Question

Part 2 of the course, session 30, recitation video - there is one critical point at (0,0) and from the discriminant we conclude that it is a saddle. I was very much disturbed, since at this point both of the second derivatives were zero. My point is this - look at the formula from which we categorized the characteristics of critical points, we divide by one of the second derivatives evaluated at the critical point. Therefore, I assume this should be rather a singularity point. I plotted it Mathematica and it turns out to be indeed a saddle, the reason for this seems to be more subtle.Thoughts?

Answer 1

** look at the formula from which we categorized the characteristics of critical points, we divide by one of the second derivatives evaluated at the critical point. ***

can you post where you see this?

Answer 2

http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-a-functions-of-two-variables-tangent-approximation-and-optimization/session-30-second-derivative-test/MIT18_02SC_L10Brds_4.png

Where a equals to one of the second derivatives evaluated at the critical point (there may be an extra factor of two, it isn´t really relevant to this problem). It states that a can´t equal zero, yet in the recitation it actually equals to zero...

Answer 3

i'll venture a guess,

i agree, the formula for categorizing critical points is not valid here since A=0

so we just go back and look at how we derived the formula in the first place,
it was from the taylor series, after setting \(f_x=f_y=0) \)and \(f_{xx} =A \) etc etc
we get
\[\Delta z = \frac{1}{2}A(\Delta x)^2 +B \Delta x \Delta y +\frac{1}{2}C (\Delta y)^2\]

at (0,0) we get A=0 B=-3 C=0
substituting
\[\Delta z = -3 \Delta x \Delta y \]

Answer 4

we can easily observe that depending on our choice for \(\Delta x\) and \(\Delta y\) we can get both negative and positive values for \(\Delta z\), thus the critical point has to be a saddle point.

Answer 5

so the (AC-B^2) formula yeilding the correct result is a co-incidence..
what do you think @phi ?

Answer 6

Adam is correct, the derivation does not cover this case. On the other hand, if either of the coefficients on the quadratic terms is zero (i.e. wxx or wyy = 0), we will get a "difference of squares" --> saddle point. If both are zero, we are left with x y, which can take both positive and negative values --> saddle point. In all of these cases, AC-B^2 becomes -B^2 which signals a saddle point. I don't know if this is a coincidence, but it does still work.

Nevertheless, good catch by Adam

btw, here is a plot of the quadratic approximation at (0,0) for the function z= x^3 -3xy+y^3