chapter 1 topics in analytic geometry

Chapter 1

Topics in Analytic Geometry

1.1 Conic Sections

Circles, ellipses, parabolas, and hyperbolas are called conic sections or conics becausethey can be obtained as intersection of a plane with a double-napped circular cone. If theplane passes through the vertex of the double-napped cone, then the intersection is a point,a pair of intersection lines, or a single line. These are called degenerate conic section.

Definitions of the Conic Sections

Definition 1.1 A parabola is the set of all points in the plane that are equidistant froma fixed line and a fixed point not on the line.

b

Vertex

b

focus

Directrix

Axis

b

The line is called the directrix of the parabola, and the point is called the focus. Aparabola is symmetric about the line that passes through the focus at right angles to thedirectrix. This line, called the axis or the axis of symmetry of the parabola, intersectsthe parabola at a point called the vertex.

Definition 1.2 An ellipse is the set of all points in the plane, the sum of whose distancesfrom two fixed points is a given positive constant that is greater than the distance betweenthe fixed points.

1

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2

The two fixed points are called the foci (plural of “focus”) of the ellipse, and themidpoint of the line segment joining the foci is called the center.

b

Focusb

Focusb

Center

b bb

Note that if the foci coincide, the ellipse reduces to a circle. For ellipse other than circle,the line segment through the foci and across the ellipse is called the major axis, and theline segment across the ellipse, through the center, and perpendicular to the major axis iscalled the minor axis. The endpoints of the major axis are called vertices.

bb

Minor axis

Major axis

bVextex b Vextex

Definition 1.3 A hyperbola is the set of all points in the plane, the difference of whosedistances from two fixed distinct points is a given positive constant that is less than thedistance between the fixed points.

The two fixed points are called the foci of the hyperbola. The midpoint of the linesegment joining the foci is called the center of the hyperbola. The line through the fociis called the focal axis, and the line through the center that is perpendicular to the focalaxis is called the conjugate axis. The hyperbola intersects the focal axis at two pointscalled the vertices. The two separate parts of a hyperbola are called the branches.

Focalaxis

Conjugateaxis

b

Focus

b

FocusVertex Vertex

Center


Associated with every hyperbola is a pair of lines, called the asymptotes of the hy-perbola. This line intersect at the center of the hyperbola and have the property that as apoint P moves along the hyperbola away from the center, the vertical distance between Pand one of the asymptotes approaches zero.

x

y

Equations of Parabolas in Standard Position

Let p denote the distance between focus and the vertex. The vertex is equidistant fromthe focus and the directrix, so the distance between the vertex and the directrix is also p;consequently, the distance between the focus and the directrix is 2p.

bp p

b

Directrix

Axis2p

2p

The equation of a parabola is simplest if the vertex is the origin and the axis of symmetryis along the x-axis or y-axis. The four possible such orientations are shown in Figure below.These are called the standard positions of a parabola, and the result equations are calledthe standard equations of a parabola.

Orientation Standard Equation

x

y

b

(p, 0)x = −p

y2 = 4px



x

y

b

(−p, 0)x = p

y2 = −4px

x

y

b(0, p)

y = −p

x2 = 4py

x

y

b

(0,−p)

y = p

x2 = −4py

To illustrate how the equations in the above Figure are obtained, we will derive theequation for the parabola with focus (p, 0) and directrix x = −p. Let P (x, y) be any pointon the parabola.

bP (x, y)

b

F (p, 0)

x = −p

y

x

D(−p, y)

Since P is equidistant from the focus and directrix, the distances PF and PD in the aboveFigure are equal; that is,

PF = PD (1.1)

From the distance formula, the distance PF and PD are

PF =√

(x− p)2 + y2 and PD =√

(x+ p)2 (1.2)


Substituting in (1.1) and squaring yields

(x− p)2 + y2 = (x+ p)2 (1.3)

and after simplifyingy2 = 4px (1.4)

The derivations of the other equations in the previous Figure are similar.

Example 1.1 Find the focus and directrix of the parabola x2 = −6y and sketch its graph.

Solution . . . . . . . . .

Example 1.2

(a) Find an equation of a parabola that has vertex at the origin, opens right, and passesthrough the point P (7,−3).

(b) Find the focus.

Solution . . . . . . . . .

Equations of Ellipses in Standard Position

It is traditional in the study of ellipse to denote the length of the major axis by 2a, thelength of the minor axis by 2b, and the distance between the foci by 2c.

b

c c

a a

b

b

b

The number a is called the semimajor axis and the number b the semiminor axis.There is a basic relationship between a, b, and c that can be obtained by examining the

sum of the distances to the foci from a point P at the end of the major axis and from apoint Q at the end of the minor axis.

bb

b

b P

Q

b

c c a− c

√b2 + c2

√b2 + c2


Because P and Q both lie on the ellipse, the sum of the distances from each of them to thefoci must be equal. Thus, we obtain

2√b2 + c2 = (a− c) + (a + c)

from which it follows that

a =√b2 + c2 (1.5)

or equivalently

c =√a2 − b2 (1.6)

From (1.5), the distance from the focus to an end of the minor axis is a, which implied thatall points on the ellipse the sum of the distance to the foci is 2a.

The equation of an ellipse is simplest if the center of the ellipse is at the origin and thefoci are on the x-axis or y-axis. The two possible such orientations are shown in Figurebelow. These are called the standard positions of an ellipse, and the result equationsare called the standard equations of an ellipse.


x

y

b

(c, 0) ab

(−c, 0)−a

b

−b

x2

a2+

y2

b2= 1

x

y

b(0,−c)

b (0, c)

b−b

a

−a

x2

b2+

y2

a2= 1

To illustrate how the equations in the above Figure are obtained, we will derive theequation for the ellipse with the foci on the x-axis.


x

y

b

F (c, 0)b

F ′(−c, 0)

bP (x, y))

Let P (x, y) be any point on the ellipse. Since the sum of the distances from P to the fociis 2a, it follows that

PF ′ + PF = 2a

so√

(x+ c)2 + y2 +√

(x− c)2 + y2 = 2a

Transposing the second radical to the right side of the equation and squaring yields

(x+ c)2 + y2 = 4a2 − 4a√

(x− c)2 + y2 + (x− c)2 + y2

and, on simplifying,√

(x− c)2 + y2 = a− c

ax

Squaring both sides again yields

x2

a2+

y2

a2 − c2= 1

which, by (1.5), can be written asx2

a2+

y2

b2= 1 (1.7)

Conversely it can be shown that any point whose coordinates satisfying (1.7) has 2a asthe sum of its distances from the foci, so that such a point is on the ellipse.

Example 1.3 Sketch the graph of the ellipse 4x2 + 18y2 = 36.

Solution . . . . . . . . .

Example 1.4 Find an equation of the ellipse with foci (0,±2) and major axis with end-points (0,±4).

Solution . . . . . . . . .


Equations of Hyperbolas in Standard Position

It is traditional in the study of hyperbola to denote the distance between the vertices by2a, the distance between the foci by 2c, and to define the quantity b as

b =√c2 − a2 (1.8)

This relationship, which can be expressed as

c =√a2 + b2 (1.9)

is pictured geometrically in Figure below.

b

cb

c

a a

bb

a

b c

The number a is called the semifocal axis of the hyperbola and the number b the semi-

conjugate axis. Moreover, for all points on a hyperbola, the distance to the farther focusminus the distance to the closer focus is 2a.

