I am struggling to understand/visualize the following graph/text. The topic is how the components of the directional derivative along u do not need to have any connection between each other.
Text: Take the plane $R^2$ and $y=(0,0)$ and define $f(r \cos( \theta), r\sin \theta) = rg(\theta)$ for any function $ g(\theta)$ that is odd, $ g(-\theta)= -g(\theta)$. Then $d_uf(0,0)= g(\theta)$ if $u=(cos(\theta),sin(\theta))$. The function f is linear along each line through the origin, but the slopes are unrelated.
Definition: $d_uf(y)=\lim_{t\to 0}(f(y+tu)-f(y))/t$. $(d_uf(y))_k$ is the kth component of $d_uf(y)$. If $f:D\to R^m$ with $D\subsetneq R^n$ is differentiable at y with differetial $df(y)$, then $d_uf$ exists at y for any u in $R^n$ and $d_uf(y)=df(y)u$.
My thoughts: I understand the idea that the directional derivative in some direction u can be calculated by the components of u in the space, and they are individual. When it comes to the example above, I do not quite follow. How does the graph above lock like? So f and g map into $R^2$. When we say $y=(0,0)$, when does f ever take this value? How is it differentiable at y? I do not see how the pieces fit together to claim that the slopes are unrelated.
New thoughts: Putting the problem to the side and coming back to it, I still do not get how f can be differentiable at y given that $(r \cos( \theta), r\sin \theta)$ can never equal $(0,0)$ unless $r=0$. What I did consider doing was to set $x = r\cos \theta$ and $y = r \sin \theta$. And so I want to get that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. I do not know if this is the right approach, but I do not get any further.
Another thing I thought about was to just directly use the definition, so I would say
$d_uf(y)=\lim_{t\to 0}(f(y+tu)-f(y))/t = \lim_{t\to 0}(f(tcos(\theta),tsin(\theta))-f(0,0))/t = \lim_{t\to 0}(tg(\theta)-f(0,0))/t$.
I do not see however what I can do about $f(0,0)$, but I imagine it is supposed to go to zero somehow, and the t will cancel, so we are left with $g(\theta)$. But my problem goes back to the issue of $f(0,0)$, or in other words $f(y)$. I do not understand this, unless we set r to zero. Upon writing this I wonder if the limit of t going to zero allows us to say $f(0,0) = \lim_{r\to 0}f(rcos(\theta),rsin(\theta)) = \lim_{r\to 0}rg(\theta)=0$. Is this correct?
Thanks!