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I am looking at the following exercise:

Show that $$\sigma (u, v) = (\text{sech } u \cos v,\text{sech } u \sin v,\tanh u)$$ is a regular surface patch for $S^2$ (it is called Mercator’s projection).

Show that meridians and parallels on $S^2$ correspond under $\sigma$ to perpendicular straight lines in the plane.

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I have shown that $\sigma_u\times\sigma_v\neq (0,0,0)$, so it is a regular surface patch for $S^2$, right?

Could you give me some hints how we could show that meridians and parallels on $S^2$ correspond under $\sigma$ to perpendicular straight lines in the plane?

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    $\begingroup$ Isn't the Meridian/parallel part quite trivial? The meridians are given by $v=const$, the parallels by $u=const$. These are clearly orthogonal in $\mathbb{R}^2$. Or am I missing something? $\endgroup$
    – Thomas
    Jan 1, 2016 at 16:24
  • $\begingroup$ At the formulation: "meridians and parallels on $S^2$ correspond under $\sigma$ to perpendicular straight lines in the plane." What is meant by the part under $\sigma$ ? @Thomas $\endgroup$
    – Mary Star
    Jan 2, 2016 at 14:03
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    $\begingroup$ This means that $\sigma$ maps straight lines in the plane to meridians and parallels. $\endgroup$
    – Thomas
    Jan 2, 2016 at 14:25
  • $\begingroup$ I see... Thanks a lot!! :-) @Thomas $\endgroup$
    – Mary Star
    Jan 2, 2016 at 14:27
  • $\begingroup$ I am looking at an other exercise related to meridians and parallels... The formulation of the exercise is "Which curves on the helicoid correspond under this isometry to the parallels and meridians of the catenoid." So do we find the parallels and the merdians of the helicoid or of the catenoid? Do we have to set $u=\text{ constant }$ and $v=\text{ constant }$ at the parametrization of the helicoid or at the parametrization of the catenoid? @Thomas $\endgroup$
    – Mary Star
    Jan 11, 2016 at 20:25

1 Answer 1

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The patch $\sigma$ is generated by rotation of curve $\alpha(u) = (\mathrm{sech}\, u,0, \tanh u)$ around $z$ axis. Therefore meridians are exactly the curves $v= const$ and parallels are the curves $u = const$.

It is clear that lines $u=const$ (parallels) and $v=const$ (meridians) are orthogonal.

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  • $\begingroup$ How do we conclude that meridians are the curves $v= const$ and parallels are the curves $u = const$ ? $\endgroup$
    – Mary Star
    Jan 1, 2016 at 23:41
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    $\begingroup$ meridians on a surface of revolution are the curves obtained by rotating the starting curve ($\alpha$ in our case) by a fixed angle ($v$ in our case), therefore they always have the equation $v = const$. Similarly, parallels are circles obtained by rotating one fixed point from the original curve and hence the equation $u=const$ (by choosing the value of $u$ you pick a point on $alpha$ and then rotate it to get the parallel). $\endgroup$
    – user26977
    Jan 2, 2016 at 10:10
  • $\begingroup$ Are the lines $u=const$ (parallels) and $v=const$ (meridians) orthogonal because we consider that the axes of $\mathbb{R}^2$ are the axis of $u$ and the axis of $v$ ? $\endgroup$
    – Mary Star
    Jan 2, 2016 at 13:50
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    $\begingroup$ exactly, we consider $\mathbb{R}^2$ with $u$ and $v$ coordinates there. $\endgroup$
    – user26977
    Jan 2, 2016 at 13:54
  • $\begingroup$ Ok... Thank you for your answer!! :-) $\endgroup$
    – Mary Star
    Jan 11, 2016 at 20:26

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