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APPENDIX A VECTORS
Vectors in Two Dimensions
A.1
D
B
u
v
C
A
FIGURE 1
Equivalent vectors
The term vector is used by scientists to indicate a quantity (such as displacement or velocity or force) that has both magnitude and direction. A vector is often represented by an arrow
or a directed line segment. The length of the arrow represents the magnitude of the vector
and the arrow points in the direction of the vector. We denote a vector by printing a letter
in boldface 共v兲 or by putting an arrow above the letter 共vl兲.
For instance, suppose a particle moves along a line segment from point A to point B.
The corresponding displacement vector v, shown in Figure 1, has initial point A (the tail)
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and terminal point B (the tip) and we indicate this by writing v 苷 AB. Notice that the vecl
tor u 苷 CD has the same length and the same direction as v even though it is in a different
position. We say that u and v are equivalent (or equal) and we write u 苷 v. The. zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.
Combining Vectors
C
B
l
Suppose a particle moves from A to B, so its displacement vector is AB. Then the particle
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changes direction and moves from B to C, with displacement vector BC as in Figure 2. The
combined effect of these displacements is that the particle has moved from A to C. The
l
l
l
resulting displacement vector AC is called the sum of AB and BC and we write
l
l
l
AC 苷 AB BC
A
FIGURE 2
In general, if we start with vectors u and v, we first move v so that its tail coincides with
the tip of u and define the sum of u and v as follows.
Definition of Vector Addition If u and v are vectors positioned so the initial point of
v is at the terminal point of u, then the sum u v is the vector from the initial
point of u to the terminal point of v.
The definition of vector addition is illustrated in Figure 3. You can see why this definition is sometimes called the Triangle Law.
u
u+v
v
v
u
FIGURE 3 The Triangle Law
u v
v+ u+
v
u
FIGURE 4 The Parallelogram Law
In Figure 4 we start with the same vectors u and v as in Figure 3 and draw another
copy of v with the same initial point as u. Completing the parallelogram, we see that
u v 苷 v u. This also gives another way to construct the sum: If we place u and v so
they start at the same point, then u v lies along the diagonal of the parallelogram with u
and v as sides. (This is called the Parallelogram Law.)
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SECTION A.1
VECTORS IN TWO DIMENSIONS
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EXAMPLE 1 Draw the sum of the vectors a and b shown in Figure 5.
a
b
SOLUTION First we translate b and place its tail at the tip of a, being careful to draw a
copy of b that has the same length and direction. Then we draw the vector a b [see
Figure 6(a)] starting at the initial point of a and ending at the terminal point of the copy
of b.
Alternatively, we could place b so it starts where a starts and construct a b by the
Parallelogram Law as in Figure 6(b).
FIGURE 5
a
a
b
a+b
a+b
b
FIGURE 6
(a)
(b)
It is possible to multiply a vector by a real number c. (In this context we call the real number c a scalar to distinguish it from a vector.) For instance, we want 2v to be the same vector as v v, which has the same direction as v but is twice as long. In general, we multiply
a vector by a scalar as follows.
Definition of Scalar Multiplication If c is a scalar and v is a vector, then the scalar
ⱍ ⱍ
multiple cv is the vector whose length is c times the length of v and whose
direction is the same as v if c 0 and is opposite to v if c 0. If c 苷 0 or v 苷 0,
then cv 苷 0.
2v
v
This definition is illustrated in Figure 7. We see that real numbers work like scaling factors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel if
they are scalar multiples of one another. In particular, the vector v 苷 共1兲v has the same
length as v but points in the opposite direction. We call it the negative of v.
By the difference u v of two vectors we mean
1
2v
u v 苷 u 共v兲
_v
_1.5v
So we can construct u v by first drawing the negative of v, v, and then adding it to u
by the Parallelogram Law as in Figure 8(a). Alternatively, since v 共u v兲 苷 u, the vector u v, when added to v, gives u. So we could construct u v as in Figure 8(b) by
means of the Triangle Law.
FIGURE 7
Scalar multiples of v
v
u
u-v
_v
u-v
v
u
FIGURE 8
Drawing u-v
(a)
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(b)
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APPENDIX A VECTORS
EXAMPLE 2 If a and b are the vectors shown in Figure 9, draw a 2b.
SOLUTION We first draw the vector 2b pointing in the direction opposite to b and twice
as long. We place it with its tail at the tip of a and then use the Triangle Law to draw
a 共2b兲 as in Figure 10.
a
_2b
a
b
a-2b
FIGURE 9
FIGURE 10
Components
For some purposes it’s best to introduce a coordinate system and treat vectors algebraically. If we place the initial point of a vector a at the origin of a rectangular two-dimensional coordinate system, then the terminal point of a has coordinates of the form 共a1, a2 兲.
(See Figure 11.) These coordinates are called the components of a and we write
a 苷 具a 1, a 2 典
We use the notation 具 a1, a2 典 for the ordered pair that refers to a vector so as not to confuse
it with the ordered pair 共a1, a2 兲 that refers to a point in the plane.
y
(a¡, a™)
a
O
FIGURE 11
x
a=ka¡, a™l
For instance, the vectors shown in Figure 12 are all equivalent to the vector
l
OP 苷 具3, 2典 whose terminal point is P共3, 2兲. What they have in common is that the
terminal point is reached from the initial point by a displacement of three units to the right
and two upward. We can think of all these geometric vectors as representations of the
l
algebraic vector a 苷 具3, 2典 . The particular representation OP from the origin to the point
P共3, 2兲 is called the position vector of the point P.
y
(4, 5)
(1, 3)
P(3, 2)
0
FIGURE 12
Representations of the vector a=k3, 2l
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x
SECTION A.1
P (a¡, a™)
position
vector of P
O
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l
Figure 13 shows a general vector a 苷 具a 1, a 2 典 with position vector OP and another repl
resentation AB of a, where the initial point is A共x 1, y1 兲 and the terminal point is B共x 2 , y2 兲.
Then we must have x 1 a 1 苷 x 2, and y1 a 2 苷 y2, so a 1 苷 x 2 x 1, and a 2 苷 y2 y1.
Thus we have the following result.
y
B(¤, fi)
VECTORS IN TWO DIMENSIONS
x
A(⁄, ›)
1
l
Given the points A共x 1, y1 兲 and B共x 2 , y2 兲, the vector a with representation AB is
a 苷 具 x 2 x 1, y2 y1 典
FIGURE 13
Representations of a=ka¡, a™l
y
(a¡+b¡, a™+b™)
a+b
a 苷 具 2 2, 1 共3兲典 苷 具4, 4 典
b™
b
The magnitude or length of the vector v is the length of any of its representations and
is denoted by the symbol v or 储 v 储. By using the distance formula to compute the length
of a segment OP, we obtain the following formula.
b¡
a
0
a™
x
b¡
The length of the vector a 苷 具 a 1, a 2 典 is
FIGURE 14
ⱍ a ⱍ 苷 sa
ca
a
ca™
a™
a¡
FIGURE 15
ⱍ ⱍ
a™
a¡
EXAMPLE 3 Find the vector represented by the directed line segment with initial point
A共2, 3兲 and terminal point B共2, 1兲.
l
SOLUTION By 1 , the vector corresponding to AB is
ca¡
2
1
a 22
How do we add vectors algebraically? Figure 14 shows that if a 苷 具a 1, a 2 典 and
b 苷 具 b 1, b 2 典 , then the sum is a b 苷 具a1 b1, a2 b2 典 , at least for the case where the
components are positive. In other words, to add algebraic vectors we add their components.
Similarly, to subtract vectors we subtract components. From the similar triangles in Figure
15 we see that the components of ca are ca1 and ca2. So to multiply a vector by a scalar we
multiply each component by that scalar.
