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Unformatted text preview: MAT235CALCULUS II SUMMER 2009 ASSIGNMENT #2 SOLUTIONS Problem 1 Planes and surfaces. a) Identify the quadric surface consisting of all points ( x,y,z ) that satisfy the equation: x + z 2 + y 2 + 6 yz = 0 . Sketch the traces that are relevant to the surface and use them to sketch the surface. Note: It is recommended that you sketch the surface with different color from the one that you sketch the axis. Solution: We can write the surface as x = ( y z ) 2 2( y + z ) 2 = 2( z y √ 2 ) 2 4( y + z √ 2 ) 2 . We let w = z y √ 2 , and v = y + z √ 2 . Note that < x,y,z > = x→ i + y→ j + z→ k = x→ i + v→ a + w→ b , where→ a = < , 1 √ 2 , 1 √ 2 > and→ b = < , 1 √ 2 , 1 √ 2 > . Observe that→ i ,→ a ,→ b are mutually orthogonal unit vectors and that→ i ×→ a =→ b . Therefore, the variables x,w and v form a new, orthogonal coordinate system. In these new coordinates, our surface is x = 2 w 2 4 v 2 . Traces: w = k , then x = 2 k 2 4 v 2 , parabolas concave on the opposite side of x axis. v = k , then x = 2 w 2 4 k 2 , parabolas concave on the same side as the x axis. x = k , 2 w 2 4 v 2 = k . If k = 0, then w = ± √ 2 v . If k 6 = 0 then the traces are hyperbolas. Therefore the surface is a hyperbolic paraboloid. For a sketch of the traces and the surface look at figure 1. b) Find the largest sphere that lies inside the tetrahedron whose vertices are the origin and the points A (1 , , 0), B (0 , 1 , 0) and C (0 , , 2). Let ( x ,y ,z ) be the center of the sphere. Since the inside of the tetrahedron lies on the octant ( x > ,y > ,z < 0), we have x > ,y > ,z < . (1) As the sphere we are looking for is the largest sphere lying inside the tetrahedron, it touches each face of the tetrahedron; its distance from each of the faces is equal to its radius r . 1 2 One can easily check that the faces of the tetrahedron are the planes, x = 0 ,y = 0 ,z = 0 and 2 x + 2 y z = 2 (which is the plane that passes through A,B and C ). The formula for the distance from a point to a plane implies r =  x  =  y  =  z  =  2 x +2 y z 2  3 . This, together with (1), yields r = x = y = z =  5 x 2  3 . It suffices now to solve the equation x =  5 x 2  3 , or 3 x =  5 x 2  . The solutions to the last equation are x = 1 and x = 1 4 and the corresponding points are (1 , 1 , 1) and ( 1 4 , 1 4 , 1 4 ). Observe that (1 , 1 , 1) satisfies the inequality 2 x + 2 y z > 2, however the origin satisfies the inequality 2 x + 2 y z < 2. Therefore, these two points lie on opposite sides of the plane 2 x +2 y z = 2; as the origin is the vertex of the tetrahedron that is opposite to the face lying in the plane 2 x +2 y z = 2, this means that (1 , 1 , 1) lies outside the tetrahedron. Therefore, the other solution, ( 1 4 , 1 4 , 1 4 ), must be the center of our sphere. Consequently, the radius of the sphere is 1 4 and its equation is ( x 1 4 ) 2 + ( y 1 4 ) 2 + ( z + 1 4 ) 2 = 1 16 . (2) Remark: The sphere (...
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This note was uploaded on 07/18/2010 for the course MATH MAT235 taught by Professor Recio during the Summer '08 term at University of Toronto Toronto.
 Summer '08
 Recio
 Calculus, Multivariable Calculus

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