What space curves can this theorem describe? We were given the following theorem in our Vector Calculus class:


  THM: For space curve $R$ which does not pass through the origin, and which has a second derivative, the following are equivalent:
  
  1) $R^{\prime\prime} \parallel R$ at all points
  
  2) $R \times R^\prime = C$ where $C$ is a constant vector
  
  3) There is a constant vector C such that either:
  
  
    a) $C \ne 0,$the curve is in the plane through 0 which is perpendicular to $C$, and the position vector $R(t)$ sweeps out area at the constant rate $\frac{\left | C \right |}{2}$
    
    or
    
    b) $C = 0$, and the curve is confined to a line through 0
  


My question is this: what curves fulfill any of these conditions? Obviously, any curve confined to a line works, as does any curve confined to an ellipse or hyperbola. Are there any other functions it could apply to, or just these conic sections?
Might as well have the curve in the $xy$ plane. Kepler, and for that matter classical electromagnetism, has acceleration $r''$ parallel to position $r,$ with magnitude proportional to $1/|r|^2.$ Gravity gives ellipses, repulsion of like electric charges gives, i suppose, hyperbola. 

Let position be $(x(t), y(t))$ and, as usual, take $r^2 = x^2 + y^2.$ The Law of Gravity is
$$  (x'', y'') = - (x,y)/(x^2 + y^2).  $$

How about The Law of Jagy,
$$  (x'', y'') = - (x,y)/(x^2 + y^2)^2.  $$ It will not give ellipses in general. On the other hand, there is one circle, $(\cos t, \sin t).$
Might as well have the curve in the $xy$ plane. Kepler, and for that matter classical electromagnetism, has acceleration $r''$ parallel to position $r,$ with magnitude proportional to $1/|r|^2.$ Gravity gives ellipses, repulsion of like electric charges gives, i suppose, hyperbola. 

Let position be $(x(t), y(t))$ and, as usual, take $r^2 = x^2 + y^2.$ The Law of Gravity is
$$  (x'', y'') = - (x,y)/(x^2 + y^2).  $$

How about The Law of Jagy,
$$  (x'', y'') = - (x,y)/(x^2 + y^2)^2.  $$ It will not give ellipses in general. On the other hand, there is one circle, $(\cos t, \sin t).$