Tag Info

15

Still not explicitly paramaterised curves, but someone may paramaterise them for me:

6

Though still not satisfying the OP's desire for smoothness of higher degree, the following construction turned out to be simpler than my initial analysis had predicted: All the curves in this family have a perimeter of $L=10$ and an area of $A=4$. The $\color{blue}{\text{blue arc}}$ has radius $r$ variyng as a function of $t$. The parameter $t$ is not ...

6

(The following example avoids piecewise definitions; but there is nothing "special" about it.) Consider the curve $\gamma$ with polar representation $$r=r_0(\phi):=2+\cos(3\phi)\ ,$$ which looks like a clover leaf, see the following figure. This curve has a length $L_0$ and encloses an area $A_0$. We now set up a perturbation of $\gamma$ in the form ...

4

Circle with two bumps Let $f$ be your standard bump function $$f(x) = \begin{cases} e^{-\frac{1}{1-x^2}}, &|x| < 1\\ 0, &\text{otherwise} \end{cases}$$ Let $g(x) = f(5x)$ (this makes your bump narrower). For $b \in [0, 1]$ (but not too close to 0), consider the curve: $$r(t) = 1 + g(t-\frac{\pi}{2}) + g(t-\frac{\pi}{2} - b\pi) \quad \text{for ... 4 As Feynman said it is "a piece of plastic for drawing smooth curves--a curly, funny-looking thing". It is used in art classes occasionally. Take ANY smooth curve that has a lowest point, draw a tangent line to the curve at that point. The line will be horizontal because the derivative there will be zero. 3 Let's make sure we understand splitting, before we talk about joining. To subdivide a Bézier curve into two, you use the deCasteljau algorithm, as illustrated in the figure below. Suppose we are given a curve defined by four control points A, P, Q, G, and a splitting parameter value u \in [0,1] (which you called t_{\text{cut}}). To split the ... 3 Take your favorite family of smooth Jordan curves parametrized by \lambda>1 say, for example x^\lambda+y^\lambda=1. These all bound area between 2 and 4. Scale them to unit area by an appropriate factor depending on \lambda. The resulting curves don't have the same perimeter, but notice that each perimeter is less than 8, say. Now apply affine ... 2 It will not be easy to come up with a "natural" explicit example, since you need a two-parameter family of curves whose lengths you can compute in an elementary way. The following example is not very sophisticated, but does the job: Take a square of side length a>0, and round off its corners using small circular arcs of radii \rho_i>0 (1\leq ... 2 Only two curves, rather than a family, and one of them is piecewise-defined, but I thought people might be able to use them as fuel for doing something better: Let$$ f(x) = \sqrt{\frac{1-x^2}{1+(x-1)^2}} $$Then we define two curves like this: Curve 1:$$ \begin{align} y &= \phantom{-}f(x) \quad \text{for } -1 \leq x \leq 1\\ \text{and}\quad y &= ...

2

If you know that the two curves you have resulted from splitting a single initial curve, then you don't need both these curves; knowing either and the cut position is enough to restore the full initial curve. Have a look at this post on Stack Overflow. It discusses how you cut a curve. What you want to do is do the reverse. To find control points $Q_i$ for ...

2

$$\textbf{v}.\textbf{r}=(\textbf{c}\times\textbf{r}).\textbf{r}=0\tag{1}$$ and $$\textbf{v}.\textbf{c}=0\tag{2}$$ Therefore $\textbf{v}\perp\textbf{r}~~\forall t$ and $\textbf{v}\perp\textbf{c}$. $$\frac{d}{dt}(\textbf{r}.\textbf{c})=\textbf{v}.\textbf{c}=0\implies \textbf{r}.\textbf{c}=constant\tag{3}$$ which is the equation of a plane orthogonal to ...

2

A French curve was a piece of plastic used by draftsmen back in the day when drawings were made by pencil and paper to make a smooth transition between things. The Feynman remark just reflects the point that the tangent is horizontal at a minimum-you probably learned that in Calculus 1. The joke is that that is there is no magic in the French curve, that ...

2

Can you show the following are equivalent? What would the reparametrizations be, explicitly? $a:[0,\sqrt{2\pi}]\to \Bbb C:t\mapsto e^{-it^2}$ $b:[0,2\pi]\to \Bbb C:t\mapsto e^{-it}$ $c:[0,2\pi]\to \Bbb C:t\mapsto e^{it}$ $d:[0,1]\to\Bbb C:t\mapsto e^{2\pi it}$ It might help to review what exactly a reparametrization is. By the way, your integrals should ...

2

Note that $\sin\left(\frac{3\pi}{2}-t\right)=-\cos t$, so we are integrating $\sqrt{2-2\cos t}$. For the integration, use the fact that $1-\cos t=2\sin^2(t/2)$. This is a version of the more familiar $\cos 2x=1-2\sin^2 x$.

2

For $(1)$ take $V_1=\Bbb R^2\backslash X$ and $V_2=U$, by the Jordan Curve Theorem $V_1$ is connected. On the other hand $V_1\cup V_2=\Bbb R^2$ hence $H^1(V_1\cup V_2)=0$. As $V_1$ and $V_2$ are connected and $H^1(V_1\cup V_2)=0$ by using the exercise5.8 it follow that $V_1\cap V_2$ is connected, but $V_1\cap V_2=U\backslash X$. For $(2)$ use same argument. ...

1

As commenters said, you are right: this is a circle More generally, if $\Gamma:[a,b]\to \mathbb C$ is a curve, and $f:[c,d]\to [a,b]$ is a strictly increasing continuous function such that $f(x)=a$ and $f(d)=b$, then the composition $\Gamma\circ f$ is a different parametrization of the same geometric object. In your case, $f(t)=t^2$ and $[c,d] ... 1 My guess for this case: The drawing was recorded as a sequence of$(x,y)$couples, using a graphical editor. The coordinates were considered as functions of an independent parameter$t$, defining$x(t)$and$y(t)$. (The simplest rule is to assign$t=\frac in$to the$i^{th}$point). Then the Fourier series mechanism allows to express these functions as ... 1 No. This is not true. However,$n(\gamma,z_0)=0$, for all$z_0$in the (unique) unbounded connected component of$\mathbb C\smallsetminus \gamma$. 1 I'll outline a short proof using only your calculation of$\gamma'(t_0)$. As you will see, it may help to do things one at a time rather than trying to write one big equation and solve for every variable simultaneously. The algebraic details I will leave for you to work out. Note that$$\frac{\sec^2 ... 1 I drew from the information provided by Semiclassical and Physicist137 (thank you for helping!) to draw out a direct solution to finding the curve connecting two points. Suppose we wanted the cycloid connecting an initial, known point$A$and a second, arbitrary point$B$. For simplicity, set$A=(0,0)$; a different initial point means a simple translation. ... 1 A cycloid can also be interpreted the equation of motion of a point in a rolling-circle. You can check here if you are not convinced. Or even prove it mathematically. Therefore, you have two parameters: the radius of the circle$r$, and the angular speed of the circle$\omega$. The angle of the point in the circle is$t\$. Then: \begin{array}{} x = ...

Only top voted, non community-wiki answers of a minimum length are eligible