Vectors
define vectors
Some physical quantities, such as temperature, length, and mass, can be specified
by a single number called a scalar. Other physical quantities, such as force and velocity,
must be characterized by a nonnegative magnitude and a direction. We call these
quantities vectors. Vectors are frequently named by boldface letters, such as v, w, or F.
Geometrically, we represent a vector with a directed line segment or an arrow.
DRAW. If we draw a vector in the x-y plane, then a vector has an initial point, P, and a
terminal point, Q. If P(1,2) and Q(4,3) , PQ represents the vector from P to Q.
The length or magnitude of the vector PQ is denoted by either PQ or PQ , with
our text using the former notation. The length of a vector PQ may be found using the
!
!
!
Distance Formula. SHOW. PQ = (x 2 " x1 ) 2 + (y 2 " y1 ) 2 . If P(1,2) and Q(4,3) , then
!
!
!
2
2
PQ = (4 "1) + (3 " 2) = 9 + 1 = 10 .
!
!
!
!
!
Two vectors are equal if they have the same length and direction. They do not
!
!
! initial and terminal points. We can
have to have the same
move a vector around the plane
if we do not change its length or direction. E.g., we can move a vector so that its initial
point is the origin. E.g., the vector PQ from P(1,2) to Q(4,3) is equal to the vector OS
from O(0,0) to S(3,1) . We call the vector OS the position vector (or radius vector) of
point P. If coordinates of P and Q are P(x1, y1 ) and Q(x 2 , y 2 ) , then position vector
equal to PQ is the vector!OS where! O(0,0) and
!
! S(x 2 " x1, y 2 " y1 ) .
vector
addition
!
!
!
The vector PQ may describe
the net!movement or displacement of an object from
!
! Q. If the
!a point P to a point
! object then
! moves from Q to R by the vector QR , we can
consider the net displacement from P to R as the vector PR . This resultant vector is
defined to!be the sum of the vectors PQ and QR , PR = PQ + QR . DRAW. We see that
QR + PQ = PQ + QR , because both sums represent the diagonal
! of a parallelogram.
Vector addition is commutative.
!
Let PQ represent the
vector
from
P(1,2)
to Q(4,3) and QR the vector from
!
! !
! Q(4,3)
! to!R(3,5) . The resultant vector PR is the vector from P(1,2) to R(3,5) ..
vector components
Earlier, we said that we can move a vector in the plane as long as its direction and
!
!
!
!
OS
magnitude do not change. Draw
from
O(0,0)
to
S(3,1)
. Now drop a perpendicular
!
!
!
!
from S to the x-axis forming a right triangle. We may consider the vector OS as the sum
of two component vectors along the x- and y-axes. We will write these components as
v x and v y . Here, v x =!3 and v y = 1. Since these components uniquely define the vector,
!
!
we can write the vector in component form as 3,1 .
!
Writing a vector in component form simplifies vector addition. In our previous
! example,!we could have
! written PQ = 3,1 and QR = "1,2 and PR = 3 "1,1+ 2 = 2,3 .
In general, if u = u1,u2 and v = v1!,v 2 , then u + v = u1 + v1,u2 + v 2 .
!
!
!
!
!
!
Let the vectors F and G represent two forces acting on an object and " is the
angle between the two forces, where F = 8N, G = 10N, " = 60° . Find the magnitude
and direction of the resultant vector. We can write F as 8,0 . To write G in component
form as G = G1,G2 , observe that G1 = G cos60° and G2 = G sin60°
! . So
1
3
G = 10cos60°,10sin60° = !10 " ,10 "
= 5,5 3 . So F+G = 8 + 5,0 + 5 3 =
2
2
!
!
! of the resultant vector,
! F+G, is equal to 132 + (5 3) 2 =
13,5 3 . The magnitude
!
!
!
!
!
!
!
!"1 ( 5 3 ) # 33.67° .
169 + 75 = 244 " 15.62 , and its angle is equal to tan
13
scalar multiplication of vectors
!
We can also multiply vectors by a scalar. For each real number k and each vector
v = v1,v 2 , we define a vector kv by the equation kv = kv1,kv 2 . For example, let u =
!
3,1 , then 2u = 6,2 and –u = "3,"1 . Geometrically, the length of the vector u is
multiplied by a factor of k and the direction of ku is the same as u if k > 0, otherwise the
direction of ku is the opposite of u if k < !