If a 苷 具 a 1, a 2 典 and b 苷 具b1, b2 典 , then
a b 苷 具a 1 b1, a 2 b2 典
a b 苷 具a 1 b1, a 2 b2 典
ca 苷 具 ca1, ca2 典
ⱍ ⱍ
EXAMPLE 4 If a 苷 具4, 3典 and b 苷 具 2, 1典 , find a and the vectors a b, a b, 3b,
and 2a 5b.
SOLUTION
ⱍ a ⱍ 苷 s4
2
32 苷 s25 苷 5
a b 苷 具4, 3典 具2, 1典 苷 具 4 2, 3 1典 苷 具 2, 4典
a b 苷 具4, 3典 具2, 1典 苷 具4 共2兲, 3 1典 苷 具6, 2典
3b 苷 3具2, 1典 苷 具3共2兲, 3共1兲典 苷 具6, 3典
2a 5b 苷 2具4, 3典 5具2, 1典 苷 具 8, 6典 具10, 5典 苷 具2, 11典
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APPENDIX A VECTORS
Properties of Vectors If a, b, and c are vectors and c and d are scalars, then
1. a b 苷 b a
2. a 共b c兲 苷 共a b兲 c
3. a 0 苷 a
4. a 共a兲 苷 0
5. c共a b兲 苷 ca cb
6. 共c d兲a 苷 ca da
7. 共cd 兲a 苷 c共da兲
8. 1a 苷 a
These eight properties of vectors can be readily verified either geometrically or algebraically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the Parallelogram Law) or as follows:
Q
a b 苷 具a 1, a 2 典 具b1, b2 典 苷 具a 1 b1, a 2 b2 典
c
苷 具b1 a 1, b2 a 2 典 苷 具b1, b2 典 具a 1, a 2 典
(a+b)+c
=a+(b+c)
b+c
We can see why Property 2 (the associative law) is true by looking at Figure 16 and
l
applying the Triangle Law several times: The vector PQ is obtained either by first constructing a b and then adding c or by adding a to the vector b c.
Two vectors play a special role. Let
a
P
苷ba
b
a+b
FIGURE 16
i 苷 具1, 0典
y
j 苷 具 0, 1典
(0, 1)
j
0
x
i
(1, 0)
These vectors i and j are called the standard basis vectors. They have length 1 and point
in the directions of the positive x- and y-axes. (See Figure 17.)
If a 苷 具 a 1, a 2 典 , then we can write
a 苷 具a 1, a 2 典 苷 具a 1, 0典 具 0, a 2 典 苷 a 1 具1, 0典 a 2 具0, 1典
FIGURE 17
a 苷 a1 i a2 j
2
Standard basis vectors
y
Thus any vector can be expressed in terms of i and j. For instance,
(a¡, a™)
a
a¡i
0
具 1, 2典 苷 i 2j
a™ j
See Figure 18 for the geometric interpretation of Equation 2 and compare with Figure 17.
x
EXAMPLE 5 If a 苷 i 2j and b 苷 4i 7j, express the vector 2a 3b in terms of
a=a¡i+a™ j
FIGURE 18
i and j.
SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have
2a 3b 苷 2共i 2j兲 3共4i 7j兲
苷 2i 4j 12i 21j 苷 14i 25 j
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SECTION A.1
Gibbs
Josiah Willard Gibbs (1839–1903), a professor
of mathematical physics at Yale College, published the first book on vectors, Vector Analysis,
in 1881. More complicated objects, called
quaternions, had earlier been invented by
Hamilton as mathematical tools for describing
space, but they weren’t easy for scientists to
use. Quaternions have a scalar part and a vector part. Gibb’s idea was to use the vector part
separately. Maxwell and Heaviside had similar
ideas, but Gibb’s approach has proved to be the
most convenient way to study space.
VECTORS IN TWO DIMENSIONS
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A unit vector is a vector whose length is 1. For instance, i and j are both unit vectors.
In general, if a 苷 0, then the unit vector that has the same direction as a is
u苷
3
a
1
a苷
a
a
ⱍ ⱍ
ⱍ ⱍ
ⱍ ⱍ
In order to verify this, we let c 苷 1兾 a . Then u 苷 ca and c is a positive scalar, so u has
the same direction as a. Also
1
ⱍ u ⱍ 苷 ⱍ ca ⱍ 苷 ⱍ c ⱍⱍ a ⱍ 苷 ⱍ a ⱍ ⱍ a ⱍ 苷 1
EXAMPLE 6 Find the unit vector in the direction of the vector 3i 4 j.
SOLUTION The given vector has length
ⱍ 3i 4 j ⱍ 苷 s3
2
共4兲2 苷 s25 苷 5
so, by Equation 3, the unit vector with the same direction is
1
5
共3i 4 j兲 苷 35 i 45 j
Applications
50°
32°
T¡
T™
100
EXAMPLE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the tensions (forces) T1 and T2 in both wires and their magnitudes.
FIGURE 19
50°
T¡
T™
50°
32°
32°
w
Vectors are useful in many aspects of physics and engineering. In Section A.3 we will see
how they describe the velocity and acceleration of objects moving along a curve. Here we
look at forces.
A force is represented by a vector because it has both a magnitude (measured in pounds
or newtons) and a direction. If several forces are acting on an object, the resultant force
experienced by the object is the vector sum of these forces.
SOLUTION We first express T1 and T2 in terms of their horizontal and vertical compo-
nents. From Figure 20 we see that
ⱍ ⱍ
ⱍ ⱍ
苷 ⱍ T ⱍ cos 32 i ⱍ T ⱍ sin 32 j
4
T1 苷 T1 cos 50 i T1 sin 50 j
5
T2
2
2
The resultant T1 T2 of the tensions counterbalances the weight w and so we must have
T1 T2 苷 w 苷 100j
FIGURE 20
Thus
(ⱍ T1 ⱍ cos 50 ⱍ T2 ⱍ cos 32) i (ⱍ T1 ⱍ sin 50 ⱍ T2 ⱍ sin 32) j 苷 100j
Equating components, we get
ⱍ ⱍ
ⱍ ⱍ
T
sin
50
ⱍ ⱍ
ⱍ T ⱍ sin 32 苷 100
Solving the first of these equations for ⱍ T ⱍ and substituting into the second, we get
T cos 50
sin 32 苷 100
ⱍ T ⱍ sin 50 ⱍ ⱍ
T1 cos 50 T2 cos 32 苷 0
1
2
2
1
1
cos 32
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A8
APPENDIX A VECTORS
So the magnitudes of the tensions are
ⱍT ⱍ 苷
1
100
⬇ 85.64 lb
sin 50 tan 32 cos 50
T cos 50
ⱍ T ⱍ 苷 ⱍ cosⱍ 32 ⬇ 64.91 lb
1
and
2
Substituting these values in 4 and 5 , we obtain the tension vectors
T1 ⬇ 55.05 i 65.60 j
A.1
Exercises
1. Are the following quantities vectors or scalars? Explain.
(a)
(b)
(c)
(d)
T2 ⬇ 55.05 i 34.40 j
The cost of a theater ticket
The current in a river
The initial flight path from Houston to Dallas
The population of the world
6. Copy the vectors in the figure and use them to draw the
following vectors.
(a) a b
(c) 12 a
(e) a 2b
(b) a b
(d) 3b
(f ) 2b a
2. What is the relationship between the point (4, 7) and the
vector 具 4, 7 典 ? Illustrate with a sketch.
b
3. Name all the equal vectors in the parallelogram shown.
A
B
a
7. In the figure, the tip of c and the tail of d are both the midpoint
of QR. Express c and d in terms of a and b.
