New answers tagged

1

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 ...


1

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) = ...


2

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.


0

As for any surface $f(x,y)=z$, the gradient vector denotes a vector in the normal direction to the surface. $\nabla (x^2 y - z^2 = 0) = 2 xy ~\hat{i}+ x^2~ \hat{j} - 2 z ~ \hat{k}$. Evaluate at $u=2,~ v=1$ and normalize to get the unit vector.


0

in parametric form your way of finding normal is correct. we use cross product of vectors tangent to u-curve and v-curve. in the non-parametric form unit normal vector is indeed normalized gradient $\frac{\nabla f}{|\nabla f|}$. so firstly we have to find non-parametric form of surface: $$x=u \\ y=v^2 \\ z=uv $$ for a curve with equation $z=f(x,y)$ the ...


1

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 $x$.


0

According to Archimedes the surface $S$ is the union of infinitesimal isosceles triangles of base length $ds$ (on the rim of the cone) and height $3\sqrt{2}$. The total rim length is $s=6\pi$, so that we arrive at $${\rm area}(S)={1\over2}\cdot6\pi\cdot3\sqrt{2}=9\pi\sqrt{2}\ .$$ In order to compute this area as a surface integral we have to produce an ...


0

The cosine of the angle between two vectors $u,v$ with respect to an inner product $\cdot$ is $$\frac{u \cdot v}{\sqrt{(u\cdot u) (v\cdot v)}}.$$ In the parameter domain you have the standard dot product, and in the surface geometry you have $u \cdot v=u^T I v$. Show that when $I$ has the form you are assuming, these two inner products produce the same angle ...


0

Fundamental Forms $$ \begin{pmatrix} E & F \\ F & G \end{pmatrix}= \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix} \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$ $$ \begin{pmatrix} e & f \\ f & g \end{pmatrix}= -\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix} \begin{pmatrix} ...


1

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.


1

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 $x^2 + y^2 + 0\times z = 4$.


2

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.


0

I think what you mean is Euler's polyhedral formula. It states that for every convex polyhedron with $V$ vertices (corners), $E$ edges and $F$ faces the formula $V-E+F=2$ holds. The same statement holds for planar graphs. So the statement for $\mathbb R^2$ is that the formula holds if your graph is a planar graph.


1

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 ...


0

Let us admit that we do not know the answer (that is to say that the smallest surface area of a cuboid for a given volume is a cube as calculus commented). You properly arrived to $$A = 2\big(lw + \frac{200}{w} + \frac{200}{l}\big)$$ Since you want $A$ to be minimum, you need to set its derivatives with respect to the variables $l,w$ equal to $0$ (think ...


2

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 ...


1

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.


2

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 ...


1

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 ...


1

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 ...


0

As far as know, the only parametric surfaces that have this nice simple property are planes. Even the intersection of two circular cylinders can not be represented as a polynomial or rational spline curve (i.e. as a NURBS curve). There are a couple of different approaches that are used in CAD systems. The first approach is to use "procedural" intersection ...


0

You can get two vectors $u$ and $v$ by taking the difference between any two pair of points. Then the cross product between these two vectors will be perpendicular to the plane. $$u\times v = \det \begin{bmatrix}i & j & k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{bmatrix}=(u_2v_3-u_3v_2)i-(u_1v_3-u_3v_1)j+(u_1v_2-u_2v_1)k$$


0

Use (K-L) and (M-L) to define two vectors in the plane. With these vectors, I can eyeball the vector that would be perpendicular. But, you could also take the cross product if you haven't developed the intuition. Alternative: $ax + by + cz = d$ is a generic plane. All three points are in the same plane. plug the points, hold $d$ constant, solve for ...


0

Observe that $(x,y,z)=(\cos t,\sin t,\sin t)$ satisfies $$\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{2})^2}+\frac{z^2}{(\sqrt{2})^2}=1\qquad\text{and}\qquad y=z$$ So, the curve lies in the intersection of an ellipsoid and a plane. Also the curve is closed since $(\cos 2\pi,\,\sin 2\pi,\,\sin 2\pi)=(\cos 0,\,\sin 0,\,\sin 0)$. Then, it looks like an ellipse.


2

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


2

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 ...


1

Nick's answer is great, but in case you aren't comfortable with pull-backs and tensor products, you can use that the first fundamental form is expressed in coordinates by $${\rm d}s^2 = E\,{\rm d}u^2 + 2F\,{\rm d}u\,{\rm d}v + G\,{\rm d}v^2,$$where $E = \langle \sigma_u,\sigma_u\rangle$, $F=\langle \sigma_u,\sigma_v\rangle$ and $G = \langle ...


2

Yes, this is the correct way to go. If the surface is the graph of a smooth function $z = f(x,y)$, then putting graph coordinates on it (as you suggest), via $\sigma(x,y) = (x,y,f(x,y))$ will give the formulas you want. The "First Fundamental Form" is just the Riemannian metric. You can get it, in coordinates, by "pulling back" the standard Euclidean metric ...


2

If a singular point of a tangent vector field $X$ means that $X(x)=0$, your assertion is not true. There exists parallelizable spheres, $S^1,S^3,S^7$ which are the only parallelizable spheres after results of Bott, Kervaire, Milnor.


14

The least you can do is 2 parametrizations. One such way of doing this is by the stereographic projection, in which a pole of the sphere is removed, and then the rest of the sphere is projected to the plane. Then the same is done by removing the opposite pole.


3

$T^2 = S^1\times S^1$ is a torus and $A = S^1\times\{1\}$ is one of the highlighted circles on torus below. $\hspace{42mm}$ $X = S^1\times[-1, 1]$ is a cylinder and $B = S^1\times\{-1, 1\}$ is the union of the boundary circles at each end. The quotient $T^2/A$ is a torus with the circle $A$ collapsed to a point; depending on whether you collapse the ...


-1

I do not know the Rubix's Twist well enough to determine possibilities and what not. However, I do know how to compute the minimum surface area of a cubic region. You have two equations (note that V is a volume constant selected beforehand): V = LWH => V/WH = L 2LW + 2LH + 2WH = P => 2V/H + 2V/W + 2WH = P P is now a variable defined by the above ...



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