(8/15/08)
Math 20C. Lecture Example Solutions for Instructors
Section 14.1, Part 2. Functions of three or more variables†
The frustum of a right circular cone with base of radius R (meters), top of radius r
(meters), and height h (meters) (Figure 1) has volume V (R, r, h) = 31 π(R2 + rR + r2 )h.
(a) What are the domain and range of V ? (b) What does V (6, 4, 9) represent and
what is its value if lengths are measured in meters? (b) What is the geometric
interpretation of V (R, 0, h)?
Example 1
FIGURE 1
Answer: (a) The domain of V consists of all points (R, r, h) in three-dimensional Rrh-space such that
R ≥ 0, r ≥ 0, h ≥ 0. • Its range is the interval [0, ∞) on a V -axis.
(b) V (9, 6, 4) is the volume of a frustum of a right circular cone with base of radius 9 meters, top of radius 6 meters,
and 4 meters high. • V (9, 6, 4) = 228π cubic meters
1
(c) V (R, 0, h) = 3
πR2 h is the volume of a right circular cone whose base has radius R and whose height is h.
(Figure A1)
Figure A1
SOLUTION FOR INSTRUCTORS:
(a) The domain of V consists of all points (R, r, h) in three-dimensional Rrh-space such that
R ≥ 0, r ≥ 0, h ≥ 0. • The range is the interval [0, ∞) on a V -axis.
(b) V (9, 6, 4) is the volume of a frustum of a right circular cone with base of radius 9 meters,
top of radius 6 meters, and 4 meters high. • V (9, 6, 4) = 31 π(92 + 6(9) + 62 )(4) = 228π cubic
meters (c) V (R, 0, h) = 13 πR2 h is the volume of a right circular cone whose base has radius R
and whose height is h. (Figure A1)
† Lecture notes to accompany Section 14.1,
Part 2 of Calculus, Early Transcendentals by Rogawski. All or most of the examples
in these notes are examples or exercises from Al Shenk’s calculus manuscript.
1
Math 20C. Lecture Example Solutions (8/15/08).
Example 2
Section 14.1, Part 2, p. 2
Describe the level surfaces of f (x, y, z) = x2 + y 2 + z 2 .
√
Answer: For c = 0, the level surface f = c is the origin. • For c > 0 it is the surface of the sphere of radius c
centered at the origin in Figure A2. • For c < 0 it is empty.
Figure A2
SOLUTION FOR INSTRUCTORS:
p
x2 + y 2 + z 2 is the distance from (x, y, z) to the origin (0, 0, 0). • For c = 0, the level surface
√
f = c consists of only one point, the origin. For c > 0 it is the surface of the sphere of radius c
centered at the origin in Figure A2. • For c < 0 it is empty.
Describe the level surfaces of g(x, y, z) = x2 + y 2 .
Example 3
Answer: The level surface g = c is the z -axis if c = 0, is the cylinder of radius
Figure A3 if c > 0, and is empty if c < 0.
√
c with the z -axis as axis in
Figure A3
SOLUTION FOR INSTRUCTORS:
Since z does not appear in the equations x2 + y 2 = c of level surfaces of g, they consist of vertical
lines, parallel to the z-axis. • x2 + y 2 is the square of the distance from (x, y, z) to the z-axis. •
If c is positive, the level surface is the cylinder in Figure A3 formed by the vertical lines through
√
the circle x2 + y 2 = c of radius c in the xy-plane. • If c = 0 the level surface is the z-axis. •
If c < 0 the level surface is empty.
Section 14.1, Part 2, p. 3
Math 20C. Lecture Example Solutions (8/15/08).
1 2
The level surface h(x, y, z) = 1 of h(x, y, z) = 91 x2 + 16
y + z 2 in Figure 2 is called an
ellipsoid because all of its cross sections are ellipses. Find equations for and draw its
cross sections (a) in the xz-plane, (b) in the yz-plane, and (c) in the xy-plane.
Example 4
A level surface of
h(x, y, z) = 19 x2 +
1 2
16 y
+ z2
FIGURE 2
1 2
x + z 2 = 1 in the xz -plane is the ellipse with x-intercepts x = ±3 and z Answer: (a) The cross section 9
intercepts z = ±1 in Figure A4a.
