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Unformatted text preview: Math 222 Prelim I Solutions Spring 2008 1. Let S be the level surface f ( x, y, z ) = x 2 + 3 y 2 + z 2 + xz = 6. (5 pts) (a) Find the tangent plane to S at the point (1 , 1 , 1). Solution . We have f = (2 x + z, 6 y, 2 z + x ), and at (1 , 1 , 1) this equals (3 , 6 , 3). The tangent plane is therefore 3( x 1) + 6( y 1) + 3( z 1) = 0. This can be simplified to x + 2 y + z = 4 if you want. (5 pts) (b) Find all points on S where the tangent plane is horizontal (parallel to the xyplane). Solution . The tangent plane is horizontal where the normal vector f is vertical, i.e., of the form (0 , , z ). Thus we have the conditions f x = 2 x + z = 0 and f y = 6 y = 0, so y = 0 and z = 2 x . Plugging y = 0 and z = 2 x into x 2 + 3 y 2 + z 2 + xz = 6 gives x 2 + 4 x 2 2 x 2 = 6, which has the solutions x = 2. Since y = 0 and z = 2 x we get the two points ( 2 , , 2 2) where the tangent plane is horizontal. (5 pts) (c) Show that the points on S where the tangent plane is vertical (parallel to a plane containing the zaxis) form an ellipse in some sloping plane. Solution . Vertical tangent planes occur where f is horizontal, of the form ( x, y, 0) so we get the condition f z = 0. This gives the plane x + 2 z = 0, or x = 2 z . Plugging x = 2 z into the equation x 2 + 3 y 2 + z 2 + xz = 6 gives 4 z 2 + 3 y 2 + z 2 2 z 2 = 3 y 2 + 3 z 2 = 6, or y 2 + z 2 = 2. This is a cylinder (with axis the xaxis), and intersecting this cylinder with the plane x = 2 z gives an ellipse in this plane. 2. Let f ( x, y ) = x 2 + y 2 2 x . (5 pts) (a) Show that the level curves of f are circles passing through the origin....
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This note was uploaded on 04/19/2010 for the course MATH 2220 taught by Professor Parkinson during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 PARKINSON
 Math, Multivariable Calculus

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