Identifying Critical Points in e^x(1-cosy): Local Extrema or Saddles?

[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.

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

 

[Ray Vickson]

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))
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

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

 

[scubakobe]

Yes, sorry forgot to include that:

f_xy=e^xsin(y)

 

[ehild]

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))
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.

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

 

[scubakobe]

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

Yes, that's where I'm having trouble. My thought is, and another post (Which I can't find now, sorry) suggested, is that e^x cannot equal zero - so x would have to be equal to 0 in order for the e^x 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.

 

[ehild]

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

ehild

 

[scubakobe]

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

ehild

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.

 

[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

 

[scubakobe]

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

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.

 

[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