Smith R., Minton R. Calculus
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10-59 CHAPTER 10
..
Review Exercises 683
Review Exercises
TRUE OR FALSE
State whether each statement is true or false and briefly explain
why. If the statement is false, try to “fix it” by modifying the given
statement to make a new statement that is true.
1. For a given curve, there is exactly one set of parametric equa-
tions that describes it.
2. In parametric equations, circles are always sketched counter-
clockwise.
3. In parametric equations, the derivative equals slope and
velocity.
4. If a point has polar coordinates (r,θ), then it also has polar
coordinates (−r,θ + π).
5. If f is periodic with fundamental period 2π , then one copy of
r = f (θ) is traced out with 0 ≤ θ ≤ 2π.
6. In polar coordinates, you can describe circles but not lines.
7. Tofindall intersections ofpolarcurvesr = f (θ)andr = g(θ ),
solve f (θ) = g(θ).
8. The focus, vertexand directrix of aparabola all lie on the same
line.
9. The equation of any conic section can be written in the form
r =
ed
e cos θ + 1
.
In exercises 1–4, sketch the plane curve defined by the para-
metric equations and find a corresponding x-y equation for the
curve.
1.
x =−1 +3cost
y = 2 +3 sin t
2.
x = 2 −t
y = 1 +3t
3.
x = t
2
+ 1
y = t
4
4.
x = cost
y = cos
2
t − 1
............................................................
In exercises 5–8, sketch the plane curves defined by the para-
metric equations.
5.
x = cos2t
y = sin 6t
6.
x = cos6t
y = sin 2t
7.
x = cos2t cos t
y = cos 2t sint
8.
x = cos2t cos 3t
y = cos 2t sin3t
............................................................
In exercises 9–12, match the parametric equations with the cor-
responding plane curve.
9.
x = t
2
− 1
y = t
3
10.
x = t
3
y = t
2
− 1
11.
x = cos2t cos t
y = cos 2t sint
12.
x = cos(t + cost)
y = cos(t + sin t)
y
x
848 4
1
1
2
3
FIGURE A
y
x
11
1
1
FIGURE B
y
x
3211
8
4
4
8
FIGURE C
y
x
0.5 10.51
0.5
1
0.5
1
FIGURE D
............................................................
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684 CHAPTER 10
..
Parametric Equations and Polar Coordinates 10-60
Review Exercises
In exercises 13 and 14, find parametric equations for the given
curve.
13. The line segment from (2, 1) to (4, 7)
14. The portion of the parabola y = x
2
+ 1 from (1, 2) to (3, 10)
............................................................
In exercises 15 and 16, find the slopes of the curves at the points
(a) t 0, (b) t 1 and (c) (2, 3).
15.
x = t
3
− 3t
y = t
2
− t + 1
16.
x = t
2
− 2
y = t + 2
............................................................
In exercises 17 and 18, parametric equations for the position of
an object are given. Find the object’s velocity and speed at time
t 0 and describe its motion.
17.
x = t
3
− 3t
y = t
2
+ 2t
18.
x = t
3
− 3t
y = t
2
+ 2
............................................................
In exercises 19–22, find the area enclosed by the curve.
19.
x = 3sin t
y = 2 cos t
20.
x = 4sin 3t
y = 3 cos 3t
21.
x = cos2t
y = sin π t
, −1 ≤ t ≤ 1
22.
x = t
2
− 1
y = t
3
− t
, −1 ≤ t ≤ 1
............................................................
In exercises 23–26, find the arc length of the curve (approximate
numerically, if needed).
23.
x = cos2t
y = sin π t
, −1 ≤ t ≤ 1
24.
x = t
2
− 1
y = t
3
− 4t
, −1 ≤ t ≤ 1
25.
x = cos4t
y = sin 5t
26.
x = sin10t
y = t
2
− 1
, −π ≤ t ≤ π
............................................................
In exercises 27 and 28, compute the surface area of the surface
obtained by revolving the curve about the indicated axis.
27.
x = t
3
− 4t
y = t
4
− 4t
, −1 ≤ t ≤ 1, about the x-axis
28.
x = t
3
− 4t
y = t
4
− 4t
, −1 ≤ t ≤ 1, about y = 2
............................................................
In exercises 29 and 30, sketch the graph of the polar equation
and find a corresponding x-y equation.
29. r = 3cosθ 30. r = 2 sec θ
............................................................
