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In order to get the arc length we must have $r$ as a function of $\theta$. But how can I find arc length of curve which is drawn between $r$ and $\theta$ if $r$ is an implicit function of $\theta$ like this:

\begin{align} \theta = \frac{a\cdot \left( b - \sqrt{r^2 - rf+p^2} \right) }{r} \end{align}

Where $f$ and $p$ are constants.

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    $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ Commented Dec 27, 2020 at 9:31
  • $\begingroup$ what is the $f$ and $p$ inside the square root? $r^2-rf+p^2$? $\endgroup$ Commented Dec 27, 2020 at 9:38
  • $\begingroup$ Everything expect r and theta is constants $\endgroup$ Commented Dec 27, 2020 at 9:42
  • $\begingroup$ @JoséCarlosSantos ok sir $\endgroup$ Commented Dec 27, 2020 at 9:43
  • $\begingroup$ You probably mean explicit. $\endgroup$ Commented Dec 27, 2020 at 10:40

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Since $ds^2=dr^2+r^2d\theta^2$, when $\theta$ is given as a function of $r$ the arc length is$$\int ds=\int\sqrt{1+r^2(d\theta/dr)^2}\;dr.$$In this case,$$\begin{align}r\frac{d\theta}{dr}&=-\frac{a}{r}\left(b+\frac{fr/2-p^2}{\sqrt{r^2-fr+p^2}}\right)\\\implies\int ds&=\int\sqrt{1+\frac{a^2}{r^2}\left(b^2+\frac{(fr/2-p^2)^2}{r^2-fr+p^2}+\frac{b(fr-2p^2)}{\sqrt{r^2-fr+p^2}}\right)}\;dr.\end{align}$$However, I suspect you may need to evaluate such an arc length numerically.

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