Changing to slanted coordinates

Question

In order to convert a partial derivative of an arbitrary function of two variables from Cartesian coordinates to polar coordinates, we can simply employ the chain rule as follows:

$$ \frac{\partial f(x,y)}{\partial r}=\frac{\partial f(x,y)}{\partial x}\cdot\frac{\partial x}{\partial r}+\frac{\partial f(x,y)}{\partial y}\cdot\frac{\partial y}{\partial r}=\cos\theta\frac{\partial}{\partial x}(f(x,y))+\sin\theta\frac{\partial}{\partial y}(f(x,y)). $$

This gives us a partial differentiation operator with respect to $r$:

$$ \frac{\partial}{\partial r}=\cos\theta\frac{\partial}{\partial x}+\sin\theta\frac{\partial}{\partial y}. $$

I want to extend this idea to the "slanted" Cartesian coordinate system, in which the vertical axis we now call the $u$-axis and what was formerly the $x$-axis is now called the $v$-axis and is at an angle of $\phi$ with the vertical $u$-axis, where $0<\phi<\pi/2$.

I want to find $\frac{\partial f(r,\theta)}{\partial u}$. By the chain rule:

$$ \frac{\partial f(r,\theta)}{\partial u}=\frac{\partial f(r,\theta)}{\partial r}\cdot\frac{\partial r}{\partial u}+\frac{\partial f(r,\theta)}{\partial \theta}\cdot\frac{\partial \theta}{\partial u}. $$

I'm having some difficulty in computing $\frac{\partial r}{\partial u}$ and $\frac{\partial \theta}{\partial u}$ in terms of the angle $\phi$.

Answer 1

First, let’s review a way to compute partial derivative operators with respect to the polar coordinates $r$ and $\theta$. If we have a function $f:(x,y)\mapsto(u,v)$, its Jacobian matrix is the matrix of partial derivatives $$J_f = {\partial(u,v)\over\partial(x,y)} = \pmatrix{{\partial u\over\partial x}&{\partial u\over\partial y}\\{\partial v\over\partial x}&{\partial v\over\partial y}}.$$ The Inverse Function Theorem tells us that at points where $\det{J_f}\ne 0$ (with suitable conditions on $f$), $f$ is locally invertible and $J_{f^{-1}} = (J_f)^{-1}$. A practical application of this result is that a way to compute the inverse partial derivatives ${\partial x\over\partial u}$ &c is to construct the Jacobian of $f$ and invert it. The other result we’ll be using is that the multivariable chain rule basically says that the Jacobian of $f\circ g$ is the product of the individual Jacobians of $f$ and $g$.

For example, for the polar-to-rectangular map $$\alpha:\pmatrix{r\\\theta}\mapsto\pmatrix{r\cos\theta\\r\sin\theta}$$ the Jacobian $J_\alpha$ is $$\pmatrix{\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta}$$ so that $$J_{f\circ\alpha} = J_f J_\beta = \pmatrix{f_x \cos\theta + f_y \sin\theta & -f_x r\sin\theta+f_y r\cos\theta},$$ as you already know from other methods.

The mapping from your “slanted” coordinates to rectangular is fairly easy to work out. It is (assuming that $\phi$ is measured clockwise) $$\beta:\pmatrix{u\\v}\mapsto\pmatrix{v\sin\phi\\u+v\cos\phi}$$ with Jacobian $$J_\alpha = \pmatrix{0&\sin\phi\\1&\cos\phi}.$$ (If the coordinate systems are related by a skew transformation instead, the matrix is slightly different, but the idea of how to use it it the same.) The partial derivatives that you’re trying to compute are, per the chain rule, the elements of the Jacobian $${\partial(r,\theta)\over\partial(u,v)} = J_\alpha^{-1}J_\beta = \pmatrix{\sin\theta&\sin(\theta+\phi)\\\frac1r\cos\theta&\frac1r\cos(\theta+\phi)}.$$ Note that the derivatives with respect to $u$ don’t depend on $\phi$ at all.