1. Dec 11, 2013

### scubakobe

1. If f(x,y)=e$^{x}(1-cos(y))$ find critical points and classify them as local maxima, local minima, or saddle points.

3. 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))
f$_{xx}$=-e$^{x}$(cos(y)-1)
f$_{yy}$=e$^{x}$(cos(y))

Knowing this, and that e$^{x}$ does not equal 0, then the critical points are periodic at 2∏n, where n is even intervals.

However, I get inconclusive when plugging it all into the second derivative test. And a quick query in WolframAlpha shows that there are indeed critical points, however no local maxima,minima or saddle points?

I also referred to another post in this form with a very similar problem, except it was (e^x)(cosy) and it was determined to have no critical points.

Any ideas on this?

Thanks,
Kobbe

2. Dec 11, 2013

### Ray Vickson

Second-order tests involve not only $f_{xx}$ and $f_{yy}$, but also the mixed-partial $f_{xy}$.

3. Dec 11, 2013

### scubakobe

Yes, sorry forgot to include that:

fxy=exsin(y)

4. Dec 11, 2013

### ehild

What do you mean with the sentence "the critical points are periodic at 2∏n"? To define a point, you have to give both x and y. The first derivatives are zero if y is equal to even number multiple of ∏, but what should be x?

ehild

5. Dec 12, 2013

### scubakobe

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 interval, but if asked about local min, max, or saddle points Wolfram states there are none.

6. Dec 12, 2013

### ehild

Well, can be x something else? What about the first derivatives if x=10 or anything, and y =2pi?

ehild

7. Dec 12, 2013

### scubakobe

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) - (f$_{xy}$(x,y))2
D(x,y)= f$_{xx}$(10,2∏)f$_{yy}$(10,2∏) - (f$_{xy}$(10,2∏))2
Which then simplifies to 0.

This means the second order test is inconclusive. However, there clearly are critical points.

8. Dec 12, 2013

### ehild

They are rather "critical lines", are they not ? The function is constant (zero) along the lines y=0, y=2pi... y=2kpi. There are no local extrema, no saddles, although the function has it minimal value zero along these lines.

ehild

9. Dec 12, 2013

### scubakobe

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.

10. Dec 12, 2013

### ehild

I do not know how these lines are called, or they have a name at all. I think you can say that all points (x, k*2pi) are critical as the first derivatives are zero, but there are no local extrema or saddles.

ehild