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 = (x2 " x1)2

+ (y2 " y1)2 . If

!

P(1,2) and

!

Q(4,3) , then

!

PQ = (4 "1)2

+ (3" 2)2

= 9 +1 = 10 . Two vectors are equal if they have the same length and direction. They do not have to have the same initial and terminal points. We can 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(x2,y2) , then position vector equal to

!

PQ is the vector

!

OS where

!

O(0,0) and

!

S(x2 " x1,y2 " y1). vector addition The vector

!

PQ may describe the net movement or displacement of an object from a point P to a point Q. If the 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 magnitude do not change. Draw

!

OS 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

!

vx and

!

vy . Here,

!

vx = 3 and

!

vy =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,v2 , then u + v =

!

u1 + v1,u2 + v2 .

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

G

!

= 10cos60°,10sin60° = 10 "1

2,10 "

3

2= 5,5 3 . So F+G

!

= 8 + 5,0 + 5 3 =

!

13,5 3 . The magnitude of the resultant vector, F+G, is equal to

!

132

+ (5 3)2

=

!

169 + 75 = 244 "15.62, and its angle is equal to

!

tan"1(5 3

13) # 33.67°.

scalar multiplication of vectors We can also multiply vectors by a scalar. For each real number k and each vector v

!

= v1,v2 , we define a vector kv by the equation

!

kv = kv1,kv2 . 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 A unit vector is a vector of length one. We can find unit vectors in any direction. Find a unit vector in the direction of u =

!

3,1 . Earlier, we found that

!

u = 10 . To find a

unit vector in the direction of u, we need to multiply u by

!

1

u, i.e.,

!

1

10

!

3,1 . A unit

vector in the direction of u is defined as

!

u

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 .

First, we find a unit vector in the direction of u, namely

!

1

10

!

3,1 . Then we multiply

this unit vector by 5 to obtain

!

5

10

!

3,1 .

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,v2 , we define the dot product (also known as the scalar product) of u and v as

!

u " v = u1v1

+ u2v2. Find the dot product of u =

!

3,1 and v =

!

"1,2 .

!

u " v = u1v1 + u2v2 = 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

shown (see exercise 9.3.75 on page 568) that

!

cos" =u # v

u v. We can use this formula to

find the angle between two vectors. Find the angle between u =

!

3,1 and v =

!

"1,2 .

First, we find the cosine of the angel between u and v, namely

!

cos" =u # v

u v=

$1

10 # 5

!

= "1

50# "0.141421. Now we find the angle

!

" :

!

" # cos$1($.0.141421) # 98.13° .

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 the line containing v, it will be a normalization (or a multiple) of v. The multiple of the

projection vector is equal to

!

u " v

v.

Find the projection of the vector u =

!

3,1 onto v =

!

"1,2 . The multiple of the

projection vector is equal to

!

u " v

v=3(#1) +1(2)

(#1)2

+ 22

=#1

5= #

1

5. So, the projection of u

onto v is the vector

!

"1

5"1,2 =

1

5,"2

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). In space, each point is associated with an ordered triple of real numbers,

!

(x,y,z) . Through a fixed point, the origin O, we draw three mutually perpendicular lines, the x-axis, 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 the form

!

(x,0,0) lie along the x-axis; points of the form

!

(0,y,0) lie along the y-axis; and 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.

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., the following equations represent planes in space:

!

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; and the third is parallel to the yz-plane. 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(x2,y2,z2), then the distance d between P and

Q is

!

d = (x2 " x1)2

+ (y2 " 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

= d2, where the center of the sphere is

the point in space

!

a,b,c( ) and the radius of the sphere is the distance d. The equation of the sphere with center at

!

P(1,2,3) and containing

!

Q("2,0,4) is

!

(x "1)2

+ (y " 2)2

+ (z " 3)2

=14 . The equation of a quadric surface (quadric surfaces are generalizations of the conic sections) have the form

!

Ax2

+ By2

+ Cz2

+ Dxy + Exz + Fyz +Gx + Hy + Iz + J = 0 .

E.g.,

!

z2

=x2

a2

+y2

b2

represents an elliptic cone.

3-D vectors To represent vectors in space, we introduce the unit vector k along the positive z-axis. 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(x2,y2,z2) is equal to the position vector

!

(x2 " x1)i + (y2 " 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 =

!

(4 " 9)i + (6 +12) j + ("4 "15)k = -5i + 18j – 19k. length, unit vectors The length (magnitude) of a 3-D vector v = ai +bj +ck is defined to be

!

v = a2

+ b2

+ c2 . E.g., the length of v = 2i - 3j – 6k is

!

v = 22

+ ("3)2

+ ("6)2

= 49 = 7.

Unit vectors are defined the same as in the plane,

!

u =v

v. Find the unit vector in

the same direction as v = 2i - 3j – 6k. We have u

!

=1

7(2i" 3j" 6k) =

2

7i"3

7j"6

7k .

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

The definition of

!

cos" =v # w

v w is the same in space as it was in the plane. Find the

angle between v = 2i + 3j – 2k and w = 3i – 4j + 5k. Since

!

cos" =#16

( 17)( 50)= #

16

850$ #.548795 , we have

!

" #123.284°.

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.

writing vectors in terms of magnitude and direction cosines 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 "#,$,% " & ).

Let v = ai + bj + ck. Notice that

!

cos" =v # i

v i=a

v,

!

cos" =v # j

v j=b

v, and

!

cos" =v # k

v k=c

v. This is equivalent to

!

a = v cos" ,

!

b = v cos" , and

!

c = v cos" .

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 =

!

