Solutions To Cylinder, Quadric Surfaces Problems

1)  A surface configuration is described by  the following analytical aspects:

A= 1,  B = 1,  C = 2,  D = 1, E = 3,  F = 1

Write the equation for the object and identify it.

Solution:

Ax ²    +    B xy  +  C y ²   +  Dx  + Ey  + F   = 0

So:  x ²    +   xy  +  2 y ²   +  x  + 3y  + 1   = 0

When graphed using MathCad:


Discloses a parabolic cylinder.


2) Describe and sketch each of the following:
a)    x ²     +  y ²   =      a ²  

b)  x ²   +  y ²  +   z ²  +  4x    - 6y  = 3  

Solutions:

a) The graph is for the cross sectional area of a circular cylinder parallel to the z-axis, and sketched below (from the graphic shown in the blog post, changed coordinates)

b)  We let z = 0 in the equation then graph the remaining equation to obtain the circular cross section for what will be a sphere (e.g. in 3D):

From this MathCad plotted graph we infer the sphere will have a center at: h = -2, k = 3,  m = 0

And radius a = 4

To satisfy the eqn:   (x - h) ²    + (y - k ) ²   + (z  - m  ) ²   =  a ² 


We check, by substituting the values, h = -2, k = 3 in the sphere equation (with z = 0, m = 0), and radius a = 4, to get:

(x - (-2)) ²   +   (y - 3)  ²  =   16

=   (x  +  2) ²   +   (y - 3)  ²  =   16

So that:   (x ²   + 4x +  4)    + (y ²  - 6y +  9)    = 16 

Then:  x ²   + 4x  + y ²  - 6y  = 16  - 9   - 4  

x ²   +  4x  + y ²  - 6y  =  3   

Which is the original equation.  

For those lacking a program like MathCad a simpler graphing program (such as Graphmat) can be used to obtain the sphere center coordinates.  This can be done by graphing a form for the original equation (again z = 0) based on completing the square, e.g. for the   x ²   +  4x  part and the  y ²  - 6y  part.  Then:

  ( x ²   + 4x  +  4)  +    ( y ²  - 6y + 9) = 3  + 9  + 4  = 16

 (Since 9 and 4 are added to each side in the process).  Then on graphing the circle in the x-y plane, e.g. (x  +  2) ²   +   (y - 3)  ²  =   16,   we obtain:

Where the reader can see by inspection the center is at (-2, 3)  This then forms the circular cross section  in xy-plane for the sphere sketched below in a (somewhat) 3 D aspect:

Where x = -6 and x = 2 form the left & right limits along the x-dimension, and y = -1 to y = 7 form the limits along the y-axis. In each case confirming the radius r = 4.

3)  Sketch (and identify)  the surface given by:  

f(x,y) =  4   -  x ²  +  y ²

This can also be written (since we are discussing a 3D surface):

z =  4   -  x ²  +  y ²

When y = 0, the surface's trace on the x-z plane is y  =    4   -  x ²    and  is a parabola that opens up in the negative z direction.   In the xy plane we see:
When x =0,  the trace on the yz plane is:  z  =    4   +  y ²    and  opens up in the positive z direction.  The result is a hyperbolic paraboloid.


When one cuts through the plane such that z1 > o  a hyperbola is obtained, e.g.

The MathCad graph of the surface is: