# Thread: Show that this curve is contained in a plane

1. ## Show that this curve is contained in a plane

Show that a curve $\displaystyle \gamma$ such that $\displaystyle \dddot{\gamma} = 0$ everywhere is contained in a plane
So of course the first thing I did was integrated it a few times to get:

$\displaystyle \gamma (t) = (a_1 t^2 + b_1 t + c_1 , a_2 t^2 + b_2 t + c_2, a_3 t^2 + b_3 t + c_3)$

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 $\displaystyle \gamma (t) = (a_1 t + b_1 u + c_1 , a_2 t + b_2 u + c_2, a_3 t + b_3 u + c_3)$
where t and u are the varying parameters. In the above integrated equation, there's 2 varying parameters, $\displaystyle t$ and $\displaystyle t^2$ so it almost fits the equation of a plane, but in this case u is restricted to u=t^2

2. ## Re: Show that this curve is contained in a plane

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...

3. ## Re: Show that this curve is contained in 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 $\displaystyle \kappa= |t'|$ is the "curvature". The unit normal is $\displaystyle n= \frac{t'}{\kappa}$. The 'binormal' is their cross product: $\displaystyle b= t\times n$. The torsion is the length of the derivative of b: $\displaystyle \tau= |b'|$.)

The fact that the torsion is a multiple of the third derivative:
$\displaystyle \tau= \frac{(\gamma'\times \gamma'')\cdot \gamma'''}{||\gamma'\cdot\gamma''||}$
(http://en.wikipedia.org/wiki/Torsion_of_a_curve)
makes this problem easy.