Partial Diff(3)

**MilK** · February 16th 2008, 06:54 AM

For this function,
$ 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

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

correct?

1 stationary points is (0,0), how can i find the other 3?

**Peritus** · February 16th 2008, 08:34 AM

correct.
The rest is pretty straight forward:

$ \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...

**MilK** · February 16th 2008, 08:36 AM

how do i get eqn 1 n 2? where does the 2(pi)k comes from?

Trio rules?

**MilK** · February 21st 2008, 08:52 AM

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

**badgerigar** · February 21st 2008, 03:47 PM

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

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

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
$2cosycosx - sin^2x = 0$

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

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

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.

**MilK** · February 23rd 2008, 12:23 AM

Originally Posted by badgerigar

All of the points satisfy the equation, but as I think you noticed they are not all in the region.

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

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

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

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

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

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?

**Peritus** · February 23rd 2008, 01:09 AM

let's examine the last two equations using the following formulas:

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

thus we get:

$\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} $

$ \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:

$ \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:

$ \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:

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

can you continue...

**MilK** · February 23rd 2008, 02:59 AM

Originally Posted by Peritus

let's examine the last two equations using the following formulas:

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

thus we get:

$\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} $

$ \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:

$ \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.
" alt="
\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?

$ \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:

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

can you continue...

**Peritus** · February 23rd 2008, 03:44 AM

substitue in one of these two equations:

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

**MilK** · February 23rd 2008, 04:01 AM

Originally Posted by Peritus

let's examine the last two equations using the following formulas:

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

thus we get:

$\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} $

$ \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:

$ \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:

$ \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:

$ \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...

**mr fantastic** · February 23rd 2008, 04:27 AM

Originally Posted by MilK

one more question...

k is an integer. The general solution of sin(A) = 0 is A = k pi, that is, $\pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$

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

**MilK** · February 23rd 2008, 06:00 AM

Originally Posted by mr fantastic

k is an integer. The general solution of sin(A) = 0 is A = k pi, that is, $\pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$

Similarly for cos(A) = 0: $A = \frac{(2k + 1)\pi}{2}$ , that is, $\pm \frac{\pi}{2}, \, \pm \frac{3\pi}{2} \, \pm \frac{5\pi}{2} .......$ . Peritus made a small mistake here by the way - it's NOT $\frac{\pi}{2}k$ since this also gives $\pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$ 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..........

**MilK** · February 24th 2008, 12:00 AM

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~~~~~~

**mr fantastic** · February 24th 2008, 12:56 AM

Originally Posted by MilK

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:

Originally Posted by Peritus

[snip]
$ \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:

$ \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:

$ \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?

**MilK** · February 24th 2008, 01:40 AM

Originally Posted by mr fantastic

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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