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You've shown that there is a connected neighborhood $U$ of the point $\ast$ of interest in the quotient space such that $U − \{ \ast \}$ has two components, but you know that no point on the plane admits such a neighborhood; therefore, the quotient space cannot be a closed surface.


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Not exactly: A flexible (but un-stretchable) wire or thread can bend isometrically into a catenary. By contrast, two-dimensional materials have internal rigidity coming from intrinsic curvature. Consequently, an ideal flexible (but un-stretchable) sheet (think of paper or a wire screen) suspended at its corners can only bend upward at the corners with ...


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HINT: What is the normal curvature at $P$ in the direction of the line? What do you conclude about the two principal curvatures at $P$?


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Any point in the annulus $U$ is uniquely of the form $(t \cos \theta, t \sin \theta)$ for some real $t \in (0,\sqrt{\pi}), \theta \in [0,2\pi).$ Map this point to the point of the cylinder $(x,y,z)=(\cos \theta, \sin \theta, \cot t^2).$ This is clearly a subset of the cylinder as it satisfies $x^2+y^2=1.$ Also, because $\theta$ ranges in $[0,2\pi),$ for any ...


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Hint: The formula for the surface area, $A_s$ is, $$A_s=2 \cdot (x^2+y^2+z^2)$$ The formula for the volume, $V$, is, $$V=x \cdot y \cdot z$$ We want the maximum volume given that the surface area is constrained to be equal to some number $S$. Taking partial derivatives, shown by adding a subscript, we should have, $$V_x=V_y=V_z=0$$ and $$0=S-2 \cdot ...


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He mentions that to specify what he means by continuity of $f^{-1}$, probably for those readers who don't have any background in topology. In general a map is called continuous if preimages of open sets are open. Therefore you need to know which sets are called open - i.e. the topology, in this case the induced topology of $S$. His way to avoid this is ...


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You should get the Klein bottle: By combining arrows $a$ and $b$ into a single red arrow, and arrows $a^{-1}$ and $b^{-1}$ into another red arrow, we get the first figure below. Now we must glue the red sides together, and blue sides together, so that the arrows line up.


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First you need to find the point(s) where you have to give the tangent plane. For the gradient you have: $$\nabla u = \left( \frac{\partial u}{\partial x},\frac{\partial u}{\partial y} \right) = \left( \frac{1}{x+1/y},\frac{-1/y^2}{x+1/y} \right)$$ And you're looking for the points $(x,y)$ where this is equal to $\left( 1,-\tfrac{16}{9}\right)$, so: ...


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What you are looking for is the function $$p(x) = \min_y f(x,y)$$ Assuming there are only minima and no maxima, we can simply say that $$p(x) = f\left(x,\left( \frac{\partial}{\partial y} f(x,y) = 0 \right)\right)$$ which evaluates to an expression that is only dependent on $x$.


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Use "oval" or "ovoid" for such curves and surfaces respectively. Make sure to provide your own definition for these words though.


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I had seen usage of terms oblate/ prolate ovaloids in American literature.


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You need to find a function, this function must be related to the parameters of the box, in this case its measures, length, height and width. Then for this role you must find the derivative and using the criteria of the first and second derivative to find the maximum and minimum.


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Well, you say you want to show that $\langle\alpha'(u),\beta(u)\rangle = 0$. But, as you go on to say, you really want to choose a different directrix curve $\tilde\alpha(u)$ so that you'll have $$\langle\tilde\alpha'(u),\beta(u)\rangle = 0\tag{$\star$}.$$ So you want to choose a scalar function $\lambda(u)$ so that, setting $\tilde\alpha(u) = ...


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There are many questions, here. I'll take stab at a few of them. The easiest way to render a set of triangles is a z-buffer algorithm. You paint each triangle into a pixel array, and you keep track of z-depth at each pixel. Pixels that are in front over-write pixels that are behind. Most graphics packages (like OpenGL or DirectX) will do this for you; all ...


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In practice, raytracers tend to decompose surfaces into triangular meshes first, because the math needed is so much simpler. However, you asked how to calculate the intersection between a line and a quadrilateral in 3D, so here goes. Let's define your ray using an unit vector $\hat{n}$ (unit referring to unit length, $\lvert\hat{n}\rvert=1$) that passes ...


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The same equation will work, as for given $x_0$ and $y_0$ on the circle, for any $z$, at the vertical of $(x_0,y_0)$, $(x_0,y_0,z)$ will be on the wall, as in the figure from Quadric Surfaces: If you really miss the $z$, try $x^2 + y^2 + 0\times z = 4$.


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Actually what you wrote is for an entire cylinder of unspecified height.But to specify where it starts and ends, the (x,y,z) surface (cylinder ) parametrization is $$ (x = 2 \cos u, y=2 \sin u, z=v ),( 0< u< 2 \pi), ( hmin < v < hmax) $$ Where the height is between two limits. Useful for extruded /prismatic surfaces.


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One way of seeing this surface is slicing it with a plane parallel to the plane $z=0$. That means setting $z=k$ and imagining what the section should look like. In your case, \begin{equation} z=k\quad\Longrightarrow\quad x^2y=3k\quad\Longrightarrow\quad y=\frac{3k}{x^2} \end{equation} so if $k>0$ the sections seen from "above" look like the function ...


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Differential geometry of curves and surfaces is very important. It is a main mathematical component of a branch of mechanical engineering called: the theory shells. Shell constructions are everywhere: airplanes, ships, rockets, cars, pressure vessels, etc. If you open any book devoted the shells theory (or, say, the Finite Element Analysis of shells), you ...



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