E
P
b
a
D
c
R
C
d
4. Write each combination of vectors as a single vector.
l
l
(a) AB BC
l
l
(c) DB AB
l
l
(b) CD DB
l l l
(d) DC CA AB
A
Q
ⱍ ⱍ ⱍ ⱍ
8. If the vectors in the figure satisfy u 苷 v 苷 1 and
ⱍ ⱍ
u v w 苷 0, what is w ?
u
B
w
v
D
C
5. Copy the vectors in the figure and use them to draw the
following vectors.
(a) u v
(c) v w
(e) v u w
u
(b) u w
(d) u v
(f ) u w v
v
9–12 Find a vector a with representation given by the directed line
l
l
segment AB. Draw AB and the equivalent representation starting at
the origin.
9. A共1, 1兲,
B共3, 2兲
10. A共4, 1兲,
11. A共1, 3兲,
B共2, 2兲
12. A共2, 1兲,
w
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B共1, 2兲
B共0, 6兲
SECTION A.1
13–16 Find the sum of the given vectors and illustrate
still air) of 250 km兾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.
geometrically.
13. 具1, 4典 ,
15. 具2, 3 典 ,
具6, 2典
具3, 4 典
14. 具3, 1典 ,
具1, 5 典
16. 具1, 2典 ,
具5, 3 典
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VECTORS IN TWO DIMENSIONS
37. A woman walks due west on the deck of a ship at 3 mi兾h. The
ⱍ ⱍ
17–24 Find a , a b, a b, 2a, and 3a 4b.
ship is moving north at a speed of 22 mi兾h. Find the speed and
direction of the woman relative to the surface of the water.
17. a 苷 具5, 12典 , b 苷 具 2, 8 典
38. Ropes 3 m and 5 m in length are fastened to a holiday decora-
18. a 苷 具 1, 2典 , b 苷 具 4, 3典
19. a 苷 具 2, 3 典 , b 苷 具 1, 4 典
20. a 苷 具 2, 1典 , b 苷 具 6, 7典
21. a 苷 i j, b 苷 i j
22. a 苷 2 i 3 j, b 苷 3i 2 j
23. a 苷 i j, b 苷 2 i j
24. a 苷 6 i, b 苷 i 2 j
tion that is suspended over a town square. The decoration has a
mass of 5 kg. The ropes, fastened at different heights, make
angles of 52 and 40 with the horizontal. Find the tension in
each wire and the magnitude of each tension.
52°
3 m
25–27 Find a unit vector that has the same direction as the given
40°
5 m
vector.
25. 具 1, 2 典
26. 具3, 5典
27. i j
39. A clothesline is tied between two poles, 8 m apart. The line
28. Find a vector that has the same direction as 具2, 4 典 but has
length 5.
29–30 What is the angle between the given vector and the positive
direction of the x-axis?
29. i s3 j
is quite taut and has negligible sag. When a wet shirt with a
mass of 0.8 kg is hung at the middle of the line, the midpoint
is pulled down 8 cm. Find the tension in each half of the
clothesline.
40. The tension T at each end of the chain has magnitude 25 N
(see the figure). What is the weight of the chain?
30. 8 i 6 j
31. If v lies in the first quadrant and makes an angle 兾3 with the
37°
37°
ⱍ ⱍ
positive x-axis and v 苷 4, find v in component form.
32. If a child pulls a sled through the snow with a force of 50 N
exerted at an angle of 38 above the horizontal, find the
horizontal and vertical components of the force.
41. Find the unit vectors that are parallel to the tangent line to the
parabola y 苷 x 2 at the point 共2, 4兲.
33. A quarterback throws a football with angle of elevation 40 and
speed 60 ft兾s. Find the horizontal and vertical components of
the velocity vector.
34–35 Find the magnitude of the resultant force and the angle it
makes with the positive x-axis.
34.
y
0
35.
20 lb
y
200 N
300 N
45°
30°
x
60°
0
x
16 lb
36. 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兾h. (This means that the direction from which
the wind blows is 45 west of the northerly direction.) A pilot is
steering a plane in the direction N60 E at an airspeed (speed in
42. (a) Find the unit vectors that are parallel to the tangent line to
the curve y 苷 2 sin x at the point 共兾6, 1兲.
(b) Find the unit vectors that are perpendicular to the tangent
line.
(c) Sketch the curve y 苷 2 sin x and the vectors in parts (a)
and (b), all starting at 共兾6, 1兲.
43. If A, B, and C are the vertices of a triangle, find
l
l
l
AB BC CA.
44. Let C be the point on the line segment AB that is twice as far
l
l
l
from B as it is from A. If a 苷 OA, b 苷 OB, and c 苷 OC, show
2
1
that c 苷 3 a 3 b.
45. (a) Draw the vectors a 苷 具 3, 2典 , b 苷 具2, 1 典 , and
c 苷 具7, 1典.
(b) Show, by means of a sketch, that there are scalars s and t
such that c 苷 sa t b.
(c) Use the sketch to estimate the values of s and t.
(d) Find the exact values of s and t.
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A10
APPENDIX A VECTORS
46. 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 苷 sa t b for suitable scalars s and t. Then give an argument using components.
47. If r 苷 具x, y 典 and r0 苷 具x 0 , y0 典 , describe the set of all points
ⱍ
ⱍ
共x, y兲 such that r r0 苷 1.
48. If r 苷 具 x, y 典 , r1 苷 具x 1, y1 典 , and r2 苷 具x 2 , y2 典 , describe the
ⱍ
ⱍ ⱍ
49. Figure 16 gives a geometric demonstration of Property 2 of
vectors. Use components to give an algebraic proof of this
fact.
50. Prove Property 5 of vectors algebraically. Then use similar
triangles to give a geometric proof.
51. Use vectors to prove that the line joining the midpoints of
two sides of a triangle is parallel to the third side and half
its length.
ⱍ
set of all points 共x, y兲 such that r r1 r r2 苷 k,
where k r1 r2 .
ⱍ
A.2
ⱍ
Vector Functions and Their Derivatives
In general, a function is a rule that assigns to each element in the domain an element in the
range. A vector-valued function, or vector function, is simply a function whose domain
is a set of real numbers and whose range is a set of vectors. This means that for every number t in the domain of a vector function r there is a unique vector denoted by r共t兲. If f 共t兲 and
t共t兲 are the components of the vector r共t兲, then f and t are real-valued functions called the
component functions of r and we can write
r共t兲 苷 具 f 共t兲, t共t兲典 苷 f 共t兲 i t共t兲 j
We use the letter t to denote the independent variable because it represents time in most applications of vector functions.
EXAMPLE 1 If
r共t兲 苷 具 ln共3 t兲, st 典
then the component functions are
f 共t兲 苷 ln共3 t兲
t共t兲 苷 st
By our usual convention, the domain of r consists of all values of t for which the expression for r共t兲 is defined. The expressions ln共3 t兲 and st are defined when 3 t 0
and t 0. Therefore the domain of r is the interval 关0, 3兲.
The limit of a vector function r is defined by taking the limits of its component functions
as follows.
If lim t l a r共t兲 苷 L, this definition is equivalent
to saying that the length and direction of the
vector r共t兲 approach the length and direction of
the vector L.
1
If r共t兲 苷 具 f 共t兲, t共t兲典 , then
具
典
lim r共t兲 苷 lim f 共t兲, lim t共t兲
tla
tla
tla
provided the limits of the component functions exist.
Equivalently, we could have used an - definition. Limits of vector functions obey the
same rules as limits of real-valued functions.
