curvature of curve in $\mathbb R^2$

Question

I'm taking a course in differential geometry this semester and I'm stuck with one of my first exercises.

Let $\alpha(s)=(x(s),y(s))$ be a curve such that $|\alpha'(s)|=1$. Prove that the curvature is given by $k(s)=|x'(s)y''(s)-x''(s)y'(s)|$.

So far we defined the curvature of such curves as $k(s)=|\alpha''(s)|$, but this doesn't really get me to the formula I'm supposed to prove. Since we haven't really done anything in the course yet, I don't think that I have to do anything really complicated here, but still I don't know what to do.

Anyone who can help me with this? Thanks.

Answer 1

Since $|\alpha'|^2=(x')^2+(y')^2$ is constant, $\alpha''$ is orthogonal to $\alpha'$. But $(-y',x')$ is also orthogonal to $\alpha'=(x',y')$, so it is parallel to $\alpha''=(x'',y'')$. What does that tell you about the inner (or dot) product between the two vectors?

Answer 2

Besides to descriptive @Harald post, we know that $v=d\alpha/dt=v\mathbf{T}$ and so $a=(dv/dt)\mathbf{T}+v^2\kappa\mathbf{N}$ and so $|v\times a|=\kappa v^3$. I assume you know the cross product which is used here.