The equation of a hyperbola is simplest if the center of the hyperbola is at the originand the foci are on the x-axis or y-axis. The two possible such orientations are shown inFigure below. These are called the standard positions of a hyperbola, and the resultequations are called the standard equations of a hyperbola.


x

y

bba

b

(−c, 0) (c, 0)

x2

a2− y2

b2= 1

x

y

b

b

b

a(0, c)

(0,−c)

y2

a2− x2

b2= 1


The derivation of these equations are similar to those already given for parabolas andellipses, so we will leave them as exercises.

A Quick Way to Find Asymptotes

They are a trick that can be used find the equations of the asymptotes of a hyperbola.They can be obtained, when needed, by replacing 1 by 0 on the right side of the hyperbolaequation, and then solving for y in terms of x. For example, for the hyperbola

x2

a2− y2

b2= 1

we would writex2

a2− y2

b2= 0 or y2 =

b2

a2x2 or y = ± b

ax

which are the equations for the asymptotes.

Example 1.5 Sketch the graph of the hyperbola 9x2 − 4y2 = 36 and showing its vertices,foci, and asymptotes.

Solution . . . . . . . . .

Example 1.6 Find the equation of the hyperbola with vertices (0,±8) and asymptotes y =±4

3x.

Solution . . . . . . . . .

Translated Conics

Equations of conic that are translated from their standard positions can be obtained byreplacing x by x − h and y by y − k in their standard equations. For a parabola, thistranslates the vertex from the origin to the point (h, k); and for ellipses and hyperbolas,this translates the center from the origin to the point (h, k).

Parabolas with vertex (h, k) and axis parallel to x-axis

(y − k)2 = 4p(x− h) [Opens right]

(y − k)2 = −4p(x− h) [Opens left]

Parabolas with vertex (h, k) and axis parallel to y-axis

(x− h)2 = 4p(y − k) [Opens up]

(x− h)2 = −4p(y − k) [Opens down]

Ellipse with center (h, k) and major axis parallel to x-axis

(x− h)2

a2+

(y − k)2

b2= 1 (a > b)


Ellipse with center (h, k) and major axis parallel to y-axis

(x− h)2

b2+

(y − k)2

a2= 1 (a > b)

Hyperbola with center (h, k) and focal axis parallel to x-axis

(x− h)2

a2− (y − k)2

b2= 1

Hyperbola with center (h, k) and focal axis parallel to y-axis

(y − k)2

a2− (x− h)2

b2= 1

Example 1.7 Find an equation of the parabola with vertex V (−4, 2) and directrix y = 5.

Solution . . . . . . . . .

Sometimes the equations of translated conics occur in expanded form, in which case weare faced with the problem of identifying the graph of a quadratic equation in x and y:

Ax2 + Cy2 +Dx+ Ey + F = 0

The basic procedure for determining the nature of such a graph is to complete the squaresof the quadratic terms and then try to match up the resulting equation with one of theforms of a translated conic.

Example 1.8 Show that the curve y = 6x2 − 12x+ 8 is a parabola.

Solution . . . . . . . . .

Example 1.9 Discuss and sketch the graph of the equation

16x2 + 9y2 + 64x− 18y − 71 = 0.

Solution . . . . . . . . .

Example 1.10 Show that the curve

9x2 − 4y2 − 54x− 16y + 29 = 0

is a hyperbola. Sketch the hyperbola and show the foci, vertices, and asymptotes in thefigure.

Solution . . . . . . . . .


1.2 Rotation of Axes; Second-Degree Equations

Quadratic Equations in x and y

We saw in Examples 1.8 to 1.10 that equations of the form

Ax2 + Cy2 +Dx+ Ey + F = 0 (1.10)

can represent conic sections. Equation (1.10) is a special case of the more general equation

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 (1.11)

which, if A, B, and C are not all zero, is called a quadratic equation in x and y. IfB = 0, then (1.11) reduces to (1.10) and the conic section has its axis or axes parallel tothe coordinate axes. However, if B 6= 0, then (1.11) contains a cross-product term Bxy,and the graph of the conic section represented by the equation has its axis or axes “tilted”relative to the coordinate axes.

Rotation of Axes

Consider the rotation of axes in Figure below.

x

y

bP (x, y)

P (x′, y′)

x′

y′

θ

α

b

(x′, 0′)b(0′, y′)

r

The axes of an xy-coordinate system have been rotated about the origin through an angle θto produce a new x′y′-coordinate system. Each point P in the plane has coordinates (x′, y′)as well as coordinates (x, y). To see how the two are related, let r be the distance from thecommon origin to the P , and let α be the angle shown in Figure. It follows that

x = r cos(θ + α), y = r sin(θ + α) (1.12)

andx′ = r cos(α), y′ = r sin(α) (1.13)

Applying the addition formulas for the sine and cosine, the relationships in (1.12) can bewritten as

x = r cos θ cosα− r sin θ sinα

y = r sin θ cosα + r cos θ sinα


and on substituting (1.13) in these equations we obtain the following relationships calledthe rotation equations :

x = x′ cos θ − y′ sin θ

y = x′ sin θ + y′ cos θ(1.14)

Example 1.11 Suppose that the axes of an xy-coordinate system are rotated through anangle of θ = 45◦ to obtain an x′y′-coordinate system. Find the equation of the curve

x2 − xy + y2 − 6 = 0

in x′y′-coordinates.

Solution . . . . . . . . .

If the rotation (1.14) are solved for x′ and y′ in terms of x and y, we obtain:

x′ = x cos θ + y sin θ

y′ = −x sin θ + y cos θ(1.15)

Example 1.12 Find the new coordinates of the point (2, 4) if the coordinate axes are rotatedthrough an angle of θ = 30◦.

Solution . . . . . . . . .

Eliminating the Cross-Product Term

In Example 1.11, we were able to identify the curve x2−xy+y2−6 = 0 as an ellipse becausethe rotation of axes eliminate the xy-term, thereby reducing the equation to a familiar form.This occurred because the new x′y′-axes were aligned with the axes of the ellipses. Thefollows theorem tells how to determine an appropriate rotation of axes to eliminate thecross-product term of a second-degree equation in x and y.

Theorem 1.1 If the equation

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 (1.16)

is such that B 6= 0, and if an x′y′-coordinate system is obtained by rotating the xy-axesthrough an angle θ satisfying

cot 2θ =A− C

B(1.17)

then, in x′y′-coordinates, Equation (1.16) will have the form

A′x′2 + C ′y′2 +D′x′ + E ′y′ + F ′ = 0

Remark

• It is always possible to satisfy (1.17) with an angle θ in the range 0 < θ < π/2. Weshall always use such a value of θ.


• The values of sin θ and cos θ needed for the rotation equations may be obtained byfirst calculating cos 2θ and then computing sin θ and cos θ from the identities

sin θ =

√

1− cos 2θ

2and cos θ =

√

1 + cos 2θ

2

Example 1.13 Discuss and sketch the graph of the equation xy = 1.

Solution . . . . . . . . .

Example 1.14 Identify and sketch the curve 41x2 − 24xy + 34y2 − 25 = 0.

Solution . . . . . . . . .

