \mathbf{C}$, then you can define a surface$\mathbf{S}(u,v)$by $$\mathbf{S}(u,v) = \mathbf{A}(u) + \mathbf{B}(v) - \mathbf{C} \quad\quad (0 \le u \le 1 \; ; \; 0 \le u \le 1)$$ Note that ... 0 Hyperboloid of 1 sheet. Saddle point is an attribute/ character of all surface points. If you hold a 3D model of it in your hands, its nomenclature is invariant by the direction of your view :) but lines of projection can have a separate name. 0 $$z=\dfrac{cr^2}{1+\sqrt{1-(1+k)c^2r^2}},$$ rationalise the denominator $$z = \frac{cr^2\left(1+\sqrt{1-(1+k)c^2r^2}\right)}{\left(1-\sqrt{1-(1+k)c^2r^2}\right)\left(1+\sqrt{1-(1+k)c^2r^2}\right)} \\ =\frac{cr^2\left(1+\sqrt{1-(1+k)c^2r^2}\right)}{(1+k)c^2r^2} \tag{1}$$ then re-write Eq.(1) as $$cr^2\sqrt{1-(1+k)c^2r^2} = z(1+k)c^2r^2 - cr^2$$ I ... 0 First part of your question Basically it's all to do with the inverse function theorem. But more explicitly: Let$\mathbf{x}:U \subset \mathbb{R}^2 \rightarrow S$be a parameterization and$q$be a point in$U$. Define$\mathbf{F}:U \times I \rightarrow \mathbb{R}^3$by $$\mathbf{F}(u,v,t):= \mathbf{x}(u,v) + t\cdot \hat{e}_3, \hspace{1in} t \in I.$$ ... 0 By simplification N comes out with a negative sign: N = - φ′′/ψ′. Since K= (LN−M^2)/(EG−F^2) = LN/EG = φ φ′′/φ^2 = φ′′/φ , do we get K φ - φ′′= 0 ?? The sign is important to choose between elliptic/hyperbolic cases using above standard Gauss definition for K. Narasimham 0 Either the question has an error, or your informant was talking entirely through their hat: after all, given any two varieties$X$and$Y$, one can get a regular map$X \rightarrow Y$by choosing any point on$Y$, and mapping$X$to that point! However, if we insert the word dominant before rational map, then the question makes sense, and indeed the answer ... 0 A differential form is sometimes written with the point as subscript, so, for example, if we denote$dS$the 2-form of the area element at a point$\gamma$we can write it also as$dS_\gamma$. Remember that a 2-form takes a couple of tangent vectors (whose area it computes) as arguments written in the brackets:$dS_\gamma(u,v)$. 1 It is much simpler to produce$S$by means of the parametric representation $${\bf f}:\quad(r,\phi)\mapsto(r\cos\phi, r\sin\phi,r)\qquad(0\leq r\leq 3, \ -\pi\leq\phi\leq\pi)\ .$$ Then $${\bf f}_r=(\cos\phi,\sin\phi,1),\quad{\bf f}_\phi=(-r\sin\phi,r\cos\phi,0), \quad {\bf f}_r\times{\bf f}_\phi=(-r\cos\phi,-r\sin\phi,r)\ .$$ As given the normal vector ... 2 What you are asking for is a special case of "Poincare duality", see also this question. Guillemin and Pollack is my favorite reference. Suppose you have a nonseparating simple loop$a$in$S$. Since$a$is nonseparating, there exists a simple loop$b$in$S$which crosses$a$transversally in exactly one point. Orienting these loops, we can assume that ... 3 By Stoke's theorem, $$\int_S (\nabla \times \mathbf{A}) \cdot \mathbf{n} \,d\sigma = \oint_C \mathbf{A} \cdot d\mathbf{r},$$ where$C = \{(x,y,z):x^2+y^2=9,z =3\}$oriented clockwise. With$\mathbf{A} = -y\mathbf{i}+x\mathbf{j}+-xyz\mathbf{k},the line integral is \eqalign{ \oint_C \mathbf{A} \cdot ... 1 Here is a drawing of the situation . The surface S is shown in tan, the yellow arrow points in the direction of increasing x . The unit normal to the surface is the vector shown in red , the plane R , x=0 is shown in white. I believe you need to develop a projection formula of the form : \int_S F \cdot \hat n dS = \int_R \frac{F \cdot \hat ... 