Page 1 Math 251 copyright Joe Kahlig 15A 1 path gives f 2y 3 y lim f 2y 3 y y 0 10y 6 10y 3 4y 6 y 3 4y 3 1 0 0 0 1 2 a fy z tan 3x y z 1 8x3 b fx 3 sec2 3x y 2 24x2 y c fz tan 3x y z ln z 3 a g 2 4 1 h3 8 12i 1 b u h2 1 2i 3 1 2 38 2 Du g 2 4 1 3 8 12 3 3 3 3 c h3 8 12i 217 d h 3 8 12i 4 the normal vector is hfx fy 1i at the point n h6 4 1i tanget plane 6 x 1 4 y 2 1 z 5 0 or 6x 4y z 9 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 x3 5y 2 z 2 50 Then n f 1 3 2 h3 30 42i parametric equation of the normal line x 1 3t y 3 30t z 2 4t 7 zy 3x4 y 2 2 2z 4 cos 4z 8 wa y 2 z 3 2at 2xyz 3 eat ateat 9 x 5 30 8 y z 11 11 11 Math 251 copyright Joe Kahlig 15A Page 2 10 fy 2xy 4y and fx y 2 16x 16 solving fy 0 gives x 2 and y 0 now set fx 0 and sub theses values in and solve for the other variable Critical points 1 0 local max 2 4 and 2 4 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 6 0 1 12 2 25 1 5 6 abs max 10 abs min 1 12
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