0.
linear combination of vectors
!
!
We can create linear combinations of two or more vectors. Let u = 3,1 and v =
!
"1,2 . Then 3u + 2v = 7,7 and u – v = 4,"1 . You'll notice that we have defined the
subtraction of two vectors, u – v as u + (-v).
We may also define a zero vector called 0. The zero vector has length zero.
!
unit vectors
!
! length one. We can find unit vectors in any direction.
A unit vector
is a vector of
Find a unit vector in the direction of u = 3,1 . Earlier, we found that u = 10 . To find a
1
1
unit vector in the direction of u, we need to multiply u by , i.e.,
3,1 . A unit
u
10
u
!
!
vector in the direction of u is defined
as .
u
!
normalization of vectors
!
!
Sometimes we want to find a vector of a particular length in a given direction; this
is called normalizing the vector. Find a vector of length 5 in the direction of u = 3,1 .
!
1
First, we find a unit vector in the direction of u, namely
3,1 . Then we multiply
10
5
!
this unit vector by 5 to obtain
3,1 .
10
!
unit vector component form
!
Two useful unit vectors are along the coordinate axes, namely i = 1,0 and j =
0,1 . Since we can write!any vector in component form as x, y , we can also write any
!
vector as a sum or difference of unit vectors, i.e., u = 3,1 is equivalent to u = 3i + j.
dot products and angles between vectors
!
Given two vectors u = u1,u2 , and v = v1,v 2 , we define the dot product (also
!
known as the scalar product) of u and v as u " v = u1v1 + u2v 2 . Find the dot product of u =
!
!
!
!
!
!
!
3,1 and v = "1,2 . u " v = u1v1 + u2v 2 = 3(#1) + 1(2) = #3 + 2 = #1. Dot products are real
numbers that can be positive, negative, or zero.
If " represents the angle between the two nonzero vectors u and v, then it can be
u# v
shown
9.3.75 on page 568) that cos" =
. We can use this formula to
!
! (see exercise
uv
find the angle between two vectors. Find the angle between u = 3,1 and v = "1,2 .
!
u# v
$1
First, we find the cosine of the angel between u and v, namely cos" =
=
uv
10 # 5
!
1
! $1 ($.0.141421)
! # 98.13° .
="
# "0.141421 . Now we find the angle " : " # cos
50
Note that two vectors are perpendicular if their dot product is equal to zero!
!
Perpendicular vectors are sometimes called orthogonal vectors.
Find a vector orthogonal to u !
= !
3,1 . Many solutions are possible, e.g., v =
1,"3 or "1,3 . All solutions are scalar multiples of 1,"3 .
projections of vectors
A projection of a vector
! u onto another vector v is the vector formed by projecting
the endpoint of u onto the line containing v. Since the projection of u onto v lies along
!
! (or a multiple) of v. The multiple of the
the line containing v, it will be a normalization
u" v
projection vector is equal to
.
v
Find the projection of the vector u = 3,1 onto v = "1,2 . The multiple of the
u " v 3(#1) + 1(2) #1
1
projection vector is equal to
. So, the projection of u
=
=
=#
2
2
v
5
5
(#1)
+
2
!
1
2
!1
!
"1,2 =
,"
.
5
5
5
Vectors in !
3-D
three-dimensional coordinate system
In the plane, each point is associated with an ordered pair of real numbers, (x, y) .
! point is associated with an ordered triple of real numbers, (x, y,z) .
In space, each
Through a fixed point, the origin O, we draw three mutually perpendicular lines, the xaxis, the y-axis, and the z-axis. On each axis, we select an appropriate scale and the
!
positive direction. The direction for the positive z-axis makes the three-dimensional
!
coordinate system right-handed.
Sketch the point P(1,2,3) .
3-D graphs
Let's locate some other points: (1,0,0),(0,2,0),(0,0,3),(1,2,0),(1,0,3),(0,2,3) . Along
with points P and O, we have plotted the points of a rectangular solid in space. Points of
! lie along the x-axis; points of the form (0, y,0) lie along the y-axis; and
the form (x,0,0)
points of the form (0,0,z) lie along the z-axis. Points of the form (x, y,0) lie in a plane
!
called the xy-plane. Points of the form (x,0,z) lie in a plane called the xz-plane. Points of
the form (0, y,z) lie in a plane called the yz-plane. In our example, the points
!
!