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SECTION A.2
VECTOR FUNCTIONS AND THEIR DERIVATIVES
EXAMPLE 2 Find lim r共t兲, where r共t兲 苷 共1 t 3 兲 i tl0
A11
sin t
j.
t
SOLUTION According to Definition 1, the limit of r is the vector whose components are
the limits of the component functions of r:
关
冋
兴
lim r共t兲 苷 lim 共1 t 3 兲 i lim
tl0
tl0
苷ij
tl0
sin t
t
册
j
(by Equation 2.4.2)
A vector function r is continuous at a if
lim r共t兲 苷 r共a兲
tla
In view of Definition 1, we see that r is continuous at a if and only if its component functions f and t are continuous at a.
There is a close connection between continuous vector functions and parametric curves.
Suppose that f and t are continuous real-valued functions on an interval I. Then the set C
of all points 共x, y兲, where
x 苷 f 共t兲
y 苷 t共t兲
y
P { f(t), g(t)}
C
r(t)=kf(t), g(t)l
x
0
FIGURE 1
C is traced out by the tip of a moving
position vector r(t).
and t varies throughout the interval I , is a parametric curve. We can think of C as being
traced out by a moving particle whose position at time t is 共 f 共t兲, t共t兲兲. If we now consider
the vector function r共t兲 苷 具 f 共t兲, t共t兲典 , then r共t兲 is the position vector of the point
P共 f 共t兲, t共t兲兲 on C. Thus any continuous vector function r defines a curve C that is traced out
by the tip of the moving vector r共t兲, as shown in Figure 1.
EXAMPLE 3 The curve defined by the vector function
y
r共t兲 苷 具 t 2 2t, t 1典 苷 共t 2 2t兲i 共t 1兲j
1
is the curve described by the parametric equations
r(t)
x 苷 t 2 2t
0
x
3
y苷t1
Here we simply take a different point of view: We regard the parabola as being traced out
by the tip of the moving vector r共t兲. (See Figure 2.)
EXAMPLE 4 Sketch the curve with vector equation r共t兲 苷 具 3 cos t, 2 sin t典 , 0 t 2.
FIGURE 2
SOLUTION The corresponding parametric equations are
x 苷 3 cos t
y 苷 2 sin t
We eliminate the parameter by observing that
冉冊 冉冊
y
x
3
2
2
y
2
2
苷 cos 2t sin2t 苷 1
So the curve is the ellipse
r(π)
FIGURE 3
0
r(0)
3
x
x2
y2
苷1
9
4
shown in Figure 3. As t increases from 0 to 2, the position vector r共t兲 rotates once
counterclockwise around the origin.
Not For Sale
A12
APPENDIX A VECTORS
Vector Equation of a Line
A line is determined when a point P0 on the line and the direction of the line are given. The
direction of a line in the plane is determined by specifying a slope or by specifying a vector that is parallel to the line. The latter case leads to the vector equation of the line. Let v
be a vector parallel to L. Let P be an arbitrary point on L and let r0 and r be the position
vectors of P0 and P (so r0 苷 OP
A0 and r 苷 OP
A in Figure 4). If a is the vector with representation P
A,
0 P as in Figure 4, then the Triangle Law for vector addition gives r 苷 r0 a.
But, since a and v are parallel vectors, there is a scalar t such that a 苷 t v. Thus
y
P¸(x¸, y¸)
a
P(x, y )
L
r¸
r
v
x
O
FIGURE 4
y
t=0
t>0
L
t<0
r¸
O
r 苷 r0 t v
2
x
which is a vector equation of L. Each value of the parameter t gives the position vector r
of a point on L. In other words, as t varies, the line is traced out by the tip of the vector r. As Figure 5 indicates, positive values of t correspond to points on L that lie on one side
of P0 , whereas negative values of t correspond to points that lie on the other side of P0 .
If the vector v that gives the direction of the line L is written in component form as
v 苷 具a, b典 , then we have tv 苷 具ta, tb 典 . We can also write r 苷 具 x, y典 and r0 苷 具x 0 , y0 典 ,
so the vector equation 2 becomes
具x, y典 苷 具x 0 ta, y0 tb 典
FIGURE 5
Two vectors are equal if and only if corresponding components are equal. Therefore we
have the scalar equations:
x 苷 x 0 at
3
y 苷 y0 bt
where t 僆 ⺢. These equations are parametric equations of the line L through the point
P0共x 0 , y0兲 and parallel to the vector v 苷 具a, b典 . Each value of the parameter t gives a point
共x, y兲 on L. Note that if v is parallel to the line, then any nonzero multiple of v is also parallel to the line and can be used to obtain a vector equation or parametric equations of the
line.
EXAMPLE 5 Find vector and parametric equations of the line that passes through the
points A共3, 2兲 and B共5, 7兲.
SOLUTION We are not explicitly given a vector parallel to the line but observe that the
l
vector v with representation AB is parallel to the line and
v 苷 具 5, 7典 具 3, 2典 苷 具2, 9典
Any nonzero multiple of v can also be used. We can use any point on the line, for
example 共3, 2兲, for the point P0 . The vector equation of the line becomes
r共t兲 苷 r0 tv 苷 具3, 2典 t 具2, 9典
or, equivalently,
r共t兲 苷 共3 2t兲i 共2 9t兲j
Parametric equations of this line are
x 苷 3 2t
y 苷 2 9t
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SECTION A.2
VECTOR FUNCTIONS AND THEIR DERIVATIVES
A13
NOTE If we restrict the parameter t as in Example 5 so that 0 艋 t 艋 1, then the parametric equations x 苷 3 ⫹ 2t, y 苷 ⫺2 ⫹ 9t represent the line segment that starts at 共3, ⫺2兲
and ends at 共5, 7兲. Different parametric representations of the same line can be obtained
by choosing P0 to be 共5, 7兲 or any other point on the line. Likewise, we can use any multiple of v, such as 具 4, 18典 , 具1, 4.5典 , or 具 ⫺6, ⫺27 典 .
Derivatives of Vector Functions
The derivative r⬘ of a vector function r is defined in much the same way as for realvalued functions:
z
r(t+h)-r(t)
Q
P
r(t)
dr
r共t ⫹ h兲 ⫺ r共t兲
苷 r⬘共t兲 苷 lim
hl0
dt
h
4
r(t+h)
C
0
x
(a) The secant vector
z
rª(t)
r(t+h)-r(t)
h
P
Q
r(t)
r(t+h)
if this limit exists. The geometric significance of this definition is shown in Figure 6. If the
l
points P and Q have position vectors r共t兲 and r共t ⫹ h兲, then PQ represents the vector
r共t ⫹ h兲 ⫺ r共t兲, which can therefore be regarded as a secant vector. If h ⬎ 0, the scalar
multiple 共1兾h兲共r共t ⫹ h兲 ⫺ r共t兲兲 has the same direction as r共t ⫹ h兲 ⫺ r共t兲. As h l 0, it
appears that this vector approaches a vector that lies on the tangent line. For this reason, the
vector r⬘共t兲 is called the tangent vector to the curve defined by r at the point P, provided
that r⬘共t兲 exists and r⬘共t兲 苷 0. The tangent line to C at P is defined to be the line through P
parallel to the tangent vector r⬘共t兲. We will also have occasion to consider the unit tangent
vector, which is
C
T共t兲 苷
x
0
(b) The tangent vector
FIGURE 6
ⱍ
r⬘共t兲
r⬘共t兲
ⱍ
The following theorem gives us a convenient method for computing the derivative of a
vector function r; just differentiate each component of r.
