Chapter 3: Functions of Several Variables
Section 3.3: Quadric Surfaces
Example 3.3.4
Put the equation 3 x2− 6 x− y+ 1=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 3.3.4(a, b) contain a graph of the surface defined by the given equation,
3 x2− 6 x− y+ 1=0
whose standard form is
y+2=⁢3x−12
obtained by completing the square in x. The standard form is the equation of a parabolic cylinder.
The point 1,−2 would be the vertex of the parabola in the xy-plane.
The level curves, drawn on the surface of the cylinder, are parabolas.
The cross sections x=c are vertical lines y=3c−12−2 in the yz-plane.
The cross sections y=c are pairs of vertical lines x=1 ±c+2/3 in the xz-plane.
x = =
Figure 3.3.4(a) Parabolic cylinder with cross sections x=c
y = =
Figure 3.3.4(b) Parabolic cylinder with cross sections y=c
Maple Solution - Interactive
Obtain the standard form
Control-drag the given equation. Press the Enter key.
Context Panel: Complete Square≻x
Select −y−2 and use the Smart Pop-Up to subtract this from both sides.
3⁢x2−6⁢x−y+1=0
= complete square
3⁢x−12−y−2=0
→add y+2 to both sides
3⁢x−12=y+2
Unfortunately, Maple 2020 is unable to draw the required graph interactively.
Maple Solution - Coded
Define f so that the graph of f=0 is a quadric surface
f≔3 x2− 6 x− y+ 1:
Complete the square and put f into standard form
Student:-Precalculus:-CompleteSquaref=0,x
−−y−2
rhs=lhs
y+2=3⁢x−12
Obtain the equivalent the surface in Figures 3.3.4(a, b)
plots:-implicitplot3df=0,x=0..2,y=−2..1,z=0..5,scaling=constrained,style=surfacecontour,axes=frame,orientation=50,75,0,tickmarks=3,4,5
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