Find curvature of spiral of archimedes

[meson0731]

Homework Statement

Find the curvature of the spiral of Archimedes r = 2θ

Homework Equations

||v x a || / ||v||^3

The Attempt at a Solution

I tried to convert the polar equation into parametric and got
x = 2θsinθ
y = 2θsinθ
z = 0

I think took the derivative of x y and z and got

x' = -2θsinθ + 2cosθ
y' = 2θcosθ + 2sinθ
z' = 0

I then put them into vectors and used the cross product to get v x a and got the magnitude to be 4θ^2. I found ||v||^3 to be 8θ^3 so I got || v x a || / ||v^3|| = (1/2θ) .

 

[Nino MMP]

I'm not sure if this is correct and I'm not sure if the curvature is supposed to be in terms of θ or r or both. Any help would be appreciated. Thank you.