Derive the formula for the Arc Length in polar coordinates I want to prove the formula:
$\frac{ds}{d\theta } =\sqrt{\rho^{2} +\left( \acute{\rho } \right)^{2}  } $
One solution is to start with rectangular coordinates
$x=\rho .\cos \left( \theta \right)  \  \  \  and\  \  \  \  y=\rho .\sin \left( \theta \right)  \  \  \  ,\  \rho =f\left( \theta \right)$
$\frac{dx}{d\theta } =\acute{\rho } .\cos \left( \theta \right)  -\rho .\  \sin \left( \theta \right)  $
$\frac{dy}{d\theta } =\acute{\rho } .\sin \left( \theta \right)  +\rho .\  \cos \left( \theta \right) $
$\frac{ds}{d\theta } =\sqrt{\left( \frac{dx}{d\theta } \right)^{2}  +\left( \frac{dy}{d\theta } \right)^{2}  } =\sqrt{\rho^{2} +\left( \acute{\rho } \right)^{2}  } \  $
Now I want to reach the same result using only polar variables.
With reference to attached figure, the arc length could be approximated by a sum of very small line segments
$\left( \Delta s\right)^{2}  =\rho^{2} +\left( \rho +\Delta \rho \right)^{2}  -2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)$
$\left( \frac{\Delta s}{\Delta \theta } \right)^{2}  =\left( \frac{\rho +\Delta \rho }{\Delta \theta } \right)^{2}  +\left( \frac{\rho }{\Delta \theta } \right)^{2}  -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)  }{\left( \Delta \theta \right)^{2}  } $
$\left( \frac{\Delta s}{\Delta \theta } \right)^{2}  =\left[ \frac{\rho +\Delta \rho }{\Delta \theta } -\frac{\rho }{\Delta \theta } \right]^{2}  +\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  }{\left( \Delta \theta \right)^{2}  } -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)  }{\left( \Delta \theta \right)^{2}  }$
$\left( \frac{\Delta s}{\Delta \theta } \right)^{2}  =\left( \frac{\Delta \rho }{\Delta \theta } \right)^{2}  +\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  }{\left( \Delta \theta \right)^{2}  } -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)  }{\left( \Delta \theta \right)^{2}  }$
Now  $\  \  \  \  \  \left( \lim_{\Delta \theta \rightarrow 0} \  \frac{\Delta \rho }{\Delta \theta } \right)^{2}  =\left( \frac{d\rho }{d\theta } \right)^{2}  =\left( \acute{\rho } \right)^{2} $
I couldn’t figure out how the remaining expression could be equal to $\rho^{2} $
Could this or a different approach lead to the same formula?


Keep the $\Delta$ terms up to second order only,
\begin{align}
  \cos (\Delta \theta) & \approx 1-\frac{(\Delta \theta)^2}{2}   \\
  (\Delta s)^2
  & \approx
  \rho^{2}+(\rho +\Delta \rho)^{2}-2\rho (\rho +\Delta \rho) \cos ( \Delta \theta ) \\
  & \approx
  [\rho^{2}-2\rho (\rho +\Delta \rho)+(\rho +\Delta \rho)^{2}]+
  \rho (\rho +\Delta \rho) (\Delta \theta)^2 \\
  & \approx
  [\rho-(\rho+\Delta \rho)]^2+\rho^2 (\Delta \theta)^2+
  \require{cancel} \cancel{\rho^2 (\Delta \rho)(\Delta \theta)^2} \\
  & \approx (\Delta \rho)^2+\rho^2 (\Delta \theta)^2 \\
\end{align}
Take $\Delta \theta \to 0$, the metric becomes
$$ds^2=d\rho^2+\rho^2 d\theta^2$$
that is
$$
\left(
  \frac{ds}{d\theta}
\right)^2=\rho^2+
\left(
  \frac{d\rho}{d\theta}
\right)^2$$
