Tag Info

Hot answers tagged

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 B├ęzier 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