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Solutions-Week6

Solutions-Week6 - Homework Week 6 Solutions 14.1.3...

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Homework Week 6 Solutions 14.1.3 Substituting ( x, y, z ) = (3 , 8 , 2) and ( x, y, z ) = (3 , - 2 , - 6) in the function, we obtain h (3 , 8 , 2) = 3 · 8 · 2 - 2 = 3 · 8 · 1 4 = 6 h (3 , - 2 , - 6) = 3 · ( - 2) · ( - 6) - 2 = - 6 · 1 36 = - 1 6 14.1.7 The function is defined if y - 2 x > 0 or y > 2 x . This is the region in the xy -plane that is above the line y = 2 x . D = { ( x, y ) : y > 2 x } x y y = 2 x y > 2 x 14.1.11 The function is defined for all x = 0. The domain is the xy -plane with the y -axis excluded. D = { ( x, y ) : x = 0 } x y x = 0 14.1.20 (a) | x | + | y | . The level curves are | x | + | y | = c , y = c - | x | , or y = - c + | x | . The graph (D) corresponds to the function with these level curves. (b) cos( x - y ). The vertical trace in the plane x = c is the curve z = cos( c - y ) in the plane x = c . These traces correspond to the graph (C). (c) - 1 1 + 9 x 2 + y 2 (e) - 1 1 + 9 x 2 + 9 y 2 . The level curves of the two functions are: - 1 1+9 x 2 + y 2 = c - 1 1+9 x 2 +9 y 2 = c 1 + 9 x 2 + y 2 = - 1 c 1 + 9 x 2 + 9 y 2 = - 1 c 9 x 2 + y 2 = - 1 - 1 c 9 x 2 + 9 y 2 = - 1 - 1 c x 2 + y 2 = - 1+ c 9 c For suitable values of c , the level curves of the function in (c) are ellipses as in (E), and the level curves of the function (e) are circles as in (A). 1
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(d) cos( x 2 ) e - 1 / ( x 2 + y 2 ) (f) cos( x 2 + y 2 ) e - 1 / ( x 2 + y 2 ) . The value of | z | is decreasing to zero as x or y are decreasing, hence the possible graphs are (B) and (F). In (f), z is constant whenever x 2 + y 2 is constant, that is, z is constant whenever ( x, y ) varies on a circle. Hence (f) corresponds to the graph (F) and (d) corresponds to (B). To summarize, we have the following matching: (a) (D) (b) (C) (c) (E) (d) (B) (e) (A) (f) (F) 14.1.22 The graph is shown in the figure: 1 2 1 0 0.5 0 0 1 0.5 - 1 - 0.5 - 0.5 - 1 - 1 y x z The horizontal trace at height c is x 2 + y = c or y = - x 2 + c . This is a parabola in the plane z = c . 5 20 10 0 2.5 0 0 5 2.5 - 5 - 2.5 - 2.5 - 5 y x z The vertical trace obtained by setting x = a is the line z = y + a 2 in the plane x = a , and the vertical trace obtained by setting y = a is the parabola z = x 2 + a in the plane y = a .
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