I take the definition of arc length of a smooth curve between $x=a$ and $x=b$ to be: $$\int^a_b \sqrt{1+f'(x)^2}dx$$
Then how do I derive the formula for polar coordinates?
And please don't use infinitesimals, (or poorly defined things). Only use rigorous arguments like epsilons and deltas.
Best Answer
If you parameterize the curve using $x=x(t)$, $y = y(t)$, $a \le t \le b$ the formula for arclength is $$L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2} \, dt.$$
You can let $r(t)$, $\theta(t)$ denote the polar coordinates of the point $(x(t),y(t))$.
Since $x(t) = r(t) \cos \theta(t)$ and $y(t) = r(t) \sin \theta(t)$ you get $$x'(t) = r'(t) \cos \theta(t) - r(t) \sin \theta(t) \theta'(t)$$and $$y'(t) = r'(t) \sin \theta(t) + r(t) \cos \theta(t) \theta'(t).$$ Square to get $$x'(t)^2 = r'(t)^2 \cos^2 \theta(t) - 2 r(t) r'(t) \cos \theta (t) \sin \theta(t) \theta'(t) + r(t)^2 \sin^2 \theta(t) \theta'(t)^2$$ $$y'(t)^2 = r'(t)^2 \sin^2 \theta(t) + 2 r(t) r'(t) \cos \theta (t) \sin \theta(t) \theta'(t) + r(t)^2 \cos^2 \theta(t) \theta'(t)^2$$ thus $$x'(t)^2 + y'(t)^2 = r'(t)^2 + r(t)^2 \theta'(t)^2$$ so that $$L = \int_a^b \sqrt{r'(t)^2 + r(t)^2 \theta'(t)^2} \, dt.$$