Chapter 3: Functions of Several Variables
Section 3.3: Quadric Surfaces
Example 3.3.3
Put the equation z2−4 x2−4y2−4 z+8 x−8 y−8=0 into standard form for a quadric surface, identify the surface, draw its graph, and discuss the nature of the level curves and plane sections.
Solution
Mathematical Solution
Figure 3.3.3(a) is a graph of the surface defined by the given equation,
z2−4 x2−4y2−4 z+8 x−8 y−8=0
whose standard form is
−x−12−y+12+z−2222=1
obtained by completing the square in x,y, and z. The standard form is the equation of a two-sheeted hyperboloid.
The point 1,−1,2 is the center of the hyperboloid.
The level curves, drawn on the surface of the hyperboloid, are circles. The cross sections x=c and y=c are hyperbolas, shown in Figure 3.3.3(b) where the slider controls the value of c. Indeed, if x=c, then the equation
z−224c2−2 c+2−y+12c2−2 c+2 =1
defines hyperbolas in the yz-plane. Likewise, the cross sections y=c are also hyperbolas, but in the xz-plane:
z−224c2+2 c+2−x−12c2+2 c+2 =1
plots:-implicitplot3d(-(x-1)^2-(y+1)^2+(z-2)^2/4-1=0,x=-3..5,y=-5..3,z=-4..8,scaling=constrained,axes=frame,orientation=[-50,60,0],style=surfacecontour,tickmarks=[4,4,6],grid=[25,25,25]);
Figure 3.3.3(a) Two-sheeted hyperboloid
x = =
Figure 3.3.3(b) Cross sections x=c
Maple Solution - Interactive
Obtain the standard form
Control-drag the given equation.
Context Panel: Manipulate Equation
Check the "Show steps stacked vertically" box.
Click the "Complete the square" button.
Add to both sides and multiply both sides as per the actions shown in the figure below.
Click the "Return Steps" button.
z2−4 x2−4y2−4 z+8 x−8 y−8=0→manipulate equation14z−22−y+12−x−12=1
Obtain the equivalent of Figure 3.3.3(a)
Context Panel: Plots≻Plot Builder≻3-D implicit plot
Set the ranges −3≤x≤5,−5≤y≤3,−4≤z≤8 style → surfacecontour
3-D Options scaling → constrained grid → [25, 25, 25] lightmodel → none
z2−4 x2−4y2−4 z+8 x−8 y−8=0→
Maple Solution - Coded
Define f so that the graph of f=0 is a quadric surface
f≔z2−4 x2−4y2−4 z+8 x−8 y−8:
Complete the square and put f into standard form
Student:-Precalculus:-CompleteSquaref=0,x,y,z
z−22−4y+12−4x−12−4=0
+4
z−22−4y+12−4x−12=4
4
14z−22−y+12−x−12=1
plots:-implicitplot3df=0,x=−3..5,y=−5..3,z=−4..8,style=surfacecontour,scaling=constrained,grid=25,25,25,axes=frame,tickmarks=4,4,6,orientation=−50,60,0
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