Let $r = r(\varphi)$ be the equation of a plane curve $\alpha$ in polar coordinates. Prove that the length of $\alpha$ is given by
$$ \int_{\varphi_1}^{\varphi_2} \sqrt{ r^2(\varphi) + \dot{r}(\varphi)^2} \, d \varphi$$
I parametrise $\alpha$ as follows
$$ \alpha(t) \enspace = \enspace |\alpha(t)| \cdot \Big( \cos \varphi(t), \sin \varphi(t) \Big)$$
And call $|\alpha(t)| = r(t)$. I then solved it this way: \begin{align} \operatorname{len}(\alpha) \enspace &= \enspace \int_a^b |\dot{\alpha}(t)| \, dt \\ &= \enspace \int_a^b \sqrt{ \dot{\alpha}_1^2(t) + \dot{\alpha}_2^2(t)} \, dt \\ &= \enspace \int_a^b \sqrt{ \dot{\varphi}^2(t) \cdot \Big[ \big( \tfrac{\partial r}{\partial \varphi}\big)^2 + r^2(\varphi) \Big] } \, dt \\ &= \enspace \int_{\varphi_1}^{\varphi_2} \sqrt{\dot{r}^2(\varphi) + r^2(\varphi)} \, d\varphi \end{align}
where I identified $\varphi(a) = \varphi_1$ and $\varphi(b) = \varphi_2$. However, I am struggling to understand one problem that comes with this:
In my approach I used $r(t) = r\big( \varphi(t) \big)$, so that the derivative is $\tfrac{d}{dt} r \big( \varphi(t) \big) = \dot{\varphi}(t) \cdot \tfrac{\partial r}{\partial \varphi}$. But am I really allowed to do this? If not, then how do I solve this? And if I am allowed to do this, why am I?
I would parametrize directly by angle $\varphi$. By recalling the general formula for a length of a curve given in the parametric form \begin{align} L=\int_{t_1}^{t_2}\sqrt{\left(\frac{\text{d}x}{\text{d}t}\right)^2+\left(\frac{\text{d}y}{\text{d}t}\right)^2}\,\text{d}t \end{align} then by considering $t\equiv\varphi$ and by substitution $x(\varphi)=r(\varphi)\cos\varphi,~y(\varphi)=r(\varphi)\sin\varphi$ one gets \begin{align} L=&\int_{\varphi_1}^{\varphi_2}\sqrt{\left(\frac{\text{d}x}{\text{d}\varphi}\right)^2+\left(\frac{\text{d}y}{\text{d}\varphi}\right)^2}\,\text{d}\varphi=\\ &\int_{\varphi_1}^{\varphi_2}\sqrt{\left( r'(\varphi)\cos\varphi-r(\varphi)\sin\varphi\right)^2+\left( r'(\varphi)\sin\varphi+r(\varphi)\cos\varphi\right)^2}\,\text{d}\varphi=\\ &\int_{\varphi_1}^{\varphi_2}\sqrt{\left(r'(\varphi)\right)^2+r^2(\varphi)}\,\text{d}\varphi \end{align} Of course you can ask how to derive the first formula, which is however only direct application of the Pythagoras theorem.