I'm trying to derive the polar functions for the Velocity Potential Function $\Phi$ from its cartesian definitions of:
\begin{align}
\frac{dΦ}{dx} &= u \\\\\frac{dΦ}{dy} &= v
\end{align}
where I believe $u$ and $v$ to be velocities in the $x$ and $y$ direction respectively.
I am trying to get to the following equation for $V_r$:
\begin{align}
\frac{dΦ}{dr} &= V_r
\end{align}
where $V_r$ is the velocity in the $r$ direction from the origin in polar coordinates. The only way I can currently complete this derrivation is by using \begin{align}
V_r = u\cos(θ)
\end{align}
with θ being the angle from the x-axis in radians. If I accept this then I can follow through with the chain rule using $r = x\cos(θ)$ and differentiating to get $\frac{dr}{dx} = \cos(θ)$.
Then it becomes \begin{align}
\frac{dΦ}{dx} = \frac{dΦ}{dr} \times \frac{dr}{dx} = \frac{dΦ}{dr}\times \frac{1}{\cos(θ)} = \frac{V_r}{\cosθ}
\end{align}
However, I have a feeling this is incorrect, and I cannot for the life of me work out how $V_r = u\cos(θ)$. Could anyone explain whether this is right and if so why?
Any help is greatly appreciated, thanks.
Best Answer
I'm going to assume you mean to use partial derivatives everywhere, otherwise there's something I'm missing.
In that case, your equation $$V_r=u\cos(\theta)$$ is, in fact, not correct. How could it be? Then your velocity would be $0$ for points in the $yy$-axis, even if $v$ were not zero. You can come up with similar arguments but it makes no sense for $V_r$ to depend only on $u$ and not $v$. These are independent velocities and $V_r$ should be some combination of both.
I think one of the main problems here lies in misuse of the chain rule, as $\Phi$ is a function of two variables, and not just one. For example, it is also not generally true that $$\frac{\partial\Phi}{\partial x}=\frac{\partial\Phi}{\partial r}\frac{\partial r}{\partial x}$$ What is correct is considering, for example, both polar variables $r$ and $\theta$ and writting $$\frac{\partial\Phi}{\partial x}=\frac{\partial\Phi}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial\Phi}{\partial \theta}\frac{\partial \theta}{\partial x}$$ as you can see here. And you can in fact use the chain rule the same way as above to determine how to differentiate $\Phi$ with respect to $r$: $$\frac{\partial\Phi}{\partial r}=\frac{\partial\Phi}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial\Phi}{\partial y}\frac{\partial y}{\partial r}$$ Since $x=r\cos(\theta)$ and $y=r\sin(\theta)$, the above becomes $$\frac{\partial\Phi}{\partial r}=u\cos(\theta)+v\sin(\theta)$$