In exercises 31–38, sketch the graph and identify all values of θ
where r 0 and a range of values of θ that produces one copy
of the graph.
31. r = 2sinθ 32. r = 2 −2sinθ
33. r = 2 −3sinθ 34. r = cos3θ + sin2θ
35. r
2
= 4sin2θ 36. r = e
cosθ
− 2cos4θ
37. r =
2
1 + 2 sinθ
38. r =
2
1 + 2 cosθ
............................................................
In exercises 39 and 40, find a polar equation corresponding to
the rectangular equation.
39. x
2
+ y
2
= 9 40. (x − 3)
2
+ y
2
= 9
............................................................
In exercises 41 and 42, find the slope of the tangent line to the
polar curve at the given point.
41. r = cos3θ at θ =
π
6
42. r = 1 −sinθ at θ = 0
............................................................
In exercises 43–48, find the area of the indicated region.
43. One leaf of r = sin5θ
44. One leaf of r = cos2θ
45. Inner loop of r = 1 −2sinθ
46. Bounded by r = 3sinθ
47. Inside of r = 1 +sinθ and outside of r = 1 + cos θ
48. Inside of r = 1 +cosθ and outside of r = 1 + sin θ
............................................................
In exercises 49 and 50, find the arc length of the curve.
49. r = 3 −4sinθ 50. r = sin4θ
............................................................
In exercises 51–53, find an equation for the conic section.
51. Parabola with focus (1, 2) and directrix y = 0
52. Ellipse with foci (2, 1) and (2, 3) and vertices (2, 0) and (2, 4)
53. Hyperbola with foci (2, 0) and (2, 4) and vertices (2, 1) and
(2, 3)
............................................................
In exercises 54–58, identify the conic section and find each ver-
tex, focus and directrix.
54. y = 3(x − 2)
2
+ 1
55.
(x + 1)
2
9
+
(y − 3)
2
25
= 1
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10-61 CHAPTER 10
..
Review Exercises 685
Review Exercises
56.
x
2
9
−
(y + 2)
2
4
= 1
57. (x − 1)
2
+ y = 4
58. (x − 1)
2
+ 4y
2
= 4
............................................................
59. Aparabolic satellite dishhas the shape y =
1
2
x
2
.Where should
the microphone be placed?
60. If a hyperbolic mirror is in the shape of the top half of
(y + 2)
2
−
x
2
3
= 1, to which point will light rays following
the path y = cx (y < 0) reflect?
In exercises 61–64, find a polar equation and graph the conic
section with focus (0, 0) and the given directrix and eccentricity.
61. Directrix x = 3, e = 0.8
62. Directrix y = 3, e = 1
63. Directrix y = 2, e = 1.4
64. Directrix x = 1, e = 2
............................................................
In exercises 65 and 66, find parametric equations for the conic
sections.
65.
(x + 1)
2
9
+
(y − 3)
2
25
= 1
66.
x
2
9
−
(y + 2)
2
4
= 1
EXPLORATORY EXERCISE
1. Sketch several polar graphs of the form r = 1 + a cos θ and
r = 1 +a sinθ using some constants a that are positive and
some that are negative, greater than 1, equal to 1 and less than
1 (for example, a =−2, a =−1, a =−1/2, a = 1/2, a = 1
and a = 2). Discuss all patterns you find.
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CHAPTER
11
Vectors and the Geometry
of Space
Auto racing at all levels has become a highly technological competition.
Formula One race cars are engineered with large wings on the back and
with undersides shaped like upside-down airplane wings, to use the air
flow to help hold the cars to the track even at speeds up to 200 mph.
The engineering of stock cars is less obvious to the untrained eye, but
is no less important. Stock car racers are challenged by, among other
things, intricate rules that severely limit the extent to which the cars can
be modified. Nevertheless, at the Bristol Motor Speedway, the track is an
oval only 0.533 mile in length and racers regularly exceed 120 miles per
hour, completing a lap in just over15 seconds. These speeds would be unsafe if the
track were not specially designed for high-speed racing. In particular, the Bristol
track is steeply banked, with a 16-degree bank on straightaways and a spectacular
36-degree bank in the corners.
As you will see in the exercises in section 11.3, the banking ofa road changes
the role of gravity.In effect,part ofthe weight ofthe car isdiverted into aforce that
helps a car make its turn safely. In this chapter, we introduce vectors and develop
the calculations needed to resolve
vectors into components. This is
a fundamental tool for engineers
designingracecarsandracetracks.