0 A simple way to do your integral is to use Stokes' theorem: the integral of\omega$over the surface of the ellipsoid is equal to the integral of$d\omega$over the enclosed volume. Since you were given that$d\omega=dx\wedge dy\wedge dz$, which is the standard volume form in$R^3$, the integral is simply plus or minus the volume enclosed in the ellipsoid. ... 0 Wrapping a rope of diameter$d$around a pipe of diameter$D$is equivalent to wrapping a ribbon of width$d$around a pipe of diameter$D+d$. For the latter problem, the length of ribbon,$L$, is whatever it takes to cover the total surface area of the pipe, hence $$dL=\pi(D+d)h$$ where$his the height of the pipe. Thus $$L=\pi\left(1+{D\over ... 0 The rope follows a helical path. Let's begin with the equation of a helix:$$\begin{align*} x &= a\cos \theta, \\ y &= a\sin \theta, \\ z &= b\theta. \end{align*}$$We now compute a and b. For a, we must consider the thickness of the rope in addition to the radius of the pipe! A rope is measured end-to-end, and when you wrap it, the ... 0 Some hints: What length of rope does it take to loop around the pipe once? (Call this L_{loop}.) How high up the pipe does this one loop cover? (Call this h.) How many loops will you need to cover the whole pipe? (Call this N. If H is the height of the pipe, then N = H/h.) Then the length of rope is L_{rope} = NL_{loop}. 2 The fraction of the triangle's area between the two planes is given by:$$ \frac{(B-z_0)^2 - (A-z_0)^2}{(z_1-z_0)(z_2-z_0)} $$In your example, z_0 = 900, z_1 = 1500, z_2 = 1200, A = 1100 and B= 1200. So the fraction of area is:$$ \begin{align} \frac{(1200 -900)^2 + (1100-900)^2}{(1500 - 900)(1200-900)} &= \frac{300^2-200^2}{600\times300}\\ ... 0 Consider a two-dimensional shape of areaA$with normal$n$. And imagine that you are looking directly at it (along$n$). You will observe it to have its true area$A$. Now imagine you rotate the shape by an angle$\theta$about some arbitrary axis so that it now points in the direction$n'$. If you are still looking along$n$then you will observe the ... 0 There exist several solutions in Matlab, as follows, they will give you both example programs and references to the underlying methods: Interpolate scattered data - MATLAB griddata, Surface Fitting using gridfit - File Exchange - MATLAB Central, Scattered data interpolation - MATLAB. The branch of mathematics is called approximation theory. The general ... 2$\Sigma$is the graph of the function$z = xe^y$. That is$(x,y,z) = (x,y, \varphi(x,y)) = G(x,y)$. So $$\int_{\Sigma}\mathbf{f}\cdot d\sigma = \int_{x}\int_{y}f(G(x,y))\cdot \left(\frac{\partial G}{\partial x}\times \frac{\partial G}{\partial y}\right)dx dy$$ What should$\varphi$be? Once you have determined that and the bounds ... 3 Let's denote the components of$f$by$f_1,f_2,f_3$. In particular,$f_3 \colon M \to \mathbb{R}$is continuous, and since$M$is compact,$f_3$attains its maximum and minimum at some points of$M$. What happens at these points? 0 Maybe define$f_x(\gamma)$instead of$f(x, \gamma)$and go with$x\mapsto \int_\Gamma f_xdS$, unless the$dS\$ part itself causes confusion.