We have the following equation $$p^2=r^2+b^2$$ Where $r$ and $\rho$ are variables. $$(\frac{dz}{dr})^2 +(\frac{dp}{dr})^2=1$$
By Using first equation for $\rho$ second equation becomes $$z(r)=b \sinh^{-1} (\frac{r}{b}) + constant $$ I understand by using above two equations we will get sine hyberbolic inverse
But now the book says eliminating r in favour of p yields the equation of the curve in p-z plane: $$ p(z) =b \cosh^{-1}(\frac{z}{b})$$
The above example is from gravity by hartle. I understand the concept of embedding. What I don't understand is math behind these equations. Like how can you use the first equation and then write the second equation as a function of $z$ to get cosine hyberbolic inverse function Please help
Assuming that the initial condition is such that $z = 0$ when $r = 0$, we have $z(r) = b\sinh^{-1}(r/b)$. We can then use $r(z) = b\sinh(z/b)$ in the equation for $p$ to eliminate $r$: $$ p^2 = b^2\sinh^2(z/b) + b^2 = b^2\cosh^2(z/b)\Longrightarrow p = b\cosh(z/b). $$ One could also do this from the differential equation directly by using the chain rule: $$ \left(\frac{dz}{dr}\right)^2 + \left(\frac{dp}{dr}\right)^2 = \left[\left(\frac{dz}{dp}\right)^2 + 1 \right]\left(\frac{dp}{dr}\right)^2 = 1\Longrightarrow \frac{dz}{dp} = \sqrt{\left(\frac{dr}{dp}\right)^2 - 1} = \frac{b}{\sqrt{p^2-b^2}} $$ which solves to $z = b\cosh^{-1}(p/b)$ using the initial condition that $z =0$ when $p = b$. So we again have $p = b\cosh(z/b)$.