Solutions to Assignment #05 –MATH 2421
Kawai
Section 8.6
[NEW] Classify the surfaces.
Surface
y2
z=4
z=
x2
r
x2 y 2
+
4
9
y2 +
z2
=1
4
Name
Additional Info?
Parabolic cylinder
Perpendicular to the
yz-plane.
Elliptic (upper) cone
Central axis = z-axis.
Hyperboloid
of ONE sheet
Central axis = y-axis.
Central axis = z-axis.
y2 +
z2
=1
4
Hyperboloid
of TWO sheets
x2 + 4y 2 +
z2
=1
4
Ellipsoid
x2
z=
y2
4
x2
9
Hyperbolic paraboloid
[NEW] Without technology, make a rough sketch on your graph paper side of the following 3D
surfaces. Orientation is obviously important.
(a) x2 + y 2 = 1: Right circular cylinder perpendicular to the xy-plane.
See below, left.
1
The radius is 1.
(b) x2 + z 2 = 1: Right circular cylinder perpendicular to the xz-plane. The radius is 1.
See previous page, right.
p
x2 + z 2 : Right circular cone opening out onto the negative y-axis (left). See
(c) y =
below, left.
(d) z = 1 x2 y 2 and z = 0 on the same set of axes.
Elliptic paraboloid, concave down. Opens out onto the negative z-axis. The vertical
shift is positive one unit in the z-axis direction. See above, right.
(#22) Use technology (Derive5) to sketch z = f (x; y) = sin (y) : This is a cylindrical wall.
This will be perpendicular to the yz-plane.
1
0.8
0.6
0.4
z
0.2
0
-0.2
-0.4
-0.6
-0.8
-5
-4
-3
-2
-1
0
1
2
3
4
5
5
4
3
2
1
0
-5
-4
-3
-2
-1
x
y
On separate xy-axes, determine the level curves for f (x; y) = sin (y) = k; for k = 0;
The level curves are the traces parallel to the xy-plane.
2
1
; 1:
2
y
8
6
When k = 0; we solve
sin (y) = 0 ) y = n ; n is any integer.
4
2
0
-8
-6
-4
-2
0
2
4
6
8
We have equally-spaced parallel lines.
x
-2
-4
-6
-8
y
8
1
When k = ; we solve
2
6
4
2
sin (y) =
0
-8
-6
-4
-2
0
2
4
6
8
1
)
2
x
-2
y=
-4
-6
6
+ 2 n or
5
+ 2 n; n is any integer.
6
We have pairs of parallel lines which repeat every 2 :
-8
y
8
6
When k = 1; we solve
4
2
0
-8
-6
-4
-2
0
2
4
6
8
sin (y) = 1 )
x
-2
y=
-4
-6
2
+ 2 n; n is any integer.
We have equally-spaced parallel lines.
-8
(#46) Assume that the xy-plane cuts through the skinniest portion of the hyperboloid of one sheet.
The origin will be at the center of that circle.
Thus, when z = 0; the circular trace has radius 200. When z = 800; the circular trace has
radius 300. Find the equation of the surface.
We should assume that the origin is the center of the hyperboloid of one sheet. Its equation
must be
x2 y 2 z 2
+ 2
= 1:
a2
b
c2
Since the level curves are circles, we must have a = b:
x2 y 2
+ 2
a2
a
3
z2
=1
c2
When z = 0; we have the circle
x2 y 2
+ 2 = 1 ) x2 + y 2 = a2
a2
a
and the radius is 200, so a = 200:
y2
x2
+
2002 2002
z2
=1
c2
When z = 800; we have
x2
y2
+
2002 2002
8002
8002
2
2
2
=
1
)
x
+
y
=
200
1
+
c2
c2
and the radius of this circle must be 300.
s
8002
8002
2002 1 + 2
= 300 ) 2002 1 + 2
c
c
1+
8002
c2
=
c2
8002
=
3002
8002
9
)
=
2
2
200
c
4
1=
= 3002
5
4
4
4
) c2 =
8002
5
5
s
r
4
4
1600
c =
(8002 ) = 800
= p :
5
5
5
The equation of the hyperboloid is
x2
y2
+
2002 2002
z2
1600
p
5
2
= 1:
Section 10.1
(#2) Describe and sketch the domain of the function f (x; y) =
3xy
:
y x2
We must have the denominator not equal to zero. When does it equal zero?
y
x2 = 0 ) y = x2
We must stay away from the parabola y = x2 : Thus, we want the all points NOT on the
parabola (dashed line). See below, left.
y
4
2
0
-4
-2
0
2
4
x
-2
-4
4
(#4) Same thing for f (x; y) =
p
1
x2
y2:
The contents of a square root must be nonnegative. Solve:
x2
1
y2
0 ) x2 + y 2
1
This region is the unit circle AND its interior. See previous page, right.
(#18) We have four traces to graph for z = f (x; y) = x2
y2:
(a) For z = 0; we have x2 y 2 = 0 in the xy-plane.
(x + y) (x y) = 0 ) y = x or y = x: We have two intersecting lines! See below,
left.
y
4
4
y
3
3
2
2
1
1
0
-4
-3
-2
0
-1
0
1
2
3
4
-4
-3
-2
-1
0
1
2
3
4
x
x
-1
-1
-2
-2
-3
-3
-4
-4
(b) For z = 1; we have the hyperbola x2
(c) For y = 0; we have
below, left.
y
z = x2
y 2 = 1: See above, right.
02 ) z = x2 : This is a parabola in the xz-plane. See
y
4
3
3
2
2
1
1
0
0
-4
-3
-2
4
-1
0
1
2
3
4
-4
-3
-2
-1
0
x
-1
-2
-2
-3
-3
-4
-4
(d) For y = 1; we have z = x2
1
2
3
4
x
-1
22 = x2
4: Vertical shift, down 4. See above, right.
5
(#24) z = f (x; y) =
x2
: See below, left.
x2 + y 2 + 1
(#26) z = f (x; y) = sin2 (x) + cos2 (y) : See above, right.
(#42) Sketch a contour plot for f (x; y) = yex = k; for k = 0; 1; 2:
The level curves have the form
yex = k ) y =
k
= ke
ex
x
:
This will be a sequence of exponential decay curves.
When k = 0; we have the line y = 0 [the x-axis].
y
4
3
2
1
0
-4
-3
-2
-1
0
1
2
3
4
x
-1
-2
-3
-4
Section 10.2
(#10) Show that
lim
(x;y)!(0;0)
2xy
= d:n:e:
x2 + 2y 2
by choosing two level curves with di¤erent k-values, both of which lead to the origin.
2
We see that y = 1x is a level curve for k = :
3
2x (x)
2x2
2
=
= ; if x 6= 0:
x2 + 2x2
3x2
3
2
Thus, if we choose this path, which leads back to the origin, we stay at z = :
3
6
4
If, instead, we choose the path y = 2x; then we see that this is a level curve for k = :
9
4x2
4
2x (2x)
= ; if x 6= 0:
=
2
2
2
9x
9
x + 2 (2x)
4
This path also leads us back to the origin, but at a di¤erent altitude/height of z = :
9
This shows that we have a jump discontinuity at the origin, and that discontinuity is not
removable.
7