advanced calculus
⑧
12.6. Cylinders and quadric surfaces
cylinders
A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass
through a given plane curve.
Sketch the the surface x
e.g. graph of z =
does notinvolve
equation y
·
·
any vertical plane with
equation y = K intersects the
graph in the curve moves
->
parallel along xz-plane
·
surface name: parabolic cylinder
rulings of cylinder parallel to
y-axis
·
are
if variable the surface is
rulings parallel to * one
(/y(z) is
missing, a
y-axis and
pass cylinder
through curve
e.g.(2 y2 + 1
=
·
circle with radius 1(z
of =k)
·
parallel to
cy plane
2
5.9.y2 z
+
1
=
radius k) NB:both these
circle with of 1(x
equations representa cylinder,
·
*
=
parallel this cylinder is
to
yz-plane Not
a circle. The trace of
·
a circle with z 0.
=
, surfaces,
quadric have the same characteristics as conic sections
A quadric surface is the graph of a second-degree equation in three variables x, y, and z. The
most general such equation is:
* Ax2
By (z2 Dxy Eyz fxz 6x Hy Iz 5 0
+
+
+ + + + +
+ + =
where A-5 are constants, butby translation and rotation can be boughtinto
I standard forms:
I
Ax2 Byz (z 2 5 0 or where C
A, B, O
· + +
=
+
=
Ax2 By2 Iz 0
+
· + =
e.g. Use traces to sketch the
quadric surface:( + YE +
=
· substitute z 0 = trace
=> in cy plane is x+
Y 1)
=
Ellipse
Horizontal in z K is:
trace plane
·
=
x2 y) 1
k2; z
k) ellipse provided 1
k 0 -> 12 4
- -
<
+ =
=
9 ↳
-> -
2<k < 2
and ellipses:
·vertical traces
parallel to
yz xz
planes are also
y2 z
+
1
=
-
k2;x kif = -
1ckx )
9 H
x2 z2 k2 if 3xk< 3
1
i( k
-
t
-
=
=
4 4
Z vertical:yz Z vertical:xz
Y Horizontal:
M xy M M
3 2
2
>x
>x
-y
C C C
- I I -
3 3 -
I I
-
2
-
3
2
W W W
combine these form ellipsoid (all in IR5:
we can traces to an 3 traces are
ellipses)
s
each trace is symmetric to each coordinate plane,
because there are only even
powers of and Z
x, y
, surface:(another def.)
TRACE
ofa
quadric
curve obtained by taking an intersection a
of
plane parallel to a coordinate plane (cy/zy/yz
plane) and the surface.
quadric
e.g. Determine the shape of the surface
[(1,4,z) G(R3: y2
+
z)
=
solution set
solution
① Find the traces where z =
K, KEIR Horizontal trace:Ellipse
vertical trace:Parabola
consider?+Y=K in a
plane (2D) <
Elliptic parabaloid
obtain solutions. know itdoes the
If
so, we no
So, we cross
not
negative Z axis.
k
If 0,x y
=
=
0
=
is the
only solution. y
a
If k >0:x2 yz
+ k
=
cy-plane
16
x2 k gets larger, ellipse
Y
- + =I AS
16k grows
I
() x2 +
y2 = j >x
L
I
(4) (r)
2
0
-
L
~
I
How does the ellipse grow? N
2
xz
plane x
y2 z
+ =
-
M
k2 = 116
k1
plane Z =
yz
-
A
N A
1 Set K
a y
=
Let x
k,
= then if k 0
=
z x2
2
1
y2
+
z
= =
I0 =
16 16
R
k 'x
=
ak 0 =
L -
>x
R
ifk 0
=
..
2 W
z
y
=
if k 1
=
↑ ·
=z
i y
+
W
c
, revision ofconic sections (10.5)
intersection a
of cone with a plane
⑪se
x2 + y2 = I a vertices: I a
92 b2
and Ib
⑭erbola xand
(*and-y:
+
y:
-
3
x -
1
=
>vertices: I a
92
assymptotes: x x
⑳bola
=
x2 OR Y2:AKI OR c2 =
dy>
b2
a = Lip
focus
Use traces to sketch z 4x2 yz
e.g.
+
=
·
substituting =
0, we
get: y2 = z
the
Thus, a
parabola parallel to
ccy-plane
·
FOr x K: =
z
412 z slice the parallel the yz-plane,
y => if we
graph with
any plane
to
+
2
=
yZ
have direction.
we a
positive parabola opening up in the upward
FOr y K:
xz/z
=
=4x + k2
parabola
= that
opens upward (steeper)
·
FOr z K:
=
cy( y2 4=
+ k ellipse
=
iff k>0
e.g.
·
fOr
Sketch
x K:
=
z
yz
=
-
FOr
x2
z K!= FOr
y K:
=
-pr
z =
yz
-
k2 k yz
=
-
x2 z x2
=
-
x2
42
a A xy xZ
yperbolic paraboloid
⑧
12.6. Cylinders and quadric surfaces
cylinders
A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass
through a given plane curve.