Chapter 2

Parametric and Polar Curves

2.1 Parametric Equation; Tangent Lines and Arc Length

for Parametric Curves

Parametric Equation

Suppose that a particle moves along a curve C in the xy-plane in such a way that x- andy-coordinates, as a functions of time, are

x = f(t), y = g(t)

We call these the parametric equations of motion for the particle and refer C as thetrajectory of the particle or the graph of the equations. The variable t is called theparameter for the equations.

x

y

b (x, y)

C

Example 2.1 Sketch the trajectory over the time interval 0 ≤ t ≤ 10 of the particle whoseparametric equations of motion are

x = t− 3 sin t, y = 4− 3 cos t

15


Solution One way to sketch the trajectory is to plot the (x, y) coordinate of points on thetrajectory at those times, and connect the points with a smooth curve.

t x y

0 0.0 1.01 -1.5 2.4

2 -0.7 5.23 2.6 7.0

4 6.3 6.05 7.9 3.16 6.8 1.1

7 5.0 1.78 5.0 4.4

9 7.8 6.710 11.6 6.5

x

y

b

b

b

b

b

b

bb

b

b

z

Example 2.2 Find the graph of the parametric equations

x = cos t, y = sin t (0 ≤ t ≤ 2π) (2.1)

Solution One way to find the graph is to eliminate the parameter t by noting that

x2 + y2 = sin2 t + cos2 t = 1

Thus, the graph is contained in the unit circle x2 + y2 = 1. Geometrically, the parametert can be interpreted as the angle swept out by the radial line from the origin to the point(x, y) = (cos t, sin t) on the unit circle. As t increases from 0 to 2π, the point traces thecircle counterclockwise, starting at (1, 0) when t = 0 and completing one full revolutionwhen t = 2π.

x

y

t

x

1

(0, 1)y

b

b(x, y)

z


Orientation

The direction in which the graph of a pair of parametric equations is traces as the parameterincreases is called the direction of increasing parameter or sometimes the orientationimposed on the curve by the equation. Thus, we make a distinction between a curve, whichis the set of points, and a parametric curve, which is a curve with an orientation. Forexample, we saw in Example 2.2 that the circle represented parametrically by (2.1) is tracedcounterclockwise as t increases and hence has counterclockwise orientation.

To obtain parametric equation for the unit circle with clockwise orientation, we canreplace t by −t in (2.1). This yields

x = cos(−t) = cos t, y = sin(−t) = − sin t (0 ≤ t ≤ 2π)

Here, the circle is traced clockwise by a point that starts at (1, 0) when t = 0 and completesone full revolution when t = 2π.

x

y

t

x

1

(0, 1)

y

b

b(x, y)

b

Example 2.3 Graph the parametric curve

x = 2t− 3, y = 6t− 7

by eliminating the parameter, and indicate the orientation on the graph.

Solution . . . . . . . . .

Expressing Ordinary Functions Parametrically

An equation y = f(x) can be expressed in parametric form by introducing the parametert = x; this yields the parametric equations

x = t, y = f(t)

For example, the portion of the curve y = cosx over the interval [−2π, 2π] can be expressedparametrically as

x = t, y = cos t (−2π ≤ t ≤ 2π)


If a function f is one-to-one, then it has an inverse function f−1. In this case the equationy = f−1(x) is equivalent to x = f(y). We can express the graph of f−1 in parametric formby introducing the parameter t = y; this yields the parametric equations

x = f(t), y = t

For example, the graph of f(x) = x5 + x+ 1 can be represented parametrically as

x = t, y = t5 + t+ 1

and the graph of f−1 can be represented parametrically as

x = t5 + t+ 1, y = t

Tangent Line to Parametric Curves

We will be concerned with curves that are given by parametric equations

x = f(t), y = g(t)

in which f(t) and g(t) have continuous first derivatives with respect to t. It can be provedthat if dx/dt 6= 0, then y is a differentiable function of x, in which case the chain rule impliesthat

dy

dx=

dy/dt

dx/dt(2.2)

This formula makes it possible to find dy/dx directly from the parametric equations withouteliminating the parameter.

Example 2.4 Find the slope of the tangent line to the curve

x = t− 3 sin t, y = 4− 3 cos t (t ≥ 0)

at the point where t = π/3.

Solution . . . . . . . . .

It follows from Formula (2.2) that the tangent line to a parametric curve will be hori-zontal at those points where dy/dt = 0 and dx/dt 6= 0, since dy/dx = 0 at such points.

At points where dx/dt = 0 and dy/dt 6= 0, the right side of (2.2) has a nonzero numeratorand a zero denominator; we will agree that the curve has infinite slope and a vertical

tangent line at such point.At points where dx/dt and dy/dt are both zero, the right side of (2.2) becomes an

indeterminate form; we call such points singular points.

Example 2.5 A curve C is defined by the parametric equations

x = t2 and y = t3 − 3t.

Find the points on C where the tangent is horizontal or vertical.


Solution . . . . . . . . .

Example 2.6 The curve represented by the parametric equations

x = t2 y = t3 (−∞ < t < +∞)

is called a semicubical parabola. The parameter t can be eliminated by cubing x andsquaring y, from which it follows that y2 = x3.

x

y

x = t2, y = t3 (−∞ < t < +∞)

The graph of this equation consists of two branches: an upper branch obtained by graphingy = x3/2 and a lower branch obtained by graphing y = −x3/2. The two branches meet atthe origin, which corresponds to t = 0 in the parametric equations. This is a singular pointbecause the derivatives dx/dt = 2t and dy/dt = 3t2 are both zero there. z

Example 2.7 Without eliminating the parameter, find dy/dx and d2y/dx2 at (1, 1) and(1,−1) on the semicubical parabola given by the parametric equations in Example 2.6

Solution . . . . . . . . .

Arc Length of Parametric Curve

The following result provides a formula for finding the arc length of a curve from parametricequations for the curve.

Arc Length Formula for Parametric Curve

If no segment of the curve represented by the parametric equations

x = x(t), y = y(t) (a ≤ t ≤ b)

is traced more than once as t increase from a to b, and if dx/dt and dy/dt are continuousfunctions for a ≤ t ≤ b, then the arclength L of the curve is given by

L =

∫ b

a

√

(

dx

dt

)2

+

(

dy

dt

)2

dt (2.3)

Example 2.8 Find the circumference of a circle of radius a form the parametric equations

x = a cos t, y = a sin t (0 ≤ t ≤ 2π)

Solution . . . . . . . . .


2.2 Polar Coordinates

Polar Coordinate Systems

A polar coordinate system in a plane consists of a fixed point O, called the pole (ororigin), and a ray emanating from a pole, called the polar axis. In such a coordinatesystem we can associate with each point P in the plane a pair of polar coordinates (r, θ),where r is the distance from P to the pole and θ is an angle from the polar axis to the rayOP .

θOOrigin

P (r, θ)

Polar axis

r

The number r is called the radial distance of P and the number θ the angular coordi-

nate (or polar angle) of P .

Example 2.9 Plot the points whose polar coordinates are given.

(a) (3, π/4) (b) (4, 2π/3) (c) (2, 5π/4) (d) (4, 11π/6)

Solution . . . . . . . . .

The polar coordinates of a point are not unique. For example, the polar coordinates

(1, 5π/3), (1,−π/3), (1, 11π/3)

all represent the same point.

5π/3

(1, 5π3)

−π/3

(1,−π3)

11π/3

(1, 11π3)

In general, if a point P has polar coordinates (r, θ), then

(r, θ + 2nπ) and (r, θ − 2nπ),

are also polar coordinates of P for any nonnegative integer n.As define above, the radial coordinate r of a point P is nonnegative, since it represents

the distance from P to the pole. However, it will be convenient to allow for negativevalues of r as well. To motivate an appropriate definition, consider the point P with polarcoordinates (3, 5π/4). We can reach this point by


• rotating the polar axis 5π/4 and then moving forward from the pole 3 units along theterminal side of the angle, or

• rotating the polar axis π/4 and then moving backward from the pole 3 units along theextension of the terminal side.

bTerminalside

Polar axis

5π/4

P (3, 5π/4) b

Terminalside

Polar axisπ/4

P (3, π/4)

This suggest that the point (3, 5π/4) might also be denoted by (−3, π/4), with minussign serving to indicate that the point is on the extension of the angle’s terminal side ratherthan on the terminal side itself.

