Midterm SOS 237 package

# Midterm SOS 237 package - MATH237 EXAM-AID SOS NIALL W...

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MATH237: EXAM-AID SOS NIALL W. MACGILLIVRAY (EDITOR); VINCENT CHAN (WRITER) October 31st, 2010 1 . ScalarFunctions ( 1 . 1 - 1 . 2 ) DEFINITION 1.1. Suppose A and B are sets. A function f is a rule that determines how a subset of A is associated with a subset of B , such that each element in A for which f is defined is sent to exactly one element in B . This subset of A for which f is defined is called the domain of f and denoted by D( f ), while the subset of B which is attained by f is called the range of f and denoted by R( f ). We typically denote f by f : A B . DEFINITION 1.2. f : R R is a scalar function, i.e. when the domain is a subset of R and the range is a subset of R . EXAMPLE 1. Define f : R 2 R by f ( x, y ) = x 2 + 2 y 2 - 2 x . Find the domain and range of f. SOLUTION. Clearly D( f ) = R 2 . For the range, notice x 2 + y 2 - 2 x x 2 - 2 x , which has a minimum at x = 1 (using what we know from 1-dimension). Then f ( x, y ) f ( - 1 , 0) = - 1. For any value z [0 , [, we can take x = 1 + 1 + z and y = 0. Indeed, this is a possible value for x we can solve for using z = x 2 - 2 x and the quadratic equation. Thus, R( f ) = [ - 1 , [. EXAMPLE 2. Define f : R 2 R by f ( x, y ) = | x | + | y | | x |-| y | . Find the domain and range of f . SOLUTION. We only need | x | 6 = | y | for f to be defined. The domain is then R 2 \ { ( x, y ) : | x | = | y |} . As for the range, notice we can obtain 1 and -1 by setting y = 0 and x = 0 respectively. Now, suppose for z R , we have f ( x, y ) = z . Then | x | + | y | = z | x |- z | y | , so that: (1 - z ) | x | = - (1 + z ) | y | We have already dealt with the case z = ± 1. Otherwise, | y | = - 1 - z 1+ z | x | . Since | x | , | y | ≥ 0, we have - 1 - z 1+ z 0. Then either both 1 - z and - 1 - z are positive, or both are negative. That is, we have z < - 1 or z > 1. 1

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W ATERLOO SOS E XAM -AID: MATH237 In either case, we can set x = 1 and y = 1 z 1+ z j x j to obtain z , so the range is (including the case z = 1 ) R ( f ) = ( 1 ; 1] [ [1 ; 1 ) . E XERCISE 1. Find the domain and range of the following functions. (a) f ( x; y ) = ln(1 j xy j ) . (b) g ( x; y ) = cos x sin y . (c) h ( x; y ) = p 1 x 2 y 2 It is difficult to sketch scalar functions from their definition alone. To help us, we look at level curves and cross-sections. D EFINITION 1.3. For f : R 2 ! R , we define the level curves of f to be the curves given by k = f ( x; y ) ; where k is a constant (in the range of f ). Sketching level curves is an exercise in solving for equations in two variables. The course notes cover three excellent examples, which are fairly standard. E XAMPLE 3. Define a function f by f ( x; y ) = x 2 + 2 y 2 2 x . Sketch the level curves and use them to sketch the surface z = f ( x; y ) . S OLUTION . From example 1, D ( f ) = R 2 and R ( f ) = [ 1 ; 1 ) . For values of k 1 , we examine x 2 + 2 y 2 2 x = k . Recall that the equation ( x h ) 2 a 2 + ( y k ) 2 b 2 = 1 determines an ellipse with major axis length 2 a , minor axis length 2 b , shifted h units right and k units up. This formula is a more general form of what we have: since there are no y terms, we will have k = 0 and since there is a 2 x term, we will want h = 1 (to get ( x 1) 2 = x 2 2 x + 1 ). Then if k 6 = 1 , x 2 + 2 y 2 2 x = k x 2 2 x + 1 + 2 y 2 = k + 1 ( x 1) 2 k + 1 + 2 y 2 k + 1 = 1 ( x 1) 2 p k + 1 2 + y 2 q k +1 2 2 = 1 : Thus, x 2 + 2 y 2 2 x = k describes an ellipse with major axis length 2 p k + 1 , minor axis length 2 q k +1 2 = p 2( k + 1) , shifted 1 unit right. If
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