5 Theorem If r共t兲 苷 具 f 共t兲, t共t兲典 苷 f 共t兲 i ⫹ t共t兲 j, where f and t are differentiable functions, then
r⬘共t兲 苷 具 f ⬘共t兲, t⬘共t兲典 苷 f ⬘共t兲 i ⫹ t⬘共t兲 j
PROOF
r⬘共t兲 苷 lim
hl0
苷 lim
hl0
苷 lim
hl0
苷
冓
1
关r共t ⫹ h兲 ⫺ r共t兲兴
h
1
关具 f 共t ⫹ h兲, t共t ⫹ h兲典 ⫺ 具 f 共t兲, t共t兲典兴
h
冓
lim
hl0
f 共t ⫹ h兲 ⫺ f 共t兲 t共t ⫹ h兲 ⫺ t共t兲
,
h
h
冔
f 共t ⫹ h兲 ⫺ f 共t兲
t共t ⫹ h兲 ⫺ t共t兲
, lim
hl0
h
h
苷 具 f ⬘共t兲, t⬘共t兲典
Not For Sale
冔
APPENDIX A VECTORS
A14
EXAMPLE 6
(a) Find the derivative of r共t兲 苷 共1 ⫹ t 3 兲 i ⫹ te⫺t j.
(b) Find the unit tangent vector at the point where t 苷 0.
SOLUTION
(a) According to Theorem 5, we differentiate each component of r:
r⬘共t兲 苷 3t 2 i ⫹ 共1 ⫺ t兲e⫺t j
(b) Since r共0兲 苷 i and r⬘共0兲 苷 j, the unit tangent vector at the point 共1, 0兲 is
y
T共0兲 苷
ⱍ
2
(1, 1)
r(1)
0
ⱍ
EXAMPLE 7 For the curve r共t兲 苷 st i ⫹ 共2 ⫺ t兲 j, find r⬘共t兲 and sketch the position vector r共1兲 and the tangent vector r⬘共1兲.
rª(1)
1
r⬘共0兲
j
苷 苷j
r⬘共0兲
1
x
SOLUTION We have
1
i⫺j
2st
r⬘共t兲 苷
FIGURE 7
Notice from Figure 7 that the tangent vector
points in the direction of increasing t. (See
Exercise 47.)
r⬘共1兲 苷
and
1
i⫺j
2
Elimination of the parameter from the equations x 苷 st, y 苷 2 ⫺ t gives y 苷 2 ⫺ x 2,
x 艌 0. In Figure 7 we draw the position vector r共1兲 苷 i ⫹ j starting at the origin and the
tangent vector r⬘共1兲 starting at the corresponding point 共1, 1兲.
Integrals of Vector Functions
The definite integral of a continuous vector function r共t兲 can be defined in much the same
way as for real-valued functions except that the integral is a vector. But then we can express
the integral of r in terms of the integrals of its component functions f and t as follows.
y
b
a
n
r共t兲 dt 苷 lim
兺 r共t*兲 ⌬t
i
n l ⬁ i苷1
冋冉兺
n
苷 lim
nl⬁
i苷1
冊 冉兺
n
f 共t*i 兲 ⌬t i ⫹
冊册
t共t*i 兲 ⌬t j
i苷1
and so
y
b
a
r共t兲 dt 苷
冉y 冊 冉y 冊
b
a
f 共t兲 dt i ⫹
b
a
t共t兲 dt j
This means that we can evaluate an integral of a vector function by integrating each component function.
We can extend the Fundamental Theorem of Calculus to continuous vector functions as
follows:
y
b
a
r共t兲 dt 苷 R共t兲]ba 苷 R共b兲 ⫺ R共a兲
where R is an antiderivative of r, that is, R⬘共t兲 苷 r共t兲. We use the notation x r共t兲 dt for indefinite integrals (antiderivatives).
Not For Sale
SECTION A.2
A15
VECTOR FUNCTIONS AND THEIR DERIVATIVES
EXAMPLE 8 If r共t兲 苷 2 cos t i sin t j, then
冉
y r共t兲 dt 苷 y 2 cos t dt
冊 冉y 冊
i
sin t dt j 苷 2 sin t i cos t j C
where C is a vector constant of integration, and
y
兾2
0
]
兾2
0
苷 2i j
Exercises
A.2
1–2 Find the domain of the vector function.
3–6 Find the limit.
3. lim 具sin t, t ln t 典
冉
20. An object is moving in the xy -plane and its position after
t2
2. r共t兲 苷
i ln共9 t 2 兲 j
t2
1. r共t兲 苷 具 st 1, s5 t 典
tl0
[
r共t兲 dt 苷 2 sin t i cos t j
4. lim
冊
t1
5. lim st 3 i 2
j
tl1
t 1
tl0
冓
e t 1 s1 t 1
,
t
t
2t
6. lim 具arctan t, e
tl
冔
典
7–12 Sketch the curve with the given vector equation. Indicate with
an arrow the direction in which t increases.
7. r共t兲 苷 具t, 2t典
9. r共t兲 苷 sin t i cos t j
11. r共t兲 苷 具 t 3 1, 2t 典
t seconds is r共t兲 苷 共t 3兲 i 共t 2 2t兲 j.
(a) Find the position of the object at time t 苷 5 .
(b) At what time is the object at the point 共1, 8兲?
(c) Does the object pass through the point 共3, 20兲?
(d) Find an equation in x and y whose graph is the path of
the object.
21. The figure shows a curve C given by a vector function r共t兲.
(a) Draw the vectors r共4.5兲 r共4兲 and r共4.2兲 r共4兲.
(b) Draw the vectors
r共4.5兲 r共4兲
0.5
r共4.2兲 r共4兲
0.2
and
(c) Write expressions for r共4兲 and the unit tangent vector T(4).
(d) Draw the vector T(4).
8. r共t兲 苷 具t 2, t 典
y
R
C
10. r共t兲 苷 i cos t j
r(4.5)
12. r共t兲 苷 具t 2, t 3 典
1
Q
r(4.2)
13–15 Find (a) a vector equation and (b) parametric equations for
P
the line passing through the pair of points.
13. 共1, 3兲 and 共2, 7兲
r(4)
14. 共3, 4兲 and 共2, 8兲
0
1
x
15. 共4, 1兲 and 共2, 5兲
22. (a) Make a large sketch of the curve described by the vector
16–18 Find (a) a vector equation, (b) parametric equations, and
(c) a Cartesian equation for the line that passes through the point P
and is parallel to the vector a.
16. P共1, 3兲,
a 苷 具 1, 2 典
18. P共2, 5兲,
a 苷 具 3, 0 典
17. P共4, 5兲,
a 苷 具2, 6典
19. An object is moving in the xy-plane and its position after
t seconds is r共t兲 苷 共t 2 2t兲 i 共t 4兲 j.
(a) Find the position of the object at time t 苷 2 .
(b) At what time is the object at the point 共15, 7兲?
(c) Does the object pass through the point 共20, 9兲?
(d) Find an equation in x and y whose graph is the path of
the object.
function r共t兲 苷 具t 2, t典 , 0 t 2, and draw the vectors
r(1), r(1.1), and r(1.1) r(1).
(b) Draw the vector r共1兲 starting at (1, 1) and compare it with
the vector
r共1.1兲 r共1兲
0.1
Explain why these vectors are so close to each other in
length and direction.
23–26 Find the domain and derivative of the vector function.
23. r共t兲 苷 具t 2, t典
24. r共t兲 苷 t sin t i t cos t j
25. r共t兲 苷 具 t 2 4, st 4 典
26. r共t兲 苷
Not For Sale
冓
t2
t
, 2
t1 t 1
冔
A16
APPENDIX A VECTORS
27–30 Find the tangent vector of unit length at the point with the
given value of the parameter t .
27. r共t兲 苷 具2t, 3t 2 典 ,
38. Let r共t兲 苷 共t 2 t 2兲 i 4t 2 j.
(a) Find a vector tangent to the curve given by r共t兲 at the
point 共2, 4兲.
