math243_exam_1_solutions

# math243_exam_1_solutions - MATH 243 SECTION 016, SPRING...

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MATH 243 SECTION 016, SPRING 2008, EXAM I SOLUTIONS Question 1: (1) Page 833 no.36; (2) Page 833 no.35; (3) Page 874 no.44; (4) Page 882 no.36. Question 2: (1) Page 849 no.38; (2) Page 857 no.32. Question 3: (1) Page 867 no.60; (2) Page 866 no.37. Question 4: (2) Page 857 no.45 (c). 1

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2 ANALYTIC GEOMETRY AND CALCULUS C .Question 1 Write inequalities or equalities to describe the region or the surface. Expla- nations on how to ﬁnd those inequalities or equalities are required. (1) The solid rectangular box in the ﬁrst octant bounded by the plane x = 1 , y = 2 and z = 3 . (4pts) Solution: Because the box lies in the ﬁrst octant, each point must comprise only nonnegative coordinates. So inequalities describing the region are 0 x 1 , 0 y 2 , 0 z 3 . (2) The half-space consisting of all points to the left of xz - plane. (4pts) Solution: This describes all points with negative y - coordinates, that is , y < 0 . (3) The surface obtained by rotating the line x = 3 y about the x - axis. (6pts) Solution: The surface is a right circular cone with vertex at (0 , 0 , 0) and axis the x - axis. For x = k 6 = 0 , the trace is a circular with center ( k, 0 , 0) and radius r = y = x 3 = k 3 . Thus the equation is ( 1 3 x ) 2 = y 2 + z 2 . (4) The surface consists of all points P such that the distance from P to the plane y = 1 is twice the distance from P to the point (0 , - 1 , 0) . (6pts) Solution: The distance from a point P = ( x, y, z ) to the plane y = 1 is | y - 1 | , so the given condition becomes | y - 1 | = 2 p
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## This note was uploaded on 04/03/2008 for the course UNKNOWN Calculus taught by Professor Junhuwu during the Spring '08 term at University of Delaware.

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math243_exam_1_solutions - MATH 243 SECTION 016, SPRING...

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