Stationary points with 2 variables

Printable View

March 1st 2010, 05:17 AM
Beard

Stationary points with 2 variables

Hi,

Locate and classify all of the stationary points of the function:

$f(x,y) = e^x(3x - y^3 + 3y)$ .

I have got as far as doing the partial derivatives for the terms but I am unsure if the partial derivatives I got are right

What I got for the derivatives so far are:

$f_{x} = 3e^x - e^xy^3$
$f_{xx} = 3e^x - e^xy^3$ <-- unsure on this one
$f_{y} = 3xe^x - 3e^xy^2 + 3$
$f_{yy} = 3xe^x - 6e^xy$
$f_{xy} = 3e^x - e^xy^3$

I am sure that after I have found these I am meant to equate them to zero (because they are stationary points) and then find x and y and then put the in one of the equations (I think its the original one if I remember correctly) and get the coordinates of the stationary points.
March 2nd 2010, 04:09 AM
CaptainBlack

Quote:

Originally Posted by Beard

Hi,

Locate and classify all of the stationary points of the function:

$f(x,y) = e^x(3x - y^3 + 3y)$ .

I have got as far as doing the partial derivatives for the terms but I am unsure if the partial derivatives I got are right

What I got for the derivatives so far are:

$f_{x} = 3e^x - e^xy^3$
$f_{xx} = 3e^x - e^xy^3$ <-- unsure on this one
$f_{y} = 3xe^x - 3e^xy^2 + 3$
$f_{yy} = 3xe^x - 6e^xy$
$f_{xy} = 3e^x - e^xy^3$

I am sure that after I have found these I am meant to equate them to zero (because they are stationary points) and then find x and y and then put the in one of the equations (I think its the original one if I remember correctly) and get the coordinates of the stationary points.

$f_x=\frac{\partial}{\partial x}\left[ 3x e^x + (-y^3+3y)e^x\right]=3e^x+3xe^x+(-y^3+3y)e^x$

CB
March 3rd 2010, 10:12 AM
Beard

I found where I was going wrong for the differentiations, I wasn't using the product rule when I did it.

I have redone the differentials using the rule and come up with this.

$f_x = 3e^x + e^x(3x - y^2 +3y)$
$f_{xx} = 6e^x + e^x(3x - y^2 +3y)$
$f_{xy} = 6 + 3x + y + y^2$
$f_y = e^x(-2y +3) + 3x - y^2 + 3y$
$f_{yy} = -2e^x - 2y +3$

I tried equating one of the to zero but I'm not getting anywhere and keep hitting dead ends. Am I able to divide by e^x on some of them?
March 3rd 2010, 12:15 PM
CaptainBlack

Quote:

Originally Posted by Beard

I found where I was going wrong for the differentiations, I wasn't using the product rule when I did it.

I have redone the differentials using the rule and come up with this.

$f_x = 3e^x + e^x(3x - y^2 +3y)$
$f_{xx} = 6e^x + e^x(3x - y^2 +3y)$
$f_{xy} = 6 + 3x + y + y^2$
$f_y = e^x(-2y +3) + 3x - y^2 + 3y$
$f_{yy} = -2e^x - 2y +3$

I tried equating one of the to zero but I'm not getting anywhere and keep hitting dead ends. Am I able to divide by e^x on some of them?

$f_y$ is still wrong

CB