4.5: Involute Curves of Space Curves

In the study of plane curves, once we defined the evolute of a curve, we considered the inverse problem of finding all curves that have a given curves as evolute. We may consider the analogous problem for space curves. If a curve Y is the evolute of a curve X, then X is said to be an "involute" of the curve Y. As in the planar case, we may define an involute by

    I(t) =X(t) +(c-s(t))T(t)
In this way, for various choices of the constant c, we obtain a one-paramater family of space curves that are involutes of a given curve.

Need to put an APPLET here! Involutes of Space Curves Demo Need to put an IMG here!

Clicking on Involute will toggle the display of the involute

    I(t) =X(t) +(a-s(t))T(t)
As in previous demos, the variable c is a constant that is to be used in the type-in curve while a is the constant that is used in the definition of the involute.

This simple demo allows you to input a curve with its domain and will show its involute.

    Look at various helices and corresponding plane curves. Now, look at various other original curves. If a curve has a planar involute, is it true that it has to be a helix?

We look at the involute of a helix.

For a circular helix, the involute will be a plane curve, which is itself an involute of a circular cross-section of the cylinder. The same will be true for any helix drawn on a cylinder over a plane curve.

A helix is defined to be a curve for which the unit tangent vectors make a fixed angle with a given unit vector W . If

    T(t) ·W=cos(a)
with a being constant, then
    T'( t)·W= 0=k(t) s'(t) P(t)· W
so
    W=cos(a)T(t) +0·P(t) +zB(t)
Since W is a unit vector, z=±sin(a) so that
    W=cos(a)T(t) ±sin(a)B(t)
Differentiating this relationship gives
    W'=0 =cos(a)k(t)s'(t)P(t)±sin( a)t(t) s'(t) P(t)
so
    t(t)k(t)=± cos(a)sin( a)
Thus we have a linear relationship between t(t) and k(t) independent of t . Conversely if we have
    t(t)=bk( t)
for all t , then if
    b=tan(a)
we may define the vector cos(a)T(t)±sin( a)B( t) and differentiate to show that it must be a constant vector, say W . This W will make a constant angle with all unit tangent vectors.

To show that an involute of a helix with axis vector W is a plane curve, we observe that

    I(t) =X(t) +(c-s(t))T(t)
so
    I'( t)=(c-s(t))k(t) s'(t) P(t)
so
    I'( t)·W=0
and
    I(t)·W =c
the condition for the involute to lie in a plane perpendicular to W .

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