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Define a function $f: \Bbb R×(0,2 \pi) \to \Bbb R^3$ (Mercator projection) by:

$$f(u,\theta) = {1 \over {\cosh\,\, u} }\begin{pmatrix}\cos\,\, \theta\\\sin\,\, \theta\\\sinh\,\, u\end{pmatrix} $$

How can I show that $f(u,\theta)$ is in the sphere $S$ (given by the equation $x^2+y^2+z^2=1$)?

Jhwana
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1 Answers1

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Hint: What happens when you compute the quantity

$$ \frac{ \cos^2 \theta + \sin^2 \theta + \sinh^2 u}{\cosh^2 u}= \frac{1 + \sinh^2 u}{\cosh^2 u} ?$$

  • $\frac{1 + \sinh^2 u}{\cosh^2 u}=1$ – Jhwana Apr 04 '13 at 22:17
  • @Fayz So this now means that....? –  Apr 04 '13 at 22:20
  • could you tell me the steps please, what should I do to get this result? – Jhwana Apr 04 '13 at 22:21
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    @Fayz Every point in the image of $f$ can be described as

    $$\left( \frac{\cos \theta}{\cosh u}, \frac{\sin \theta}{\cosh u},\frac{\sinh u}{\cosh u} \right). $$

    Now what does it mean to say that the image of $f$ lands on the sphere?

     –  Apr 04 '13 at 22:31