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The following picture shows a family of surfaces, one for each value of \(a \in [0,1]\). When \(a = 0\), we have the catenoid, parametrised by \[ p_0 : \mathbb{R}^2 \rightarrow \mathbb{R}^3, \quad p_0(u,v) = (\cosh(v) \cos(u), \cosh(v) \sin(u), v) \] and when \(a = 1\) we have the helicoid, parametrised by \[ p_1 : \mathbb{R}^2 \rightarrow \mathbb{R}^3, \quad p_1(u,v) = (\sinh(v) \cos(u), \sinh(v) \sin(u), u) . \] In Problem Class 3, we showed that the map \(f : \mathbb{R}^3 \rightarrow \mathbb{R}^3\) given by \[ f(x, y, z) = \left( \frac{x \sqrt{x^2 + y^2 + 1}}{\sqrt{x^2 + y^2}} , \frac{y \sqrt{x^2 + y^2 + 1}}{\sqrt{x^2 + y^2}} , \text{arccosh}(\sqrt{x^2 + y^2 + 1}) \right) \] defines an isometry from the helicoid to the catenoid. The picture below shows that this isometry can be seen by deforming helicoid into the catenoid, which you can do by moving the slider to change the value of \(a\).
