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Thread: Partial Diff(3)

  1. #1
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    Partial Diff(3)

    For this function,
    $\displaystyle
    z= sin^2xcosy + sin^2ycosx

    $

    determine the 4 stationary points in the region 0=< x <pi ; 0=< y <pi

    I got these two eqn after partial diff for x and y

    $\displaystyle
    sinx(2cosxcosy - sin^2y) = 0$
    $\displaystyle
    siny(2cosycosx - sin^2x) = 0

    $

    correct?

    1 stationary points is (0,0), how can i find the other 3?
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  2. #2
    Senior Member Peritus's Avatar
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    correct.
    The rest is pretty straight forward:

    $\displaystyle

    \begin{gathered}
    \left\{ \begin{gathered}
    2\cos x\cos y - \sin ^2 y = 0 \hfill \\
    2\cos y\cos x - \sin ^2 x = 0 \hfill \\
    \end{gathered} \right. \Rightarrow \sin ^2 y = \sin ^2 x \hfill \\
    \Leftrightarrow \left( {\sin x - \sin y} \right)\left( {\sin x + \sin y} \right) = 0 \hfill \\
    1.\quad x = y + 2\pi k \hfill \\
    2.\quad x = - y + 2\pi k \hfill \\
    \end{gathered}

    $

    now substitute these results in the above equations and choose the solutions in the given range...
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  3. #3
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    how do i get eqn 1 n 2? where does the 2(pi)k comes from?

    Trio rules?
    Last edited by MilK; Feb 16th 2008 at 07:21 PM.
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  4. #4
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    i got these 4 points (0,0) (pi,pi) (pi/2, pi) (pi,pi/2)
    are these correct?

    i think there is smthing wrong somewhere because i can't determine the 4 stationary points in the region 0=< x <pi ; 0=< y <pi

    the answer stated 1 point is min, 2 are max and 1 is saddle.

    Can anyone show me how to get? thanks
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  5. #5
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    i got these 4 points (0,0) (pi,pi) (pi/2, pi) (pi,pi/2)
    are these correct?
    All of the points satisfy the equation, but as I think you noticed they are not all in the region.

    For $\displaystyle sinx(2cosxcosy - sin^2y) = 0$ we need either sin x = 0 or $\displaystyle 2cosxcosy - sin^2y = 0$.

    For $\displaystyle siny(2cosycosx - sin^2x) = 0$ we need either sin y = 0 or $\displaystyle 2cosxcosy - sin^2x = 0$

    Now you need to set up a pair of equations for each possible combination that will make both equations work:

    sin x = 0
    sin y = 0

    sin x = 0
    $\displaystyle 2cosycosx - sin^2x = 0$

    $\displaystyle 2cosxcosy - sin^2y = 0$
    sin y = 0

    $\displaystyle 2cosxcosy - sin^2y = 0$
    $\displaystyle 2cosycosx - sin^2x = 0$

    Have a go at solving each of the pairs. The last one is a bit more difficult, but Peritus has already given you a hint.
    Last edited by badgerigar; Feb 21st 2008 at 03:48 PM. Reason: Whoops, I should really make sure I am cutting and pasting the right latex
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  6. #6
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    Quote Originally Posted by badgerigar View Post
    All of the points satisfy the equation, but as I think you noticed they are not all in the region.

    For $\displaystyle sinx(2cosxcosy - sin^2y) = 0$ we need either sin x = 0 or $\displaystyle 2cosxcosy - sin^2y = 0$.

