Or you could just say that if the third derivative of your curve is 0 everywhere, then the function must be a quadratic, which has a parabola as its graph. Clearly as a 2 dimensional curve it must lie on a plane...
So of course the first thing I did was integrated it a few times to get:Show that a curve such that everywhere is contained in a plane
where the a, b and c's are constants.
Now how do I continue from there?
EDIT:
Actually I think I have it,
The parametric equation of a plane is
where t and u are the varying parameters. In the above integrated equation, there's 2 varying parameters, and so it almost fits the equation of a plane, but in this case u is restricted to u=t^2
Or you could just say that if the third derivative of your curve is 0 everywhere, then the function must be a quadratic, which has a parabola as its graph. Clearly as a 2 dimensional curve it must lie on a plane...
If you have taken, or are taking, differential geometry (this looks like an elementary differential geometry problem to me) you could just show that if the third derivative of the position vector is the 0 vector then the torsion is 0.
(If t is the unit tangent vector at any point on the curve, then t' is perpendicular to the curve. It length is the "curvature". The unit normal is . The 'binormal' is their cross product: . The torsion is the length of the derivative of b: .)
The fact that the torsion is a multiple of the third derivative:
(http://en.wikipedia.org/wiki/Torsion_of_a_curve)
makes this problem easy.