Let $x = u\cos(v)$, and $y = u\sin(v)$, and assume $f(u,v)$ is given. Determine $f_x$ and $f_y$ in terms of $u, v, f_u$, and $f_v$.

Question

Let $x = u \cos(v)$ and $y = u \sin(v)$, and assume $f(u, v)$ is given. Determine $f_x$ and $f_y$ in terms of $u$, $v$, $f_u$, and $f_v$.

I thought of chain rule like $$ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} + \frac{df}{dv} \cdot \frac{dv}{dx}. $$ But in order to find $du/dx$, I have to find $u$ in terms of $v$ and $x,y$. Is there anyway to solve this?

Answer 1

Since you have the cartesian coordinates $x$ and $y$ in terms of the polar coordinates $u$ and $v$, calculate those derivatives and take reciprocals: $$ \frac{dx}{du} = \cos v \quad\implies\quad \frac{du}{dx} = \frac{1}{\cos v} $$ Hopefully, you can finish from here.

Answer 2

Firstly, you should be dealing in partial derivatives, as $f_x=\tfrac{\partial f(u,v)}{\partial x}$, and $f_y=\tfrac{\partial f(u,v)}{\partial y}$ .

As such, the Jacobian Inverse Function Theorem applies.

$$\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{bmatrix}^{-1}$$

Finally, apply this to the chain rule in matrix form.$$\begin{bmatrix}\dfrac{\partial f(u,v)}{\partial x}&\dfrac{\partial f(u,v)}{\partial y}\end{bmatrix}=\begin{bmatrix}\dfrac{\partial f(u,v)}{\partial u}&\dfrac{\partial f(u,v)}{\partial v}\end{bmatrix}\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}\end{bmatrix}$$

All that's left is to apply the differentiations, and matrix operations.