Curvature of the Archimedean spiral in polar coordinates

Question

Let's take a look on the Archimedean spiral. The parametric equation is:

$$c : \mathbb R \to \mathbb R^{2} \,;\, c(t) := (t\cos(t), t\sin(t))$$

The goal of the exercise is to compute the curvature of the spiral in polar coordinates. What I've tried? I convert the parametric equation into polar coordinates:

$$x(\varphi) = r(\varphi)\cos(\varphi)$$

$$y(\varphi) = r(\varphi)\sin(\varphi)$$ Is that correct so far?

I know how to compute the curvature with the parametric equation, that's not the point. But how I can do it with the polar coordinates?

Thanks in advance

Answer 1

Let $r$, $\varphi\colon [a,b]\to\mathbb R$ be $C^{\infty}$. Applying the definition of the curvature to $\gamma=r\cdot(\cos\mathrel\circ\varphi,\sin\mathrel\circ\varphi)$, a lengthy calculation shows that $$\kappa=\frac{r^2\dot\varphi^3+2\dot r^2\dot\varphi-r\ddot r\dot\varphi+r\dot r\ddot\varphi}{(\dot r^2+r^2\dot\varphi^2)^{3/2}}$$

In your case: $r(t)=\varphi(t)=t$, hence $\kappa(t)=\frac{t^2+2}{\sqrt{1+t^2}^3}$.

Answer 2

Your parametric equation in polar is $r=t, \theta=t$

Answer 3

$$r=θ \\ dr/dθ=1 \\ ds=ρdθ \\ (rdθ)2+(dr)2=(ds)2 \\ (rdθ)2+(dr)2=(ρdθ)2 \\ (r)2+(dr/dθ)2=(ρ)2 \\ (r)2+(1)2=(ρ)2 \\ ρ=\sqrt{r2+1}=\sqrt{θ2+1}$$