Midterm SOS 237 package

Midterm SOS 237 package

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Unformatted text preview: MATH237: EXAM-AID SOS NIALL W. MACGILLIVRAY (EDITOR); VINCENT CHAN (WRITER) 1. October 31st, 2010 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 : R2 → R by f (x, y ) = x2 + 2y 2 − 2x. Find the domain and range of f. SOLUTION. Clearly D(f ) = R2 . For the range, notice x2 +y 2 −2x ≥ x2 − 2x, 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 = x2 − 2x and the quadratic equation. Thus, R(f ) = [−1, ∞[. EXAMPLE 2. Define f : R2 → R by f (x, y ) = |x |+ |y | |x|−|y | . Find the domain and range of f . SOLUTION. We only need |x| = |y | for f to be defined. The domain is then R2 \ {(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 |x|. Since |x|, |y | ≥ 0, 1+z 1−z we have − 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 In either case, we can set . and WATERLOO SOS E XAM -AID: MATH237 to obtain , so the range is (including the case ) E XERCISE 1. Find the domain and range of the following functions. (a) (b) (c) 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 , we define the level curves of to be the curves given by . . where is a constant (in the range of ). 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 the surface . S OLUTION. From example 1, . Recall that the equation by . Sketch the level curves and use them to sketch and . For values of , we examine determines an ellipse with major axis length , minor axis length , shifted units right and units up. This formula is a more general form of what we have: since there are no terms, we will have and since there is a term, we will want (to get ). Then if , Thus, describes an ellipse with major axis length , minor axis length , shifted 1 unit right. If , we get just the single point , and this is the exceptional level curve. We can thus generate the image of , and we get a surface called an elliptical paraboloid. 2 WATERLOO SOS E XAM -AID: MATH237 E XERCISE 2. Sketch the level curves and use them to sketch the surface . D EFINITION 1.4. For , we define the cross-sections of or to be the curves given by where are constants. They correspond to the intersection of the surface with the vertical planes respectively. , E XAMPLE 4. Define a function by yield , . Sketch the cross-sections. while the cross-sections yield S OLUTION. The cross-sections . Sketching these for E...
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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