This chapter represents a
crossroadsfromtheprimarilytwo-
dimensional world of first-year
calculus to the three-dimensional
world of many important scien-
tific and engineering problems.
Therestofthecalculuswedevelop
in this book builds directly on the
basic ideas developed here.
687
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688 CHAPTER 11
..
Vectors and the Geometry of Space 11-2
11.1 VECTORS IN THE PLANE
Todescribe the velocity of a moving object, we must specify both its speed and the direction
in which it’s moving. Acceleration and force also have both a size (e.g., speed) and a
direction. We represent such a quantity graphically as a directed line segment, that is, a line
segment with a specific direction (indicated by an arrow).
P (initial point)
Q (terminal point)
PQ
FIGURE 11.1
Directed line segment
We denote the directed line segment extending from the point P (the initial point)to
the point Q (the terminal point)by
−→
PQ. (See Figure 11.1.) We refer to the length of
−→
PQ
as its magnitude, denoted
−→
PQ. Mathematically, we consider all directed line segments
with the same magnitude and direction to be equivalent, regardless of the location of their
initial point and we use the term vector to describe any quantity that has both a magnitude
and a direction. We should emphasize that the location of the initial point is not relevant;
only the magnitude and direction matter. In other words, if
−→
PQ is the directed line segment
from the initial point P to the terminal point Q, then the corresponding vector v represents
−→
PQ as well as every other directed line segment having the same magnitude and direction
as
−→
PQ. In Figure 11.2, we indicate three vectors that are all considered to be equivalent,
even though their initial points are different. In this case, we write
a = b = c.
It is often helpful to think of vectors as representing some specific physical quantity.
For instance, when you see the vector
−→
PQ, you might imagine moving an object from the
initial point P to the terminal point Q. In this case, the magnitude of the vector would
represent the distance the object is moved and the direction of the vector would point from
the starting position to the final position.
a
b
c
y
x
FIGURE 11.2
Equivalent vectors
In this text, we usually denote vectors by boldface characters such as a, b and c,as
seen in Figure 11.2. Since you will not be able to write in boldface, you should use the
arrow notation (e.g.,
−→
a ). When discussing vectors, we refer to real numbers as scalars.
It is very important that you begin now to carefully distinguish between vector and scalar
quantities. This will save you immense frustration both now and as you progress through
the remainder of this text.
Look carefully at the three vectors shown in Figure 11.3a. If you think of the vector
−→
AB as representing the displacement of a particle from the point A to the point B, notice
that the end result of displacing the particle from A to B (corresponding to the vector
−→
AB),
followed by displacing the particle from B to C (corresponding to the vector
−→
BC) is the
same as displacing the particle directly from A to C, which corresponds to the vector
−→
AC
(called the resultant vector). We call
−→
AC the sum of
−→
AB and
−→
BC and write
−→
AC =
−→
AB +
−→
BC.
To add two vectors, we locate the initial point of one at the terminal point of the other
and complete the parallelogram, as indicated in Figure 11.3b. The vector lying along the
diagonal, with initial point at A and terminal point at C is the sum
−→
AC =
−→
AB +
−→
AD.
C
A
B
AC
→
BC
→
AB
→
C
D
A
B
AC
→
AD
→
BC
→
AB
→
FIGURE 11.3a
Resultant vector
FIGURE 11.3b
Sum of two vectors
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11-3 SECTION 11.1
..
Vectors in the Plane 689
Asecond basic arithmetic operation for vectorsis scalar multiplication. Ifwemultiply
a vector u by a scalar (a real number) c > 0, the resulting vector will have the same
direction as u, but will have magnitude cu. On the other hand, multiplying a vector u by
a scalar c < 0 will result in a vector with opposite direction from u and magnitude |c|u.
(See Figure 11.4.)
u
3u
2u
Qu
a
y
x
A(a
1
, a
2
)
a
a
2
a
1
O
a
2
a
1
FIGURE 11.4
Scalar multiplication
FIGURE 11.5
Position vector a =a
1
, a
2
Since the location of the initial point is irrelevant, we typically draw vectors with their
initial point located at the origin. Such a vector is called a position vector. Notice that the
terminal point of a position vector will completely determine the vector, so that specifying
the terminal point will also specify the vector. For the position vector a with initial point at
theorigin and terminal point at the point A(a
1
, a
2
)(see Figure 11.5), we denote the vectorby
a =
−→
OA =a
1
, a
2
.