(1,0,0),(0,2,0),(0,0,0),(1,2,0) lie in the xy-plane.
!
!
!
!
onto v is the vector "
!
All points obeying the equation z = 3 (e.g., (0,0,3),(1,0,3),(0,2,3),(1,2,3) ) lie in a
plane parallel to and three units above the xy-plane.
In general, equations of planes in " 3 are of the form ax + by + cz + d = 0 , where a,
b, c, and d are real numbers. E.g.,
equations represent planes in space:
! the following
!
x + 3y + 2z = 6, y + z = 5, and x = 4 . The first has intercepts (6,0,0), (0,2,0), and (0,0,3);
the second is parallel to the x-axis;
parallel to the yz-plane.
! and the third is !
The distance between points in space is given
by an extension of the distance
formula in the plane. If P(x1, y1,z1 ) and Q(x 2 , y 2 ,z2 ) , then the distance d between P and
!
Q is d = (x 2 " x1 ) 2 + (y 2 " y1 ) 2 + (z2 " z1 ) 2 . E.g., the distance d between P(1,2,3) and
!
!
!
!
!
!
!
Q("2,0,4) is d = ("2 "1) 2 + (0 " 2) 2 + (4 " 3) 2 = 9 + 4 + 1 = 14 . The equation of a
!
!
sphere has the form (x " a) 2 + (y " b) 2 + (z " c) 2 = d 2 , where the center of the sphere is
! d. The equation of
the point in space ( a,b,c ) and the radius of the sphere is the distance
the !
sphere with center at P(1,2,3) and containing Q("2,0,4) is
2
2
(x "1) 2 + (y
! " 2) + (z " 3) = 14 .
The
(quadric surfaces are generalizations of the
! equation of a quadric 2surface
2
conic sections) !
have the form Ax + By !+ Cz 2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0 .
x2 y2
E.g., z 2 = 2 + 2 represents an elliptic cone.
a
b
3-D vectors
!
To represent vectors in space, we introduce the unit vector k along the positive zaxis. If v is a vector initial point at the origin O and terminal point at P(a,b,c) , then v =
ai + bj + ck. The scalars a, b, and c are called the components of the vector v. We can
also write the vector v in component form as a,b,c . We call a vector whose initial
point is the origin a position vector.
!
We can move vectors around in space, too. The vector v from P(x1, y1,z1 ) to
Q(x 2 , y 2 ,z2 ) is equal to the position vector (x 2 " x1 )i + (y 2 " y1 ) j + (z2 " z1 )k . E.g., the
!
vector v from P(1,2,3) and Q("2,0,4) is equal to ("2 "1)i + (0 " 2) j + (4 " 3)k = -3i – 2j
+ k.
!
We multiply and add vectors in space the same as we do in the plane.
Find 2v – 3w where v = 2i!+ 3j – 2k and w = 3i – 4j + 5k. 2v – 3w =
! – 19k.
(4 "!
9)i + (6 + 12) !
j + ("4 "15)k = -5i + 18j
length, unit vectors
The length (magnitude) of a 3-D vector v = ai +bj +ck is defined to be
v = a 2 + b 2 + c 2 . E.g., the length of v = 2i - 3j – 6k is v = 2 2 + ("3) 2 + ("6) 2 = 49
= 7.
v
Unit vectors are defined the same as in the plane, u = . Find the unit vector in
v
!
1
2 3 6
the same direction as v = 2i - 3j – 6k. We have u = (2i " 3j " 6k) = i " j " k .
7
7 7 7
dot products
!
!
The definition of a dot product of two vectors v = ai + bj + ck and w = di +ej +fk
is extended to space as v " w = ad + be + cf . For example, the dot product of v = 2i + 3j –
2k and w = 3i – 4j + 5k is v " w = 2(3) + 3(#4) + (#2)(5) = 6 #12 #10 = #16 .
angle between vectors, orthogonal vectors
v#w
The definition
of cos" =
is the same in space as it was in the plane. Find the
!
vw
!
angle between v = 2i + 3j – 2k and w = 3i – 4j + 5k. Since
#16
16
cos" =
=#
$ #.548795 , we have " # 123.284°.
( 17)( 50)
850
!
Two vectors are orthogonal in space iff their dot product is zero. E.g., the
vectors v = 2i + 3j – 2k and w = 3i – 4j - 3k are orthogonal.
