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Local Maxima, Minima, Saddles

  1. Dec 11, 2013 #1
    1. If f(x,y)=e[itex]^{x}(1-cos(y))[/itex] 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[itex]_{x}[/itex]=-e[itex]^{x}[/itex](cos(y)-1)
    f[itex]_{y}[/itex]=e[itex]^{x}[/itex](sin(y))
    f[itex]_{xx}[/itex]=-e[itex]^{x}[/itex](cos(y)-1)
    f[itex]_{yy}[/itex]=e[itex]^{x}[/itex](cos(y))

    Knowing this, and that e[itex]^{x}[/itex] 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. jcsd
  3. Dec 11, 2013 #2

    Ray Vickson

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    Second-order tests involve not only ##f_{xx}## and ##f_{yy}##, but also the mixed-partial ##f_{xy}##.
     
  4. Dec 11, 2013 #3
    Yes, sorry forgot to include that:

    fxy=exsin(y)
     
  5. Dec 11, 2013 #4

    ehild

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    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
     
  6. Dec 12, 2013 #5
    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.
     
  7. Dec 12, 2013 #6

    ehild

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

    ehild
     
  8. Dec 12, 2013 #7
    Yes, x can be anything as long as the y=2∏. For both the f[itex]_{x}[/itex] and f[itex]_{y}[/itex].

    So the x can range from 0[itex]\rightarrow[/itex]∞, 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[itex]_{xx}[/itex](x,y)f[itex]_{yy}[/itex](x,y) - (f[itex]_{xy}[/itex](x,y))2
    D(x,y)= f[itex]_{xx}[/itex](10,2∏)f[itex]_{yy}[/itex](10,2∏) - (f[itex]_{xy}[/itex](10,2∏))2
    Which then simplifies to 0.

    This means the second order test is inconclusive. However, there clearly are critical points.
     
  9. Dec 12, 2013 #8

    ehild

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    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
     
  10. Dec 12, 2013 #9
    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.
     
  11. Dec 12, 2013 #10

    ehild

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    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
     
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