## extrema

tough one
find all critical points of
f(x, y)=e^x(1-cos y)

and classify these critical points.
 Recognitions: Homework Help For a critical point to occur, both partial derivatives must be equal to zero at that point. So the first steps would be to find the partial derivatives, and then to solve them for zero. Try that first.
 \\f(x,y) = e^x(1-\cos y) \\ \\\frac{\partial f}{\partial x} = e^x(1-\cos y),\ \ \frac{\partial f}{\partial y} = e^x\sin y \\ \\ e^x(1-\cos y) = 0 \\ \cos y = 1 \\ y = 2k\pi,\ k\in\mathbb{Z} \\ \\ e^x\sin y = 0 \\ \sin y = 0 \\ y = k\pi,\ k\in\mathbb{Z} \\ \\\mbox{critical points along the lines }\ y = 2k\pi,\ k\in\mathbb{Z} \\\mbox{which coincidentaly, is also the lines at which f(x,y) = 0} \\ is this right

Recognitions:
Homework Help

## extrema

You forgot the actual LaTeX tags lol! Thats kind of hard to read right now, can u put them in? [ tex ] and [ /tex ] without the spaces in between.
 don't worry i can't work out how to use this thing bit complex anyways im pretty sure it's right i just don't know how to classify the critical points. All the critical points of the function have positive second partial derivitive in y, and zero second partial derivitive in x, and zero second partial derivitive in x and y. what kind of classification is given to that?
 Recognitions: Homework Help Try graphing it to visualize it a bit. Remember that its either only: a maxima, minima or a saddle point.
 Recognitions: Gold Member Science Advisor Staff Emeritus How about, instead of using formulas you think about what ex(1- cos(y)) looks like in the vicinity of a point y= 2k$\pi$? In particlar, what does 1- cos(y) look like there? I also want to point out that 1) this is clearly homework 2) you showed NO attempt to do this in your original post.