MA261-A Calculus III
2006 Fall
Homework 1 Solutions
Due 9/8/2006 8:00AM
9.1 #4 What are the projections of the point (2; 3; 5) on the xy-, yz-, and xz-planes? Draw a
rectangular box with the origin and (2; 3; 5) as opposite vertices and with its faces parallel
to the coordinates planes. Label all vertices of the box. Find the length of the diagonal
of the box.
[Solution]
The projection of the point (2; 3; 5) on the xy-plane is (2; 3; 0).
The projection of the point (2; 3; 5) on the yz-plane is (0; 3; 5).
The projection of the point (2; 3; 5) on the xz-plane is (2; 0; 5).
The length of the diagonal is the distance from the origin to (2; 3; 5), which equals
q
(2
0)2 + (3
0)2 + (5
0)2 =
p
38.
9.1 #6 (a) What does the equation x = 4 represent in R2 ? What does it represent in R3 ?
(b) What does the equation y = 3 represent in R3 ? What does z = 5 represent? What
does the pair y = 3; z = 5 represent? In other words, describe the set of points
(x; y; z) such that y = 3 and z = 5. Illustrate with s sketch.
[Solution]
(a) The equation x = 4 represent in R2 :
y
5
4
-5
-4
-3
-2
-1
The equation x = 4 represent in R3 :
1
0
1
2
3
4
5
x
2
10
5
-10
-5
0
0 0
5
10
x
-5
-10
-5
y
5
10
-10
(b) The equation y = 3 represent in R3 :
5
10
-10
-5
-10
0
00
-5
x
5
10
y
5
-5
10
-10
The equation z = 5 represent in R3 :
6.0
5.5
z
5.0
4.5
-4
-4
-2
4
4.0
0 0
x2
-2
y
2
4
For the pair y = 3; z = 5, we have the form (x; 3; 5). So, the graph looks like
3
6.0
5.5
z
5.0
4.5
-4
4.0
2.0 0
2
4
-2
2.5
x
3.0
3.5
y
4.0
9.1 #10 Find an equation of the sphere with center (2; 6; 4) and radius 5. Describe its intersection
with each of the coordinate planes.
[Solution]
By the general equation of a sphere, it’s
2)2 + (y
(x
( 6))2 + (z
4)2 = 52 ,
which is
(x
2)2 + (y + 6)2 + (z
4)2 = 25.
The intersection point of xy-plane looks like (x; y; 0). The sphere equation becomes
(x 2)2 + (y + 6)2 + (0 4)2 = 25, that is (x 2)2 + (y + 6)2 = 9. It is a circle in
xy-plane with center (2; 6; 0) and radius 3.
The intersection point of xz-plane looks like (x; 0; z). The sphere equation becomes
(x 2)2 + (0 + 6)2 + (z 4)2 = 25, that is (x 2)2 + (y + 6)2 = 11. So, no points of
(x; 0; z) satisfy our sphere equation. No intersection with xz-plane.
The intersection point of yz-plane looks like (0; y; z). The sphere equation becomes
(0 2)2 + (y + 6)2 + (z 4)2 = 25, that p
is (y + 6)2 + (z 4)2 = 21. It is a circle in
yz-plane with center (0; 6; 4) and radius 21.
9.1 #14 Show that the equation
4x2 + 4y 2 + 4z 2
8x
16y = 1
represents a sphere, and …nd its center and radius.
[Solution]
We have
4x2 + 4y 2 + 4z 2 8x 16y = 1
=)
x2 + y 2 + z 2
2x
4y =
1
4
x2
4y + z 2 =
1
4
=)
2x + y 2
4
=)
x
2
2x +
2
2
2
+y
2
4y +
4
2
2
1
+z = +
4
=)
x2
2x + 1 + y 2
1)2 + (x
2
+
4
2
2
21
4
p !2
21
.
2
4y + 4 + z 2 =
=)
(x
2
2
2
2)2 + (z
0)2 =
Therefore, it is a sphere with center (1; 2; 0) and radius
p
21
.
2
9.1 #16 Find an equation of a sphere if one of its diameters has end-points (2; 1; 4) and (4; 3; 10).
