Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f(x,y)= \ln(x^2 + y^2 +1)$

The partial derivatives of $f(x,y)$ being:

$f_x(x,y) = \frac{2x}{x^2+y^2+1}$ and $f_y(x,y)=\frac{2y}{x^2+y^2+1}$

Setting each partial derivative equal to zero, adding the two linear equations two, and solving for $y$, I get:

$y = - x$

In this problem Finding the $x$ and $y$ values such that the partial derivatives are zero simultaneously, there was only one ordered pair. In the problem before us, I solved it the same way I did in the problem given in the link Does the solution above imply that there are an infinite amount of $(x,y)$ pairs, such that there partial derivatives are zero? I ask, because this particular problem is an even number, and the answer key provides no answers for such questions.

share|improve this question
    
No, you get $x = y = 0$. –  vonbrand Mar 10 '13 at 16:04
add comment

1 Answer

up vote 4 down vote accepted

Hint: Since the denominators of these partial derivatives are never zero, then the partial derivatives are equal to zero precisely when their numerators are.

share|improve this answer
    
Yes, that is what I did. Then I had simultaneous equations, $2y = 0$ and $2x = 0$. When I added the two equations together, I got $2x +2y = 0 \implies y = -x$. –  Mack Mar 10 '13 at 14:42
1  
As in the other problem, you've added your equations together unnecessarily. You should only add two equations together if it eliminates variables. In this case, your variables are all but solved for. $2y=0$ implies $y=0$ and $2x=0$ implies $x=0$. Thus, $(0,0)$ is the solution point. –  Cameron Buie Mar 10 '13 at 14:45
    
Oh, I see. Thank you, Cameron! –  Mack Mar 10 '13 at 14:47
1  
P.S.: If you do add equations together to eliminate variables, don't forget to back-substitute at the end, so you can solve for all of the variables! –  Cameron Buie Mar 10 '13 at 14:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.