Sketch the the surface x
e.g. graph of z =
does notinvolve
equation y
·
·
any vertical plane with
equation y = K intersects the
graph in the curve moves
->
parallel along xz-plane
·
surface name: parabolic cylinder
rulings of cylinder parallel to
y-axis
·
are
if variable the surface is
rulings parallel to * one
(/y(z) is
missing, a
y-axis and
pass cylinder
through curve
e.g.(2 y2 + 1
=
·
circle with radius 1(z
of =k)
·
parallel to
cy plane
2
5.9.y2 z
+
1
=
radius k) NB:both these
circle with of 1(x
equations representa cylinder,
·
*
=
parallel this cylinder is
to
yz-plane Not
a circle. The trace of
·
a circle with z 0.
=
, surfaces,
quadric have the same characteristics as conic sections
A quadric surface is the graph of a second-degree equation in three variables x, y, and z. The
most general such equation is:
* Ax2
By (z2 Dxy Eyz fxz 6x Hy Iz 5 0
+
+
+ + + + +
+ + =
where A-5 are constants, butby translation and rotation can be boughtinto
I standard forms:
I
Ax2 Byz (z 2 5 0 or where C
A, B, O
· + +
=
+
=
Ax2 By2 Iz 0
+
· + =
e.g. Use traces to sketch the
quadric surface:( + YE +
=
· substitute z 0 = trace
=> in cy plane is x+
Y 1)
=
Ellipse
Horizontal in z K is:
trace plane
·
=
x2 y) 1
k2; z
k) ellipse provided 1
k 0 -> 12 4
- -
<
+ =
=
9 ↳
-> -
2<k < 2
and ellipses:
·vertical traces
parallel to
yz xz
planes are also
y2 z
+
1
=
-
k2;x kif = -
1ckx )
9 H
x2 z2 k2 if 3xk< 3
1
i( k
-
t
-
=
=
4 4
Z vertical:yz Z vertical:xz
Y Horizontal:
M xy M M
3 2
2
>x
>x
-y
C C C
- I I -
3 3 -
I I
-
2
-
3
2
W W W
combine these form ellipsoid (all in IR5:
we can traces to an 3 traces are
ellipses)
s
each trace is symmetric to each coordinate plane,
because there are only even
powers of and Z
x, y
, surface:(another def.)
TRACE
ofa
quadric
curve obtained by taking an intersection a
of
plane parallel to a coordinate plane (cy/zy/yz
plane) and the surface.
quadric
e.g. Determine the shape of the surface
[(1,4,z) G(R3: y2
+
z)
=
solution set
solution
① Find the traces where z =
K, KEIR Horizontal trace:Ellipse
vertical trace:Parabola
consider?+Y=K in a
plane (2D) <
Elliptic parabaloid
obtain solutions. know itdoes the
If
so, we no
So, we cross
not
negative Z axis.
k
If 0,x y
=
=
0
=
is the
only solution. y
a
If k >0:x2 yz
+ k
=
cy-plane
16
x2 k gets larger, ellipse
Y
- + =I AS
16k grows
I
() x2 +
y2 = j >x
L
I
(4) (r)
2
0
-
L
~
I
How does the ellipse grow? N
2
xz
plane x
y2 z
+ =
-
M
k2 = 116
k1
plane Z =
yz
-
A
N A
1 Set K
a y
=
Let x
k,
= then if k 0
=
z x2
2
1
y2
+
z
= =
I0 =
16 16
R
k 'x
=
ak 0 =
L -
>x
R
ifk 0
=
..
2 W
z
y
=
if k 1
=
↑ ·
=z
i y
+
W
c
, revision ofconic sections (10.5)
intersection a
of cone with a plane
⑪se
x2 + y2 = I a vertices: I a
92 b2
and Ib
⑭erbola xand
(*and-y:
+
y:
-
3
x -
1
=
>vertices: I a
92
assymptotes: x x
⑳bola
=
x2 OR Y2:AKI OR c2 =
dy>
b2
a = Lip
focus
Use traces to sketch z 4x2 yz
e.g.
+
=
·
substituting =
0, we
get: y2 = z
the
Thus, a
parabola parallel to
ccy-plane
·
FOr x K: =
z
412 z slice the parallel the yz-plane,
y => if we
graph with
any plane
to
+
2
=
yZ
have direction.
we a
positive parabola opening up in the upward
FOr y K:
xz/z
=
=4x + k2
parabola
= that
opens upward (steeper)
·
FOr z K:
=
cy( y2 4=
+ k ellipse
=
iff k>0
e.g.
·
fOr
Sketch
x K:
=
z
yz
=
-
FOr
x2
z K!= FOr
y K:
=
-pr
z =
yz
-
k2 k yz
=
-
x2 z x2
=
-
x2
42
a A xy xZ
yperbolic paraboloid