In general, the terminal side of the angle θ + π is the extension of the terminal side ofθ, we define negative radial coordinates by agreeing that

(−r, θ) and (r, θ + π)

to be polar coordinates for the same point.

Relationship Between Polar and Rectangular Coordinates

Frequently, it will be useful to superimpose a rectangular xy-coordinate system on top ofa polar coordinate system, making the positive x-axis coincide with the polar axis. Thenevery point P will have both rectangular coordinates (x, y) and polar coordinates (r, θ).

x

y

θ

r y

x

P (r, θ) = P (x, y)

From the above Figure, these coordinates are related by the equations

x = r cos θ, y = r sin θ (2.4)


These equation are well suited for finding x and y when r and θ are known. However, tofind r and θ when x and y are known, we use equations

r2 = x2 + y2, tan θ =y

x(2.5)

Example 2.10 Find the rectangular coordinates of the point P whose polar coordinates are(4, 2π/3).

Solution . . . . . . . . .

Example 2.11 Find polar coordinates of the point P whose rectangular coordinates are(1,−1).

Solution . . . . . . . . .

Graphs in Polar Coordinates

We will now consider the problem of graphing equations in r and θ, where θ is assumed tobe measured in radians. Some examples of such equations are

r = 1, θ = π/4, r = θ, r = sin θ, r = cos 2θ

Example 2.12 Sketch the graph of

(a) r = 1 (b) θ = π/4

Solution . . . . . . . . .

Example 2.13 Sketch the graph of r = 2 cos θ in polar coordinates by plotting points.

Solution . . . . . . . . .

Example 2.14 Sketch the graph of r = cos 2θ in polar coordinates.

Solution Instead of plotting points, we will use the graph of r = cos 2θ in rectangularcoordinates to visualize how the polar graph of this equation is generated. This curve iscalled a four-petal rose.


y = cos 2θ

π2

π 3π2

2π

θ

r

1

−1

r varies from1 to 0 as θvaries from0 to π/4.

r varies from0 to −1 as θvaries fromπ/4 to π/2.

r varies from−1 to 0 as θvaries fromπ/2 to 3π/4.

r varies from0 to 1 as θvaries from3π/4 to π.

r varies from1 to 0 as θvaries fromπ to 5π/4.

r varies from0 to −1 as θvaries from5π/4 to 3π/2.

r varies from−1 to 0 as θvaries from3π/2 to 7π/4.

r varies from−1 to 0 as θvaries from7π/4 to 2π.


Symmetry Tests

Observe that the polar graph r = cos 2θ in above Figure is symmetric about the x-axis andthe y-axis. This symmetry could have been predicted from the following theorem.

Theorem 2.1 (Symmetry Tests).

(a) A curve in polar coordinates is symmetric about the x-axis if replacing θ by −θ in itsequation produces an equivalent equation.

(b) A curve in polar coordinates is symmetric about the y-axis if replacing θ by π − θ inits equation produces an equivalent equation.

(c) A curve in polar coordinates is symmetric about the origin if replacing θ by θ+ π, orreplacing r by −r in its equation produces an equivalent equation.

0

π/2

(r, θ)

(r,−θ)

(a)

0

π/2(r, π − θ) (r, θ)

(b)

0

π/2

(r, θ)

(r, θ + π)or

(−r, θ)

(c)

Example 2.15 Show that the graph of r = cos 2θ is symmetric about the x-axis and y-axis.

Solution . . . . . . . . .

Families of Lines and Rays Through the Pole

For any constant θ0, the equationθ = θ0 (2.6)

is satisfied by the coordinates of the form P (r, θ0), regardless of the value of r. Thus, theequation represents the line that passes through the pole and makes an angle of θ0 with thepolar axis.

0

π/2

θ

θ = θ0


Families of Circles

We will consider three families of circles in which a is assumed to be a positive constant:

r = a r = 2a cos θ r = 2a sin 2θ

• The equation r = a represents a circle of radius a, centered at the pole

• The equation r = 2a cos θ represents a circle of radius a, centered on the x-axis andtangent to the y-axis at the origin.

• The equation r = a represents a circle of radius a, centered on the y-axis and tangentto the x-axis at the origin.

0

π/2

ar = a

0

π/2

b

(a, π)b

(a, 0)

r = −2a cos θ r = 2a cos θ

0

π/2

b (a,−π2)

b (a, π2)

r = 2a sin θ

r = −2a sin θ

Example 2.16 Sketch the graphs of the following equations in polar coordinates.

(a) r = 4 cos θ (b) r = −5 sin θ (c) r = 3

Solution . . . . . . . . .

Families of Rose Curves

In polar coordinates, equations of the form

r = a sinnθ or r = a cosnθ

in which a > 0 and n is a positive integer represent families of flower-shaped curves calledroses. The rose consists of n equally spaced petals of radius a if n is odd and 2n equallyspaced petals of radius a if n is even.


r = a sin 2θ r = a sin 3θ r = a sin 4θ

r = a sin 5θ r = a sin 6θ

r = a cos 2θ r = a cos 3θ r = a cos 4θ

r = a cos 5θ r = a cos 6θ

Families of Cardioids and Limacons

Equations with any of the four forms

r = a± b sin θ r = a± b cos θ

in which a > 0 and b > 0 represent polar curves called limacons. There are four possibleshapes for a limacon that can be determined from the ratio a/b. If a = b (the case a/b = 1),then the limacons is called a cardioids because of its heart-shaped appearance.


a/b < 1

Limacon withinner loop

a/b = 1

Cardioid

1 < a/b < 2

Dimpled limacon

a/b ≥ 2

Convex limacon

Example 2.17 Sketch the graph of the equation

r = 2(1− cos θ)

in polar coordinates.

Solution . . . . . . . . .

Families of Spirals

A spiral is a curve that coils around a central points. The most common example is thespiral of Archimedes, which has an equation of the form

r = aθ (θ ≥ 0) or r = aθ (θ ≤ 0)

In these equations, θ is in radians and a is positive.

Example 2.18 Sketch the graph of r = θ (θ ≥ 0) in polar coordinates by plotting points.

Solution . . . . . . . . .

2.3 Tangent Lines, Arc Length, and Area for Polar

Curves

Tangent Lines to Polar Curves

Our first objective in this section is to find a method for obtaining slopes of tangent linesto polar curves of the form r = f(θ) in which r is a differentiable function of θ. A curve ofthis form can be expressed parametrically in terms of parameter θ by substituting f(θ) forr in the equations x = r cos θ and y = r sin θ. This yields

x = f(θ) cos θ, y = f(θ) sin θ


from which we obtain

dx

dθ= −f(θ) sin θ + f ′(θ) cos θ = −r sin θ +

dr

dθcos θ

dy

dθ= f(θ) cos θ + f ′(θ) cos θ = r cos θ +

dr

dθsin θ

Thus, if dx/dθ and dy/dθ are continuous and if dx/dθ 6= 0, then y is a differentiable functionof x, and Formula (2.2) with θ in place of t yields

dy

dx=

dy/dθ

dx/dθ=

r cos θ + sin θdr

dθ

−r sin θ + cos θdr

dθ

(2.7)

Example 2.19 Find the slope of the tangent line to the cardioid r = 1 + sin θ at the pointwhere θ = π/3.

Solution . . . . . . . . .

Example 2.20 Find the points on the cardioid r = 1− cos θ at which there is a horizontaltangent line, a vertical tangent line, or a singular point.

Solution . . . . . . . . .

Theorem 2.2 If the polar curve r = f(θ) passes through the origin at θ = θ0, and ifdr/dθ 6= 0 at θ = θ0, then the line θ = θ0 is tangent to the curve at the origin.

Arc Length of a Polar Curve

If no segment of the polar curve r = f(θ) is traced more than once as θ increases from α toβ, and if dr/dθ is continuous for α ≤ θ ≤ β, then the arc length L from θ = α to θ = β is

L =

∫ β

α

√

[f(θ)]2 + [f ′(θ)]2 dθ =

∫ β

α

√

r2 +

(

dr

dθ

)2

dθ (2.8)

Example 2.21 Find the arc length of the spiral r = eθ between θ = 0 and θ = π.

Solution . . . . . . . . .

Example 2.22 Find the total arc length of the cardioid r = 1 + cos θ.

Solution . . . . . . . . .


Area in Polar Coordinates

0

θ = α

θ = β

r = f(θ)

R

Area Problem in Polar Coordinates

Suppose that α and β are angles that satisfy the condition

α < β ≤ α + 2π

and suppose that f(θ) is continuous and nonnegative for α ≤ θ ≤ β. Find the area of theregion R enclosed by the polar curve r = f(θ) and the rays θ = α and θ = β.

Area in Polar Coordinates

If α and β are angles that satisfy the condition

α < β ≤ α + 2π

and if f(θ) is continuous and either nonnegative or nonpositive for α ≤ θ ≤ β, then thearea A of the region R enclosed by the polar curve r = f(θ) (α ≤ θ ≤ β) and the linesθ = α and θ = β is

A =

∫ β

α

1

2[f(θ)]2dθ =

∫ β

α

1

2r2 dθ (2.9)

The hardest part of applying (2.9) is determining the limits of integration. This can bedone as follows:

Step 1 Sketch the region R whose area is to be determined.

Step 2 Draw an arbitrary “radial line” from the pole to the boundary curve r = f(θ).

Step 3 Ark, “Over what intervals must θ vary in order for the radial line to sweep out theregion R?”

Step 4 The answer in Step 3 will determine the lower and upper limits of integration.


Example 2.23 Find the area of the region in the first quadrant that is within the cardioidr = 1− cos θ.

Solution . . . . . . . . .

Example 2.24 Find the entire area within the cardioid of Example 2.23.

Solution . . . . . . . . .

Example 2.25 Find the area enclosed by the rose curve r = cos 2θ.

Solution . . . . . . . . .

Example 2.26 Find the area of the region that is inside of the cardioid r = 4+4 cos θ andoutside of the circle r = 6.

Solution . . . . . . . . .

Chapter 3

Three-Dimensional Space; Vectors

3.1 Rectangular Coordinates in 3-Space; Spheres

Rectangular Coordinate Systems

To begin, consider three mutually perpendicular coordinate lines, called the x-axis, they-axis, and the z-axis, positioned so that their origin coincide.

O

x

y

z

The three coordinate axes form a three-dimensional rectangular coordinate system (orCartesian coordinate system) The point of intersection of the coordinate axes is calledthe origin of the coordinate system.

The coordinate axes, taken in pairs, determine three coordinate planes : the xy-plane, the xz-plane, and the yz-plane, which divide space into eight octants. To eachpoint P in 3-space corresponds to ordered triple of real numbers (a, b, c) which measureits directed distances from the three planes. We call a, b, and c the x-coordinate, y-coordinate, and z-coordinate of P , respectively, and we denote the point P by (a, b, c)or by P (a, b, c).

The following facts about three-dimensional rectangular coordinate systems:

31


Region Descriptionxy-plane Consists of all points of the form (x, y, 0)xz-plane Consists of all points of the form (x, 0, z)yz-plane Consists of all points of the form (0, y, z)x-axis Consists of all points of the form (x, 0, 0)y-axis Consists of all points of the form (0, y, 0)z-axis Consists of all points of the form (0, 0, z)

Distance in 3-Space; Spheres

Recall that in 2-space the distance d between the points P1(x1, y1) and P2(x2, y2) is

d =√

(x2 − x1)2 + (y2 − y1)2

The distance formula in 3-space has the same form, but it has a third term to account forthe added dimension. The distance between the points P1(x1, y1, z1) and P2(x2, y2, z2) is

d =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

Example 3.1 Find the distance d between the points (2,−3, 4) and (−3, 2,−5).

Solution . . . . . . . . .

Recall that the standard equation of a circle in 2-space that has center (x0, y0) andradius r is

(x− x0)2 + (y − y0)

2 = r2

Analogously, standard equation of the sphere in 3-space that has center (x0, y0, z0)and radius r is

(x− x0)2 + (y − y0)

2 + (z − z0)2 = r2 (3.1)

If the terms in (3.1) are expanded and like terms are collected, then the resulting equationhas the form

x2 + y2 + z2 +Gx+Hy + Iz + J = 0 (3.2)

Example 3.2 Find the center and radius of the sphere

x2 + y2 + z2 − 10x− 8y − 12z + 68 = 0

Solution . . . . . . . . .

In general, completing the squares in (3.2) produces an equation of the form

(x− x0)2 + (y − y0)

2 + (z − z0)2 = k2

• If k > 0, then the graph of this equation is a sphere with center (x0, y0, z0) and radius√k.

• If k = 0, then the sphere has radius zero, so the graph is the single point (x0, y0, z0).


• If k < 0, the equation is not satisfied by any values of x, y, and z, so it has no graph.

Theorem 3.1 An equation of the form

x2 + y2 + z2 +Gx+Hy + Iz + J = 0

represents a sphere, a point, or has no graph.

3.2 Vectors

Many physical quantities such as area, length, mass, and temperature are completely de-scribed once the magnitude of the quantity is given. Such quantities are called scalar.Other physical quantities, called vectors are not completely determined until both magni-tude and a direction are specified.

Vectors can be represented geometrically by arrows in 2-space or 3-space: the directionof the arrow specifies the direction of the vector and the length of the arrow describes itsmagnitude. The tail of the arrow is called the the initial point of the vector, and the tipof the arrow the terminal point.

We will denote vectors with lowercase boldface type such as a, k, v, w, and x. Whendiscussing vectors, we will refer to real numbers as scalars. Scalar will be denoted bylowercase italic type such as a, k, w, and x.

Two vectors, v and w, are considered to be equal (also called equivalent) if theyhave the same length and same direction, in which case we write v = w. Geometrically,two vectors are equal if they are translations of one another; thus, the three vectors in thefollowing figure are equal, even though they are in different positions.

If the initial point of v is A and the terminal point is B, then we write v =−→AB when

we want to emphasize the initial and terminal points.

bA

bB

If the initial and terminal points of a vector coincide, then the vector has length zero;we call this the zero vector and denote it by 0. The zero vector does not have a specificdirection, so we will agree that it can be assigned any convenient direction in a specificproblem.


Definition 3.1 If v and w are vectors, then the sum v +w is the vector from the initialpoint of v to the terminal point of w when the vectors are positioned so the initial point ofw is at the terminal point of v.

v

w

v +w

Definition 3.2 If v is a nonzero vector and k is a nonzero real number (a scalar), thenthe scalar multiple kv is defined to be the vector whose length is |k| times the length ofv and whose direction is the same as that of v if k > 0 and opposite to that of v if k < 0.We define kv = 0 if k = 0 or v = 0.

v

2v

1

2v

(−1)v

(−3

2)v

Observe that if k and v are nonzero, then the vectors v and kv lie on the same line iftheir initial points coincide and lies on parallel or coincident lines if they do not. Thus, wesay that v and kv are parallel vectors. Observe also that the vector (−1)v has the samelength as v but is oppositely directed. We call (−1)v the negative of v and denote it by−v. In particular, −0 = (−1)0 = 0.