    For $\displaystyle siny(2cosycosx - sin^2x) = 0$ we need either sin y = 0 or $\displaystyle 2cosxcosy - sin^2x = 0$

    Now you need to set up a pair of equations for each possible combination that will make both equations work:

    sin x = 0
    sin y = 0 // point(0,0)

    sin x = 0
    $\displaystyle 2cosycosx - sin^2x = 0$ // point (0,pi/2)

    $\displaystyle 2cosxcosy - sin^2y = 0$ //point (pi/2,0)
    sin y = 0

    $\displaystyle 2cosxcosy - sin^2y = 0$ // struck, don't understand hint
    $\displaystyle 2cosycosx - sin^2x = 0$

    Have a go at solving each of the pairs. The last one is a bit more difficult, but Peritus has already given you a hint.
    how to get x= y+2(pi)k
    x= -y+2(pi)k?
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  7. #7
    Senior Member Peritus's Avatar
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    let's examine the last two equations using the following formulas:

    $\displaystyle \begin{gathered}
    \sin x - \sin y = 2\cos \left( {\frac{{x + y}}
    {2}} \right)\sin \left( {\frac{{x - y}}
    {2}} \right) \hfill \\
    \sin x + \sin y = 2\sin \left( {\frac{{x + y}}
    {2}} \right)\cos \left( {\frac{{x - y}}
    {2}} \right) \hfill \\
    \end{gathered} $

    thus we get:


    $\displaystyle \begin{gathered}
    1.\quad 2\cos \left( {\frac{{x + y}}
    {2}} \right)\sin \left( {\frac{{x - y}}
    {2}} \right) = 0 \hfill \\
    \Leftrightarrow \frac{{x + y}}
    {2} = \frac{\pi }
    {2}k\quad \quad \quad \Leftrightarrow \frac{{x - y}}
    {2} = \pi k \hfill \\
    \Leftrightarrow x = - y + \pi k\quad \quad \quad \Leftrightarrow x = y + 2\pi k \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    2.\quad 2\sin \left( {\frac{{x + y}}
    {2}} \right)\cos \left( {\frac{{x - y}}
    {2}} \right) = 0 \hfill \\
    \Leftrightarrow \frac{{x + y}}
    {2} = \pi k\quad \quad \quad \Leftrightarrow \frac{{x - y}}
    {2} = \frac{\pi }
    {2}k \hfill \\
    \Leftrightarrow x = - y + 2\pi k\quad \quad \Leftrightarrow x = y + \pi k \hfill \\
    \end{gathered} $

    so we've got the following solutions:

    $\displaystyle
    \left\{ \begin{gathered}
    x = \pm y + 2\pi k \hfill \\
    x = \pm y + \pi k \hfill \\
    \end{gathered} \right.
    $

    after substituting the second solution in one of the first two equations I've given you we get:

    $\displaystyle
    \begin{gathered}
    - 2\cos ^2 y - \sin ^2 y = 0 \hfill \\
    \Leftrightarrow \cos ^2 y = - 1 \hfill \\
    \end{gathered} $

    thus this solution is obviously illegal:
    substituting x = -y or x = y gives the same result namely:

    $\displaystyle
    \begin{gathered}
    \Leftrightarrow 3\cos ^2 x = 1 \hfill \\
    \Leftrightarrow \cos x = \pm \sqrt {\frac{1}
    {3}} \hfill \\
    \end{gathered}
    $

    can you continue...
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  8. #8
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    Quote Originally Posted by Peritus View Post
    let's examine the last two equations using the following formulas:

    $\displaystyle \begin{gathered}
    \sin x - \sin y = 2\cos \left( {\frac{{x + y}}
    {2}} \right)\sin \left( {\frac{{x - y}}
    {2}} \right) \hfill \\
    \sin x + \sin y = 2\sin \left( {\frac{{x + y}}
    {2}} \right)\cos \left( {\frac{{x - y}}
    {2}} \right) \hfill \\
    \end{gathered} $

    thus we get:


    $\displaystyle \begin{gathered}
    1.\quad 2\cos \left( {\frac{{x + y}}
    {2}} \right)\sin \left( {\frac{{x - y}}
    {2}} \right) = 0 \hfill \\
    \Leftrightarrow \frac{{x + y}}
    {2} = \frac{\pi }
    {2}k\quad \quad \quad \Leftrightarrow \frac{{x - y}}
    {2} = \pi k \hfill \\
    \Leftrightarrow x = - y + \pi k\quad \quad \quad \Leftrightarrow x = y + 2\pi k \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    2.\quad 2\sin \left( {\frac{{x + y}}
    {2}} \right)\cos \left( {\frac{{x - y}}
    {2}} \right) = 0 \hfill \\
    \Leftrightarrow \frac{{x + y}}
    {2} = \pi k\quad \quad \quad \Leftrightarrow \frac{{x - y}}
    {2} = \frac{\pi }
    {2}k \hfill \\
    \Leftrightarrow x = - y + 2\pi k\quad \quad \Leftrightarrow x = y + \pi k \hfill \\
    \end{gathered} $

    so we've got the following solutions:

