# How to derive the transformation that maps general curvilinear coordinates to their cartesian image.

Assume an arbitrary orthogonal curvilinear coordinate system: $$\mathbf θ \in \mathrm A \subset \mathbb R ^2 \leftrightarrow \mathbf x \in \mathbb R ^2$$

Notice the following notation that I use: $$\mathrm D \mathbf x ( \mathbf θ ) \equiv \begin{bmatrix} \frac{\partial θ_1}{\partial x_1} & \frac{\partial θ_1}{\partial x_2} \\ \frac{\partial θ_2}{\partial x_1} & \frac{\partial θ_2}{\partial x_2} \\ \end{bmatrix}$$

$$\mathrm d \mathbf θ \equiv \begin{bmatrix} \mathrm d θ_1 \\ \mathrm d θ_2 \\ \end{bmatrix}$$

I believe the fundamental theorem of calculus ensures that: $$\int_{\mathbf q_0} ^{\mathbf θ} \mathrm D \mathbf x ( \mathbf q ) \mathrm d \mathbf q = \mathbf x (\mathbf θ)$$

In other words: $$\int_{ q_{1_{0}}} ^{ θ_1} \frac{\partial \mathbf x}{\partial q_1} \mathrm d q_1 \, + \, \int_{ q_{2_{0}}} ^{ θ_2} \frac{\partial \mathbf x}{\partial q_2} \mathrm d q_2 = \mathbf x (\mathbf θ)$$

With linearization based on Taylor series expansion: $$\frac{\partial \mathbf x}{\partial θ_1} \ θ_1 + \frac{\partial \mathbf x}{\partial θ_2} \ θ_2 = \mathrm D \mathbf x ( \mathbf θ ) \ \mathbf θ = \mathbf x (\mathbf θ)$$

But the foregoing result is wrong. I looked into a math textbook and found the following result, presented without proof or justification: $$\frac{\partial \mathbf x}{\partial θ_1} \ θ_1 / \left \Vert \frac{\partial \mathbf x}{\partial θ_1} \right \Vert + \frac{\partial \mathbf x}{\partial θ_2} \ θ_2 / \left \Vert \frac{\partial \mathbf x}{\partial θ_2} \right \Vert = \mathrm D \mathbf x ( \mathbf θ ) \left ( \mathrm D ^\mathtt T \mathbf x ( \mathbf θ ) \ \mathrm D \mathbf x ( \mathbf θ ) \right )^{-1/2} \ \mathbf θ = \mathbf x (\mathbf θ)$$

If somebody would show how to derive the matrix operation that correctly transforms the curvilinear coordinates into the cartesian coordinates, then my question will become resolved. It would also be helpful to indicate precisely where there exists error(s) in my attempted derivation.

Here is a valid procedure, using cylindrical coordinates. Let $\mathbf r = \left ( \begin{matrix} r & \theta & z\end{matrix} \right)$ and $\mathbf x (\mathbf r) = \left(\begin{matrix} r ~ \cos \theta & r ~ \sin \theta & z \\ \end{matrix} \right).$
Then $$\mathrm D \mathbf x ( \mathbf r ) \left ( \mathrm D ^\mathtt T \mathbf x ( \mathbf r ) \ \mathrm D \mathbf x ( \mathbf r ) \right )^{-1/2} = \left(\begin{matrix} \hat {\mathbf h}_r & \hat {\mathbf h}_\theta & \hat {\mathbf h}_z \\ \end{matrix} \right) = \left[\begin{matrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$$ where $\hat {\mathbf h}_i$ is a unit basis vector parallel to the curve of constant parameter $i$.
$\mathbf x (\mathbf r) = \left[\begin{matrix} \hat {\mathbf h}_r & \hat {\mathbf h}_\theta & \hat {\mathbf h}_z \\ \end{matrix} \right] \left[\begin{matrix} \cos \theta & \sin \theta & 0 \\ - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \left[\begin{matrix} r ~ \cos \theta \\ r ~ \sin \theta \\ z \\ \end{matrix} \right] = r~ \hat {\mathbf h}_r + 0 ~ \hat {\mathbf h}_\theta + z ~ \hat {\mathbf h}_z$.
There exists no linear transformation to map the cylindrical coordinates $\mathbf r$ to their cartesian image $\mathbf x (\mathbf r )$. The person who made the original post asked for "the matrix operation that correctly transforms the curvilinear coordinates into the cartesian coordinates", which is obviously absurd.