Vector subtraction is defined in terms of addition and scalar multiplication by

v −w = v + (−w)

In the special case where v = w, their difference is 0; that is,

v + (−v) = v− v = 0

Vectors in Coordinate Systems

If a vector v is positioned with its initial point at the origin of the rectangular coordinatesystem, then the terminal point will have coordinates of the form (v1, v2) or (v1, v2, v3),depending on whether the vector is in 2-space or 3-space. We call these coordinates thecomponents of v, and we write v in component form using the bracket notation

v = 〈v1, v2〉 or v = 〈v1, v2, v3〉In particular, the zero vectors in 2-space and 3-space are

v = 〈0, 0〉 and v = 〈0, 0, 0〉respectively.


x

y

v

b (v1, v2)b (v1, v2, v3)

v

x

y

z

Theorem 3.2 Two vectors are equivalent if and only if their corresponding components areequal.

For example,〈a, b, c〉 = 〈−2, 3, 5〉

if and only if a = −2, b = 3, and c = 5.

Arithmetic Operations on Vectors

Theorem 3.3 If v = 〈v1, v2〉 and w = 〈w1, w2〉 are vectors in 2-space and k is any scalar,then

v +w = 〈v1 + w1, v2 + w2〉v −w = 〈v1 − w1, v2 − w2〉

kv = 〈kv1, kv2〉

Similarly, if v = 〈v1, v2, v3〉 and w = 〈w1, w2, w3〉 are vectors in 3-space and k is any scalar,then

v +w = 〈v1 + w1, v2 + w2, v3 + w3〉v−w = 〈v1 − w1, v2 − w2, v3 − w3〉

kv = 〈kv1, kv2, kv3〉

Example 3.3 If v = 〈−1, 2,−3〉 and w = 〈2, 0,−4〉, then

v +w =

3v =

−2w =

w − 3v =

Vectors With Initial Point Not At The Origin

Suppose that P1(x1, y1) and P2(x2, y2) are points in 2-space and we interested in finding

the components of the vector−−→P1P2. As illustrated in the following figure, we can write this

vector as −−→P1P2 =

−−→OP2 −

−−→OP1 = 〈x2, y2〉 − 〈x1, y1〉 = 〈x2 − x1, y2 − y1〉


Ox

y

−−→OP1

−−→OP2

−−−→P1P2

bP1(x1, y1)

bP2(x2, y2)

Thus, we have shown that the components of the vector−−→P1P2 can be obtained by subtracting

the coordinates of its initial point from the coordinates of its terminal point. Similarcomputations hold in 3-space, so we have established the following result.

Theorem 3.4 If−−→P1P2 is a vector in 2-space with initial point P1(x1, y1) and terminal point

P2(x2, y2), then−−→P1P2 = 〈x2 − x1, y2 − y1〉

Similarly, if−−→P1P2 is a vector in 3-space with initial point P1(x1, y1, z1) and terminal point

P2(x2, y2, z2), then−−→P1P2 = 〈x2 − x1, y2 − y1, z2 − z1〉

Example 3.4 In 2-space the vector from P1(3, 2) to P2(−1, 4) is

−−→P1P2 = 〈−1− 3, 4− 2〉 = 〈−4, 2〉

and in 3-space the vector from A(1,−2, 0) to B(−3, 1, 2) is

−→AB = 〈−3− 1, 1− (−2), 2− 0〉 = 〈−4, 3, 2〉 z

Rules of Vector Arithmetic

Theorem 3.5 For any vectors u, v, and w and any scalars k and `, the following rela-tionships hold:

(a) u+ v = v + u (e) k(`u) = (k`)u

(b) (u+ v) +w = u+ (v +w) (f) k(u+ v) = ku+ kv

(c) u+ 0 = 0+ u = u (g) (k + `)u = ku+ `u

(d) u+ (−u) = 0 (h) 1u = u

Norm of a Vector

The distance between the initial and terminal points of a vector v is called the length, thenorm, or the magnitude of v and is denoted by ‖v‖. This distance does not change if


the vector is translated. The norm of a vector v = 〈v1, v2〉 in 2-space is given by

‖v‖ =√

v21+ v2

2

and the norm of a vector v = 〈v1, v2, v3〉 in 3-space is given by

‖v‖ =√

v21+ v2

2+ v2

3

Example 3.5 Find the norm of v = 〈4,−2〉, and w = 〈−1, 3, 5〉.

Solution . . . . . . . . .

For any vector v and scalar k, the length of kv must be |k| times the length of v; thatis,

‖kv‖ = |k|‖v‖Thus, for example,

‖5v‖ = |5|‖v‖ = 5‖v‖‖ − 3v‖ = | − 3|‖v‖ = 3‖v‖‖ − v‖ = | − 1|‖v‖ = ‖v‖

This applies to vectors in 2-space and 3-space.

Unit Vectors

A vector of length 1 is called a unit vector. In an xy-coordinate system the unit vectorsalong the x-axis and y-axis are denoted by i and j, respectively; and in xyz-coordinatesystem the unit vectors along the x-axis, y-axis and z-axis are denoted by i, j, and k,respectively.

x

y

j

(0, 1)

i (1, 0)

x

z

y

i j

k

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

Thus,

i = 〈1, 0〉, j = 〈0, 1〉 In 2-space

i = 〈1, 0, 0〉, j = 〈0, 1, 0〉, k = 〈0, 0, 1〉 In 3-space


Every vector in 2-space is expressible uniquely in terms of i and j, and every vector in3-space is expressible uniquely in terms of i, j, and k as follows:

v = 〈v1, v2〉 = 〈v1, 0〉+ 〈0, v2〉 = v1〈1, 0〉+ v2〈0, 1〉 = v1i+ v2j

v = 〈v1, v2, v3〉 = v1〈1, 0, 0〉+ v2〈0, 1, 0〉+ v3〈0, 0, 1〉 = v1i + v2j + v3k

Example 3.6

2-space 3-space

〈3,−4〉 = 3i− 4j 〈2, 3,−5〉 = 2i+ 3j− 5k

〈5, 0〉 = 5i+ 0j = 5i 〈0, 0, 3〉 = 3k

〈0, 0〉 = 0i+ 0j = 0 〈0, 0, 0〉 = 0i+ 0j+ 0k = 0

(3i− 2j) + (i+ 4j) = 2i+ 2j (2i− j+ 3k) + (3i+ 2j− k) = i + j+ 2k

3(2i− 4j) = 6i− 12j 2(3i+ 4j− k) = 6i+ 8j− 2k

‖3i+ 4j‖ =√32 + 42 = 5 ‖2i− j+ 3k‖ =

√

22 + (−1)2 + 32 =√14

‖v1i+ v2j‖ =√

v21+ v2

2‖〈v1, v2, v3〉‖ =

√

v21+ v2

2+ v2

3

Normalizing a Vector

A common problem in applications is to find a unit vector u that has the same directionas some given nonzero vector v. This can be done by multiplying v by the reciprocal of itslength; that is,

u =1

‖v‖ v =v

‖v‖is a unit vector with the same direction as v.

The process of multiplying v by the reciprocal of its length to obtain a unit vector withthe same direction is called normalizing v.

Example 3.7 Find the unit vector that has the same direction as

v = 2i− j− 2k.

Solution . . . . . . . . .