    $\displaystyle
    \left\{ \begin{gathered}
    x = \pm y + 2\pi k \hfill \\ ----> where does the k comes from?
    x = \pm y + \pi k \hfill \\
    \end{gathered} \right.
    $

    after substituting the second solution in one of the first two equations I've given you we get: <-------lost... sub which eqn with which?

    $\displaystyle
    \begin{gathered}
    - 2\cos ^2 y - \sin ^2 y = 0 \hfill \\
    \Leftrightarrow \cos ^2 y = - 1 \hfill \\
    \end{gathered} $

    thus this solution is obviously illegal:
    substituting x = -y or x = y gives the same result namely:

    $\displaystyle
    \begin{gathered}
    \Leftrightarrow 3\cos ^2 x = 1 \hfill \\
    \Leftrightarrow \cos x = \pm \sqrt {\frac{1}
    {3}} \hfill \\
    \end{gathered}
    $

    can you continue...
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  9. #9
    Senior Member Peritus's Avatar
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    substitue in one of these two equations:

    $\displaystyle
    \begin{gathered}
    2\cos x\cos y - \sin ^2 y = 0 \hfill \\
    2\cos y\cos x - \sin ^2 x = 0 \hfill \\
    \end{gathered}
    $
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  10. #10
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    Quote Originally Posted by Peritus View Post
    let's examine the last two equations using the following formulas:

    $\displaystyle \begin{gathered}
    \sin x - \sin y = 2\cos \left( {\frac{{x + y}}
    {2}} \right)\sin \left( {\frac{{x - y}}
    {2}} \right) \hfill \\
    \sin x + \sin y = 2\sin \left( {\frac{{x + y}}
    {2}} \right)\cos \left( {\frac{{x - y}}
    {2}} \right) \hfill \\
    \end{gathered} $

    thus we get:


    $\displaystyle \begin{gathered}
    1.\quad 2\cos \left( {\frac{{x + y}}
    {2}} \right)\sin \left( {\frac{{x - y}}
    {2}} \right) = 0 \hfill \\
    \Leftrightarrow \frac{{x + y}}
    {2} = \frac{\pi }
    {2}k\quad \quad \quad \Leftrightarrow \frac{{x - y}}
    {2} = \pi k \hfill \\
    \Leftrightarrow x = - y + \pi k\quad \quad \quad \Leftrightarrow x = y + 2\pi k \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    2.\quad 2\sin \left( {\frac{{x + y}}
    {2}} \right)\cos \left( {\frac{{x - y}}
    {2}} \right) = 0 \hfill \\
    \Leftrightarrow \frac{{x + y}}
    {2} = \pi k\quad \quad \quad \Leftrightarrow \frac{{x - y}}
    {2} = \frac{\pi }
    {2}k \hfill \\
    \Leftrightarrow x = - y + 2\pi k\quad \quad \Leftrightarrow x = y + \pi k \hfill \\
    \end{gathered} $

    so we've got the following solutions:

    $\displaystyle
    \left\{ \begin{gathered}
    x = \pm y + 2\pi k \hfill \\
    x = \pm y + \pi k \hfill \\
    \end{gathered} \right.
    $

    where does the k comes from?
    after substituting the second solution in one of the first two equations I've given you we get:

    $\displaystyle
    \begin{gathered}
    - 2\cos ^2 y - \sin ^2 y = 0 \hfill \\
    \Leftrightarrow \cos ^2 y = - 1 \hfill \\
    \end{gathered} $

    thus this solution is obviously illegal:
    substituting x = -y or x = y gives the same result namely:

