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20Chw3

# 20Chw3 - the ones above would also give you 2/2 14.2 14(6...

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MATH 20C graded homework 3. (18+2 points) 14.1 # 28 (6 points): Describe the vertical and horizontal traces of f ( x, y ) = 9 - x 2 - y 2 and sketch the graph. Solution: The horizontal trace at height z = c has equation x 2 + y 2 = 9 - c. If c < 9, this is a circle of radius 9 - c , so that as c decreases, the radius increases. If c = 9, this is a point (with coordinates (0 , 0 , 9)). There are no traces when c > 9. The vertical trace in the plane x = a is the parabola z = (9 - a 2 ) - y 2 . It opens downward with vertex at ( a, 0 , 9 - a 2 ). The vertical trace in the plane y = b is the parabola z = (9 - b 2 ) - x 2 . It opens down with vertex at (0 , b, 9 - b 2 ). [The sketch of the surface should look like a paraboloid opening downwards with vertex at (0 , 0 , 9).] About the grading this week: I’m looking for a detailed description of the traces. Pictures are worth a thousand words, so brief sketches of the traces indicating the radii of level curves or vertices of parabolas got you 2/2. Detailed verbal explanations like

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Unformatted text preview: the ones above would also give you 2/2. 14.2 # 14 (6 points): Evaluate the limit or determine that it does not exist. lim ( x,y ) → (0 , 0) x 2 + y 2 1 + y 2 . Solution: The function x 2 + y 2 1 + y 2 is continuous everywhere in the xy-plane because it is a rational function whose denominator 1 + y 2 is never zero. The limit can be evaluated by substitution lim ( x,y ) → (0 , 0) x 2 + y 2 1 + y 2 = 1 = 0 . 14.2 # 34 (6 points): Show that lim ( x,y ) → (0 , 0) y 2 x 2 + y 2 does not exist. Solution : Let f ( x,y ) = y 2 x 2 + y 2 . Approaching (0 , 0) along the line y = mx , lim ( x,mx ) → (0 , 0) f ( x,mx ) = lim x → m 2 x 2 x 2 + m 2 x 2 = m 2 1 + m 2 . Since the limit of f ( x,y ) depends on the direction from which ( x,y ) approaches (0 , 0), then the limit does not exist....
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20Chw3 - the ones above would also give you 2/2 14.2 14(6...

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