275- Constant curvature

The text we are using for Calculus III introduces the notion of unit tangent vector, principal unit normal vector, and curvature, for smooth curves vec r(t), and it also mentions that circles and lines are planar curves of constant curvature. (A curve is planar if its image is contained in a plane, i.e., if it describes a two-dimensional trajectory.)

Surprisingly, though, the text does not explain (not even in the exercises) that circles and lines are the only smooth planar curves of constant curvature. The argument for this is simple enough, so I will show it here:

Recall that if vec r(t) is a smooth curve, we set

displaystylevec T=frac{dvec r}{ds}

where s denotes arc length. We can define the curvature kappa of vec T as

displaystylekappa=left|frac{dvec T}{ds}right|.

Assume that kappa is constantly equal to zero. Then frac{dvec T}{ds}=vec 0, so vec T is constant, so vec r=svec T+vec r_0 for some constant vector vec r_0. This is the parametric equation of a line, so we have shown that the only curves with zero curvature are straight lines. Notice that for this we did not need to assume that vec r is a planar curve. From now on we may assume that this is the case and that veckappa is constant and different from zero. 

In terms of t, we can equivalently compute vec T as 

displaystylevec T=frac{dvec r/dt}{|dvec r/dt|}=frac{vec v}{|vec v|}.

We then set vec N to be the unit vector in the direction of frac{dvec T}{ds}, so

displaystylekappavec N=frac{dvec T}{ds}

or, equivalently in terms of t,

displaystylevec N=frac{dvec T/dt}{|dvec T/dt|},

so displaystyle vec T'=kappa|vec v|vec N. Recall that vec Ncdotvec T=0, since vec Tcdotvec T=1 is constant, so its derivative is zero.

Since vec Ncdotvec N=1 is also constant, again we have that vec Ncdot vec N'=0. Since vec T is also perpendicular to vec N and we are assuming that vec r moves on a plane, this means that vec N' is a multiple of vec T,

vec N'(t)=a(t)vec T(t),

where a(t) is some scalar function. To compute a, notice that vec N'cdot vec T=avec Tcdotvec T=a. But also vec Tcdot vec N=0, so vec T'cdotvec N+vec Tcdotvec N'=0, or a=-vec T'cdotvec N. Since vec T'=kappa|vec v|vec N, we finally find that 

a=-kappa|vec v|mbox{ and }vec N'=-kappa|vec v|vec T.

Recall also that the osculating circle to the curve vec r at the point t is the circle of the same curvature as vec r at t and centered in the direction of the vector vec N at t, so its center is given by 

displaystyle vec c(t)=vec r(t)+frac1{kappa}vec N(t),

since the curvature of a circle is the inverse of its radius. In particular, since kappa is constant, all the osculating circles to vec r have the same radius. Now we prove that vec c is constant (from the assumptions that vec r is planar and kappa is constant). To see this, compute vec c,'(t) to find

displaystyle vec c,'(t)=vec r,'(t)+frac1{kappa}vec N'(t)=vec v-frac{kappa|vec v|}{kappa}vec T,

or vec c,'=vec v-|vec v|vec T=vec 0

But this completes the proof, because then displaystyle|vec r(t)-vec c|=frac1{kappa}, since |vec N|=1, i.e., the curve vec r lies on a circle of radius 1/kappa. {sf QED}

On the other hand, there are curves of constant curvature other than lines and circles, if the curve is not confined to a plane. The helix vec r(t)=cos(t)vec i+sin(t)vec j+tvec k is an example.


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