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14 6. Directional Derivatives & The Gradient Vecter of : The directional derivative Def inition...

14 6. Directional Derivatives & The Gradient Vecter of : The directional derivative Def inition g(x g) (a,b) in af unction , & the glass direction P n Gifyis functionge a three variables , a b, in of the direction a unit vector =Cu , 2 , us) is eie , Hatho btho 0 < thog-flabc Difla , , b, c= 1 5 limit provided the exists * Note : Diff Theoremon of afta differentin tonal functi o any unit veeter : 2 , 102) and Dif(x y) = = , ExamplesLetf(File a < f -3) g =. , = 2) = 2 fx 2(n(+ + x) 2) 2() ( gy -if(z = = (r = + : , = + 24n8(i) Lyndigt is 8x 2xy 3 fx(1 ,0 2) 0 unit vector in direction = = = , (Y y+ ) = - - + 2 2) 1 gy gy(l = = X ,0 , = fz (( , 0 , 2) 4 22 f = = = -is i , 2) 0(5) 1(5) a() = Dif(1 + ·: ,0, = + Definitiona ifunction a if ·. (a 1 b) Definition : A is called point a critical point he Vagifandforn · Example find : critical of of the points each the following functions. f(x y) yz al , = xz - fx So = 2x ((x y) (0 0) = , , Jy 2y = a f(x y) , = - jx* Ex + dxy y + - x 2x fx Gfy = ↓ + =. y = - - gy ax zy = - = - x(x = 2x - b) (0 , 0) (4 , 8) , - 2, 6) (x y) = x(x 4)(x 2) - , , + = 0 8x2 = - - , xXx = 0 - > O X= 0 - = y x - = 8 2 y = 2 - = - 4 x y = - a) +y f(x y) xy x yz , = - x - y) 0 2x 1 y y = = or + fx y zxy y2 y(1 2x = 0 0 = - - = = - = 2y + (y) x = 0x 0 or )x() = x= x = 2yx - 0 - = gy X - - = = 10 , 0) = (0 1 % ERM (2y + 1 , 2x)) (2y + 1 , 0) = = 7 - (0 , 2 x + 1) = > Definition:TheSecondPrivatiTest:Suoa are continous b) b) D D(a ,b) fxx(a fyy(a 3) (gxy( - , = = , , is localminga If Exx1 a) and 3k 0 , then lab) a D < 0 , - x 0) = Think 2 = +y & 10 1 # b) IfDo and fthismaxof 9) ythendepointa a) If x* Dro , then the test is inconclusive 2= + * y 2= - xt - y z = xt - ya (local min & (0 , 0) (local maxd(00) (suddle & 2010 12 (or1R3) Definition Aregit : ion D in is closed pointit of boundary contains all its Definition Aregion Din (espisboundedpri : a closed not closed bounded : red d Definition functionfof 2varimbles : A hasan abo a D; (x,y) flaib) all in in this case is called an absolute , value of f Dat D We say f absolute max on. has an min on a regionin (nib) if f(x , ) - f(a , b) for (x,y) flaib) all in D ; this case , is called an absolute min value of fAbson D Closed & Bounded Method for finding & Max Min on : Region ① Find all critical points of finside D ② Find the extreme values ofandf on the boundary of D (by parametrizing the boundary using single-var cale ( optimization O Compare of fo points obtained the values the in and Steps 1 52. The largest is the abso max the smallest is abs of f on D the min.. Example of f(xy) X+y find abs maximin 2x : the = the closed G filled-in of & triangular region on vertices -2) is he point(s) & (2 , 0) , 10 , 2) g(0 , which these occur (110) ① fx = 2x - 2 = 0 =x = 1 (x , y) = Jy zy = = 0 y = = 0 - = X 2 y - o ②x - +82 O = 0 - 2 (0 , 0 , 37(01-2) ② f(0 y) y , = , 10 2) , ③ j'(0 y) = 2y 0 y = = 0 = , 2 - x , 0x - y = x) x+ ( x) xx[0 2) 10 , 2) f(x , , 2- = g'(x , 2 - x) = 2x 2x + 2x - 2(2 x) - - 4 - - 2 2 = 0 = x= z 4x - 4 = x12 y 2 0 = x - , 10 -2 2) 2x50 27(2 , -2) x= (x - , f(x = , x , 0 z('z -2) j(x 2) 2x + 2(x 2) = - 2 = 0 = , x x - - = , ③ fl0 flomaxf(, 0) = 0 = = fl MAfz gi = = 0 f(IO 14 8 Lagrange Multipliers. Method the to a abs of constraint Multipliers To object Lagrange : fi n glxy a , 2) = A follow the steps : ① Find (x 2) such that all points , y , z) x g(x z) f(x = = y y , , , , y , z) g(x , = F CD is called Lagrange 2) Multiplier of the (x , y, (this the amount the represents by which absolute external value with to K changes (the respect slight in allowance) a increase constraint ② (This up) : Find all points (x17 2) rarely comes 1 the u) & which constraint gly on = , y , 5 Vg(x z) = , y , ③ Compare of f values at obtained the the points in Steps 192. The largest is the abs max value & of f He smallest is the abs min value on k g(x y z) = , , Example : find lagrange Use abso multipliers to of max/min Circle occur. x2 + values y gy onea Optimize g(x y) x+ f(x(1y) y on 8 = y , = x - , = f(x y) y) (2x = 1) zy) g(x = , = , - = , , 1 x2x 3 2x ( = = = ① x = g(x y) S g(xy) k - y) = , =2 , = - 1 = x yy = x x+ y g(x g = = , x) Ex zyx= + = 5 - = x 2x2 5 y - = = =2 y x= = = = (2 , -2) (-2 , 2) ② Note that Eg(x y) , =

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