! and direction cosines
writing vectors in terms of magnitude
!
A nonzero vector v can be described by specifying its magnitude and its direction
angles " , " , and " , where " is the angle between v and i, " is the angle between v and j,
and " is the angle between v and k ( 0 " #, $ , % " & ).
v#i a
v# j b
= , cos " =
= , and
Let v = ai + bj + ck. Notice that cos" =
v
i
v
v
j
v
! !
!
!
!
v#k c
!
!
cos " =
= . This is equivalent
to a = v cos" , b = v cos " , and c = v cos " .
vk v
Therefore, we can write v = ai !
+ bj + ck = v cos!
" i + v cos # j + v cos " k . This form
of the vector is called its magnitude and direction cosine form.
!
E.g., write the vector v!= "3 i + 2j –!6k in terms of magnitude
and direction
!
3
2
2
2
angles. DRAW. We see that v =! a + b + c = 9 + !
4 + 36 = 49 = 7 , cos" = # ,
7
2
6
cos " = , and cos " = #! . So, " = cos#1(# 3 7) $ 115.4° , " = cos#1(2 7) $ 73.4°, and
7
7
#1
! . Therefore, v = 7cos115.4°i + 7cos 73.4°j
+ 7cos149.0°k .
" = cos (#6 7) $ 149.0°
!
cross-products
!
!
!
! If v = ai + bj + ck and w = di + ej + fk, we define the cross product of v and w to
be v x w = (bf " ce)i + (cd " af )j + (ae " bd)k . Cross products are only defined for
!
! of two 3-D vectors is another vector, whereas the dot
3-D vectors. The cross products
product is a scalar (number).
E.g., let v = 2i - j + 3k and w = 7j - 4k, then v x w =
!
("1# 4 " 3# 7)i + (3# 0 " 2 # "4)j + (2 # 7 " "1# 0)k = -17i + 8j + 14k.
geometric interpretations of cross-products
The cross product, v x w, of two vectors, v and w, has some important properties.
First, the cross product v x w is orthogonal to both v and w. Therefore, v x w = !
17i + 8j + 14k is orthogonal to both v = 2i - j + 3k and w = 7j - 4k. Show that the dot
products (v x w) " v and (v x w) " w are both equal to zero.
Second, the length of the cross product v x w, v x w , is equal to the area of the
parallelogram having v and w as adjacent sides. Therefore, the area of the parallelogram
having!sides v = 2i - !
j + 3k and w = 7j - 4k is equal to
-17i + 8j +14k = 17 2 + 8 2 + 14 2 = 289 +!64 + 196 = 549 " 23.43 .
!
!
!
!
!
!
vector-valued functions
A vector-valued function F assigns to each number t a unique vector
F(t) = f 1 (t)i + f 2 (t)j + f 3 (t)k , where f1, f 2 , and f 3 are real-valued functions of t.
E.g., the graph of F(t) = (3 " t)i + (2t)j + ("4 + 3t)k is the collection of all points
(x, y,z) with x = 3 " t, y = 2t, and z = "4 + 3t for all t. These points include the point
P (3,0,"4) and all points aligned with the vector v = -i + 2j + 3k from P. In other words,
!
the graph of F is a line in space.
!
Polar functions and their graphs
! polar coordinate system
In the rectangular coordinate system, we identify points in the plane by their xand y-coordinates. We can also identify points in the plane by using their distance from
the origin and their angle from the positive x-axis. Today, we will study functions
defined using this new way of identifying points in the plane.
Let’s begin by defining the polar coordinate system. We begin by drawing a halfline (or ray) from a fixed point called the pole (think origin); the half-line is called the
polar axis (think positive x-axis). Let P be any point in the plane. If we draw the line
segment from the pole to point P, this line segment has some distance r and makes an
angle " with the polar axis. We can label the point P as (r," ) . This means any point may
be identified by a distance from the pole and by an angle from the polar axis.
However, the angle " is not unique for P, e.g., P (r," + 2#k) would also identify
the same point P. Negative values of r are possible, and r is referred to as a directed
!
distance. E.g., we could identify our point P as! ("r,# + $ ) . Since the angle choice
depends on whether r is positive or negative, the angle is referred to as a directed angle.
!
Thus, a point in the plane may have an infinite!number of polar coordinates.
"
7#
#
Plot the points P (2, ) , Q (0, " )!
, R (1," ) , and S ("2, ) .
