I guess that's true, however in my classes so far we've only referred to points, and not "critical lines."
So the answer would be no critical points? The 2∏ intervals are simply a recurring minimum value, but not necessarily a local minimum.
Yes, x can be anything as long as the y=2∏. For both the f_{x} and f_{y}.
So the x can range from 0\rightarrow∞, but the only time the first partials will equal 0 is when y=2∏. Or even multiples, as I had said.
So here's the second order test...
D(x,y)= f_{xx}(x,y)f_{yy}(x,y) -...
Yes, that's where I'm having trouble. My thought is, and another post (Which I can't find now, sorry) suggested, is that ex cannot equal zero - so x would have to be equal to 0 in order for the ex to become 1.
A graph in WolframAlpha suggests a saddle point; and it does confirm 2∏ as the...
1. If f(x,y)=e^{x}(1-cos(y)) find critical points and classify them as local maxima, local minima, or saddle points.
The Attempt at a Solution
I found the partials and mixed partial for the second derivative test as follows:
f_{x}=-e^{x}(cos(y)-1)
f_{y}=e^{x}(sin(y))...