Deformation of Helicoid to Catenoid I am working on an exercise from Tapp "Differential Geometry of Curves and Surfaces" (Ex 4.56) about the deformation of the Helicoid to the Catenoid. We are given for $U = \{ (\theta, t) \in \mathbb{R}^2 \mid -\pi < \theta < \pi \}$, define a parameterisation $\sigma_s:U \to \mathbb{R}^3$
$$\sigma_s(\theta,t) = (\cos s) (c \cosh t \cos \theta, c \cosh t \sin \theta, ct) + (\sin s) (c \sinh t \cos\theta, c \sinh t \sin \theta, c \theta)$$
For $s = 0$ this surface is a catenoid and $s = \pi/2$ is a helicoid. To show isometry between these two surfaces I want to show that the first fundamental form is invariant to $s$. My approach is to take the partials:
$$\sigma_{s,\theta} = \cos(s)(-c\cosh(t)\sin(\theta),c\cosh(t)\cos(\theta), 0) \\+ \sin(s)(-c\sinh(t)\sin(\theta),c\sinh(t)\cos(\theta), c)$$
$$\sigma_{s,\theta} = \cos(s)w_1 + \sin(s)w_2$$
$$\sigma_{s,t} = \cos(s)(c\sinh(t)\cos(\theta),c\sinh(t)\sin(\theta), c) \\+ \sin(s)(c\cosh(t)\cos(\theta),c\cosh(t)\sin(\theta), 0)$$
$$\sigma_{s,t} = \cos(s)w_3 + \sin(s)w_4$$
Now computing the coefficients we have:
$$E = \langle \cos(s)w_1 + \sin(s)w_2, \cos(s)w_1 + \sin(s)w_2 \rangle $$
$$E = \cos^2(s)\langle w_1, w_1 \rangle +2\sin(s)\cos(s)\langle w_1, w_2\rangle + \sin^2(s) \langle w_2, w_2 \rangle $$
We have:
$$\langle w_1, w_1 \rangle = c^2\cosh^2(t)\sin^2(\theta) +c^2\cosh^2(t)\cos^2(\theta) $$
$$ = c^2\cosh^2(t)(\sin^2(\theta) + \cos^2(\theta))$$
$$ = c^2\cosh^2(t)$$
and:
$$ \langle w_1, w_2 \rangle = c^2\cosh(t)\sinh(t)\sin^2(\theta) + c^2\cosh(t)\sinh(t)\cos^2(\theta) $$
$$= c^2\cosh(t)\sinh(t)(\sin^2(\theta) + \cos^2(\theta))$$
$$= c^2\cosh(t)\sinh(t)$$
and:
$$\langle w_2, w_2 \rangle = c^2\sinh^2(t)\sin^2(\theta) +c^2\sinh^2(t)\cos^2(\theta) + c^2 $$
$$ = c^2\sinh^2(t)(\sin^2(\theta) + \cos^2(\theta)) + c^2 $$
$$ = c^2\sinh^2(t) + c^2 $$
$$ = c^2(1 + \sinh^2(t)) $$
$$ = c^2\cosh^2(t)$$
Plugging in:
$$E = \cos^2(s)c^2\cosh^2(t) +2\sin(s)\cos(s)c^2\cosh(t)\sinh(t) + \sin^2(s) c^2\cosh^2(t) $$
$$ = (\sin^2(s) +\cos^2(s))c^2\cosh^2(t) +2\sin(s)\cos(s)c^2\cosh(t)\sinh(t)$$
$$ = c^2\cosh^2(t) +2c^2\sin(s)\cos(s)\cosh(t)\sinh(t)$$
Now the first term looks correct in that it aligns with the first coefficient in the helicoid first fundamental form but the second term definitely depends on $s$ and it seems to me it should be zero. Where have I gone wrong? Is my approach correct? Is there a simpler way?
 A: The fundamental form of your equation is:
$$
g = \frac{c^2}2 \begin{pmatrix}
1+\cosh 2t+\sin 2s\sinh 2t & \sin 2s \\
\sin 2s & 1+\cosh 2t+\sin 2s\sinh 2t
\end{pmatrix}.
$$
It is not invariant of $s$. However, since $g$ depends only on $\sin 2s$, it is the same when $s=0$ and $s=\pi/2$.
Your approach is correct in general. But you could just have shown that $g(w_1,w_3) = g(w_2,w_4)$
Requested edit. When $s=0$, $(\sigma_\theta, \sigma_t)=(w_1, w_3)$. When $s=\pi/2$, $(\sigma_\theta, \sigma_t)=(w_2, w_4)$.
We need to show that $g(w_1,w_3) = g(w_2,w_4)$ or
$$
\begin{pmatrix}
w_1^2 & w_1\cdot w_3 \\ w_1\cdot w_3 & w_3^2
\end{pmatrix}
=\begin{pmatrix}
w_2^2 & w_2\cdot w_4 \\ w_2\cdot w_4 & w_4^2
\end{pmatrix}
$$
It can be simplified further if you consider three orthonormal vectors $e=(\cos\theta, \sin\theta,0)$, $n=e_\theta = (-\sin\theta, cos\theta, 0)$ and $z=(0,0,1)$. Then:
$$
w_1 = n\cosh t, \qquad w_3 = e\sinh t + z
$$
It's obvious now that $w_1\cdot w_3=0$ and $w_1^2=w_3^2=\cosh^2t$.
For $g(w_2, w_4)$:
$$
w_2 = n \sinh t + z,\qquad w_4 = e\cosh t,
$$
it's also true.