4
6
6
Converting between polar and rectangular coordinates
See figure on p. 577. Sometimes we want to convert from polar to rectangular
!
coordinates or vice versa.
!
!
!
To convert from polar to rectangular coordinates,
given a point (r," ) use
x = r cos" or y = r sin " .
To convert from rectangular to polar coordinates, given a point (x, y) use
y
x 2 + y 2 = r 2 or tan " = .
!
x
!
!
"
#
3
1. Convert (2, ) to rectangular coordinates. x = r cos" = 2cos = 2( ) = 3 .
6
6
2
!
#
1
!
y = r sin " = 2sin
= 2( ) = 1. So (x, y) = ( 3,1) .
6
2
2
2
2
2
2.!Convert (1, -1) to polar coordinates.
! r = x + y = (1) + ("1) = 2 .
#1
tan " =
= #1 and " = tan#1 (#1)!, since the point (x, y) is in quadrant IV and the arctan is
1
$
#
only defined for quadrants I and IV the!angle " = # . So (r," ) = ( 2," ) . Other
4
4
3#
7"
!
possibilities are (" 2, ) and ( 2, ) .
4
4
!
!
!
!
!
distance formula for polar coordinates
See figure 6 on p. 579. Let's find the distance between two points in the polar
coordinate system. We can make a triangle by drawing the line segment connecting the
two points. Then by using the Law of Cosines, we can find the distance between the two
points. d 2 = r12 + r22 " 2r1r2 cos(#2 " #1 ) .
Find the distance between the points (4, " ) and (3,5 " 3) .
d = 4 2 + 32 " 2(4)(3) cos(5 # 3" # ) = 16 + 9 " 24 cos(2 # 3) = 25 " 24("1 2) = 37
!
polar functions
!
!coordinate system are usually written as
Recall that functions in the rectangular
!
y = f(x). Similarly, functions in the polar coordinate system are written as r = f( " ), i.e.,
the directed distance of a point is a function of the angle " . Examples of polar functions
are: r = " , r = 1, r = sin " , r = 1 + sin " , r = 4/(1 + sin " ), r = d/cos( " – " ).
graphs of polar functions
!
Where graphs of rectangular functions are drawn from left to right on the plane,
!
graphs of polar
drawn counterclockwise from the polar axis.
! functions are
!
!
!
! !
Let's draw the graph of r = sin " .
Let’s begin by making a table of values.
"
0 " /6 " /3 " /2 2 " /3 5 " /6 " 7 " /6 4 " /3 3 " /2 5 " /3 11 " /6 2 "
0 1 2 !3 2 1
3 2 1 2 0 "1 2 " 3 2 "1 " 3 2 "1 2 0
r
Plot these points on the polar coordinate systems and sketch a smooth curve
connecting
graph
!
! these
! points.
!
!Notice
! the !
! is drawn
! counterclockwise
!
!
! as "! increases from
"!
" <!" !
0 to 2 !
radians.
Notice
when
< 2 " . What
effect
! ! that r was
! !
! did this have
!
!
! negative
!
on the graph? It retraced its graph.
When did the graph pass through the pole (origin)? At (0, 0), (0, " ) and (0, 2 " ).
!
The graph passes through the pole (origin) when r = 0.
!
!
!
!
Sometime it is helpful to determine the values of " for which r is a maximum.
#
Looking at our table of values, for which values of " is r a maximum?
" =! . What is
!
2
"
the maximum value of r? (maximum value of !
r = sin = 1) Could we have predicted
2
!
#
this from the function? Yes, sin " is a maximum when " = !
. What if our function had
2
3#
been r = 1 - 2sin " ? Then r would have
. What is this
! a maximum value when " =
2
!
3#
maximum value of r? (maximum value of r =!
1" 2sin
= 1" 2("1) = 3 )
2
!polar equations of circles
! of a circle. See figure 7 p.
We can use the distance formula to find an equation
580. If (r," ) is an arbitrary point on the circle of radius a with center (r0 ,"0 ) , then
!
a 2 = r 2 + r02 " 2rr0 cos(# " #0 ) is an equation of this circle. If the center of the circle is the
origin, then r0 = 0 and a 2 = r 2 (or r = ±a ).
!
!
!
!
$
$
$
12 = r 2!+ 2 2 " 2(2)r!cos(# " ) % 1 = r 2 + 4 " 4r cos(# " ) % r 2 " 4r cos(# " ) = "3
2
2
2
!
