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 November 13th, 2016, 06:55 PM #1 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 Find points of maximum slope on surface I got this surface: $\displaystyle f(x,y) = 3/(1 + x^2 + y^2)$ I want to find the points that have maximum slope. I think I'm suppose to create the function (aka. the speed at given point): $\displaystyle g = sqrt((f'x)^2 + (f'y)^2)$ And find where g'x = 0 g'y = 0 Issue is, the expression is gets so complex I have no idea what to do. g can be simplified to g^2 thus getting rid of the squareroot. $\displaystyle f'x = -6x/(1 + x^2 + y^2)^2$ $\displaystyle f'y = -6y/(1 + x^2 + y^2)^2$ $\displaystyle g^2 = 36*(x^2 + y^2)/(1 + x^2 + y^2)^4$ Now this may look simple but I have no idea what to do next. What I want to do is get $\displaystyle (g^2)'x = 0$ $\displaystyle (g^2)'y = 0$ but those equations are so messy nobody can understand where the maximum lies. I am getting nowhere on this subject, I need desperate help!
 November 13th, 2016, 08:50 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,505 Thanks: 1374 Note the problem is symmetric in $x$ and $y$ I'd write $f=\dfrac{3}{1+r^2}$ $f^\prime=\dfrac{df}{dr}$ Now find critical point in one variable as usual. set $f^{\prime \prime}=\dfrac{d^2f}{{dr}^2}=0$ and solve for $r$

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