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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 = < , 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 (...
View Full Document

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.

Page1 / 15

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online