[Solution]
Since this diameter has end-points (2; 1; 4) and (4; 3; 10), the midpoint of the line
segment with endpoints (2; 1; 4) and (4; 3; 10) is the center of this sphere and the radius
is the half of the length of this line segment.
The midpoint is 2+4
; 1+3
; 4+10
= (3; 2; 7). The length of the line segment is
2
2
2
q
p
(4 2)2 + (3 1)2 + (10 4)2 = 2 11.
p
So, the radius is 11. Therefore, the equation is
2
p
(x 3)2 + (x 2)2 + (z 7)2 =
11 = 11.
9.1 #28 Describe in words the region of R3 represented by the inequality
x2 + y 2 + z 2 > 2z.
[Solution]
x2 + y 2 + z 2 > 2z
is equivalent to
x2 + y 2 + z 2 2z > 0.
Consider x2 + y 2 + z 2 2z = 0. This turns into x2 + y 2 + z 2 2z + 1 = 1, which is,
x2 + y 2 + (z 1)2 = 12 , a sphere with center (0; 0; 1) and radius 1. Note that
x2 + y 2 + z 2
2z > 0
is also equivalent to
x2 + y 2 + (z 1)2 > 1.
So, our region is the space outside the sphere with center (0; 0; 1) and radius 1, not
including the sphere.
9.1 #32 Write inequalities to describe the solid upper hemisphere of the sphere of radius 2 center
at the origin.
[Solution]
First, the equation of the sphere of radius 2 center at the origin is
x2 + y 2 + z 2 = 22 .
The upper hemisphere is sitting above the xy-plane. So, we should have z
0.
5
It’s solid. So, we need everything inside the sphere. This implies that x2 + y 2 + z 2 22 .
Combining these two conditions, we have x2 + y 2 + z 2 4 and z 0 to describe the
solid upper hemisphere of the sphere of radius 2 center at the origin.
9.1 #34 Consider the points P such that the distance from P to A ( 1; 5; 3) is twice the distance
from P to B (6; 2; 2). Show that the set of all such points is a sphere, and …nd its center
and radius.
[Solution]
Assume that P (x; y; z). The distance from P to A is
q
jP Aj = (x ( 1))2 + (y 5)2 + (z 3)2 .
The distance from P to B is
q
jP Bj = (x
6)2 + (y
2)2 + (z
We know that jP Aj = 2 jP Bj. This implies
q
q
2
2
2
(x ( 1)) + (y 5) + (z 3) = 2 (x
( 2))2 .
6)2 + (y
2)2 + (z
( 2))2
=)
(x
( 1))2 + (y
5)2 + (z
3)2 = 4 (x
6)2 + (y
2)2 + (z
( 2))2
=)
x2 + 2x + y 2
10y + z 2
6z + 35 = 4x2
48x + 4y 2
16y + 4z 2 + 16z + 176
=)
3x2
50x + 3y 2
=)
x2
=)
25
3
50
x+
3
x2
25
3
=
50
x + y2
3
2
2
+ (1) +
x
2
Therefore, the center is
+ (y
25
; 1;
3
6
22
y + z2 + z =
3
3
141
3
2
+ y2
11
3
=)
25
3
6y + 3z 2 + 22z + 141 = 0
2y + (1)2 + z 2 +
11
3
2
2
47
p !2
332
2 83
=
=
.
9
3
p
with the radius 32 83.
11
1)2 + z +
3
11
3
22
z+
3
2
9.2 #22 If a child pulls a sledthrough the snow with a force of 50N exerted at an angle of 38
above the horizontal, …nd the horizontal and vertical components of the force.
[Solution]
The horizontal component is 50 cos 38 and the vertical component is 50 sin 38 .
6
9.2 #24 Velocities have both direction and magnitude and thus are vectors. The magnitude of a
velocity vector is called speed. Suppose that a wind is blowing from the direction N45 W
at a speed of 50 km/h. (This means that the direction from which the wind blow is 45
west of the northerly direction.) A pilot is steering a plane in the direction N60 E at an
airspeed (speed in still air) of 250 km/h. The true course, or track, of the plane is the
direction of the resultant of the velocity vectors of the plane and the wind. The ground
speed of the plane is the magnitude of the resultant. Find the true course and the ground
speed of the plane.
