# Prove that curve is a straight line iff curvature vanishes.

For my class in classical differential geometry I have to solve the following problem:

Let $$\alpha$$ be a regular curve in $$\mathbb{R}^2$$. Proof that:

a) $$\quad$$ $$\alpha$$ is a straight line if and only if the curvature vanishes.

b) $$\quad$$ If $$\dot{\alpha}$$ and $$\ddot{\alpha}$$ are linearly dependent, then $$\alpha$$ is a straight line.

Conisder a). At first this problem seemed easy to me. Because if the curvature is defined via

$$\ddot{\alpha}(t) = \kappa(t) \cdot n(t) \qquad \qquad (1)$$

where $$n(t)$$ is the unit vector perpendicular to the tangent vector, then it is easy to see that $$\kappa = 0$$ implies $$\ddot{\alpha} = 0$$ and therefore $$\alpha$$ is a straight line. However, I realsied, that the above equation is derived by using the fact

$$\big< \dot{\alpha} , \dot{\alpha} \big> = 1 \quad \Longrightarrow \quad \big< \dot{\alpha}, \ddot{\alpha} \big> = 0$$

in the case of $$\alpha$$ being parameterised by arc length. Since my given $$\alpha$$ is not parameterised by arc length, I cannot assume equation (1) to hold and therefore cannot assume my $$\ddot{\alpha}$$ to be perpendicular to the tangent vector. My conclusion would be wrong then. Any ideas on how to solve this problem for general curves $$\alpha$$?

b) I have no idea on how to show this. Any tips here?

• If you can prove $(a)$ then $(b)$ follows immediately. If $\alpha'(t)$ and $\alpha''(t)$ are linearly dependent for all $t\in \mathbb{R}$ we know that $\alpha'(t) \times \alpha''(t)=\vec{0}$ for all $t$ and so $\kappa(t)=\frac{\|\alpha'(t) \times \alpha''(t)\|}{\|\alpha'(t)\|}^3=0$. Note the denominator in the previous expression for $\kappa(t)$ never vanishes by our assumption that $\alpha$ is a regular curve.
– user801306
Commented Mar 13, 2021 at 19:39
• I forgot to mention that the curve lies in $\mathbb{R}^2$. I updated my question accordingly. Commented Mar 13, 2021 at 19:46
• Set $\alpha(t)=\big(x(t),y(t)\big)$. If $\alpha'(t),\alpha''(t)$ are linearly dependent, then $$x'(t)y''(t)-y'(t)x''(t)=\det\Big(\begin{smallmatrix} x'(t) & x''(t)\\ y'(t) & y''(t) \end{smallmatrix}\Big)=0$$ Hence $$\kappa(t)=\frac{x'(t)y''(t)-y'(t)x''(t)}{\Big(\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2}\Big)^3}=0$$ for all $t$. Also, you can assume WLOG that you're parameterizing w.r.t arc length since you're curve $\alpha$ is regular.
– user801306
Commented Mar 13, 2021 at 20:16

Bearing in mind that any regular curve can be parametrized by the arc length then

a) if $$\alpha$$ is a straight line, i.e., $$\alpha(t)=\text{p}+\text{v}t$$ and so $$k(t)=||\alpha''(t)||=||0||=0$$.

If $$k(t)=0$$ then $$\alpha''(t)=0$$. If you calculate the antiderivative twice you obtain $$\alpha'(t)=\text{v}$$ and finally $$\alpha(t)=\text{p}+\text{v}t$$ for some $$\text{p,v}\in\mathbb{R^2}$$.

b) As you saw $$\langle\alpha',\alpha''\rangle=0$$ then $$\alpha''$$ is null vector or $$\alpha'$$ is perpendicular to $$\alpha''$$, but as they are linear dependent then $$\alpha''=0$$ and by a) it can only be a straight line.

Hint for $$(a)$$ : WLOG Assume $$\alpha(0)=0$$ (Curvature is invariant under isometries) and try looking at $$\gamma(s)=(\int_0^scos(\theta(u))du, \int_0^ssin(\theta(u))du)$$.