student lecture 31 quadric surfaces

Upload: uploadingperson

Post on 14-Oct-2015

7 views

Category:

Documents


0 download

DESCRIPTION

.

TRANSCRIPT

  • LECTURE 31

    QUADRIC SURFACES

    When a standard quadric surface

    x2

    a2 y

    2

    b2 z

    2

    c2= 1

    is rotated in space mixed terms xy, xz and yz appear in the equation. It is possible to expressthe resulting quadratic form in matrix form xTAx = 1 where A is a symmetric matrix. Ananalysis of the eigenvectors and eigenvalues of A will reveal both the structure and the principalaxes of the surface.

    Consider the curve 2x24xy+5y2 = 54. The presence of the mixed term xy indicatesthat this is a standard object (ellipse or hyperbola) which has been tilted to some degreeso that its major and minor axes no longer point in the x and y directions. To under-stand the curve we need to apply a specific transformation which untilts the curve intostandard form. Our first step is to rewrite the quadratic form in terms of matrices.

    Claim: 2x2 4xy + 5y2 = 54 is equivalent to ( x y )(

    2 22 5

    )(x

    y

    )= 54

    Proof:

    This gets us over into the arena of matrices where the theory of eigenvalues andeigenvectors may be brought into play! Observe that by its nature of construction, thematrix A will be symmetric and thus its eigenvectors will be naturally orthogonal to eachother. We now undertake a complete eigen-analysis of A.

    1

  • 2

  • So the eigenvalues of A are 1 and 6 with associated unit eigenvectors

    25

    15

    and

    1

    5

    25

    respectively. Observe that the eigenvectors are indeed orthogonal.These di-

    rections X =

    25

    15

    and Y =

    1

    5

    25

    actually form the principal axes of the

    curve. That

    is the curve sits properly on the eigenvectors of A. We will prove this formally in the nextlecture. Furthermore in the {X, Y } system the equation of the curve is 1X2 + 6Y 2 = 54.(Note the use of the eigenvalues). We can now identify the curve as an ellipse whoseclosest point to the origin is 3 units in the Y direction. Thus the closest points to the

    origin are

    3

    5

    65

    in the {x, y} system. The transformation which interrelates the

    two coordinate systems is

    (x

    y

    )= P

    (X

    Y

    )where P is the usual matrix of eigenvectors

    25

    15

    15

    25

    . Sketch:

    In summary:The quadratic form xTAx where A is a symmetric matrix has principal axes given by

    the orthogonal eigenvectors of A and the associated quadratic curves and quadric surfacesmay be transform into standard objects with the eigenvalues as coefficients.

    3

  • Example 1 Express the equation of the surface

    x2 + 2y2 + 2z2 + 4xy 4xz + 6yz = 30

    in terms of its principal axes X , Y and Z and hence determine the nature of the surface.

    Find an orthogonal matrix P implementing the transformation through

    xy

    z

    = P

    XY

    Z

    .

    Deduce the shortest distance from the surface to the origin and the

    xy

    z

    coordinates

    of these closest point(s).

    The equation may be written as(x y z

    ) 1 2 22 2 32 3 2

    xy

    z

    = 30

    Now the matrix A =

    1 2 22 2 32 3 2

    has eigenvalues -3, 3 and 5 with associated eigen-

    vectors

    11

    1

    , 21

    1

    and

    01

    1

    . Observe the orthogonality of the eigenvectors!

    4

  • P =

    13

    26

    0

    13

    16

    12

    13

    16

    12

    , 3X2 + 3Y 2 + 5Z2 = 30

    Hyperboloid of one sheet with closest distance to the origin of6 at

    03

    3

    We will start the next lecture with a formal proof of these algorithms and results.

    31You can now do Q 91 and 92

    5

    skaeLine