(b) Find parametric equations for the tangent line to the curve
at the point 共2, 4兲.
t苷1
t 苷 兾6
28. r共t兲 苷 2 sin t i 3 cos t j,
29. r共t兲 苷 共1 t 兲 i t j,
3
2
30. r共t兲 苷 具 t cos t, t sin t典 ,
t苷1
39– 42 Evaluate the integral.
t 苷 兾4
31. r共t兲 苷 具 t 2, t 2 1 典 ,
32. r共t兲 苷 具t 2, t 3 典,
t 苷 1
41.
y 共e
34. r共t兲 苷 e i e
t
j,
t苷0
35. r共t兲 苷 e i e j,
t苷0
t
2t
t
t 苷 兾4
36. r共t兲 苷 共1 cos t兲 i 共2 sin t兲 j,
t 苷 兾6
2
37. Let r共t兲 苷 共t t兲 i j.
t
(a) Find a vector tangent to the curve given by r共t兲 at the
point where t 苷 2 .
(b) Find parametric equations for the tangent line to the curve
at the point where t 苷 2 .
r(t+h)-r(t)
h
rª(t)
Q
冊
4
2t
i
j dt
1 t2
1 t2
42.
y 共cos t i sin t j兲 dt
0
43. Find r共t兲 if r共t兲 苷 t 2 i 4t 3 j and r共0兲 苷 j.
45.
d
关 f 共t兲 u共t兲兴 苷 f 共t兲 u共t兲 f 共t兲 u共t兲
dt
46.
d
关u共 f 共t兲兲兴 苷 f 共t兲 u共 f 共t兲兲
dt
47. Show that the tangent vector to a curve defined by a vector
function r共t兲 points in the direction of increasing t. [Hint: Refer
to Figure 6 and consider the cases h 0 and h 0 separately.]
In this section we show how the ideas of derivatives and tangent vectors can be used in
physics to study the motion of an object, including its velocity and acceleration, along a
curve.
Suppose a particle moves so that its position vector at time t is r共t兲. Notice from Figure 1
that, for small values of h, the vector
1
P
r(t)
r(t+h)
C
FIGURE 1
i ln t j兲 dt
冉
y
Curvilinear Motion: Velocity and Acceleration
y
O
t
1
40.
45– 46 If f is a real-valued function and u is a vector-valued function, where f and u are both differentiable, prove the differentiation formula.
2
A.3
0
44. Find r共t兲 if r共t兲 苷 sin t i cos t j and r共0兲 苷 i j.
t苷1
33. r共t兲 苷 sin t i 2 cos t j,
共16t 3 i 9t 2 j兲 dt
y
31–36
(a) Sketch the plane curve with the given vector equation.
(b) Find r共t兲.
(c) Sketch the position vector r共t兲 and the tangent vector r共t兲 for
the given value of t.
1
39.
x
r共t h兲 r共t兲
h
approximates the direction of the particle moving along the curve r共t兲. Its magnitude measures the size of the displacement vector per unit time. The vector 1 gives the average
velocity over a time interval of length h and its limit is the velocity vector v共t兲 at time t :
2
v共t兲 苷 lim
hl0
r共t h兲 r共t兲
苷 r共t兲
h
Thus the velocity vector is also the tangent vector and points in the direction of the tangent
line.
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SECTION A.3
A17
CURVILINEAR MOTION: VELOCITY AND ACCELERATION
ⱍ
ⱍ
The speed of the particle at time t is the magnitude of the velocity vector, that is, v共t兲 .
As in the case of one-dimensional motion, the acceleration of the particle is defined as the
derivative of the velocity:
a共t兲 苷 v共t兲 苷 r共t兲
EXAMPLE 1 The position vector of an object moving in a plane is given by
r共t兲 苷 t 3 i t 2 j. Find its velocity, speed, and acceleration when t 苷 1 and illustrate
geometrically.
y
SOLUTION The velocity and acceleration at time t are
v(1)
v共t兲 苷 r共t兲 苷 3t 2 i 2t j
a(1)
a共t兲 苷 r共t兲 苷 6t i 2 j
(1, 1)
0
FIGURE 2
x
and the speed is
ⱍ v共t兲 ⱍ 苷 s共3t
兲 共2t兲2 苷 s9t 4 4t 2
2 2
When t 苷 1, we have
v共1兲 苷 3 i 2 j
a共1兲 苷 6 i 2 j
ⱍ v共1兲 ⱍ 苷 s13
These velocity and acceleration vectors are shown in Figure 2.
The vector integrals that were introduced in Section 10.4 can be used to find position vectors when velocity or acceleration vectors are known, as in the following example.
EXAMPLE 2 A moving particle starts at an initial position r共0兲 苷 具1, 0典 with initial
velocity v共0兲 苷 i j. Its acceleration is a共t兲 苷 4t i 6t j. Find its velocity and position
at time t.
SOLUTION Since a共t兲 苷 v共t兲, we have
v共t兲 苷 y a共t兲 dt 苷 y 共4t i 6t j兲 dt
苷 2t 2 i 3t 2 j C
To determine the value of the constant vector C, we use the fact that v共0兲 苷 i j. The
preceding equation gives v共0兲 苷 C, so C 苷 i j and
v共t兲 苷 2t 2 i 3t 2 j i j
苷 共2t 2 1兲 i 共3t 2 1兲 j
Since v共t兲 苷 r共t兲, we have
r共t兲 苷 y v共t兲 dt
苷 y 关共2t 2 1兲 i 共3t 2 1兲 j兴 dt
苷
( 23 t 3 t) i 共t 3 t兲 j D
Putting t 苷 0, we find that D 苷 r共0兲 苷 i, so
r共t兲 苷
( 23 t 3 t 1) i 共t 3 t兲 j
Not For Sale
A18
APPENDIX A VECTORS
In general, vector integrals allow us to recover velocity when acceleration is known and
position when velocity is known:
t
v共t兲 苷 v共t0兲 y a共u兲 du
t0
t
r共t兲 苷 r共t0兲 y v共u兲 du
t0
If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion. The vector version of this law states that if, at any time t, a
force F共t兲 acts on an object of mass m producing an acceleration a共t兲, then
F共t兲 苷 ma共t兲
The angular speed of the object moving with
position P is 苷 d兾dt, where is the angle
shown in Figure 3.
EXAMPLE 3 An object with mass m that moves in a circular path with constant angular
speed has position vector r共t兲 苷 a cos t i a sin t j. Find the force acting on the
object and show that it is directed toward the origin.
v共t兲 苷 r共t兲 苷 a sin t i a cos t j
SOLUTION
y
a共t兲 苷 v共t兲 苷 a 2 cos t i a 2 sin t j
P
Therefore Newton’s Second Law gives the force as
¨
0
x
F共t兲 苷 ma共t兲 苷 m 2共a cos t i a sin t j兲
Notice that F共t兲 苷 m 2 r共t兲. This shows that the force acts in the direction opposite to
the radius vector r共t兲 and therefore points toward the origin (see Figure 3). Such a force
is called a centripetal (center-seeking) force.
