Solution to Set 1, due Friday April 2
(1) √
(Section
4e) Determine
the distance between (1 +
√ 1.1, Problem √
√
5, − 2, 2) and (1 − 5, 2, −1).
Solution: distance between given points
q
√
√
√
√
= [(1 + 5) − (1 − 5)]2 + (− 2 − 2)2 + (2 − (−1))2
q √
√
√
√
= (2 5)2 + (2 2)2 + 32 = 20 + 8 + 9 = 37
(2) (Section 1.1, Problem 7) Sketch and identify the graph of 2x +
z = 2.
Solution: On the x−z-plane, we get the straight line 2x+z = 2
with intercept 2 and slope −2. Since equation does not involve
y at all, the surface is obtained by parallel translation of this
straight line along y-axis to obtain a plane:
z
(0,0,2)
z+2x=2
Plane
y
(1,0,0)
x
Figure 1. Sketching 2x + z = 2
(3) (Section 1.1, No. 13) Identify and describe the set of points
(x, y, z) satisfying the relation −1 ≤ z ≤ 1.
Solution: z ≤ 1 is the set of all points (x, y, z) below plane
z = 1 (which is parallel to the x − y plane), while z ≥ −1 is the
set of all points above plane z = −1. Therefore, −1 ≤ z ≤ −1
is the interesection of these two sets; i.e. all points between
planes z = −1 and z = +1:
1
2
z
(0,0,1)
z=1
Plane
In between two planes
y
−1 < z < 1
x
(0,0,−1)
z=−1 plane
Figure 2. Set of points (x, y, z) satisfying −1 ≤ z ≤ 1
is in between or on the two planes z = 1 and z = −1.
(4) (Section 1.1, No. 15) Identify the region described by (x − 1)2 +
(y + 1)2 + (z + 3)2 > 81.
Solution: Since the expression on the left is the distance square
between point (x, y, z) and (1, −1, −3) and this is less than 92 , it
follows that the set of points {(x, y, z)} satisfying above relation
lies outside the sphere of radius 9 centered at (1, −1, −3).
(5) (Section 1.1, NO. 18) Find the radius of smallest sphere centered
at (3, −8, 6) that contains the origin.
Solution. The √
answer is the distance
of this point from the
√
origin, which is 32 + 82 + 62 = 109.
(6) (Section 1.2, Problem 2) Sketch the graph of g(x, y) = cos y in
3 − D.
Solution: Since equation does not involve x at all, we draw
the curve z = cos y in the y − z plane and translate the figure
along the z-axis. This leads to:
3
z
z
z= cos y
y
y
x
x
z=−1 plane
Figure 3. Sketch of graph of z = cos y
Figure 4. Computer graph of z = cos y, tilted to show
the 3-D structure
(7) Section 1.2, Problem 5. Sketch the graph of the function q(x, y) =
−x − 2y for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
Solution: In the y − z plane, where x = 0, we obtain z =
q(0, y) = −2y, a straightline through the origin with slope −2.
In the x − z plane, where y = 0, we have z = q(x, 0) = −x,
which is also another straight line passing through the origin
with slope -1. Also, the intersection of the surface with the
plane x = 1 results in curve z = −1 − 2y and the intersection of
the surface with the plane y = 1 results in the curve z = −2−x.
We obtain a plane as sketched below:
4
z
(0,0,0)
(0,1,0)
z=−2y
(1,0,0) z=−x
x
plane
(0,1,−2)
(1,0,−1) z=−x−2y
z=1−2y
y
z=−2−x
(1,1,−3)
Figure 5. Sketch of graph of q(x, y) = −x − 2y for
0 ≤ x ≤ 1, 0 ≤ y ≤ 1
(8) Section 1.2, Problem 6. Sketch the graph of the function r(x, y) =
2 + x − y.
Solution: For z = r(x, y) = 2 + x − y, note the intersection
with the y − z plane gives z = 2 − y, which is a straight line
with slope −1 and intercept 2. On the x − z-plane, we obtain
z = r(x, 0) = 2 + x a straightline with slope 1 and intercept 2.
z
z=2+x
plane
z=2−y
+x
z=2−y
y
x
Figure 6. Sketch of graph of r(x, y) = 2 + x − y
(9) Section
1.2, Problem 7. Sketch the graph of the function s(x, y) =
p
1
2 + y2.
x
2
Solution: On the y − z plane, the intersection curve is that of
part of the straight line z = y2 for y ≥ 0. Since the height z =
s(x, y) only depends on the distance from the z-axis, the surface
under question is the surface of revolution around z-axis which
generates a cone as shown. Also, note that the intersection
5
ofpsurface with the plane z = k of the surface results in k =
1
x2 + y 2 and so x2 + y 2 = 4k 2 , which is a circle..
