Towards the origin it gives the size and direction of

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towards the origin It gives the size and direction of the “steepest ascent” of z
where D is called ‘the discriminant’ of f . B2. 15 critical or turning points f has a ‘critical’ or ‘turning’ point wherever f = 0 (that is, all first partial derivatives of f are zero) A critical point r c of f ( r ) is a: maximum if f decreases in all directions minimum if f increases in all directions saddle if f increases in some directions and decreases in others (as r moves away from r c ) For a critical point at the origin : If f = 0 at r = 0 (that is, 0 is a critical point) then f = A + 1 2 f xx x 2 + f xy xy + 1 2 f yy y 2 + · · · provided f xx , f xy and f yy are evaluated at r c = 0 . If 1 2 f xx x 2 + f xy xy + 1 2 f yy y 2 can be factorised then f ( x, y ) has the form f = f (0 , 0) + 1 2 f yy ( y ax )( y bx ) + · · · which would describe a saddle at ( x, y ) = (0 , 0) . x y = B ± B 2 4 AC 2 A A B C Factorising 1 2 f xx x 2 + f xy xy + 1 2 f yy y 2 requires that D = f 2 xy 4( 1 2 f xx )( 1 2 f yy ) = f 2 xy f xx f yy > 0 ( i.e. B 2 4 AC > 0 ) B2. 16 derivative test in 2 dimensions If f ( x, y ) has a critical point at ( x, y ) = ( x c , y c ) , where f x = f y = 0 , the discriminant D = f xx f yy f 2 xy helps to identify a maximum, minimum or saddle: D < 0 = saddle D > 0 = maximum if f xx < 0 or f yy < 0 D > 0 = minimum if f xx > 0 or f yy > 0 The discriminant is the determinant of the Hessian matrix D = f xx f xy f xy f yy = f xx f yy f xy 2 (which is easily remembered) If D = 0 the derivative test does not help. Instead, we should examine changes near the point. Example : z = x 3 y 2
B2. 17 contour lines and f Contours of a function f ( r ) are defined by f ( r ) = const. (for di ff erent values of the constant) in 2 dimensions: for r R 2 contours are paths in 3 dimensions: for r R 3 contours are surfaces Examples : y = x , 2 y = 3 x f = ( y x )(2 y + 3 x ) f < 0 f > 0 f y x y x y = x 2 y = 3 x f 1 1 f f f = 2 y 2 + xy 3 x 2 f = 2 y 2 2 x 2 contours cross at a saddle contours encircle a max/min f = y 6 x, 4 y + x f = 4 x, 2 y f xx f xy f xy f yy = 6 1 1 4 = 25 f xx f xy f xy f yy = 4 0 0 2 = 8 f has the direction of fastest increase in f ( r ) it is orthogonal to the contours of f | f | varies inversely with distance between contours B2. 18 finding critical points Critical points of f ( r ) are found where f = 0 Example : Find the critical points of z = 1 2 x 2 x + 1 3 y 3 y and draw a rough sketch of its contours. rough sketch : remember contours can only cross at a saddle-point they encircle a maximum or a minimum (1, 1) y x (1, 1)
B2. 19 another critical point example Example : Find all critical points of f ( x, y ) = cos x sin y for 0 x 2 π and 0 y 2 π . Draw a rough sketch of the contours of f ( x, y ) . Critical points are where f x = f y = 0 , i.e. ( x, y ) (0 , 1 2 π ) ( π , 1 2 π ) (2 π , 1 2 π ) (0 , 3 2 π ) ( π , 3 2 π ) (2 π , 3 2 π ) D 1 1 1 1 1 1 f xx 1 1 1 type saddle min. saddle max. saddle max.

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