Vectors Determined by Length and Angle

If v is a nonzero vector with its initial point at the origin of an xy-coordinate system, andif θ is the angle from the positive x-axis to the radial line through v, then the x-componentof v can be written as ‖v‖ cos θ and the y-component as ‖v‖ sin θ;


x

y

θ

‖v‖ cos θ

‖v‖

‖v‖ sin θ

and hence v can be expressed in trigonometric form as

v = ‖v‖〈cos θ, sin θ〉 or v = ‖v‖ cos θi+ ‖v‖ sin θj (3.3)

In the special case of a unit vector u this simplifies to

u = 〈cos θ, sin θ〉 or u = cos θi + sin θj (3.4)

Example 3.8

(a) Find the vector of length 3 that makes an angle of π/3 with the positive x-axis.

(b) Find the angle that the vector v = i−√3j makes with the positive x-axis.

Solution . . . . . . . . .

Vectors determined by length and a vector in the same direction

It is a common problem in many applications that a direction in 2-space or 3-space isdetermined by some known unit vector u, and it is interest to find the components of avector v that has the same direction as u and some specified length v. This can be doneby expressing v as

v = ‖v‖uand then reading off the components of ‖v‖u.

Example 3.9 Find the vector v of length√5 and has the same direction as the vector from

the point A(0, 0, 4) to the point B(2, 5, 0).

Solution . . . . . . . . .

Resultant of Two Concurrent Forces

The effect that a force has on an object depends on the magnitude and direction of theforces and the point at which it is applied. If two forces F1 and F2 are applied at the samepoint on an object, then the two forces have the same effect on the object as the single forceF1 + F2 applied at the point.


F1

F2

F1 + F2

Physicists and engineers call F1+F2 the resultant of F1 and F2, and they say that theforces F1 and F2 are concurrent to indicate that they are applied at the same point.

Example 3.10 Suppose that two forces are applied to an eye bracket, as show in Figurebelow. Find the magnitude of the resultant and the angle θ that it makes with the positivex-axis.

x

y

30◦40◦ ‖F1‖ = 200N

‖F2‖ = 300N

b

Solution . . . . . . . . .

3.3 Dot Product; Projection

Definition of the Dot Product

Definition 3.3 If u = 〈u1, u2〉 and v = 〈v1, v2〉 are vectors in 2-space, then the dot

product of u and v is written as u · v and is defined as

u · v = u1v1 + u2v2

Similarly, if u = 〈u1, u2, u3〉 and v = 〈v1, v2, v3〉 are vectors in 3-space, then their dotproduct is defined as

u · v = u1v1 + u2v2 + u3v3

Note that the dot product of two vectors is a scalar. For example,

〈4,−3〉 · 〈3, 2〉 = (4)(3) + (−3)(2) = 6

〈1, 2,−3〉 · 〈4,−1, 2〉 = (1)(4) + (2)(−1) + (−3)(2) = −4


Algebraic Properties of the Dot Product

Theorem 3.6 If u, v, and w are vectors in 2-space or 3-space and k is a scalar, then:

(a) u · v = v · u(b) u · (v +w) = u · v + u ·w(c) k(u · v) = (ku) · v = u · (kv)(d) v · v = ‖v‖2

(e) 0 · v = 0

Angle Between Vectors

Suppose that u and v are nonzero vectors in 2-space or 3-space that are positioned so theirinitial points coincide. We define the angle between u and v to be the angle θ determinedby the vectors that satisfies the condition 0 ≤ θ ≤ π.

θ

v

u

θ

v

u b

θ

vu θ

v

u

Theorem 3.7 If u and v are nonzero vectors in 2-space or 3-space, and if θ is the anglebetween them, then

cos θ =u · v

‖u‖‖v‖ (3.5)

Example 3.11 Find the angle between

(a) u = 〈4,−3,−1〉 and v = 〈−2,−3, 5〉

(b) u = −4i+ 5j + k and v = 2i+ 3j− 7k

(c) u = i− 2j+ 2k and v = −3i + 6j− 6k

Solution . . . . . . . . .

Example 3.12 Find the angle ABC if A = (1,−2, 3), B = (2, 4,−6), and C = (5,−3, 2).

θ

A(1,−2, 3)

B(2, 4,−6) C(5,−3, 2)

Solution . . . . . . . . .


Interpreting the Sign of the Dot Product

It will often be convenient to express Formula (3.5) as

u · v = ‖u‖‖v‖ cos θ (3.6)

which expresses the dot product of u and v in terms of the lengths of these vectors andthe angle between them. Since u and v are assumed to be nonzero vectors, this version ofthe formula make it clear that the sign of u · v is the same as the sign of cos θ. Thus, wecan tell from the dot product whether the angle between two vectors is acute or obtuse orwhether the vectors are perpendicular.

θ

v

u

u · v > 0

θ

v

u

u · v < 0

bθ

v

u

u · v = 0

Direction Angles

In both 2-space and 3-space the angle between a nonzero vector v and the vectors i, j,and k are called the direction angles of v, and the cosines of these angles are called thedirection cosines of v. Formulas for the direction cosines of a vector can be obtainedform Formula (3.5). For example, if v = v1i+ v2j+ v3k, then

cosα =v · i

‖v‖‖i‖ =v1‖v‖ , cos β =

v · j‖v‖‖j‖ =

v2‖v‖ , cos γ =

v · k‖v‖‖k‖ =

v3‖v‖

α

β

v

ix

j

y

α

βγ

v

i

j

k

x

y

z

Theorem 3.8 The direction cosines of a nonzero vector v = v1i+ v2j+ v3k are

cosα =v1‖v‖ , cos β =

v2‖v‖ , cos γ =

v3‖v‖


The direction cosines of a vector v = v1i + v2j + v3k can be computed by normalizingv and reading off the components of v/‖v‖, since

v

‖v‖ =v1‖v‖ i +

v2‖v‖ j+

vk‖v‖ k = (cosα)i+ (cos β)j+ (cos γ)k

Moreover, we can show that the direction cosines of a vector satisfy the equation

cos2 α+ cos2 β + cos2 γ = 1

Example 3.13 Find the direction angles of the vector v = 4i− 5j+ 3k.

Solution . . . . . . . . .

Decomposing Vectors into Orthogonal Components

Our next objective is to develop a computational procedure for decomposing a vector intosum of orthogonal vectors. For this purpose, suppose that e1 and e2 are two orthogonalunit vectors in 2-space, and suppose that we want to express a given vector v as a sum

v = w1 +w2

so that w1 is a scalar multiple of e1 and w2 is a scalar multiple of e2.

v

e1 w1

e2

w2

That is, we want to find scalars k1 and k2 such that

v = k1e1 + k2e2 (3.7)

We can find k1 by taking the dot product of v with e1. This yields

v · e1 = (k1e1 + k2e2) · e1= k1(e1 · e1) + k2(e2 · e1)= k1‖e1‖2 + 0 = k1

Similarly,

v · e2 = (k1e1 + k2e2) · e2 = k1(e1 · e2) + k2(e2 · e2) = 0 + k2‖e2‖2 = k2

Substituting these expressions for k1 and k2 in (3.7) yields

v = (v · e1)e1 + (v · e2)e2 (3.8)


In this formula we call (v · e1)e1 and (v · e2)e2 the vector components of v along e1 ande2, respectively; and we call v · e1 and v · e2 the scalar components of v along e1 ande2, respectively.

If θ denote the angle between v and e1, then

v · e1 = ‖v‖ cos θ and v · e2 = ‖v‖ sin θ

and the decomposition (3.7) can be expressed as

v = (‖v‖ cos θ)e1 + (‖v‖ sin θ)e2 (3.9)

Example 3.14 Let

v = 〈2, 3〉, e1 =

⟨

1√2,1√2

⟩

, and e2 =

⟨

− 1√2,1√2

⟩

Find the scalar components of v along e1 and e2 and the vector components of v along e1and e2.