    $\displaystyle
    \begin{gathered}
    \Leftrightarrow 3\cos ^2 x = 1 \hfill \\
    \Leftrightarrow \cos x = \pm \sqrt {\frac{1}
    {3}} \hfill \\
    \end{gathered}
    $

    can you continue...
    one more question...
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  11. #11
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    Quote Originally Posted by MilK View Post
    one more question...
    k is an integer. The general solution of sin(A) = 0 is A = k pi, that is, $\displaystyle \pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$

    Similarly for cos(A) = 0: $\displaystyle A = \frac{(2k + 1)\pi}{2}$, that is, $\displaystyle \pm \frac{\pi}{2}, \, \pm \frac{3\pi}{2} \, \pm \frac{5\pi}{2} .......$. Peritus made a small mistake here by the way - it's NOT $\displaystyle \frac{\pi}{2}k$ since this also gives $\displaystyle \pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$ and obviously cos is NOT 0 for those values ....
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    k is an integer. The general solution of sin(A) = 0 is A = k pi, that is, $\displaystyle \pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$

    Similarly for cos(A) = 0: $\displaystyle A = \frac{(2k + 1)\pi}{2}$, that is, $\displaystyle \pm \frac{\pi}{2}, \, \pm \frac{3\pi}{2} \, \pm \frac{5\pi}{2} .......$. Peritus made a small mistake here by the way - it's NOT $\displaystyle \frac{\pi}{2}k$ since this also gives $\displaystyle \pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$ and obviously cos is NOT 0 for those values ....
    so the eqn i get shld be this?
    x=y+(pi)k/2
    x=-y+(pi)k/2
    x=y+(pi)k
    x=-y+(pi)k

    after sub these eqn into 2cosxcosy-sin^2y, i can't simplifiy it and get the ans..........
    Last edited by MilK; Feb 23rd 2008 at 05:15 PM.
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  13. #13
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    Hi

    after working out the solution i got

    y = 54.73 x = 2(pi)+54.73 ------------for x=2(pi) +y eqn

    the value or x is out of range
    their is one simliar ans with c value out of range

    others are -ve points........

    wrong somewhere?

    Help~~~~~~
    Last edited by MilK; Feb 24th 2008 at 12:14 AM. Reason: ~~~~~~~
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  14. #14
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    Quote Originally Posted by MilK View Post
    Hi

    after working out the solution i got

    y = 54.73 x = 2(pi)+54.73 ------------for x=2(pi) +y eqn

    the value or x is out of range
    their is one simliar ans with c value out of range

    others are -ve points........

    wrong somewhere?

    Help~~~~~~
    The solutions have already been given to you:

    Quote Originally Posted by Peritus View Post
    [snip]
    $\displaystyle
    \left\{ \begin{gathered}
    x = \pm y + 2\pi k \hfill \\
    x = \pm y + \pi (2k + 1) \hfill \\
    \end{gathered} \right.
    $
    (Second one edited by Mr F)

    after substituting the second solution in one of the first two equations I've given you we get:

    $\displaystyle
    \begin{gathered}
    - 2\cos ^2 y - \sin ^2 y = 0 \hfill \\
    \Leftrightarrow \cos ^2 y = - 1 \hfill \\
    \end{gathered} $

    thus this solution is obviously illegal:
    substituting x = -y or x = y gives the same result namely:

    $\displaystyle
    \begin{gathered}
    \Leftrightarrow 3\cos ^2 x = 1 \hfill \\
    \Leftrightarrow \cos x = \pm \sqrt {\frac{1}
    {3}} \hfill \\
    \end{gathered}
    $
    [snip]
    Where in the above solution are you stuck?
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  15. #15
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    Quote Originally Posted by mr fantastic View Post
    The solutions have already been given to you:



    Where in the above solution are you stuck?
    cos x = +/- sqrt of (1/3)
    which gives x = +/- 0.7593

    since x is not -ve, this gives x= 0.7593 and sub this value into eqn will give a y value above the limit of the question.....
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