For an arbitrary level curve, $f(x,y) = c$, $c$ a constant, can we say that the slope of the level curve is $-\frac{f_x}{f_y}$? I am getting this equation from saying that, for a level curve, $f_x(a,b)dx+f_y(a,b)dy=0$?
If what I have above is true, can someone explain the intuition for the equation $f_x(a,b)dx+f_y(a,b)dy=0$ to me? It seems to me as if it says that the rate of change of $x$ times the change in $x$ and the rate of change of $y$ times the change in $y$ have to cancel out, which makes sense since we want to stay on our level curve, but I am not sure if this intuition is valid for infinitesimals. Also, perhaps there is better intuition for the derivatives than rates of change?
All I have found cleary relating to this so far is that The equation for the tangent to a level curve here seems similar, but I do not know if it works for infinitesimals.
EDIT: Also, assume $c \neq 0, f_x (a,b) \neq 0, f_y (a,b) \neq 0$. Similarly, they cannot be $\pm \infty$.
Best Answer
Here is perhaps another way that I have found to think about this.
We want the change in $f(x,y)$, $-\frac{f_x}{f_y}$ which I will call $df$. We can use the partial derivatives of $f$ to write a linear approximation of $f$ near some point on the level curve, $(a,b)$. This linear approximation would be: $$L(i,s) = f(a,b) + [f_x(a,b)](i-a)+[f_y(a,b)](s-b)$$ This linear approximation is very good very close (infinitesimally close) to $(a,b)$ (this fact and how to find the linear approximation/why it is written as above can be found in calculus books or through a google search). Therefore, an infinitesimally small movement along the level curve, $df$ (a first order change) can be written as $$L(i + di,s + ds) - L(i,s) =0$$ because $df =0$ (since we are on a level curve).
Some algebra will give us $[f_x(a,b)]dx+[f_y(a,b)]dy =0$ (i replaced di and ds with dx and dy since they are movements in the same directions), which is the slope of the linear approximation and thus the slope of the level curve.