"3i + 2j – 6k in terms of magnitude and direction

angles. DRAW. We see that

!

v = a2

+ b2

+ c2

= 9 + 4 + 36 = 49 = 7 ,

!

cos" = #3

7,

!

cos" =2

7, and

!

cos" = #6

7. So,

!

" = cos#1(#3 7) $115.4°,

!

" = cos#1(2 7) $ 73.4°, and

!

" = cos#1(#6 7) $149.0°. Therefore, v =

!

7cos115.4°i + 7cos73.4°j + 7cos149.0°k . 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 3-D vectors. The cross products of two 3-D vectors is another vector, whereas the dot 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 = 172+ 82

+142= 289 + 64 +196 = 549 " 23.43.

vector-valued functions A vector-valued function F assigns to each number t a unique vector

!

F(t) = f1(t)i+ f2(t)j+ f3(t)k , where

!

f1, f2, and f3 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 x- and 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 half-line (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.

Plot the points P

!

(2,"

6), Q

!

(0,") , R

!

(1,"#

4) , and S

!

("2,7#

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 = rcos" or

!

y = rsin" . To convert from rectangular to polar coordinates, given a point (x, y) use

!

x2

+ y2

= r2 or

!

tan" =y

x.

1. Convert

!

(2,"

6) to rectangular coordinates.

!

x = rcos" = 2cos#

6= 2(

3

2) = 3 .

!

y = rsin" = 2sin#

6= 2(

1

2) =1. So (x, y) =

!

( 3,1).

2. Convert (1, -1) to polar coordinates.

!

r = x2

+ y2

= (1)2

+ ("1)2

= 2 .

!

tan" =#1

1= #1 and

!

" = tan#1(#1), since the point (x, y) is in quadrant IV and the arctan is

only defined for quadrants I and IV the angle

!

" = #$

4. So

!

(r,") =

!

( 2,"#

4) . Other

possibilities are

!

(" 2,3#

4) and

!

( 2,7"

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.

!

d2

= r12

+ r22"2r1r2 cos(#2 "#1) .

Find the distance between the points

!

(4,") and

!

(3,5" 3) .

!

d = 42

+ 32"2(4)(3)cos(5# 3" #) = 16+9"24cos(2# 3) = 25"24("1 2) = 37

polar functions Recall that functions in the rectangular coordinate system are usually written as 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 functions are drawn counterclockwise from the polar axis. 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

!

" r

!

0

!

1 2

!

3 2

!

1

!

3 2

!

1 2

!

0

!

"1 2

!

" 3 2

!

"1

!

" 3 2

!

"1 2

!

0 Plot these points on the polar coordinate systems and sketch a smooth curve connecting these points. Notice the graph is drawn counterclockwise as

!

" increases from 0 to 2

!

" radians. Notice that r was negative when

!

"<

!

" < 2

!

" . What effect did this have 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?

!

" =#

2. What is

the maximum value of r? (maximum value of

!

r = sin"

2=1) Could we have predicted

this from the function? Yes,

!

sin" is a maximum when

!

" =#

2. What if our function had

been r = 1 - 2sin

!

" ? Then r would have a maximum value when

!

" =3#

2. What is this

maximum value of r? (maximum value of

!

r =1" 2sin3#

2=1" 2("1) = 3)

polar equations of circles We can use the distance formula to find an equation of a circle. See figure 7 p. 580. If

!

(r,") is an arbitrary point on the circle of radius a with center

!

(r0,"0) , then

!

a2

= r2

+ r02"2rr0 cos(# "#0) is an equation of this circle. If the center of the circle is the

origin, then

!

r0

= 0 and

!

a2

= r2 (or

!

r = ±a).

Find a polar equation of the circle with radius 1 and center

!

(2,"

2) .

!

12

= r2

+22"2(2)r cos(# "

$

2)%1= r

2+ 4 " 4r cos(# "

$

2)% r

2" 4r cos(# "

$

2) = "3

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

value (both positive and negative). Draw the line

!

" = #$

4.

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(" #$) .

IF

!

r cos(" #$

4) =1 is the polar equation of a line, then

1. Find the polar coordinates of the point N.

!

1,"

4

#

$ %

&

' (

2. Find the polar coordinates of the point on the line where

!

" =#

2.

!

2,"

2

#

$ %

&

' (

3. Sketch the line.

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:

!

x = rcos"

!

y = rsin"

!

x2

+ y2

= r2

!

tan" =y

x.

Convert

!

x2

+ y2

=16 to polar form. Substitute for x and y.

!

(rcos")2 + (rsin")2 = r2(cos

2" + sin2") = r

2(1) =16 ; r = 4. This is the graph of a circle

centered at the pole with radius 4.

Convert

!

r =2

1" sin# to rectangular form. Converting from polar to

rectangular form is not as easy. Multiply both sides by (1 – sin

!

" ).

!

r " rsin# = 2 . Since r sin

!

" = y, we have r – y = 2 or r = y + 2. Squaring both sides,

!

r2

= (y + 2)2

= y2

+ 4y + 4 . Substituting for

!

r2 , we have

!

x2

+ y2

= y2

+ 4y + 4 . Subtracting

!

y2 from both sides, we

have

!

x2

= 4y + 4 = 4(y +1). This is the graph of a parabola with vertex (0, -1).

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.

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. 588.

A polar graph is symmetric across the origin if

!

f "( ) = r and

!

f "( ) = #r . 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 polar functions. We can then sketch their graphs by plotting a few key points and 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 Example 2 on p. 587. 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 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# . Polar functions with lemniscate graphs (figure eight or two-leaved rose) are of the form

!

r2 = a2

sin 2"( ) or cos 2"( )( ) . See Example 4 on p. 590.