Chapter 3: Functions of Several Variables
Section 3.3: Quadric Surfaces
Example 3.3.12
Put the equation 4x2+2y2−z2−8 x+4 y+4 z−2=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
Figures 3.3.12(a, b) each contains a graph of the surface defined by the given equation,
4x2+2y2−z2−8 x+4 y+4 z−2=0
whose standard form is
x−12+y+1222− z−2222=1
obtained by completing the square in x,y and z. The standard form is the equation of a single-sheeted hyperboloid with center 1,−1,2.
The level curves, drawn on the surface of the cone, are the ellipses
x−12c2−4 c+8/4+y+12c2−4 c+8/2=1
The cross sections x=c and y=c are hyperbolas, shown in Figures 3.3.12(a, b) where the sliders control the values of c. Indeed, if x=c, then the equation
y+122 c 2−c−z−224 c 2−c=1
defines hyperbolas in the yz-plane, seen in Figure 3.3.12(a). Likewise, the cross sections y=c are the hyperbolas
x−121−2 c−c2/2−z−2221−2 c−c2=1
defined in the xz-plane, and shown in Figure 3.3.12(b).
x = =
Figure 3.3.12(a) Single-sheeted hyperboloid with cross sections x=c
y = =
Figure 3.3.12(b) Single-sheeted hyperboloid with cross sections y=c
Maple Solution - Interactive
Obtain the standard formb
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.
4x2+2y2−z2−8 x+4 y+4 z−2=0→manipulate equation−14z−22+12y+12+x−12=1
Obtain the equivalent of the surfaces in Figures 3.3.12(a, b)
Context Panel: Plots≻Plot Builder≻3-D implicit plot
Set the ranges −1≤x≤3,−4≤y≤2,−2≤z≤6 style → surfacecontour
3-D Options≻grid → [25, 25, 25] scaling → constrained lightmodel → none
4x2+2y2−z2−8 x+4 y+4 z−2=0→
Maple Solution - Coded
Define f so that the graph of f=0 is a quadric surface
f≔4x2+2y2−z2−8 x+4 y+4 z−2:
Complete the square and put f into standard form
Student:-Precalculus:-CompleteSquaref=0,x,y,z
−z−22+2y+12+4x−12−4=0
−−4
−z−22+2y+12+4x−12=4
4
−14z−22+12y+12+x−12=1
Obtain the equivalent of the surfaces drawn in Figures 3.3.12(a, b)
plots:-implicitplot3df=0,x=−3..5,y=−6..4,z=−4..8,scaling=constrained,axes=frame,orientation=−50,60,0,style=surfacecontour,tickmarks=4,6,8,grid=15,15,15
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