1 2
(b) The cross section 16
y + z 2 = 1 in the yz -plane is the ellipse with y -intercepts y = ±4 and z -intercepts
z = ±1 in Figure A4b.
1 2
x2 + 16
y = 1 in the xy -plane is the ellipse with x-intercepts x = ±3 and y -intercepts
(c) The the cross section 1
9
y = ±4 in Figure A4c.
z
1 2
9x
+ z2 = 1
2
z
1 2
16 y
+ z2 = 1
2
3
Figure A4a
Figure A4b
+
1 2
16 y
=1
2
4 y
x
1 2
9x
y
2
x
Figure A4c
SOLUTION FOR INSTRUCTORS:
1 2
(a) The xz-plane has the equation y = 0. • Its intersection with 91 x2 + 16
y + z 2 = 1 is the
2
2
ellipse 19 x + z = 1 with x-intercepts x = ±3 and z-intercepts z = ±1 in Figure A4a.
1 2
(b) The yz-plane has the equation x = 0. • Its intersection with 19 x2 + 16
y + z 2 = 1 is the
2
1 2
ellipse 16 y + z = 1 with y-intercepts y = ±4 and z-intercepts z = ±1 in Figure A4b.
1 2
y + z 2 = 1 is the
(c) The xy-plane has the equation z = 0. • Its intersection with 19 x2 + 16
1 2
1 2
ellipse 9 x + 16 y = 1 with x-intercepts x = ±3 and y-intercepts y = ±4 in Figure A4c.
Math 20C. Lecture Example Solutions (8/15/08).
Section 14.1, Part 2, p. 4
Describe the level surfaces h = c of the function of Example 4 (a) with c > 1 and
(b) with 0 < c < 1.
Example 5
Answer: (a) The level surfaces h = c with c > 1 are geometrically similar ellipsoids outside the ellipsoid h = 1
of Figure 2. (b) The level surfaces h = c with 0 < c < 1 are geometrically similar ellipsoids inside the ellipsoid
h = 1 of Figure 2.
SOLUTION FOR INSTRUCTORS:
See the answer.
Figure 3 is a cross-sectional view of level surfaces of the Van Allen belts of cosmic
radiation that surround the earth. (To visualize the level surfaces imagine that the
curves as drawn are rotated around the north-south axis of the earth.) The radiation is
measured in counts per second and the scale shows the distance in earth radii (≈ 4000
miles). At approximately what distances from the equator is the radiation the greatest?
Example 6
FIGURE 3
Answer: The radiation is greatest at approximately 0.5 earth radii and at approximately 2.5 earth radii from the
equator.
SOLUTION FOR INSTRUCTORS:
See the answer.
Example 7
What is the domain of P (w, x, y, z) =
p
1 − w 2 − x2 − y 2 − z 2 ?
Answer: The domain consists of all “points” (w, x, y, z ) in wxyz -space such that w 2 + x2 + y 2 + z 2 ≤ 1.
SOLUTION FOR INSTRUCTORS:
p
P (w, x, y, z) = 1 − w2 − x2 − y 2 − z 2 is defined where 1 − w2 − x2 − y 2 − z 2 ≥ 0. •
The domain consists of all “points” (w, x, y, z) in wxyz-space such that w2 + x2 + y 2 + z 2 ≤ 1.
Example 8
On July 21, 2006, Ford Motor stock sold for $6.19 per share, Hewlett Packard stock
sold for $31,80 per share, Motorola stock sold for $20.60 per share, Pepsi Cola stock
sold for $62.48 per share, and Yahoo stock sold for $25.27 per share. Give a formula for
the cost C(x1 , x2 , x3 , x4 , x5 ) on that day of x1 shares of Ford Motor stock, x2 shares
of Hewlett Packard stock, x3 shares of Motorola stock, x4 shares of Pepsi Cola stock,
and x5 shares of Yahoo stock. What is the domain of this function?
Answer: C (x1 , x2 , x3 , x4 , x5 ) = 6.19x1 + 31.80x2 + 20.60x3 + 62.48x4 + 25.27x5 dollars. • The domain is
the set of “points” (x1 , x2 , x3 , x4 , x5 ), such that x1 , x2 , x3 , x4 , and x5 are all nonnegative integers
SOLUTION FOR INSTRUCTORS:
See the answer.