We call a
1
and a
2
the components of the vector a;a
1
is the first component and a
2
is
the second component. Be careful to distinguish between the point (a
1
, a
2
) and the position
vector a
1
, a
2
. Note from Figure 11.5 that the magnitude of the position vector a follows
directly from the Pythagorean Theorem. We have
a=
a
2
1
+ a
2
2
. (1.1)
Magnitude of a vector
Notice that it follows from the definition that for two position vectors a =a
1
, a
2
and
b =b
1
, b
2
, a = b if and only if their terminal points are the same, that is if a
1
= b
1
and
a
2
= b
2
. In other words, two position vectors are equal only when their corresponding
components are equal.
y
x
A(a
1
, a
2
)
B(b
1
, b
2
)
C(a
1
b
1
, a
2
b
2
)
O
FIGURE 11.6
Adding position vectors
To add two position vectors,
−→
OA =a
1
, a
2
and
−→
OB =b
1
, b
2
, we draw the position
vectorsin Figure 11.6andcompletetheparallelogram,asbefore.FromFigure11.6,wehave
−→
OA +
−→
OB =
−→
OC.
Writingdownthepositionvectorsintheircomponentform,wetakethisasourdefinition
of vector addition:
a
1
, a
2
+b
1
, b
2
=a
1
+ b
1
, a
2
+ b
2
. (1.2)
Vector addition
So, to add two vectors, we simply add the corresponding components. For this reason, we
say that addition of vectors is done componentwise. Similarly, we define subtraction of
vectors componentwise, so that
a
1
, a
2
−b
1
, b
2
=a
1
− b
1
, a
2
− b
2
. (1.3)
Vector subtraction
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690 CHAPTER 11
..
Vectors and the Geometry of Space 11-4
We give a geometric interpretation of subtraction later in this section.
Recall that if we multiply a vector a by a scalar c, the result is a vector in the same
direction as a (for c > 0) or the opposite direction as a (for c < 0), in each case with
magnitude |c|a. We indicate the case of a position vector a =a
1
, a
2
and scalar multiple
c > 1inFigure11.7aandfor0 < c < 1inFigure11.7b.Thesituationfor c < 0isillustrated
in Figures 11.7c and 11.7d.
y
x
C(ca
1
, ca
2
)
A(a
1
, a
2
)
O
FIGURE 11.7a
Scalar multiplication (c > 1)
For the case where c > 0, notice that a vector in the same direction as a, but with
magnitude |c|a, is the position vector ca
1
, ca
2
, since
ca
1
, ca
2
=
(ca
1
)
2
+ (ca
2
)
2
=
c
2
a
2
1
+ c
2
a
2
2
=|c|
a
2
1
+ a
2
2
=|c|a.
y
x
C(ca
1
, ca
2
)
A(a
1
, a
2
)
O
y
x
C(ca
1
, ca
2
)
A(a
1
, a
2
)
O
y
x
C(ca
1
, ca
2
)
A(a
1
, a
2
)
O
FIGURE 11.7b
Scalar multiplication (0 < c < 1)
FIGURE 11.7c
Scalar multiplication (c < −1)
FIGURE 11.7d
Scalar multiplication (−1 < c < 0)
Similarly, if c < 0, you can show that ca
1
, ca
2
is a vector in the opposite direction from a,
withmagnitude|c|a.Forthisreason,wedefinescalarmultiplicationofpositionvectorsby
Scalar multiplication
ca
1
, a
2
=ca
1
, ca
2
, (1.4)
for any scalar c. Further, notice that this says that
ca=|c|a. (1.5)
EXAMPLE 1.1 Vector Arithmetic
For vectors a =2, 1 and b =3, −2, compute (a) a + b, (b) 2a, (c) 2a + 3b,
(d) 2a − 3b and (e) 2a −3b.
Solution (a) From (1.2), we have
a + b =2, 1+3, −2=2 +3, 1 −2=5, −1.
(b) From (1.4), we have
2a = 22, 1=2 ·2, 2 · 1=4, 2.
(c) From (1.2) and (1.4), we have
2a + 3b = 22, 1+33, −2=4, 2+9, −6=13, −4.
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11-5 SECTION 11.1
..
