Problem on chain rule and change of coordinates

Let $f$ be a function of cartesian coordinates $x,y$. It is possible to express $f$ in terms of polar coordinates $r,\theta$ by $f(r,\theta) = f\bigl(x(r,\theta), y(r, \theta)\bigr)$.

(a) What are the expressions for $x(r, \theta)$ and $y(r, \theta)$? That is, write down the $x$ and $y$ coordinates of a point with polar coordinates $r, \theta$.

(b) Use the chain rule to express $\begin{pmatrix}\partial_r f \\ \partial_\theta f \end{pmatrix}$ as a matrix times $\begin{pmatrix}\partial_x f \\ \partial_y f \end{pmatrix}$.

Solution
(a) Recall that in polar coordinates
Polar coordinates

$$x(r,\theta)= r \cos \theta, \qquad y(r,\theta) = r \sin \theta.$$
Chain rule with functions of several variables

(b) We directly apply the chain rule to compute $\partial_r$ and $\partial_\theta$ of $f \Bigl( x(r,\theta), y(r,\theta) \Bigr)$:
$$
\begin{align}
\partial_r f &= \partial_x f \ \partial_r x + \partial_y f \ \partial_r y \\
\partial_\theta f &= \partial_x f \ \partial_\theta x + \partial_y f \ \partial_\theta y
\end{align}
$$
Matrix multiplication

Writing these equations in matrix form, we get
$$
\begin{pmatrix} \partial_r f \\ \partial_\theta f \end{pmatrix} = \begin{pmatrix} \partial_r x & \partial_r y \\ \partial_\theta x & \partial_\theta y\end{pmatrix} \begin{pmatrix} \partial_x f \\ \partial_y f \end{pmatrix}.
$$
Partial derivatives

With the explicit formulas $x(r,\theta) = r \cos \theta$ and $y(r,\theta) = r \sin \theta$, we can compute
\begin{align}
\partial_r x &= \cos \theta \\
\partial_r y &= \sin \theta \\
\partial_\theta x &= -r \sin \theta \\
\partial_\theta y &= r \cos \theta \\
\end{align}
We end up with
$$
\begin{pmatrix} \partial_r w \\ \partial_\theta w \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ -r \sin \theta & r \cos \theta \end{pmatrix} \begin{pmatrix} \partial_x w \\ \partial_y w \end{pmatrix}.
$$

Problem on chain rule and change of coordinates

Solution

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