erte - 4(a f x(2 1 = 2 f y(2 1 =-5 f(2 1 = 1 so z = 1 2...

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MATH 241H - EXAM 1 , September 29, 2003 - Solutions 1. (a) a × b (b) ( a · b / a · a ) a (c) ( k a k / k b k ) b 2. The plane is given by (3 , - 2 , 0) + s (2 , 1 , 1) + t (1 , - 1 , - 3) so P 0 = (3 , - 2 , 0) is a point and a = (2 , 1 , 1) and b = (1 , - 1 , - 3) are vectors in the plane. To get a normal vector we take n = a × b = ( - 2 , 7 , - 3). Thus the equation is - 2( x - 3) + 7( y + 2) - 3( z - 0) = 0. 3. (a) Plug in to get f (1 , 0) = 2. (b) Doesn’t exist because limits are different from different directions. For example lim x 0 f ( x, 0) = 2 and lim y 0 f (0 , y ) = 3. (c) No because limit does not exist. (d) f ( x, 0) = 2 if x 6 = 0 and f (0 , 0) = 3. Thus f ( x, 0) is not continuous at x = 0 so f x (0 , 0) does not exist. (e) f (0 , y ) = 3 for all y so f y (0 , 0) = 0.
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Unformatted text preview: 4. (a) f x (2 , 1) = 2 , f y (2 , 1) =-5 , f (2 , 1) = 1 so z = 1 + 2( x-2)-5( y-1). (b) f (2 . 01 , . 99) ∼ 1 + 2(2 . 01-2)-5( . 99-1) = 1 . 07. 5. ( a ) Df ( x, y ) = 2 x-y 2-2 xy 2 y 1 3 ( b ) D ( f ◦ g )(3 ,-1) = Df (2 , 1) Dg (3 ,-1) = 3-4 2 1 3 ± 1-2 2-1 ² = -5-2 4-2 7-5 6. (a) ∇ f (1 , 2 ,-1) = (3 ,-12 , 12) so D v f (1 , 2 ,-1) = (3 ,-12 , 12) · (1 / 3 ,-2 / 3 , 2 / 3) = 17. (b) 3( x-1)-12( y-2) + 12( z + 1) = 0....
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This note was uploaded on 04/04/2010 for the course MATH 113 taught by Professor Staff during the Spring '08 term at Maryland.

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