FIGURE 3
EXAMPLE 4 A projectile is fired with angle of elevation and initial velocity v0. (See
Figure 4.) Assuming that air resistance is negligible and the only external force is due to
gravity, find the position function r共t兲 of the projectile. What value of maximizes the
range (the horizontal distance traveled)?
y
v¸
a
SOLUTION We set up the axes so that the projectile starts at the origin. Since the force
0
x
d
FIGURE 4
due to gravity acts downward, we have
F 苷 ma 苷 mt j
ⱍ ⱍ
where t 苷 a ⬇ 9.8 m兾s2. Thus
a 苷 t j
Since v共t兲 苷 a, we have
v共t兲 苷 tt j C
where C 苷 v共0兲 苷 v0 . Therefore
r共t兲 苷 v共t兲 苷 tt j v0
Integrating again, we obtain
r共t兲 苷 12 tt 2 j t v0 D
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SECTION A.3
CURVILINEAR MOTION: VELOCITY AND ACCELERATION
A19
But D 苷 r共0兲 苷 0, so the position vector of the projectile is given by
r共t兲 苷 12 tt 2 j t v0
3
ⱍ ⱍ
If we write v0 苷 v0 (the initial speed of the projectile), then
v0 苷 v0 cos i v0 sin j
and Equation 3 becomes
r共t兲 苷 共v0 cos 兲t i [共v0 sin 兲t 12 tt 2 ] j
The parametric equations of the trajectory are therefore
If you eliminate t from Equations 4, you will see
that y is a quadratic function of x. So the path
of the projectile is part of a parabola.
4
x 苷 共v0 cos 兲t
y 苷 共v0 sin 兲t 12 tt 2
The horizontal distance d is the value of x when y 苷 0. Setting y 苷 0, we obtain t 苷 0
or t 苷 共2v0 sin 兲兾t. The latter value of t then gives
d 苷 x 苷 共v0 cos 兲
2v0 sin v02 共2 sin cos 兲
v02 sin 2
苷
苷
t
t
t
Clearly, d has its maximum value when sin 2 苷 1, that is, 苷 兾4.
EXAMPLE 5 A projectile is fired with muzzle speed 150 m兾s and angle of elevation 45
from a position 10 m above ground level. Where does the projectile hit the ground, and
with what speed?
SOLUTION If we place the origin at ground level, then the initial position of the projectile
is (0, 10) and so we need to adjust Equations 4 by adding 10 to the expression for y.
With v 0 苷 150 m兾s, 苷 45, and t 苷 9.8 m兾s2, we have
x 苷 150 cos共兾4兲t 苷 75s2 t
y 苷 10 150 sin共兾4兲t 12 共9.8兲t 2 苷 10 75s2 t 4.9t 2
Impact occurs when y 苷 0, that is, 4.9t 2 75s2t 10 苷 0. Solving this quadratic
equation (and using only the positive value of t), we get
t苷
75s2 s11,250 196
⬇ 21.74
9.8
Then x ⬇ 75s2共21.74兲 ⬇ 2306, so the projectile hits the ground about 2306 m away.
The velocity of the projectile is
v共t兲 苷 r共t兲 苷 75s2 i (75s2 9.8t) j
So its speed at impact is
ⱍ v共21.74兲 ⱍ 苷 s(75s2 )
2
(75s2 9.8 ⴢ 21.74)2 ⬇ 151 m兾s
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A20
A.3
APPENDIX A VECTORS
Exercises
1. The figure shows the path of a particle that moves with
position vector r共t兲 at time t.
(a) Draw a vector that represents the average velocity of the
particle over the time interval 2 t 2.4.
(b) Draw a vector that represents the average velocity over the
time interval 1.5 t 2.
(c) Write an expression for the velocity vector v(2).
(d) Draw an approximation to the vector v(2) and estimate the
speed of the particle at t 苷 2.
10. If an arrow is shot upward on the moon with a velocity of
15 i 40 j m兾s , its position after t seconds is given by
r共t兲 苷 15t i 共40t 0.83t 2 兲 j
(a)
(b)
(c)
(d)
(e)
Find the velocity of the arrow after 1 s.
Find the velocity of the arrow when t 苷 a .
When will the arrow hit the moon?
With what velocity will the arrow hit the moon?
With what speed will the arrow hit the moon?
11. A satellite completes one orbit of the earth along the equator
y
at an altutude of 1000 km every 1 h 46 min. Find the velocity,
speed, and acceleration of the satellite at any time. [Note: The
earth’s radius is approximately 6600 km.]
r(2.4)
2
r(2)
1
r(1.5)
0
1
12. A satellite completes one orbit of the earth along the equator
at an altutude of 2000 km every 2 h 8 min. Find the velocity,
speed, and acceleration of the satellite at any time. [Note: The
earth’s radius is approximately 6600 km.]
x
2
13–14 Find the velocity and position vectors of a particle that
has the given acceleration and the specified initial velocity and
position.
2– 4 The vector function r共t兲 represents the position of a particle at
time t . Find the velocity, speed, and acceleration at the given value
of t .
2. r共t兲 苷 具 4 cos t, 3 sin t典 ,
t苷2
6. r共t兲 苷 具 2 t, 4st 典,
r共0兲 苷 0
15. a共t兲 苷 2 i t j,
v共0兲 苷 i j,
16. a共t兲 苷 t 2 i cos 2t j,
v共0兲 苷 j,
r共0兲 苷 i j
r共0兲 苷 i
17. What force is required so that a particle of mass m has the posi-
tion function r共t兲 苷 t 3 i t 2 j ?
t苷1
7. r共t兲 苷 3 cos t i 2 sin t j ,
8. r共t兲 苷 e t i e 2t j ,
v共0兲 苷 i j,
(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position.
; (b) Use a calculator or computer to graph the path of the particle.
t苷2
given position function. Sketch the path of the particle and draw
the velocity and acceleration vectors for the specified value of t.
1
14. a共t兲 苷 10 j,
r共0兲 苷 0
15–16
t苷1
5–8 Find the velocity, acceleration, and speed of a particle with the
5. r共t兲 苷 具 2 t 2, t 典,
v共0兲 苷 i j,
t 苷 兾3
3. r共t兲 苷 共t 4 7t兲 i 共t 2 2t兲 j,
4. r共t兲 苷 st 2 5 i t j,
13. a共t兲 苷 2 j,
18. A force with magnitude 20 N acts in the positive y -direction on
an object with mass 4 kg. The object starts at the origin with
initial velocity v共0兲 苷 i j. Find its position function and its
speed at time t .
t 苷 兾3
t苷0
19. A projectile is fired with an initial speed of 200 m兾s and
9. If a ball is thrown into the air with a velocity of 10 i 30 j ft兾s ,
its position after t seconds is given by
20. Rework Exercise 19 if the projectile is fired from a position
r共t兲 苷 10t i 共30t 16t 2 兲 j
100 m above the ground.
(a) Find the velocity of the ball when t 苷 1 .
(b) What is the speed of the ball when t 苷 1 ?
;
Graphing calculator or computer required
angle of elevation 60. Find (a) the range of the projectile,
(b) the maximum height reached, and (c) the speed at impact.
21. A ball is thrown at an angle of 45 to the ground. If the ball
lands 90 m away, what was the initial speed of the ball?
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SECTION A.3
22. A gun is fired with angle of elevation 30. What is the
muzzle speed if the maximum height of the shell is 500 m?
23. A gun has muzzle speed 150 m兾s. Find two angles of eleva-
tion that can be used to hit a target 800 m away.
24. A batter hits a baseball 3 ft above the ground toward the cen-
ter field fence, which is 10 ft high and 400 ft from home
plate. The ball leaves the bat with speed 115 ft兾s at an angle
50 above the horizontal. Is it a home run? (In other words,
does the ball clear the fence?)
; 25. Water traveling along a straight portion of a river normally
flows fastest in the middle, and the speed slows to almost
zero at the banks. Consider a long stretch of river flowing
north, with parallel banks 40 m apart. If the maximum water
speed is 3 m兾s, we can use a quadratic function as a basic
model for the rate of water flow x units from the west bank:
3
f 共x兲 苷 400
x共40 x兲.