2
z
x2 + y 2 = 4 k
z=y/2
2
y
x
Figure 7. Sketch of graph of s(x, y) =
1
2
p
x2 + y 2
(10) Section 1.2, Problem 22. Describe and sketch the relationship
between the graphs of the following pairs of equations: (a) z =
ex , (b) z = 14 ex − 2.
Solution: Note for z = ex does not involve y so the surface
is obtained by translating the curve z = ex on the x − z plane
along the y-axis and we get the shape shown below. Note case
b. shape is recovered from case a. by replacing x by x − ln 4
(noting ex−ln 4 = 14 ex ) and z by z + 2. So, we shift the shape in
case a. along the x axis by ln 4 and along the negative z axis
by −2 to obtain graph for case b.
z
z=e x
z =e x/4 −2
y
x
Figure 8. Sketch of graphs of z = ex and z = 41 ex − 2
6
(11) Section 1.3, Problem 3: Identify and sketch the surface x2 −
2y 2 + z = 0.
Solution: Equation implies z = 2y 2 − x2 . This is of the form
z = Ay 2 + Bx2 , with A > 0, B < 0. From discussion in class,
this is a saddle. To sketch this, note the following
i. Intersection of surface with x = 0 gives z = 2y 2– parabola
opening upwards going through (0, 0, 0). Intersection of surface with x = +1 gives z = 2y 2 − 1; again a parabola but
shifted downwards by -1 with lowest point at (1, 0, −1). Similar
parabola from intersection with x = −1 plane. We sketch this
as shown:
z
Saddle
z=2y
2
2
z=−x
y
2
z=2 y −1
x
Figure 9. Sketching z = 2y 2 − x2
Figure 10. Computer assisted 3-D plot of z = 2y 2 − x2
7
(12) (Section 1.3, No. 7) Identify and sketch the graph corresponding
to
9x2 + 4y 2 + 36z 2 − 18x + 16y = 0
Solution: Completing the square, we note
9(x2 − 2x + 1) + 4(y 2 + 4y + 4) + 36z 2 = 9 + 16 = 25
So,
(x − 1)2 (y + 2)2
z2
+
+
=1
25/9
25/4
(25)/36
Get ellipsoid centered at (1, -1, 0) with semi-major minor axes
lengths 5/3, 5/2 and 5/6 respectively. The sketch is given below.
z
y
(1,−2,0) center of ellipsoid
x
Figure 11. Sketching of ellipsoid 9x2 + 4y 2 + 36z 2 −
18x + 16y = 0
Figure 12. Computer plot of ellipsoid 9x2 +4y 2 +36z 2 −
18x + 16y = 0. Never mind the egg-shell breaking
8
(13) (Section 1.3, problem 9) Analyze and sketch the graph of the
hyperboloid with two sheets
x2 y 2 z 2
− 2 − 2 + 2 =1
a
b
c
2
2
2
Solution: We rewrite equation as zc2 = 1 + xa2 + yb2
i. On the x − z plane, i.e. y = 0, the intersection curve:
2
z2
− yb2 = 1–a pair of hyperbola centered at (0, 0, 0) with axis
c2
along the z-axis.
ii. On the y − z plane, i.e. x = 0, the intersection curve:
2
z2
− xa2 = 1–a pair of hyperbola centered at (0, 0, 0) with axis
c2
along z-axis.
iii. Intersection of surface with z = k for |k| > c: f rack 2 c2 −1 =
2
x2
+ y – ellipse centered at (0,0,0) with major and minor axis
a2p b2
p
a k 2 /c2 − 1 and a k 2 /c2 − 1 respectively. Note no intersec2
tion curve possible when kc2 − 1 < 0, i.e. |k| < c.
z
y
x
2
Figure 13. Sketch of hyperboloid − xa2 −
y2
b2
+
z2
c2
=1
9
2
Figure 14. Computer plot of hyperboloid − xa2 −
z2
= 1 for a = 2, b = 3, c = 1
c2
y2
b2
+
(14) (Section 1.3, problem 13). Identify and sketch x2 + y 2 − 2z 2 =
−5.
2
2
Solution: Note we can solve for z to obtain z 2 = x2 + y2 + 25
or
x2 y 2
z2
= 1+
+
5/2
5
5
p
5/2,
This√
is a special
case
of
Problem
9
given
above
with
c
=
√
a = 5, b = 5. So, the sketch of the plot is roughly as given
in last figure. Note the interesection curves on the x−z plane is
2
z2
= 1 + y5 and the intersection curves
the pair of hyperbolae 5/2
on the x − y plane is again a pair of hyperbolae
z2
5/2
=1+
x2
.
5