Page 1 Math 251 copyright Joe Kahlig 15A 1 path gives f x 3x2 lim f x 3x2 x 0 12x4 12x2 x2 9x4 1 9x2 0 0 1 0 2 a fx 5 cos 5x y z 12x2 y b fy sin 5x z y z 1 4x3 c fz sin 5x y z ln y 3 a g 3 2 1 h6 4 24i 1 b u h4 4 2i 6 4 2 44 4 Du g 3 2 1 6 4 24 6 6 6 3 c h6 4 24i 628 d h 6 4 24i 4 the normal vector is hfx fy 1i at the point n h12 10 1i tanget plane 12 x 2 10 y 1 1 z 9 0 or 12x 10y z 25 5 since we know f 1 2 3 then dx x 1 3 1 0 3 dy y 1 9 p 2 0 1 let z f x y x y 3 1 1 and fx 1 2 fx p 3 6 2 x y 3y 2 fy p and fy 1 2 2 2 x y3 dz fx dx fy dy thus dz 0 15 f 1 3 1 9 f 1 2 dz 3 0 15 2 85 6 Let f x y z x2 2y 2 z 3 10 Then n f 1 1 2 h2 4 12i parametric equation of the normal line x 1 2t y 1 4t z 2 12t 7 zy 3x4 y 2 2 2z 4 sec2 4z 8 wa 2xyz 3 2at 3xy 2 z 2 ateat eat Math 251 copyright Joe Kahlig 15A 9 x Page 2 5 30 8 y z 11 11 11 10 fx 2xy 4x and fy x2 16y 16 solving fx 0 gives x 0 and y 2 now set fy 0 and sub theses values in and solve for the other variable Critical points 0 1 local max 4 2 and 4 2 saddle points 11 step 1 find fx and fy and solve for any critical points in the interior you get the point 1 5 note this in not in the region so ignore it Boundary 1 x asis x x and y 0 for 0 x 2 g x f x 0 3x2 x g 0 6x 1 set equal to zero and solve has a critical value at x 1 6 Boundary 2 vertical line x 2 x 2 y y for 0 y 8 g y f 2 y 10 y since g 0 1 no critical values Boundary 3 x x and y 4x for 0 x 2 g x f x 4x x2 3x take deriva and find critical value x 1 5 critical points function value 0 0 0 2 0 10 2 8 2 1 12 1 6 0 2 25 1 5 6 abs max 10 abs min 1 12
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