Let $f(x,y)=\frac{x^2+y^2} x$ In each point of the circle $x^2+y^2-2y=0$ calculate the directional derivative 
Let $f(x,y)=\frac{x^2+y^2} x$
a) In each point of the circle $x^2+y^2-2y=0$, what is the value of the directional derivative with respect to a vector $(a,b) \in R^2$

My approach:
the circle is $x^2+(y-1)^2=1$ and can be parametrized as $(cos \theta,sin \theta +1)$
On the other hand $\nabla f(x,y)=(\frac {x^2-y^2}{x^2},\frac {2y} x)$ , if $(x,y)$ is on the circle then $x=cos \theta$ and $y=sin \theta +1$
And $\frac{∂f}{∂v}(x,y)=\nabla f(x,y) \cdot v $, $v=(a,b)$.
I think I should substitute  $x=cos \theta$ and $y=sin \theta +1$ in $\nabla f(x,y)$ and do the product is it right?

b) When you are on a point of that circle, in which direction should you move to make $f$ not to change (note it depends on the point).

My approach:
I compute the level curves and for $f(x,y)=k$ I get $(x-\frac k 2 )^2 +y^2=\frac {k^2}4$ which is a circle with center $(\frac k 2,0)$ and radius $\frac k 2$.
Is it correct if for a point in the circle a) I found a level curve that pass through that point  f$? In that case, how can I find that level curve and how can i get the direction?

c) On the same circle and for each point , in which direction should you move to get the maximum variation of f?

It is almost the same problem of b) and calculate $\nabla f$
 A: $$\nabla f(x,y)=\left(\frac{x^2-y^2}{x^2},\frac{2y}{x}\right)$$
equals
$$\left(2y\frac{1-y}{x^2},\frac{2y}{x}\right)$$
for $(x,y)$ on the circle. Thus the directional derivative with respect to $(a,b)$ is given by
$$2y\left(a\frac{1-y}{x^2}+\frac{b}{x}\right).$$
Note that $y=0$ would imply that $x=0$ and $f$ is not defined in $(0,0)$. Hence we may assume that $y\neq 0$. In this case the directional derivative is zero if and only if
$$a\frac{1-y}{x^2}+\frac{b}{x}=0\Rightarrow a(1-y)+bx=0.$$
Setting $a=1$ we get $b=\frac{y-1}{x}$.
A: We have that
$$x^2+y^2-2y=0 \iff x^2+(y-1)^2=1$$
which can be parametrized by
$$(x,y)=(\cos \theta, \sin \theta +1) \implies \hat v=(-\sin \theta, \cos \theta), \,\theta\in[0,2\pi)$$
and therefore
$$\nabla f(x,y) =\left(\frac{x^2-y^2}{x^2}, \frac{2y}{x}\right)=\left(\frac{-2\sin^2\theta-2\sin \theta}{\cos^2 \theta}, \frac{2\sin \theta +2}{\cos \theta}\right)$$
from which we can evaluate
$$\frac{∂f}{∂\hat v}(x,y)=\nabla f(x,y) \cdot \hat v=\frac{2\sin^3\theta+2\sin^2 \theta}{\cos^2 \theta}+ 2\sin \theta +2=$$
$$=(2\sin \theta+2)\left(\tan^2 \theta+1\right)=2y\left(\frac{(y-1)^2}{x^2}+1\right)$$
which is the answer for point “a”.
For point “b” and “c” recall the geometrical meaning of the gradient and refer to


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*Find a unit vector that minimizes the directional derivative at a point
