There is another slick little interpretation of the curvature. This one is both easier and harder to see: easier in that the argument for it is much shorter and doesn't use quite such an intricate argument; harder in that it lies so close to the surface it is just harder to notice.
As usual, we start with some $C^2$ regular curve $\gamma$ in $\mathbb{R}^3$, and we assume that it is parametrized by arclength with arclength parameter $s$. At this point, it can seem a trivial observation, but the important thing is that the tangent vector $T(s)$ has length which is always equal to $1$.
Now we step back and consider the assignment $\alpha: s \to T(s)$ as a function on its own terms, and let $\gamma$ fade into the background for a minute. For each input $s$, $T(s)$ is a vector in $\mathbb{R}^3$. Usually, one visualizes each of these vectors as attached to the corresponding points $\gamma(s)$, but we will just take them all based at the origin of $\mathbb{R}^3$. Because $||T(s)|| = 1$, the curve $\alpha$ lies on the unit sphere!
This curve $\alpha$ is called the (tangent) spherical image of $\gamma$.
Note that there is no reason to believe that $\alpha$ is parametrized by arclength. The parameter $s= s_{\gamma}$ is adapted to $\gamma$, not to $\alpha$. Still, since $\gamma$ is $C^2$, we see that $\alpha$ is a $C^1$ curve and so it makes sense to measure its length via the usual arclength construction with an integral:
\[ \mathrm{len}(\alpha) = \int_a^b ||\alpha'(s)|| \, ds \]
We have one more small observation to make, and again it seems nearly trivial once you see that it is important. If we now think further about what the construction means for $\alpha$, we see that
\[ ||\alpha'(s)|| = ||T'(s)|| = \kappa(s). \]
So this means that the length of $\alpha$ is really the total curvature of $\gamma$.
Of course, we can do better than that. The above means that the ratio of the arclength parameters on $\alpha$ and $\gamma$ is $\kappa$:
\[ \dfrac{ds_{\alpha}}{ds_{\gamma}} = \kappa .\]
Now, the classical language of taking ratios of infinitesimal quantities has a bad rap in some circles. We can shore this up with differential forms, but it is better to slightly change the explanation, because it give a better geometric insight.
Imagine we fix a particular value $s_0$ of the parameter and consider some very small neighborhood $I_{\varepsilon} = (s_0-\varepsilon, s_0+\varepsilon)$ around it. In particular, we wish to choose $\varepsilon$ small enough that we can assume that $\kappa$ is essentially constant on the interval. Then we find that the length of $\gamma(I_{\varepsilon})$ is
\[ \mathrm{len}(\gamma(I_{\varepsilon})) = \int_{s_0-\varepsilon}^{s_0+\varepsilon} ||\gamma'(s)|| \, ds = 2\varepsilon ,\]
and the length of the spherical image of this same interval is
\[ \mathrm{len}(\alpha(I_{\varepsilon})) = \int_{s_0-\varepsilon}^{s_0+\varepsilon} ||\alpha'(s)|| \, ds \approx 2\varepsilon \cdot \kappa(s_0) . \]
This means that we can study the ratio of these lengths as $\varepsilon \to 0$:
\[ \dfrac{ \mathrm{len}(\gamma(I_{\varepsilon})) }{\mathrm{len}(\alpha(I_{\varepsilon}))} \to \kappa(s_0). \]
So the curvature $\kappa(s)$ is the ratio of length of the spherical image to the length of the original curve over infinitesimally small intervals centered at $s$.
This can be seen by looking at some pictures for very simple examples: consider two planar circles, one of very large radius, and the other of very small radius. Over a short length of the curve, the large circle has a tangent that moves just a little. So its spherical image traces a very short curve. Thus the ratio is very small, and we get a small value of curvature.
Over the same short length of curve, the tangent of the small circle moves quite a lot. The spherical image is then a longish curve, and the ratio ends up rather large. This corresponds to a large value of the curvature.
