# Tag Info

4

It's not true (even if you disregard constant functions). It is possible to define a $C^\infty$ function $\gamma$ from $[0,1]$ into $\mathbb R^2$ such that $\gamma(t) = [0,0]$ for infinitely many $t$ but $\gamma$ is not constant on any interval, and $\gamma([0,1])$ is the union of infinitely many loops each containing $[0,0]$. But for your $\rho$, this ...

2

Do you perhaps mean Lissajous curves?

2

By definition, the arc-length of a curve $\gamma: [a,b] \to \Bbb R^n$ is: $$L[\gamma] = \int_a^b \|\gamma'(t)\|\,{\rm d}t.$$ Here, $\gamma: [0,2\pi] \to \Bbb R^3$ is given by $\gamma(t) = (t,\cos t, \sin t)$, so $\gamma'(t) = (1,-\sin t, \cos t)$ and so: $$\|\gamma'(t)\| = \sqrt{2} \implies L[\gamma] = \int_0^{2\pi}\sqrt{2}\,{\rm dt} = 2\sqrt{2} \pi.$$ ...

1

Making a paper model of the cylinder in question and then flattening it out you will see that the curve appears as diagonal of a square with side length $2\pi$. Its length therefore is $\sqrt{2}\cdot2\pi$.

1

The following way to visualize the curve works in this situation, but it is not reliable in general. If you look at the $yz$-plane, the curve traces out a circle of length $2\pi$. If you look in the $x$-axis, the curve traces out a line of length $2\pi$. If you think of the final curve as a combination of these two curves, you could "apply" the ...

1

Edited, based on comments (1) In my copy of the book, property (b) includes the constraint that $\Gamma$ is convex (2) Your counterexample does not work because you are not using the correct parametrization. In your parametrization $\gamma(t), \gamma(t+\pi)$ are not on opposite sides of the circle. For a circle of radius $R$ shifted by $\epsilon$, a ...

1

Even if $\gamma$ is injective, it's not true. Here's a counterexample: $$\gamma(t) = \begin{cases} (-e^{1/t},0), & t<0,\\ (0,0), &t=0,\\ \left( e^{-1/t},\ e^{-1/t}\sin \frac{1}{t}\right), & t>0. \end{cases}$$ This is $C^1$ (in fact $C^\infty$) and injective, but there's no $C^1$ reparametrization with nonvanishing right-hand derivative at ...

1

A partial answer. Call a point $(x,y)$ ordinary if $2x$ is not an odd integer, $2y$ is not an odd integer, $x+y$ is not an integer, and $x-y$ is not an integer. It's not hard to see that if $(x,y)$ is ordinary then it has a unique nearest lattice point and a unique second-nearest lattice point. So if $(x,y)$ is ordinary then the circle centered at $(x,y)$ ...

1

Suppose the plane is given as $$Ax + By + Cz = 0$$ and the point $P$ is $(a, b, c)$. THen the ray from $P$ normal to the plane consists of points of the form $$R(t) = (a, b, c) + t(A, B, C)$$ For what $t$ is $R(t)$ on the plane? We must have  A (a + tA) + B(b + tB) + C(c + tC) = 0 \\ aA + bB + cC + t(A^2 + B^2 + C^2) = 0 \\ t = -\frac{aA +bB + ...

1

HINT: Suppose we write $C$ as a graph over the tangent line at $p$, so we represent $C$ as $y=f(x)$, with $f(0)=0$ and $f'(0)=0$. Do you know (or can you derive) a formula for $\kappa$ in terms of derivatives of $f$? Can you express $2h/d^2$ in terms of $f(x)$ and find the limit as $x\to 0$?

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