Slope of a paraboloid given a point and angle I am trying to compute the following for a personal coding endeavor. Suppose we have a function $f:\mathbb{R}^2\rightarrow\mathbb{R}^1$ defined by $f(x,y)=x^2+y^2$. I first would like to check that my approach for the following are correct:
Slope at the origin in the +$x$ direction is $0$. Slope at $(0,12)$ in the +$y$ direction is the partial derivative with respect to $y$ which is $24$. Slope at $(-3,0)$ in the -$x$ direction is the partial derivative with respect to $x$ which is $-6$.
Now here is the piece that I can't seem to solve. At point $(a, b)$ on the paraboloid $f(x,y)$, looking in the direction of $\theta$ on the $xy$-plane, what is the slope of the function?
I've tried finding a unit vector ($\hat{n}$) $\langle \cos(\theta), \sin(\theta) \rangle$ to define a normal to the plane (the purpose of trying to find this plane is that it will "slice" a parabola out of the curve, making the slope calculation a one-dimensional derivative), but I can't seem to get much further than defining the plane as
{$\vec{r}|(\vec{r}-p_0)\cdot\hat{n}=0$} with $p_0$ being the offset $(a,b)$.
From the little internet roaming I have done all I have come across is that the slope of a plane given a unit normal vector is the $z$-coordinate of that vector (here), but I don't see how this applies to my situation.  Is there a solution using the plane that I am trying to find, or just a more elegant approach in general?
Edit: For any derivatives in the result, please leave them in their derivative form (rather than computing them)—for the sake of applicability to other functions.
 A: From $z = f(x, y) = x^2 + y^2 $
Moving along the line $ p_1 =  (x_1,y_1) =   (x_0, y_0) + t u = p_0 + t u $ , where $u$ is a unit vector, the value of the function, when linearized, is
$ f(p_1) = f(p_0 + t u) = f(p_0) + t \nabla f \cdot u $
Therefore, the slope of the curve when moving along $u$ is
$ \text{Slope} = \dfrac{ f(p_1) - f(p_0) }{t} = \nabla f \cdot u $
And we have
$\nabla f =\begin{bmatrix} f_x \\ f_y \end{bmatrix} = \begin{bmatrix} 2 x_0 \\ 2 y_0 \end{bmatrix} $
and
$u = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} $
So
$\text{Slope} = 2 x_0 \cos \theta + 2 y_0 \sin \theta $
We can also use the tangent plane at $p_0$ whose equation is
$f_x (x-x_0) + f_y (y-y_0) - (z-z_0) =0$
And intersect it with the plane
$ -u_y (x - x_0) + u_x (y-y_0) =0 $
The intersection is along the direction vector:
$(f_x,f_y,-1) \times (-u_y, u_x,0)= (u_x,u_y, f_x u_x+f_y u_y)$
The slope is given by
$ \dfrac{ f_x u_x + f_y u_y }{\sqrt{ u_x^2 + u_y^2 }} = f_x u_x + f_y u_y $
And this last expression is simply $2 x_0 \cos \theta + 2 y_0 \sin \theta $
