Surface swept out by tangent vector

Question

This question is from Novikov and Fomenko's Modern Geometry Part I. I quote the question here (problem 6 in Exercise 8.4) in full:

Let $S$ denote the surface swept out (i.e. "generated") by the tangent vector to a given curve with curvature $k(l)$. Prove that if the curve is twisted, but in such a way as to preserve the curvature $k(l)$, then the metric on the surface $S$ is also preserved.

I am not able to comprehend (or even visualize) what the first line means. If I understand it correctly, the tangent vector refers to the curve $\frac{d\vec{r}}{dt}$, where $ \vec{r} \equiv \vec{r}(t)$ is the given curve. But I am not able to understand what surface this corresponds to; in particular what surface is being "swept out" by the said vector?

I believe that if I can somehow write down the equation for the surface S, the other things will be straightforward, including the form of the metric (and possibly using the Ferrer-Srenet formulae).

Answer 1

I try to answer this question visually, focussing on "What is the are swept out by the tangent vector?"

Let's take a look at some function, say $f(x)=x^3$:

Now what is the tangent vector? You can think of a tangent line as a line that touches the graph at some point $P(x,f(x))$ and has the same slope as the graph of $f(x)$ at that point $P$. Since the derivative represents the slope, we can find a tangent line by taking the derivative $f'(x)$.

Now we can take this point $P$ and let its $x$ (and accordingly its $y$) coordinate change while plotting the tangent:

Now you could imagine that the line is you pencil and leaves some sort of color at every point that it has touched. This is the area swept out by this line.

A tangent vector is similar, but instead of being an infinitely long line, it is just a vector. So the area swept out by it will be finite.

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I made these graphics using GeoGebra. Unfortunately, due to my limited skills, I was not able to draw a tangent vector. The process is described here, but didn't work for me.

Answer 2

Is this the question in full, because as it stands it doesn't make much sense.

Take for example a simple, the circle. Then the tangent lines sweep out the plane minus the interior of this circle and the metric on this surface, by observation, is flat.

If the curve is a helix, then we obtain a helicoid. This surface is also flat.

For a general curve, general we obtain a ruled surface. This is a surface where through every point there is a straight line. Rules surfaces are always flat. In fact, Catalan proved in 1842 that the helicoid and the plane were the only minimal surfaces in 3d that were also ruled surfaces.

The question also doesn't explain what we are to understand by 'the metric ... is preserved'. Normally this phrase would be used if the surface undergoes some kind of transformation - but here, there is none in sight. Hence my ???

Answer 3

As I interpret it, the other answers are not incorrect, but the question says nothing about the curvature of the surface - it may very well be flat. The curvature of the curve is $k(l)$. You might want to use a FS frame to parametrize the surface somehow. The surface inherits a metric from the ambient space of the curve.

Now consider changing the curve $r$ to a close-by twisted version $\tilde r$, where the curvature $k$ is the same at each point. I think the question asks to show the metric is the same at each point as well.

I don't know the details of the answer yet, but I would maybe even try something like $\tilde r_\delta = r + \delta s$ where $\delta>0$ is small and $s$ is a vector field along the curve $r$, and require that $\frac{dk}{d\delta}\rvert_{\delta=0}=0$, if it is necessary.