# 12_6 - becomes y = z 2 So the equation of the revolution...

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12.6 21-28 Match the equation with its graph (labeled I-VIII). Give reasons for your choices. 21. VII. Because x 2 + 9 z 2 = 1 is an ellipse lying in the xz -plane. 22. IV. Because 9 x 2 + z 2 = 1 is an ellipse lying in the xz -plane. 23. II. Because x 2 ° y 2 = 1 is a hyperbola lying in the xz -plane which passes through (1 ; 0 ; 0) : 24. III. Because ° x 2 ° z 2 = 1 is an empty set. 25. VI. Because y = 2 x 2 and y = z 2 are parabolas lying in the xy -plane and the yz -plane, respectively. 26. I. Because y 2 = x 2 is a cross lying in the xy -plane. 27. VIII. Because the graph is independant of y: 28. V. Because y = x 2 ; y = ° z 2 are parabolas lying in the xy -plane and the yz -plane, respectively. Moreover, they open to opposite directions. 41. See the graph in page A114. 43. When we rotate the parabola y = x 2 about the y -axis with 90 ° , it becomes
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Unformatted text preview: becomes y = z 2 . So the equation of the revolution surface is y = x 2 + z 2 : 46. Let P = ( x; y; z ) . The distance from P to the x-axis is p y 2 + z 2 . The distance from P to the yz-plane is x . So the equation is p y 2 + z 2 = 2 x; i.e. 4 x 2 & y 2 & z 2 = 0 49. It su¢ ces to check c + 2( b & a ) t = ( b + t ) 2 & ( a + t ) 2 and c & 2( b + a ) t = ( b & t ) 2 & ( a + t ) 2 : This comes immediately from the relation c = b 2 & a 2 : 50. Multiply both sides of x 2 + 2 y 2 & z 2 + 3 x = 1 with 2 ; we have 2 x 2 + 4 y 2 & 2 z 2 + 6 x = 2 : Subtract 2 x 2 + 4 y 2 & 2 z 2 & 5 y = 0 from it, we get 6 x + 5 y = 2 ; which lies in the xy-plane. 1...
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