Differential Geometry of Curves

Let $\alpha (s)$ be a closed plane curve. Define the diameter $d_\alpha$ of $\alpha (s)$ to be

Now assume that the curvature $k(s) \geq 1$ for all $s$.

Let $\alpha$ be a regular $C^\infty$ curve in $\mathbb {R}^{3}$ with nonvanishing curvature. Suppose the normal vector $N(t)$ is proportional to the position vector; that is, $N(t) = c(t)\alpha (t)$ for all $t$, where $c$ is a smooth function. Determine all such curves.

A surface $S \subset \mathbb {R}^{3}$ is called triply ruled if at every $p \in S$ we can find three open line segments $L_1, L_2, L_3$ lying in $S$ such that $L_1 \cap L_2 \cap L_3 = \{ p\}$. Determine all triply ruled surfaces.

Let $\gamma : (a,b) \to \mathbb {R}^{3}$ be smooth with $\left|\gamma '\right| = 1$ and curvature $k$ and torsion $\tau$, both nonvanishing. Denote the Frenet frame by $\{ T, N, B\}$. Assume that there exists a unit vector $a \in \mathbb {R}^{3}$ with

Show that if $\gamma$ is a geodesic on the cone $z = \sqrt{x^2+y^2}$, $(x,y) \in \mathbb {R}^{2} \setminus \{ (0,0)\}$, then $\gamma$ intersects itself at most a finite number of times.

Let $\gamma : (a,b) \to \mathbb {R}^{3}$ be a $C^\infty$ curve parameterized by arc length, with curvature and torsion $k(s)$ and $\tau (s)$. Assume $k(s) \neq 0$, $\tau (s) \neq 0$ for all $s \in (a,b)$, and let $T$ and $N$ denote the unit tangent and normal vectors to $\gamma$. The curve $\gamma$ is called a Bertrand curve if there exists a regular curve $\bar\gamma : (a,b) \to \mathbb {R}^{3}$ such that for each $s \in (a,b)$ the normal lines of $\gamma$ and $\bar\gamma$ at $s$ are equal. In this case, $\bar\gamma$ is called the Bertrand mate of $\gamma$ and we can write $\bar\gamma (s) = \gamma (s) + rN(s)$ for some $r = r(s) \in \mathbb {R}$. (Note that $s$ might not be an arc length parameter for $\bar\gamma$.)

Let $x(s)$ be a curve in $\mathbb {R}^{3}$ parameterized by arc-length. Assume that $\tau (s) \neq 0$ and $k'(s) \neq 0$ for all $s$. Show that a necessary and sufficient condition for $x(s)$ to lie on a sphere is that

Let $M \subset \mathbb {R}^{3}$ be the torus obtained by rotating the circle $\{ (0, y, z) : (y-2)^2 + z^2 = 1\}$ around the $z$-axis, and let $c(t) = (2\cos t, 2 \sin t, 1)$ ("top circle"). Is this curve a geodesic on $M$? Explain without long computations.

Let $\gamma : [0,1] \to \mathbb {R}^{3}$ be a $C^\infty$ curve with $\left|\gamma '\right| = 1$ and non-vanishing curvature. Assume the torsion $\tau \equiv 0$.

Let $f : \mathbb {R} \to \mathbb {R}$ be positive and smooth. Let $M$ be the surface in $\mathbb {R}^{3}$ obtained by rotating the graph $\{ (x, f(x)) : x \in \mathbb {R}\}$ of $f$ in the $xz$ plane about the $x$ axis. Characterize in terms of $f$ the set of $x$ such that $\dfrac {1}{f(x)}$ is a principal curvature of $M$ at $(x, 0, f(x))$.

Let $\alpha (s) \subset \mathbb {R}^{3}$ be a smooth curve parameterized by arc-length. Assume that the position vector $\alpha (s)$ is always a linear combination of the binormal and normal vectors $B(s), N(s)$ of $\alpha (s)$. Show that $\alpha (s)$ does not pass through $O \in \mathbb {R}^{3}$.

Let $M^2 \subset \mathbb {R}^{3}$ be the cylinder $x^2+y^2=1$. Suppose the curve $\alpha (s) \subset M^2$ is parameterized by arclength.

Let $\Gamma$ be a two-dimensional distribution in $\mathbb {R}^{3}$ defined by

Call a normal vector field $v$ along a space curve $\gamma$ parallel if $v'$ is always tangent to $\gamma$.

Let $M$ be a surface in $\mathbb {R}^{3}$ and let $P$ be a plane. Suppose $M$ and $P$ intersect orthogonally. Show that the intersection curve (parameterized by arclength) is a geodesic on $M$.

Let $\alpha (s)$ be a $C^2$ curve in $\mathbb {R}^{3}$ parameterized by arclength. Suppose that for some function $f(s)$, $\alpha ''(s) = f(s)\alpha (s)$. What can you deduce about $f(s)$, $\alpha (s)$?

Suppose $\gamma (t)$ parameterizes a space curve with curvature function $\rho (t)$. Define a new curve $\bar\gamma$ by setting, for each $t \in \mathbb {R}$, $\bar\gamma (t) := c\gamma (t/c)$, where $c \in \mathbb {R}$ is an arbitrary fixed constant. Derive the curvature function for this new curve.

Let $\alpha : (a,b) \to \mathbb {R}^{3}$ be a smooth curve with non-vanishing curvature.

Let $\alpha$ and $\beta$ be two regular curves in $\mathbb {R}^{3}$. The curve $\beta$ is called an involute of $\alpha$ if for all $t$, $\beta (t)$ lies on the line tangent to $\alpha$ at $\alpha (t)$ and $\left\langle \alpha '(t), \beta '(t) \right\rangle = 0$. Show that every involute of a generalized helix $\alpha$ is a plane curve. (Recall that $\alpha$ is a generalized helix if for some constant vector $u \neq 0$, $\left\langle u, \alpha '(s) \right\rangle = \text{const}$, where $s$ is arclength for $\alpha$.)

Suppose that the unit normal vector to a surface $M \subset \mathbb {R}^{3}$ is constant along a regular curve $\alpha \subset M$. Deduce that in this case, $\alpha$ is an asymptotic curve, that a plane contains it, and that the Gauss curvature of $M$ vanishes at each point of $\alpha$.

[Note: A curve in a surface is called asymptotic if its acceleration is everywhere tangent to that surface.]

Sketch the closed regular plane curve $\beta : [-\pi , \pi ] \to \mathbb {R}^{2}$ having $\beta (0) = (0,0)$ and $\beta '(0) = (1,0)$, if $\beta$'s curvature function $k(s)$ ($s =$ arclength) is odd, satisfies

and has the following graph. (Include an explanation with your sketch.)

Let $X \subset \mathbb {R}^{2}$ be a connected one-dimensional real analytic submanifold, not contained in a line. Prove that not every tangent line to $X$ is bitangent -- that is, it is not the case that for all $p \in X$ there exists $q \neq p$ such that the tangent line to $X$ at $p$ equals the tangent line to $X$ at $q$ as lines in $\mathbb {R}^{2}$.