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November 13th, 2016, 06:55 PM   #1
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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!
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November 13th, 2016, 08:50 PM   #2
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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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