Proof: Curvature Zero -> Motion along a line

[Thomas_]

Homework Statement

Proof that, if a particle moves along a space curve with curvature 0, then its motion is a along a line.

Homework Equations

$$K=\frac{||r'(t)\times r''(t)||}{(||r'(t)||)^3}$$
(curvature of a space curve)

The Attempt at a Solution

Assume the curve is smooth, so r'(t) cannot be the zero vector. The numerator must be 0. I evaluate the cross product (set it to 0), and get the following equations.

$$g'(t)h''(t) = h'(t)g''(t)$$
$$f'(t)h''(t) = h'(t)f''(t)$$
$$f'(t)g''(t) = g'(t)f''(t)$$

Here I don't know what to do to get to the equation of a line.

Thank you in advance.

 

[Dick]

The easiest way to see this to take the curve parametrization so that |r'(t)|=1. If you differentiate r'(t).r'(t) you see that r'(t).r''(t)=0 (so r' and r'' are perpendicular). Curvature=0 tells you also that r'(t)xr''(t)=0 (so r' and r'' are parallel). What does that tell you about r''(t)?

 

[Thomas_]

Thank you very much for your answer.

That would mean that r''(t) is the 0-vector. Which means that there is no acceleration, no change of direction, motion should be straight.

However, my professor told me that this is not formal enough. I tried to arrive at a similar conclusion by using ||a x b|| = ||a|| ||b|| sin(theta).

Is there no way to formally arrive "back" at the parameterized equation of a line by using the result I already have?

 

[Dick]

If r''(t) is zero then r'(t) is a constant. So r'(t)=r'(0) for all t. Integrating r'(t) to get r(t) then gives you r(t)=r(0)+r'(0)*t, right? That isn't formal enough? It looks messy to try to argue starting with what you have to the conclusion r(t) is a line.