CURVILINEAR MOTION: VELOCITY AND ACCELERATION
A21
(a) A boat proceeds at a constant speed of 5 m兾s from a point
A on the west bank while maintaining a heading perpendicular to the bank. How far down the river on the opposite bank will the boat touch shore? Graph the path of the
boat.
(b) Suppose we would like to pilot the boat to land at the
point B on the east bank directly opposite A. If we maintain a constant speed of 5 m兾s and a constant heading,
find the angle at which the boat should head. Then graph
the actual path the boat follows. Does the path seem
realistic?
26. Another reasonable model for the water speed of the river in
Exercise 25 is a sine function: f 共x兲 苷 3 sin共 x兾40兲. If a
boater would like to cross the river from A to B with constant
heading and a constant speed of 5 m兾s, determine the angle
at which the boat should head.
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A22
ANSWERS TO VECTOR EXERCISES
Answers to Vector Exercises
A.4
EXERCISES A.1
N
EXERCISES A.2
PAGE A8
1. (a) Scalar
(b) Vector (c) Vector (d) Scalar
l
l l
l l
l l
l
3. AB DC, DA CB, DE EB, EA CE
5. (a)
(b)
u
u+v
v
(c)
w
v
y
≈+¥=1
1
x
0
x
1
u
_v
u-v
v+w
(e)
1
y=2x
0
(d)
5. 2 i ⫹ 2 j
9.
y
w
u+w
u
PAGE A15
N
3. 具 0, 0典
1. [1, 5]
7.
(f)
u
u
w
v
_w
11.
_v
y
x= 81 Á+1
2
u-w-v
v+u+w
0
7. c 2 a 2 b, d 2 b 2 a
1
1
1
9. a 具4, 1 典
x
2
1
11. a 具3, 1 典
y
y
A(_1, 3)
B(3, 2)
B(2, 2)
A(_1, 1)
a
0
0
x
13. 具5, 2 典
x
a
15. 具5, 1 典
y
13. (a) r共t兲 苷 具 1, 3典 ⫹ t具⫺3, 4典
(b) x 苷 1 ⫺ 3t, y 苷 3 ⫹ 4t
15. (a) r共t兲 苷 具4, ⫺1 典 ⫹ t具⫺6, 6典
(b) x 苷 4 ⫺ 6t, y 苷 ⫺1 ⫹ 6t
17. (a) r共t兲 苷 具⫺4, 5典 ⫹ t具⫺2, 6典
(b) x 苷 ⫺4 ⫺ 2t, y 苷 5 ⫹ 6t
(c) y 苷 ⫺3x ⫺ 7
19. (a) 共8, 6兲
(b) t 苷 3
(c) No
(d) x 苷 y 2 ⫺ 6y ⫹ 8
21. (a)
(b), (d)
y
k2, 3l
k6, _2l
r(4.2)-r(4)
0.2
R
C
k3, _4l
k_1, 4l
1
k5, _1l
k5, 2l
Q
r(4.2)
y
r(4.2)-r(4)
x
0
r(4.5)-r(4)
0.5
r(4.5)-r(4)
r(4.5)
17. 13, 具3, 4典, 具7, 20 典, 具10, 24 典, 具7, 4 典
19. s13 , 具3, 1 典, 具1, 7 典, 具4, 6 典, 具10, 7 典
21. s2 , 2 i, 2 j, 2 i 2 j, 7 i j
0
R
C
P
r(4)
r(4.5)
1
Q
1
x
r(4.2)
23. s2, 3 i, i 2 j, 2 i 2 j, 11 i j
P
25. 具 1兾s5, 2兾s5 典
29.
33.
35.
37.
39.
41.
45.
27. 共i j兲兾s2
31. 具 2, 2s3 典
60
⬇ 45.96 ft兾s, ⬇38.57 ft兾s
100 s7 ⬇ 264.6 N, ⬇139.1
s493 ⬇ 22.2 mi兾h, N8W
T1 ⬇ 196 i 3.92 j, T2 ⬇ 196 i 3.92 j
共i 4 j兲兾s17
43. 0
y
(a), (b)
(d) s 97 , t 117
sa
a
0
b
r(4)
0
(c) r⬘共4兲 苷 lim
hl0
r共4 ⫹ h兲 ⫺ r共4兲
r⬘共4兲
; T共4兲 苷
h
r⬘共4兲
ⱍ
23. 共⫺⬁, ⬁兲 , r⬘共t兲 苷 具2t, 1典
27.
具 1兾s10, 3兾s10 典
29.
31. (a), (c)
tb
1
ⱍ
25. 关4, ⬁兲, r⬘共t兲 苷 具 2t, 1兾共2st ⫺ 4 兲 典
具 3兾s13, 2兾s13 典
(b) r⬘共t兲 苷 具 1, 2t 典
y
c
x
T(4)
(_3, 2)
rª(_1)
r(_1)
47. A circle with radius 1, centered at 共x0, y0 兲
0
Not For Sale
x
x
A 23
ANSWERS TO VECTOR EXERCISES
33. (a), (c)
(b) r⬘共t兲 苷 cos t i ⫺ 2 sin t j
y
2
” œ„
, œ„2 ’
2
15. (a) r共t兲 共t 2 t 1兲 i (b)
π
rª” 4 ’
0
π
13. v共t兲 i 共2t 1兲j, r共t兲 t i 共t 2 t兲j
( 16 t 3 t 1) j
8
x
r” 4 ’
(1, 1) t=0
1
35. (a), (c)
17.
19.
21.
25.
rª(0)
(1, 1)
r(0)
0
x
20
1
(b) r⬘共t兲 苷 2e 2t i ⫹ e t j
y
F共t兲 6mt i 2m j
(a) ⬇3535 m
(b) ⬇1531 m
(c) 200 m兾s
30 m兾s
23. ⬇10.2, ⬇79.8
(a) 16 m
(b) ⬇23.6 upstream
12
37. (a) 5 i ⫺ 2 j
(b) x 苷 6 ⫹ 5t, y 苷 1 ⫺ 12 t, t 苷 0
39. 4 i ⫺ 3 j
41. e t i ⫹ 共t ln t ⫺ t兲j ⫹ C
1
43. 3 t 3 i ⫹ 共t 4 ⫹ 1兲j
1
0
0
_4
EXERCISES A.3
PAGE A20
N
1. (a), (b)
(c) lim
hl0
(d)
y
(a)
r共2 ⫹ h兲 ⫺ r共2兲
h
y
v(2)
(b)
r(2.4)
r(2.4)
2
r(2)
2
r(2)
1
r(1.5)
1
r(1.5)
x
0
1
x
2
3. 11 i ⫹ 4 j, s137 , 12 i ⫹ 2 j
5. v共t兲 苷 具⫺t, 1 典
ⱍ
y
v(2)
a共t兲 苷 具⫺1, 0 典
v共t兲 苷 st 2 ⫹ 1
(_2, 2)
ⱍ
a(2)
0 x
7. v共t兲 苷 ⫺3 sin t i ⫹ 2 cos t j
a共t兲 苷 ⫺3 cos t i ⫺ 2 sin t j
v共t兲 苷 s5 sin 2 t ⫹ 4
ⱍ
ⱍ
y
v” π3 ’
(0, 2)
a” π3 ’
0
9. (a) 10 i ⫺ 2 j
3
” 2 , œ„
3’
(3, 0)
x
(b) s104 ft兾s
11. v共t兲 ⬇ 具⫺27,030 sin 3.56t, 27,030 cos 3.56t典,
v ⬇ 27,030 km兾h,
ⱍ ⱍ
a共t兲 ⬇ 具⫺96,131 cos 3.56t, ⫺96,131 sin 3.56t典
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40
_12
4