How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?

Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of that curve.

Definition:- Attach a string to a point on a curve. Extend the string so that it is tangent to the curve at the point of attachment. Then wind the string up, keeping it always taut. The locus of points traced out by the end of the string is called the involute of the original curve,

My argument:- Suppose $\Gamma$ is the given curve. Attach a string to a point on $\Gamma$. Extend the string so that it is tangent to the curve at the point of attachment. Then wind the string up, keeping it always taut. The locus of points traced out by the end of the string is called the involute of the $\Gamma$. Let this involute be $\Delta.$ To find the evolute, we find centre of curvature to at each point of $\Delta$. Joining these points. How the evolute of $\Delta=\Gamma$? Please explain


3 Answers 3


If $\alpha(t)$ is a regular curve arbitrarily parametrized by $t$, not necessarily the arc-length of $\alpha$, and with nowhere vanishing curvature $\kappa_\alpha(t)$, the involute of $\alpha(t)$ originating at the point $\alpha(t_0)$ is defined to be the curve

$\beta(t) = \alpha(t) - \dfrac{\dot \alpha(t)}{\vert \dot \alpha(t) \vert} \displaystyle \int_{t_0}^t \vert \dot \alpha(u) \vert \;du; \tag 1$

that is, $\beta(t)$ is the point $\alpha(t)$ translated by the distance

$\displaystyle \int_{t_0}^t \vert \dot \alpha(u) \vert \; du \tag 2$

in the direction of the unit vector

$-\dfrac{\dot \alpha(t)}{\vert \dot \alpha(t) \vert} \tag 3$

tangent to $\alpha(t)$ at $t$; the presence of the "$-$" sign in (1), (3) allows the analogy of the "stretched string", often used to describe the involute, to go forward: (1) describes, in mathematical terms, the free end of a taught string of length (2) having been "peeled off" of the curve $\alpha(t)$ from the point $\alpha(t_0)$ in the direction of the point $\alpha(t)$; the presence of the "$-$" sign in (1) corresponds to the fact that as the string is peeled off towards $\alpha(t)$ from $\alpha(t_0)$, the distance of $\beta(t)$ from $\alpha(t)$ increases in the opposite direction.

These niceties of geometric visualization aside, equation (1) may be written in somewhat simpler terms, first via the observation that

$\dfrac{\dot \alpha(t)}{\vert \dot \alpha(t) \vert} = T_\alpha(t) \tag 4$

is the unit tangent vector field to $\alpha(t)$, and second by means of the fact that the integral (2) is in fact the arc-length along $\alpha(t)$ 'twixt $\alpha(t_0)$ and $\alpha(t)$:

$s - s_0 = \displaystyle \int_{t_0}^t \vert \dot \alpha(u) \vert \; du, \tag 5$

where $s_0$ is arbitrarily assigned to $\alpha(t_0)$ (relative to some unspecified reference point on $\alpha(t)$); by means of (4) and (5) we may write (1) in terms of arc-length parametrization along $\alpha(t)$ as

$\beta(s) = \alpha(s) - T_\alpha(s)(s - s_0). \tag 6$

We wish to find the evolute of $\beta(s)$; to this end we find its tangent vector

$\dot \beta(s) = \dot \alpha(s) - \dot T_\alpha(s)(s - s_0) - T_\alpha(s)$ $= T_\alpha(s) - \dot T_\alpha(s)(s - s_0) - T_\alpha(s) = -\kappa_\alpha(s)N_\alpha(s)(s - s_0), \tag 7$

where $\kappa_\alpha(s)$ and $N_\alpha(s)$ are the curvature and unit normal vector field to $\alpha(s)$; in deriving (7) we have called upon the Frenet-Serret equation

$\dot T_\alpha(s) = \kappa_\alpha(s) N_\alpha(s). \tag 8$

It should be borne in mind that though $s$ is the arc-length along $\alpha(s)$, it is not in general the arc-length $s'$ along $\beta(s)$; therefore we may not assume that

$\vert \dot \beta(s) \vert = 1, \tag 9$


$\dfrac{ds'(s)}{ds} = \vert \dot \beta(s) \vert, \tag{9.5}$

and to find the unit tangent vector $T_\beta(s)$ to $\beta(s)$ we must normalize (7); we have

$\vert \dot \beta(s) \vert = \vert -\kappa_\alpha(s) N_\alpha(s)(s - s_0) \vert = \kappa_\alpha(s)(s - s_0) ,\tag{10}$

since $\kappa_\alpha(s) > 0$ and we may take $s - s_0 > 0$ by assuming, without loss of generality, that $t > t_0$ in (5); then

