a) Show that the curvature of the planar graph r = f($\theta$) at a general point is
$\kappa(\theta)$ = $\frac{[2f'^2(\theta) + f^2(\theta) - f(\theta)f''(\theta)]}{[(f'^2(\theta) + f^2(\theta))^{3/2}]}$
b) Find the curvature of the curve r = a(1 - cos $\theta$).
I have been looking in my textbook for a solution to this problem and I have not come across anything too helpful. I have seen this formula in my book but nothing providing me with a gateway to finding a solution. Any help would be much appreciated on a) and/or b).
So I guess your curvature formula is $$\kappa = \frac{\|{\bf r}'(t)\times{\bf r}''(t)\|}{\|{\bf r}'(t)\|^3} $$
There are several and this should work. First note to go from polar to cartesian we have $${\bf r}(t) = \langle r\cos(t),r\sin(t) \rangle $$
So for you $${\bf r}(t) = \langle f(t)\cos(t),f(t)\sin(t),0 \rangle $$
Now just compute ${\bf r}'(t)$ and ${\bf r}''(t)$ (you'll need the product rule). It goes something like $${\bf r}'(t) = \langle f'(t)\cos(t) + f(t)(-\sin(t)), \ldots \rangle $$
In part (b) just apply the formula from part (a).
So you have all these pieces above like $f(\theta)$ and $f'(\theta)$. These pieces are
$$f(\theta) = a(1 - \cos(\theta))$$ and $$f'(\theta) = a\sin(\theta)$$