Calculus Calculus Math Forum

 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 $r$ Tags find, maximum, points, slope, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Addez123 Calculus 4 November 12th, 2016 06:51 AM Hatschi1 Algebra 3 August 17th, 2016 09:39 AM bigmyk2k Calculus 1 June 10th, 2016 06:21 PM cafegurl Algebra 2 April 29th, 2010 07:48 AM 486dx66 Algebra 1 February 10th, 2010 11:03 PM

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