How could we calculate the signed curvature?

Question

We have that $$\epsilon (s)=\gamma (s)+\frac{1}{\kappa_s(s)}\cdot n_s(s)$$

We regard $\epsilon$ as the parametrization of a new curve, called the evolute of $\gamma$ (if $\gamma$ is any regular plane curve, its evolute is defined to be that of a unit-speed reparametrization of $\gamma$).

Assume that $\dot \kappa_s (s) \neq 0$ for all values of $s$ (a dot denoting $\frac{d}{ds}$ ), say $\dot \kappa_s >0$ for all $s$ (this can be achieved by replacing $s$ by $−s$ if necessary).

Show that the arc-length of $\epsilon$ is $\frac{-1}{\kappa_s(s)}$ (up to adding a constant), and calculate the signed curvature of $\epsilon$.

Show also that all the normal lines to $\gamma$ are tangent to $\epsilon$ (for this reason, the evolute of $\gamma$ is sometimes described as the ‘envelope’ of the normal lines to $\gamma$).

Show that in particular case of the cycloid ( $a > 0$ is a constant),

$$\gamma (t) = a(t − \sin t, 1 − \cos t), 0 <t< 2\pi ,$$

the evolute is

$$\epsilon (t) = a(t + \sin t, −1 + \cos t), $$

and that after a suitable re-parametrization $\epsilon$ can be obtained from $\gamma$ by a translation of the plane.

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I have shown that the arc-length of $\epsilon$ is $\frac{-1}{\kappa_s(s)}$ up to adding a constant.

Could you give me a hint how we could calculate the signed curvature of $\epsilon$ ?

Also how could we show that all the normal lines to $\gamma$ are tangent to $\epsilon$ ?

Answer 1

This is exercise $2.2.7$ in Pressley's Elementary Differential Geometry (which has solution sketches in the back of the text). I've expanded a bit on the solution provided.

Hints:

For signed curvature, what is the definition of signed curvature in terms of your signed normal and tangent? Remember that you're trying to find $\kappa_{s_{\epsilon}}$.

For the normal line part, what does it mean for a line to be tangent to $\epsilon$? What does a normal line at a specific point $s_0$ look like? We calculated the tangent to $\epsilon$ in the first part.

(More complete answer below)

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$$\epsilon (s)=\gamma (s)+\frac{1}{\kappa_s(s)}n_s(s)$$

$\textbf{1.}$ We show that the arc length of $\epsilon$ is $\frac{-1}{\kappa_s(s)}$ up to a constant. First off, I'm going to forget about the $s$ notation-wise so assume everything is a function of $s$ unless stated otherwise.

Well, we take the derivative of $\epsilon$ with respect to $s$ to get the arc length: $$\dot{\epsilon} = \dot{\gamma} + \frac{1}{\kappa_s}(-\kappa_s \mathbf{t})-\frac{\dot{\kappa_s} \mathbf{n_s}}{\kappa_s^2}=-\frac{\dot{\kappa_s} \mathbf{n_s}}{\kappa_s^2}$$

(recall $\mathbf{\dot{t}} = \kappa_s \mathbf{n_s} \implies \mathbf{\dot{n_s}} = -\kappa_s \mathbf{t}$ for a unit-speed curve since $\mathbf{n_s} \cdot \mathbf{t} = 0$ so $\mathbf{\dot{n_s}} \cdot \mathbf{t} + \mathbf{n_s} \cdot \mathbf{\dot{t}} = 0 \implies \mathbf{\dot{n_s}} \cdot \mathbf{t}= - \kappa_s(\mathbf{n_s \cdot n_s}) = -\kappa_s \implies \mathbf{\dot{n_s}}(\mathbf{t}\cdot \mathbf{t}) = \mathbf{\dot{n_s}} = -\kappa_s \mathbf{t}$)

Now, the arc length is given by $$u=\int \| \dot{\epsilon} \| \,ds = \int \frac{\dot{\kappa_s}}{\kappa_s^2} \,ds = -\frac{1}{\kappa_s} + C$$ Note that the second equality holds since we assumed $\dot{\kappa_s} > 0$.

$\textbf{2.}$ We calculate the signed curvature $\kappa_{s_{\epsilon}}.$ Recall the signed curvature is the rate at which the tangent vector rotates. In particular, $$\mathbf{\dot{t}}_{\epsilon} = \kappa_{s_{\epsilon}}\mathbf{n_s}_{\epsilon}$$ In this case, we take the tangent vector to be $\mathbf{t}_{\epsilon}=-\mathbf{n_s}$. Rotating the tangent vector counterclockwise by $-\pi/2$ gives us our signed unit normal. In particular, the signed normal is just $\mathbf{n_s}_{\epsilon}=\mathbf{t}$. Now, $$\frac{d (-\mathbf{n_s})}{\,du} = \kappa_s \mathbf{t} \frac{ds}{du} = \kappa_s \mathbf{t} \frac{\kappa_s^2}{\dot{\kappa_s}}= \frac{\kappa_s^3}{\dot{\kappa_s}}\mathbf{t}= \kappa_{s_{\epsilon}}\mathbf{n_s}_{\epsilon}$$ In particular, since the derivative of the tangent vector is the signed curvature times the signed unit normal, dotting the derivative of the tangent vector with the signed unit normal gives the result. That is, take the dot product of the above expression with $\mathbf{t}$ to get the signed curvature of $\epsilon$: $$\frac{\kappa_s^3}{\dot{\kappa_s}}$$

