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Let $f(x,y)$ be a real-valued function on a domain $D$ in $\mathbb{R}^2$, and let $(x_s, y_s)$ be a saddle point of $f(x,y)$ in $D$. That is to say, \begin{align} \frac{\partial f}{\partial x}(x_s, y_s) =&0, \\ \frac{\partial f}{\partial y}(x_s, y_s) =&0, \end{align} but that $f$ does not take the maximum or minimum value at $(x_s,y_s)$. We can implicitly define the level curves through the saddle point by the condition, \begin{equation} f(x,y) = f(x_s, y_s). \tag{1} \label{eq: 1} \end{equation} Let us write the explicit forms of these level curves as \begin{align} y =& h_i(x), &i=&1,\dots,N, \tag{2} \label{eq: 2} \end{align} where $N$ is the number of the contour lines passing through the saddle point. How can I calculate the derivative, \begin{align} \frac{d h_i}{dx}(x_s,y_s)&,& i=&1,\dots,N, \tag{3} \label{eq: 3} \end{align} i.e., the slopes of the level curves at the saddle point?

I am confused because it looks like a $0/0$ indefinite form. Please note that by taking derivative of the both sides of eq. (\ref{eq: 1}) with respect to $x$, \begin{equation} \frac{\partial f}{\partial x}(x,y) + \frac{\partial f}{\partial y}(x,y) \frac{dy}{dx} = 0, \tag{4} \label{eq: 4} \end{equation} and if $\partial f/\partial y \not=0$, \begin{equation} \frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}(x,y)}{\frac{\partial f}{\partial y}(x,y)}. \tag{5} \label{eq: 5} \end{equation} However, at the saddle point, \begin{align} \frac{\partial f}{\partial x}(x_s,y_s) =& 0, \tag{6} \label{eq: 6} \\ \frac{\partial f}{\partial y}(x_s,y_s) =& 0, \tag{7} \label{eq: 7} \end{align} and therefore it looks that as $(x,y) \rightarrow (x_s,y_s)$, \begin{equation} \frac{dy}{dx} \rightarrow \frac{0}{0}. \tag{8} \label{eq: 8} \end{equation}

For some examples of $f(x,y)$, I can determine the level curves $\{h_i(x)\}$ explicitly, but still have difficulty to calculate $dh_i/dx$ at $(x_s,y_s)$ as it takes the $0/0$ indefinite form.


EDITED 2016-07-21 07:40 GMT+2 I replaced all occurrences of contour lines by level curves and added definition of a saddle point.

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  • $\begingroup$ Ah so here comes in the Hessian. $\endgroup$
    – norio
    Jul 21, 2016 at 5:27
  • $\begingroup$ What are contour lines, in your understanding? Lines of equal $f$? Am I right that you require the set of saddles to form a continuous line, of which the point $(x_s,y_s)$ is an element? $\endgroup$ Jul 21, 2016 at 15:09
  • $\begingroup$ @keenPenguin Thank you for the comment. Yes, I meant by 'contour line' a line on which $f$ is constant. However, I now realized that I should refer to such a line as 'level curve' since 'contour line' may mean any line on a 2- or higher dimensional space. About the saddle, I was thinking it as an isolated point. By a saddle point, I meant a point where $\partial f/\partial x =0$ and $\partial f/\partial y = 0$ but where $f$ is neither maximum or minimum. At this point, I think two or more level curves cross each other. $\endgroup$
    – norio
    Jul 21, 2016 at 17:19

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At the saddle point, you need to expand up to the second order and the Taylor development essentially becomes

$$f-f_0=ax^2+2bxy+cy^2,$$ (a parabolic hyperboloid) which you can factor as the product of two lines.

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