I'm trying to prove that if $\vec{x}:I\rightarrow\mathbb{R}^2$ is a curve parametrized by arc length and $\theta(t)$ is the angle between the tangent line to $\vec{x}$ at point $t$ and the $x$ axis, then $\kappa=\theta'$, where $\kappa$ denotes the curvature.
I know that for a curve in $\mathbb{R}^2$, the curvature is given by
$\kappa=\frac{1}{|\vec{x}'|^3}det(\vec{x}'\vec{x}'')$.
I also know that the tangent line to $\vec{x}$ at point $\alpha\in\mathbb{R}$ is given by $y(t)=\vec{x}(\alpha)+t\vec{x}'$.
I can picture the problem, but can't write the solutions. Any ideas?
Best Answer
Assume a planar curve (for simplicity function) $y=f(x)$, i.e. $\overline{x} (t)=(t,f(t))$, then $$ k=\frac{\left(\dot{\overline{x}},\ddot{\overline{x}}\right)}{|\dot{\overline{x}}|^3} $$ or equivalently $$ k=\frac{y''}{\left(1+(y')^2\right)^{3/2}} $$ Suppose that $\phi$ is the angle between the tangent line $(\epsilon)$ and the $x$ axis. We have $\frac{dy}{dx}=\tan \phi$ or equivalently $\phi=\arctan\left(\frac{dy}{dx}\right)$. Taking derivatives in both parts of the last equation we have $$ \frac{d\phi}{dx}=\frac{y''}{1+(y')^2} $$ But $$ \frac{ds}{dx}=\sqrt{1+(y')^2} $$ Hence $$ k=\frac{y''}{1+(y')^2}\frac{1}{\sqrt{1+(y')^2}}=\frac{\frac{d\phi}{dx}}{\frac{ds}{dx}}=\frac{d\phi}{ds}. $$