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ProblemSet2 - MAT235-CALCULUS II SUMMER 2009 ASSIGNMENT#2...

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MAT235-CALCULUS 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 = < 0 , 1 2 , 1 2 > and -→ b = < 0 , - 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 , 0), B (0 , 1 , 0) and C (0 , 0 , - 2). Let ( x 0 , y 0 , z 0 ) be the center of the sphere. Since the inside of the tetrahedron lies on the octant ( x > 0 , y > 0 , z < 0), we have x 0 > 0 , y 0 > 0 , z 0 < 0 . (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
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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 0 | = | y 0 | = | z 0 | = | 2 x 0 +2 y 0 - z 0 - 2 | 3 . This, together with (1), yields r = x 0 = y 0 = - z 0 = | 5 x 0 - 2 | 3 . It suffices now to solve the equation x 0 = | 5 x 0 - 2 | 3 , or 3 x 0 = | 5 x 0 - 2 | . The solutions to the last equation are x 0 = 1 and x 0 = 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 ( x - 1) 2 + ( y - 1) 2 + ( z + 1) 2 = 1 touches the four planes corresponding to the faces of the tetrahedron from outside. Problem 2 Vector functions and space curves.
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