As usual, we start with some $C^2$ regular curve $\gamma$ in $\mathbb{R}^3$, and we assume that it is parametrized by arclength with arclength parameter $s$. At this point, it can seem a trivial observation, but the important thing is that the tangent vector $T(s)$ has length which is always equal to $1$.
Now we step back and consider the assignment $\alpha: s \to T(s)$ as a function on its own terms, and let $\gamma$ fade into the background for a minute. For each input $s$, $T(s)$ is a vector in $\mathbb{R}^3$. Usually, one visualizes each of these vectors as attached to the corresponding points $\gamma(s)$, but we will just take them all based at the origin of $\mathbb{R}^3$. Because $||T(s)|| = 1$, the curve $\alpha$ lies on the unit sphere!
This curve $\alpha$ is called the (tangent) spherical image of $\gamma$.
Note that there is no reason to believe that $\alpha$ is parametrized by arclength. The parameter $s= s_{\gamma}$ is adapted to $\gamma$, not to $\alpha$. Still, since $\gamma$ is $C^2$, we see that $\alpha$ is a $C^1$ curve and so it makes sense to measure its length via the usual arclength construction with an integral:
\[ \mathrm{len}(\alpha) = \int_a^b ||\alpha'(s)|| \, ds \]
We have one more small observation to make, and again it seems nearly trivial once you see that it is important. If we now think further about what the construction means for $\alpha$, we see that
\[ ||\alpha'(s)|| = ||T'(s)|| = \kappa(s). \]
So this means that the length of $\alpha$ is really the total curvature of $\gamma$.
Of course, we can do better than that. The above means that the ratio of the arclength parameters on $\alpha$ and $\gamma$ is $\kappa$:
\[ \dfrac{ds_{\alpha}}{ds_{\gamma}} = \kappa .\]
Now, the classical language of taking ratios of infinitesimal quantities has a bad rap in some circles. We can shore this up with differential forms, but it is better to slightly change the explanation, because it give a better geometric insight.
Imagine we fix a particular value $s_0$ of the parameter and consider some very small neighborhood $I_{\varepsilon} = (s_0-\varepsilon, s_0+\varepsilon)$ around it. In particular, we wish to choose $\varepsilon$ small enough that we can assume that $\kappa$ is essentially constant on the interval. Then we find that the length of $\gamma(I_{\varepsilon})$ is
\[ \mathrm{len}(\gamma(I_{\varepsilon})) = \int_{s_0-\varepsilon}^{s_0+\varepsilon} ||\gamma'(s)|| \, ds = 2\varepsilon ,\]
and the length of the spherical image of this same interval is
\[ \mathrm{len}(\alpha(I_{\varepsilon})) = \int_{s_0-\varepsilon}^{s_0+\varepsilon} ||\alpha'(s)|| \, ds \approx 2\varepsilon \cdot \kappa(s_0) . \]
This means that we can study the ratio of these lengths as $\varepsilon \to 0$:
\[ \dfrac{ \mathrm{len}(\gamma(I_{\varepsilon})) }{\mathrm{len}(\alpha(I_{\varepsilon}))} \to \kappa(s_0). \]
So the curvature $\kappa(s)$ is the ratio of length of the spherical image to the length of the original curve over infinitesimally small intervals centered at $s$.
This can be seen by looking at some pictures for very simple examples: consider two planar circles, one of very large radius, and the other of very small radius. Over a short length of the curve, the large circle has a tangent that moves just a little. So its spherical image traces a very short curve. Thus the ratio is very small, and we get a small value of curvature.
Over the same short length of curve, the tangent of the small circle moves quite a lot. The spherical image is then a longish curve, and the ratio ends up rather large. This corresponds to a large value of the curvature.
Plotted here you see the original curve in red, with the image of interval $I_{\varepsilon}$ thicker, and the tangent spherical image plotted as a thicker curve on the unit circle. To see the effect, try changing the values of 'rad' and 'eps'. The number 'rad' is the radius of $\gamma$, and hence is the reciprocal of its curvature. The number 'eps' is the arclength of the original curve covered by the interval $I_{\varepsilon}$.
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