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Stationary points of $z(x, y)=cosxe^-x^2-y^2$

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3

1

$begingroup$

This is a problem from one of past exams from my course.

Consider a function $z(x, y)=cosxe^-x^2-y^2$.

Show that stationary points of $z$ lie along the $x$-axis, and satisfy $tanx=k(x)$. Give an expression for $k(x)$.

This one I managed to do, I got that for $z_x=z_y=0$ we have $y=0$ and $tanx=-2x$.

Determine the nature of these stationary points of $z(x,y)$.

This one I have problems with. I know the conditions for maxima, minima etc. in two variables ($D=z_xxz_yy-z_xy^2$ test), but I got into some horrible algebraic mess when trying to determine $D$.

Is there any "cleverer" method to determine the nature of the stationary points?

For what it's worth, $z_x = -e^-x^2-y^2(2xcosx+sinx)$ and $z_y = -2ycosxe^-x^2-y^2.$

calculus multivariable-calculus stationary-point

share|cite|improve this question

edited Mar 22 at 10:14

VanDerWarden

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean singular points? Instead of stationary?
  $endgroup$
  – dmtri
  Mar 22 at 10:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri As far as I know, points such that $fracpartialzpartialx=0$ and similarly for $y$ are called stationary points, see for instance personal.maths.surrey.ac.uk/S.Zelik/teach/calculus/…
  $endgroup$
  – VanDerWarden
  Mar 22 at 10:25
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

This is a problem from one of past exams from my course.

Consider a function $z(x, y)=cosxe^-x^2-y^2$.

Show that stationary points of $z$ lie along the $x$-axis, and satisfy $tanx=k(x)$. Give an expression for $k(x)$.

This one I managed to do, I got that for $z_x=z_y=0$ we have $y=0$ and $tanx=-2x$.

Determine the nature of these stationary points of $z(x,y)$.

This one I have problems with. I know the conditions for maxima, minima etc. in two variables ($D=z_xxz_yy-z_xy^2$ test), but I got into some horrible algebraic mess when trying to determine $D$.

Is there any "cleverer" method to determine the nature of the stationary points?

For what it's worth, $z_x = -e^-x^2-y^2(2xcosx+sinx)$ and $z_y = -2ycosxe^-x^2-y^2.$

calculus multivariable-calculus stationary-point

share|cite|improve this question

edited Mar 22 at 10:14

VanDerWarden

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean singular points? Instead of stationary?
  $endgroup$
  – dmtri
  Mar 22 at 10:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri As far as I know, points such that $fracpartialzpartialx=0$ and similarly for $y$ are called stationary points, see for instance personal.maths.surrey.ac.uk/S.Zelik/teach/calculus/…
  $endgroup$
  – VanDerWarden
  Mar 22 at 10:25
  
  

  
  
  
  

add a comment | 

$begingroup$

This is a problem from one of past exams from my course.

Consider a function $z(x, y)=cosxe^-x^2-y^2$.

Show that stationary points of $z$ lie along the $x$-axis, and satisfy $tanx=k(x)$. Give an expression for $k(x)$.

This one I managed to do, I got that for $z_x=z_y=0$ we have $y=0$ and $tanx=-2x$.

Determine the nature of these stationary points of $z(x,y)$.

This one I have problems with. I know the conditions for maxima, minima etc. in two variables ($D=z_xxz_yy-z_xy^2$ test), but I got into some horrible algebraic mess when trying to determine $D$.

Is there any "cleverer" method to determine the nature of the stationary points?

For what it's worth, $z_x = -e^-x^2-y^2(2xcosx+sinx)$ and $z_y = -2ycosxe^-x^2-y^2.$

calculus multivariable-calculus stationary-point

share|cite|improve this question

edited Mar 22 at 10:14

VanDerWarden

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

$endgroup$

share|cite|improve this question

edited Mar 22 at 10:14

VanDerWarden

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

share|cite|improve this question

edited Mar 22 at 10:14

VanDerWarden

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

asked Mar 22 at 10:10

VanDerWardenVanDerWarden

636831

1 Answer
1

active

oldest

votes

2

$begingroup$

At any stationary point $(x,0)$ we have
$$eqalign
left[matrixz_xx&z_xycr z_xy&z_yyright]&=
left[
beginarraycc
e^-x^2 left(left(4 x^2-3right) cos (x)+4 x sin (x)right) & 0 \
0 & -2 e^-x^2 cos (x) \
endarray
right]cr
&=-cos(x)e^-x^2left[matrix3+4x^2&0cr0&2right]

$$
where in the last step, I used $tan(x)=-2x$. Now, we have a local minimum
on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)<0$ and a local maximum on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)>0$.

share|cite|improve this answer

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881

$endgroup$

add a comment | 

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2

$begingroup$

At any stationary point $(x,0)$ we have
$$eqalign
left[matrixz_xx&z_xycr z_xy&z_yyright]&=
left[
beginarraycc
e^-x^2 left(left(4 x^2-3right) cos (x)+4 x sin (x)right) & 0 \
0 & -2 e^-x^2 cos (x) \
endarray
right]cr
&=-cos(x)e^-x^2left[matrix3+4x^2&0cr0&2right]

$$
where in the last step, I used $tan(x)=-2x$. Now, we have a local minimum
on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)<0$ and a local maximum on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)>0$.

share|cite|improve this answer

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881

$endgroup$

add a comment | 

2

$begingroup$

At any stationary point $(x,0)$ we have
$$eqalign
left[matrixz_xx&z_xycr z_xy&z_yyright]&=
left[
beginarraycc
e^-x^2 left(left(4 x^2-3right) cos (x)+4 x sin (x)right) & 0 \
0 & -2 e^-x^2 cos (x) \
endarray
right]cr
&=-cos(x)e^-x^2left[matrix3+4x^2&0cr0&2right]

$$
where in the last step, I used $tan(x)=-2x$. Now, we have a local minimum
on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)<0$ and a local maximum on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)>0$.

share|cite|improve this answer

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881

$endgroup$

add a comment | 

$begingroup$

At any stationary point $(x,0)$ we have
$$eqalign
left[matrixz_xx&z_xycr z_xy&z_yyright]&=
left[
beginarraycc
e^-x^2 left(left(4 x^2-3right) cos (x)+4 x sin (x)right) & 0 \
0 & -2 e^-x^2 cos (x) \
endarray
right]cr
&=-cos(x)e^-x^2left[matrix3+4x^2&0cr0&2right]

$$
where in the last step, I used $tan(x)=-2x$. Now, we have a local minimum
on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)<0$ and a local maximum on every stationary point $(x,0)$ with $tan(x)=-2x$ and $cos(x)>0$.

share|cite|improve this answer

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881

$endgroup$

share|cite|improve this answer

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881

answered Mar 22 at 13:13

Omran KoubaOmran Kouba

24.8k13881