!
"
2
Find a polar equation of the circle with radius 1 and center (2, ) .
polar equations of lines
Let's consider polar equations of lines. We have two cases:
1. The line passes through the pole (origin)
2. The line does not pass through the pole
If a line passes through the pole and makes an angle "0 with the positive x-axis,
then the polar equation of the line is " = "0 . Since r is not related to " , r can assume any
$
! 4
value (both positive and negative). Draw the line " = # .
See fig. 10 p. 581. For a line that does not pass through
the pole, we draw a
!
!
perpendicular line segment from the pole to the line L at some point N. Let's say the
polar coordinates of N are (d," ) . Let point
! P, (r," ) , be an arbitrary point on the line L.
Then in the right triangle ONP, we have cos(" # $ ) = d r or d = r cos(" # $ ) .
$
!4
IF r cos(" # ) =1 is the polar equation of a line, then
!
# &
! %1, " (
!
1. Find the polar coordinates
of the point N.
$ 4'
!
2. Find the polar coordinates of the point on the line where " =
3. Sketch the line.
"&
# #
. % 2, (
2 $
2'
!
converting polar functions to rectangular and vice
! versa
!
Some graphs are more easily written in polar form than rectangular
form, r = e" .
Other graphs are more easily written in rectangular form, e.g., y = 2x + 1. Sometimes in
calculus we want to convert from one form to another to simplify our calculations. To
convert functions, we use the same four formulas we saw earlier:
!
y
y = r sin "
x = r cos"
x 2 + y 2 = r 2 tan " = .
x
Convert x 2 + y 2 = 16 to polar form. Substitute for x and y.
(r cos " ) 2 + (r sin " ) 2 = r 2 (cos 2 " + sin 2 " ) = r 2 (1) = 16 ; r = 4. This is the graph of a circle
! centered at the
! 4.
! pole with radius
!
2
to rectangular form. Converting from polar to
!Convert r =
1" sin #
rectangular form is not as easy. Multiply both sides by (1 – sin " ). r " r sin # = 2 . Since r
sin " = y, we have r – y = 2 or r = y + 2. Squaring both sides, r 2 = (y + 2) 2 = y 2 + 4 y + 4 .
Substituting for r 2 , we have x 2 + y 2 = y 2 + 4 y + 4 . Subtracting y 2 from both sides, we
!
have x 2 = 4 y + 4 = 4(y + 1) . This is the graph of a parabola
! ! with vertex (0, -1).
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symmetry tests for polar coordinates
Graphs in polar coordinates may be symmetric about the x-axis, the y-axis, or the
origin like graphs in rectangular coordinates.
A polar graph is symmetric across the x-axis if r = f (" ) = f (#" ) . This is similar
to our earlier definition of even functions that are symmetric across the x-axis. This
definition makes sense, since the points ( r," ) and ( r,"# ) are reflections of each other
across the x-axis.
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A polar graph is symmetric across the y-axis if f ("# ) = "r or equivalently
f (" # $ ) = r . A polar graph may only satisfy one of these tests. Read Example 3 on p.
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588.
A polar graph is symmetric across the origin if f (" ) = r and f (" ) = #r .
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spirals, cardioids, limacons, and lemniscates
Polar graphs can assume many interesting shapes. Common types of polar graphs
are spirals, cardioids, limacons, and lemniscates. We can recognize these graphs by their
! by plotting!a few key points and
polar functions. We can then sketch their graphs
applying symmetry tests, where possible.
Polar functions with spiral graphs are of the form r = f (" ) , where f is a
monotonically increasing or decreasing function. E.g., r = " 2 , " # 0 . Spiral functions are
not symmetric.
Polar functions with cardioid graphs are of the form r = a ± asin " (or cos " ) . See
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Example 2 on p. 587.
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Polar functions with limacon graphs are of the form r = b ± asin " (or cos " ) . If
a = b , we have a cardioid. If a > b , then we have a limacon with an inner loop. See
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Example 3 on p. 589. If a < b , then we have a limacon with a dimple or a kidney-shaped
graph. Draw r = 3 " 2cos# .
! eight or two-leaved rose) are of the
Polar functions with lemniscate graphs (figure
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2
2
form r = a sin
!(2" ) (or cos(2" )) . See Example 4 on p. 590.
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