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Critial points

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Q4(b)
    http://www.physics.hku.hk/~phys1315/doc/MPI_HW3.pdf" [Broken]

    2. Relevant equations



    3. The attempt at a solution
    After I set fx and fy =0
    I get cosx=cosy.
    How to solve it?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 19, 2011 #2

    lanedance

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    so I'm assuming f is a function of 2 variables... well what do you know about the cos function?
     
  4. Oct 19, 2011 #3
    Sorry for my stupid mistake, I have forgotten to upload the question.
    Please see question 4(b).

    I know about the cos function.
    After solving, it should be x=2*Pi*n+y.
    However, the question only need to find critial pt within 0<y,x<Pi
     
  5. Oct 19, 2011 #4

    lanedance

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    ok so what do you get for fx and fy?
     
  6. Oct 19, 2011 #5
    fx=cos(x+y)+cos(x)
    fy=cos(x+y)+cos(y)
     
  7. Oct 19, 2011 #6

    HallsofIvy

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    Ys, since both derivatives are 0, you can set them equal to each other and reduce, that is cos(x+y)+ cos(x)= cos(x+y)+ cos(y) reduces to cos(x)= cos(y), but then you still need the equations themselves.

    It might help you to know that cos(x+y)= cos(x)cos(y)- sin(x)sin(y). If cos(x)= cos(y) then [itex]sin(x)= \pm sin(y)[/itex]
     
  8. Oct 19, 2011 #7
    I don't understand this statement...
     
  9. Oct 19, 2011 #8

    HallsofIvy

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    Do you know that [itex]cos^2(x)+ sin^2(x)= 1[/itex]? From that [itex]cos(x)= \pm\sqrt{1- sin^2(x)}[/itex]. That means that cos(x)= cos(y) gives [itex]\sqrt{1- sin^2(x)}= \pm\sqrt{1- sin^2(y)}[/itex]. Squaring both sides, [itex]1- sin^2(x)= 1- sin^2(y)[/itex] so that [itex]sin^2(x)= sin^2(y)[/itex] and, taking the square root again- [itex]sin(x)= \pm sin(y)[/itex].

    My purpose in saying that was after using the identity I gave you to reduce cos(x+y)+ cos(x)= 0 to [itex]cos(x)cos(y)- sin(x)sin(y)+ cos(x)= 0[/itex] you can use the fact that cos(x)= cos(y) and either sin(x)= sin(y) or sin(x)= -sin(y) to reduce to an equation in x only:
    if sin(x)= sin(y), then [itex]cos^2(x)- sin^2(x)+ cos(x)= 0[/itex] or
    if sin(x)= -sin(y), then [itex]cos^2(x)+ sin^2(x)+ cos(x)= 0[/itex]

    Now use [itex]sin^2(x)= 1- cos^2(x)[/itex] to reduce those to equations in cos(x) only.
     
  10. Oct 19, 2011 #9
    Oh no. How complicated the answer is!
    I know these trigonometric identities but I never think of I should use them like this.

    I would like to ask what is the problem if I solve cos(x)=cos(y) directly?
     
  11. Oct 19, 2011 #10

    HallsofIvy

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    You can't. That is a single equation in two variables- there are an infinite number of x, y pairs that satisfy it. y= x= any number is a solution.
     
  12. Oct 19, 2011 #11

    lanedance

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    i think we may be over complicating it a bit, so rather than equating fx and fy, how about noticing the following

    fx=cos(x+y)+cos(x)=0
    gives
    cos(x)=-cos(x+y)

    similarly for fy
    fy=cos(x+y)+cos(y)=0
    gives
    cos(y)=-cos(x+y)

    so it is true that for the above to be satisfied requires cos(x)=cos(y), though that only means fx=fy, not fx=fy=0, which is the overall goal.

    however it will be instructive to first consider cos(x)=cos(y), to solve this, consider a graph of cos(x), clearly it repeats on every 2.pi, so if y is s solution, so is y+2.pi

    Now confine x to (-pi,pi), there will be 2 solutions to cos(x)=cos(y), clearly y=x and using the fact cos is symmetric cos(y)=cos(-y), gives y=-x.

    Thus all the solutions are of the form:
    y = x+2.pi
    y = -x+2.pi

    Now going back to the main requirement for fx=fy=0:
    cos(x)=-cos(x+y)
    cos(y)=-cos(x+y)

    can you do a similar process to find allowable (x,y) pairs?
     
  13. Oct 25, 2011 #12
    Sorry that I didn't notice your reply.
    We can subs y = x+2.pi into cos(x)=-cos(x+y)
    By solving we get pi and pi/3

    Is it correct?
     
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