How I express the Jacobian matrix between xy-plane and cylinder?

Solution 1:

Let's use coordinates $u,v$ in the domain and $\theta,z$ in the image. As you've said, it's natural to use $\frac{\partial}{\partial u},\frac{\partial}{\partial v}$ as a basis for the tangent space of the domain (at every point) and, similarly, $\frac{\partial}{\partial \theta},\frac{\partial}{\partial z}$ as a basis for the tangent space of the image (at every point).

Note that we parametrize the cylinder $x^2+y^2=R^2$ by $(R\cos\theta,R\sin\theta,z)$, and so $\dfrac{\partial}{\partial\theta} = R(-\sin\theta,\cos\theta,0)$ and $\dfrac{\partial}{\partial z} = (0,0,1)$.

With your mapping $F(u,v) = \big(R\cos(\frac uR),R\sin(\frac uR),v)$, we find \begin{align*} \frac{\partial F}{\partial u} = dF_{(u,v)}\left(\frac{\partial}{\partial u}\right) &= \big(-\sin(\frac uR),\cos(\frac uR), 0\big) = \frac1R\frac{\partial}{\partial\theta} \\ \frac{\partial F}{\partial v} = dF_{(u,v)}\left(\frac{\partial}{\partial v}\right) &= (0,0,1) = \frac{\partial}{\partial z}. \end{align*} Thus, the matrix representation of $dF_{(u,v)}$ in terms of the given bases is $$\left[\begin{matrix} \frac1R & 0 \\ 0 & 1 \end{matrix}\right].$$ (It's hard to tell here, but remember that the columns of this matrix tell us how the respective basis vectors map.)