$\textbf{3.}$ We show that all normal lines to $\gamma$ are tangent to $\epsilon$.

Well, let's look at a point on the normal line at $\gamma(s_0)$ for some arbitrary $s_0$. It looks like $\gamma(s_0) + C\mathbf{n_s}(s_0)$ for some $C$. Since $\epsilon(s_0) = \gamma(s_0) + \frac{1}{\kappa_s(s_0)}\mathbf{n_s}(s_0)$, the intersection occurs when $C=\frac{1}{\kappa_s(s_0)}$. Well, we calculated the tangent of $\epsilon$ at $s_0$: $$\dot{\epsilon}(s_0)=-\frac{\dot{\kappa_s(s_0)} \mathbf{n_s}(s_0)}{\kappa_s^2(s_0)}$$ so that the tangent there is parallel to $\mathbf{n_s}(s_0)$.

$\textbf{4.}$ Regarding the evolute of the cycloid, this is just a computation, with a lot of the steps highlighted above. Regarding the reparameterization, consider $\tilde{t} = t + \pi$.

Answer 2

We have such cycloids as heirarchial or as in a generation, a way of looking at it in differential geometry. The material is included in several texts. Rather than repeat it, I try to explain in a simpler context, as necessary information is already given. Introduced here is another cycloid $\omega $ to see their continuity in a generational pattern.

$\omega, \epsilon , \gamma $ ( like grandfather, father, son) are cycloids evolved as derived figure from $\omega$ downwards as involutes, i.e,

Involute of $\omega $ is $\epsilon$, involute of $ \epsilon $ is $\gamma.$

Likewise in the other direction we have..

Evolute of $\gamma$ is $\epsilon$, evolute of $ \epsilon $ is $\omega.$

If $s$ is arc length and $R$ is radius of curvature ( reciprocal of $ \kappa(s)) $,

$$ s(\omega) = R (\epsilon ) $$

and

$$ s(\epsilon) = R (\gamma) $$

The above supposedly clarifies that calculating evolute of $\gamma$ for $\epsilon $ and finding involute of $\epsilon $ for $\gamma$ are the same, whichever is easier to calculate could be chosen.

In each case,

$$ s = R = 1/\kappa(s)\, ! $$

reckoned from cusp to any point. At hump top point curvature is $1/ 4a$ . In case we reckon $ s$ from the mid-point,

$$ s^2 + R^2 = (4a)^2 $$

proceeding similarly with limits of integration changed, usual length of arc calculation shows.

$$ s = 8 a \sin^2 (t/4) $$

The sign of curvature is always positive for hump downwards configuration. (The sign gets positive for prolate/curtate trochoids only. You can see the cycloid cusp at ground contact becoming smooth with derivatives curving up for these cases).

These are integrated/reckoned from cusp to middle. Since you say "upto a constant length" you understand the datum points. I.e., if $\gamma$ is involute of $ \epsilon $ , then $ \epsilon $ is one among a parallel set of evolute curves of constant normal distance separation along $n$.(Bertrand curve here, Bertrand surfaces if in 3-D equidistant normals.)

The first question you have yourself answered. Note that it is a property of any involute/evolute pair and is not peculiar to the cycloids.

The second question about hint is straight-forward when the natural/intrinsic equations of $ \gamma, \epsilon $ are seen together:

$$ s_{\gamma}=R_{\gamma} \tag{1} $$ $$ s_{\epsilon}=R_{\gamma} ; s_{\gamma}=R_{\epsilon} ; \tag{2} $$ $$ s_{\epsilon}=R_{\epsilon} \tag{3} $$

( In natural equation we have these geometric quantities conserved by isometry, as products of same first fundamental form in two dimensions.)

It is seen that the two are identical except for Euclidean rotations and displacements...( A fact that could be generalized to hypo/hyper cycloids from its natural equation).

EDIT1:

It is seen that $ t_1 \rightarrow t + \pi $ and euclidean motion (translation) , $x_1 \rightarrow x + \pi$ , $ y_1 \rightarrow y -2a, $ we reach to the second cycloid $ \gamma \rightarrow \epsilon. $

(Another fact not directly related is that the normal is bisected by x-axis.)