$T_\beta(s) = \dfrac{\dot \beta(s)}{\vert \dot \beta(s) \vert} = \dfrac{-\kappa_\alpha(s)N_\alpha(s)(s - s_0)}{\kappa_\alpha(s)(s - s_0)} = -N_\alpha(s), \tag{11}$

from which we may compute the unit normal $N_\beta(s)$ to $\beta(s)$; indeed,

$\dot T_\beta(s) = -\dot N_\alpha(s) = \kappa_\alpha(s)T_\alpha(s), \tag{12}$

where we have used the Frenet-Serret equation complementing (8),

$\dot N_\alpha(s) = -\kappa_\alpha(s)T_\alpha(s); \tag{13}$

then in accord with (9.5),

$\kappa_\beta(s)N_\beta(s) = \dfrac{dT_\beta(s)}{ds'} = \dfrac{ds}{ds'}\dfrac{\dot T_\beta(s)}{ds} = \dfrac{\dot T_\beta(s)}{\vert \dot \beta(s) \vert} = \dfrac{\kappa_\alpha(s)T_\alpha(s)}{\kappa_\alpha(s)(s - s_0)} = \dfrac{T_\alpha(s)}{s - s_0}, \tag{14}$

from which

$\kappa_\beta(s) = \vert \kappa_\beta(s) \vert \vert N_\beta(s) \vert = \vert \kappa_\beta(s) N_\beta(s) \vert = \left \vert \dfrac{T_\alpha(s)}{s - s_0} \right \vert = \dfrac{\vert T_\alpha(s) \vert}{s - s_0} = \dfrac{1}{s - s_0}; \tag{15}$

thus (14) yields

$N_\beta(s) = T_\alpha(s), \tag{16}$


$\dfrac{1}{\kappa_\beta(s)} N_\beta(s) = (s - s_0)T_\alpha(s); \tag{17}$

the evolute of $\beta(s)$ is thus

$\beta(s) + \dfrac{1}{\kappa_\beta(s)}N_\beta(s)$ $= \beta(s) + (s - s_0)T_\alpha(s) = \alpha(s) - (s - s_0)T_\alpha(s) + (s - s_0)T_\alpha(s) = \alpha(s), \tag{18}$

our original curve, parametrized now by its arc-length $s$.


The set is identical upto rotation. All curves are the same and can be simply obtained by rotation

These involutes are also called *curves of constant width d *

$ \theta = $ Total Released length/ base circle radius

The set of all curves $\Gamma$ is formed by a variable extension say $d=\Delta n $ along normal of curve which is same as the tangent taut string length segment.

For same $d$ different choice of initial boundary point on the cusp of involute where you chose to unwind the taut string produces another curve as shown The width $d$ exactly equals arc length released out of contact from the base circle.

For 2D involute case the differential relation/ equation of all the involute family (primes differentiation with respect to $\theta$ polar angle) is

$$ r. \cos \psi = r_{base\, circle};\, \psi^{'}= \cot^2 \psi ;\, r^{'}= r \cot \psi $$

This can be derived by considering polar differential triangles.

Plots are integrands of above ODE where primes are with respect to polar angle $\angle XOP$.Curves are made with $45^{\circ} $ start polar angle difference. They are all same except for the arbitrary constant of integration.

In the example you gave the circle is $\Gamma$ whose evolute of its involute is the same circle so it is confusing due to symmetry...

In an example of filial relations.. your father'son can be you or your other sibling, who is a member of your family.. this applies to differential relations also.

In conclusion the Evolute of an Involute of a curve Γ is Γ itself upto a constant of integration associated with the independent variable in the defining differential equation.

Involte Normal Distance Constant

I shall give another simpler example where involutes and evolutes are same cycloids but placed differently.

enter image description here

The involutes of bottom cycloid $\Gamma$ is sketched at top, call it Invo_cycloid. The evolute of Invo cycloid again happens to be the sibling continuation of $ \Gamma,$ extending to the right, is it not?

If you wish to verify the cycloids, DE is

$$ \dfrac{ y y^{''}}{1+y^{'2}}=\dfrac12 $$

where primes denote differentiation with variable on x-axis. The plots of above taut strings (that always double as tangents to $\Gamma$ and as normals to their involute).


HINT: Write down an explicit parametrization and compute, using the Frenet formulas. It might help to characterize the involute as follows: $\beta$ is the involute of $\alpha$ if $\beta(t)$ lies on the tangent line to $\alpha$ at $\alpha(t)$ and their tangent vectors are orthogonal at the corresponding points.


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