Vectors in the Plane 691
(d) From (1.3) and (1.4), we have
2a − 3b = 22, 1−33, −2=4, 2−9, −6=−5, 8.
(e) Finally, from (1.1), we have
2a − 3b = −5, 8 =
√
25 + 64 =
√
89.
Observe that if we multiply any vector (with any direction) by the scalar c = 0, we get
a vector with zero length, the zero vector:
0 =0, 0.
Further, notice that this is the only vector with zero length. (Why is that?) The zero vector
also has no particular direction. Finally, we define the additive inverse −a of a vector a in
the expected way:
−a =−a
1
, a
2
=(−1)a
1
, a
2
=−a
1
, −a
2
.
Notice that this says that the vector −a is a vector with the opposite direction as a and since
−a=(−1)a
1
, a
2
= |−1|a=a,
−a has the same length as a.
DEFINITION 1.1
Two vectors having the same or opposite direction are called parallel. The zero
vector is considered parallel to every vector.
It then follows that two (nonzero) position vectors a and b are parallel if and only if
b = ca, for some scalar c. In this event, we say that b is a scalar multiple of a.
EXAMPLE 1.2 Determining When Two Vectors Are Parallel
Determine whether the given pair of vectors is parallel: (a) a =2, 3 and b =4, 5,
(b) a =2, 3 and b =−4, −6.
Solution (a) Notice that from (1.4), we have that if b = ca, then
4, 5=c2, 3=2c, 3c.
For this to hold, the corresponding components of the two vectors must be equal. That
is, 4 = 2c (so that c = 2) and 5 = 3c (so that c = 5/3). This is a contradiction and so, a
and b are not parallel.
(b) Again, from (1.4), we have
−4, −6=c2, 3=2c, 3c.
In this case, we have −4 = 2c (so that c =−2) and −6 = 3c (which again leads us to
c =−2). This says that −2a =−4, −6=b and so, 2, 3 and −4, 6 are parallel.
We denote the set of all position vectors in two-dimensional space by
V
2
={x, y|x, y ∈ R}.
You can easily show that the rules of algebra given in Theorem 1.1 hold for vectors in V
2
.
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692 CHAPTER 11
..
Vectors and the Geometry of Space 11-6
THEOREM 1.1
For any vectors a, b and c in V
2
, and any scalars d and e in R, the following hold:
(i) a + b = b + a (commutativity)
(ii) a + (b + c) = (a +b) +c (associativity)
(iii) a + 0 = a (zero vector)
(iv) a +(−a) = 0 (additive inverse)
(v) d(a +b) = da + db (distributive law)
(vi) (d + e)a = da +ea (distributive law)
(vii) (1)a = a (multiplication by 1) and
(viii) (0)a = 0 (multiplication by 0).
b
a
a b
y
x
O
FIGURE 11.8
b + (a − b) = a
PROOF
We prove the first of these and leave the rest as exercises. By definition,
a + b =a
1
, a
2
+b
1
, b
2
=a
1
+ b
1
, a
2
+ b
2
Since addition of real
numbers is commutative.
=b
1
+ a
1
, b
2
+ a
2
=b + a.
Notice that using the commutativity and associativity of vector addition, we have
b + (a − b) = (a −b) +b = a + (−b + b) = a +0 = a.
From our graphical interpretation of vector addition, we get Figure 11.8. Notice that this
now gives us a geometric interpretation of vector subtraction.
For any two points A(x
1
, y
1
) and B(x
2
, y
2
), observe from Figure 11.9 that the vector
−→
AB corresponds to the position vector x
2
− x
1
, y
2
− y
1
.
y
A(x
1
, y
1
)
B(x
2
, y
2
)
x
x
1
y
1
y
2
x
2
x
2
x
1
y
2
y
1
AB
→
FIGURE 11.9
Vector from A to B
EXAMPLE 1.3 Finding a Position Vector
Find the vector with (a) initial point at A(2, 3) and terminal point at B(3, −1) and
(b) initial point at B and terminal point at A.
Solution (a) We show this graphically in Figure 11.10a. Notice that
−→
AB =3 −2, −1 −3=1, −4.
x
y
1, 4
A(2, 3)
B(3, 1)
22
2
4
2
4
AB
→
x
y
1, 4
A(2, 3)
B(3, 1)
22
2
4
2
4
BA
→
FIGURE 11.10a
−→
AB =1, −4
FIGURE 11.10b
−→
BA =−1, 4