[Solution]
A vector represents the direction N45 W and the magnitude 50 is ( 50 sin 45 ; 50 cos 45 ).
A vector represents the direction N60 E and the magnitude 250 is (250 sin 60 ; 250 cos 60 ).
The true course is to add them together, which is
( 50 sin 45 + 250 sin 60 ; 50 cos 45 + 250 cos 60 )
because you can put the initial point of the …rst vector topthe end point
of the
second vector
p
p
2
3
2
50 2 + 250 2 ; 50 2 + 250 12 =
to form the resultant. So, the true course is
p
p
p
p
125+25
2
p
p
125 3 25 2; 125 + 25 2 . The angle is tan = 125
, which implies that =
3 25 2
tan
1
p
125+25
2
p
p
=
125 3 25 2 q
ground speed is
0:72458 = 41:514 . This tells us the direction is E41:514 N. The
p
p 2
p 2
125 3 25 2 + 125 + 25 2 = 241:93.
9.2 #28 The tension T at each end of the chain has magnitude 25N. What is the weight of the
chain?
[Solution]
Each side contributes 25 sin 37 magnitude of the force in the vertical direction. So,
the weight is 2 25 sin 37 = 30:091.
9.2 #30 Suppose that a and b are nonzero vectors that are not parallel and c is any vector in the
plane determined by a and b. Give a geometric argument to show that c can be written
as c = sa + tb for suitable scalars s and t. Then give an argument using components.
[Solution]
(Skip)
9.2 #32 Suppose a vector a makes angles , and with the positive x-, y-, and z-axes, respectively. Find the components of a and show that
cos2
+ cos2
+ cos2
= 1.
[Solution]
Consider the vector a and the x-axe. We can draw a line from the end point of the
vector a to the x-axe which forms a right angle with the x-axe. Now, you have a rightangle triangle with the hypotenuse is a and the adjacent side is on the x-axe with the
angle in between these two sides. So, the length of the adjacent side can be calculated
by jaj cos . This is the x-coordinate of a. Similarly, we can …nd the y- and z-coordinates
of a. We conclude that a = hjaj cos ; jaj cos ; jaj cos i.
7
The length of a is
This implies that
=)
=)
q
jaj = (jaj cos )2 + (jaj cos )2 + (jaj cos )2 .
jaj2 = (jaj cos )2 + (jaj cos )2 + (jaj cos )2
jaj2 = jaj2 cos
cos2
+ jaj2 cos + jaj2 cos
+ cos2
+ cos2
= 1.
9.2 #38 Suppose the three coordinate planes are all mirrored and a light ray given by the vector
a = ha1 ; a2 ; a3 i …rst strikes the xz-plane, as shown in the …gure. Use the fact that the
angle of incidence equals the angle of re‡ection to show that the direction of the re‡ected
ray is given by b = ha1 ; a2 ; a3 i. Deduce that, after being re‡ected by all three mutually
perpendicular mirrors, the resulting ray is parallel to the initial ray.
[Solution]
Firstable, you draw a line to represent xz-plane. Then, draw the vector a toward a
point on the line, namely O. Starting from O, draw your re‡ection vector b. Now you
have a re‡ection picture on hand.
Remember that vectors are "positionless". So, you can move the vector a starting from
O. Now, you have two vectors a and b both starting from O and pointing outward from
xz-plane and xz-plane is sitting in the middle of them.
According to the picture we have now, the x-coordinates of a and b should be the same
and the z-coordinates of a and b shoule be the same too because when you project a
and b into xz-plane, you reach the same point. So, the only di¤erence is y-coordinate
and a and b are sitting in the exactly opposite position to the y-axe. So, b shoule be
ha1 ; a2 ; a3 i.
Similarly, the same happens to the xy- and yz-planes. Thus, we can conclude that the
resulting vector is h a1 ; a2 ; a3 i which is parallel to the initial ray.