Solution . . . . . . . . .

Example 3.15 A rope is attached to a 100-lb block on a ramp that is inclined at an angleof 30◦ with the ground.

30◦

How much force does the block exert against the ramp, and how much force must be appliedto the rope in a direction parallel to the ramp to prevent the block from sliding down theramp?

Solution . . . . . . . . .

Orthogonal Projections

The vector components of v along e1 and e2 in (3.8) are also called the orthogonal projectionsof v on e1 and e2 and are denoted by

proje1v = (v · e1)e1 and proje2v = (v · e2)e2

In general, if e is a unit vector, then we define the orthogonal projection of v on e tobe

projev = (v · e) e (3.10)


The orthogonal projection of v on an arbitrary nonzero vector b can be obtained by nor-malizing b and then applying Formula (3.10); that is,

projbv =

(

v · b

‖b‖

)(

b

‖b‖

)

which can be rewritten as

projbv =v · b‖b‖2 b (3.11)

Moreover, if we subtract projbv from v, then the resulting vector

v − projbv

will be orthogonal to b; we call this the vector component of v orthogonal to b.

v

b projbv

v − projbv

Acute angle between v and b

v v − projbv

bprojbv

Obtuse angle between v and b

Example 3.16 Find the orthogonal projection of v = i + j + k on b = 2i + 2j, and thenfind the vector component of v orthogonal to b.

Solution . . . . . . . . .

Work

Recall that we define the work W done on the object by a constant force of magnitude Facting in the direction of motion over the distance d to be

W = Fd = force × distance (3.12)

If we let F denote a force vector of magnitude ‖F‖ = F acting in the direction of motion,then we can write (3.12) as

W = ‖F‖dMoreover, if we assume that the object moves along a line from point P to point Q, then

d = ‖−→PQ‖, so that the work can be expressed entirely in vector form as

W = ‖F‖‖−→PQ‖

The vector−→PQ is called the displacement vector for the object.

In the case where a constant force F is not in the direction of motion, but rather makesan angle θ with the displacement vector, then we define the work W done by F to be

W = (‖F‖ cos θ)‖−→PQ‖ = F · −→PQ (3.13)


b

P

F‖F‖

bQ

Work = ‖F‖‖−→PQ‖

b

P

F‖F‖

‖F‖ cos θθ

bQ

Work = (‖F‖ cos θ)‖−→PQ‖

Example 3.17 A force F = 8i + 5j in pound moves an object from P (1, 0) to Q(7, 1),distance measured in feet. How much work is done?

Solution . . . . . . . . .

Example 3.18 A wagon is pulled horizontally by exerting a constant force of 10 lb on thehandle at an angle of 60◦ with the horizontal. How much work is done in moving the wagon50 ft?

Solution . . . . . . . . .

3.4 Cross Product

Determinants

Before we define the cross product, we need to define the notion of determinant.

Definition 3.4 The determinant of a 2× 2 matrix of real number is defined by∣

∣

∣

∣

a1 a2b1 b2

∣

∣

∣

∣

= a1b2 − a2b1.

Definition 3.5 The determinant of a 3 × 3 matrix of real number is defined as a com-bination of three 2× 2 determinants, as follows:

∣

∣

∣

∣

∣

∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣

∣

∣

∣

∣

∣

= a1

∣

∣

∣

∣

b2 b3c2 c3

∣

∣

∣

∣

− a2

∣

∣

∣

∣

b1 b3c1 c3

∣

∣

∣

∣

+ a3

∣

∣

∣

∣

b1 b2c1 c2

∣

∣

∣

∣

(3.14)

Note that Equation (3.14) is referred to as an expansion of the determinant along

the first row .

Cross Product

We now turn to the main concept in this section.

Definition 3.6 If u = 〈u1, u2, u3〉 and v = 〈v1, v2, v3〉 are vectors in 3-space, then thecross product u× v is the vector defined by

u× v =

∣

∣

∣

∣

u2 u3

v2 v3

∣

∣

∣

∣

i−∣

∣

∣

∣

u1 u3

v1 v3

∣

∣

∣

∣

j +

∣

∣

∣

∣

u1 u2

v1 v2

∣

∣

∣

∣

k (3.15)


or, equivalently,

u× v = (u2v3 − u3v2)i− (u1v3 − u3v1)j+ (u1v2 − u2v1)k (3.16)

Observe that the right side of Formula (3.15) can be written as

u× v =

∣

∣

∣

∣

∣

∣

i j ku1 u2 u3

v1 v2 v3

∣

∣

∣

∣

∣

∣

(3.17)

Example 3.19 Let u = 〈1, 2, 3〉 and v = 〈4, 5, 6〉. Find (a) u× v and (b) v × u.

Solution . . . . . . . . .

Algebraic Properties of the Cross Product

• The cross product is defined only for vectors in 3-space, whereas the dot product isdefined for vectors in 2-space and 3-space.

• The cross product of two vectors is a vector, whereas the dot product of two vectorsis a scalar.

The main algebraic properties of the cross product are listed in the following theorem.

Theorem 3.9 If u, v, and w are any vectors in 3-space and k is any scalar, then:

(a) u× v = −(v × u)

(b) u× (v +w) = (u× v) + (u×w)

(c) (u+ v)×w = (u×w) + (v ×w)

(d) k(u× v) = (ku)× v = u× (kv)

(e) u× 0 = 0× u = 0

(f) u× u = 0

The following cross products occur so frequently that it is helpful to be familiar withthem:

i× j = k j× k = i k× i = jj× i = −k k× j = −i i× k = −j

(3.18)

These results are easy to obtain; for example,

i× j =

∣

∣

∣

∣

∣

∣

i j k1 0 00 1 0

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

0 01 0

∣

∣

∣

∣

i−∣

∣

∣

∣

1 00 0

∣

∣

∣

∣

j+

∣

∣

∣

∣

1 00 1

∣

∣

∣

∣

k

However, rather than computing these cross products each time you need them, you canuse the diagram in Figure below.


j

k

i

Geometric Properties of the Cross Product

The following theorem shows that the cross product of two vectors is orthogonal to bothfactors.

Theorem 3.10 If u and v are vectors in 3-space, then:

(a) u · (u× v) = 0 (u× v is orthogonal to u)

(b) v · (u× v) = 0 (u× v is orthogonal to v)

Example 3.20 Let u and v be vectors as in Example 3.19. Show that u× v is orthogonalto both u and v.

Solution . . . . . . . . .

Theorem 3.11 Let u and v be nonzero vectors in 3-space, and let θ be the angle betweenthese vectors when they are positioned so their initial points coincide.

(a) ‖u× v‖ = ‖u‖‖v‖ sin θ

(b) The area A of the parallelogram that has u and v as adjacent sides is

A = ‖u× v‖ (3.19)

(c) u × v = 0 if and only if u and v are parallel vectors, that is, if and only if they arescalar multiples of one another.

Example 3.21 Find the area of the parallelogram with two adjacent sides formed by thevectors u = 〈1, 2,−2〉 and v = 〈3, 0, 1〉.

Solution . . . . . . . . .

Example 3.22 Find the area of the triangle that is determined by the points P1(2, 2, 0),P2(−1, 0, 2), and P3(0, 4, 3).

Solution . . . . . . . . .


Scalar Triple Products

If u = 〈u1, u2, u3〉, v = 〈v1, v2, v3〉, and w = 〈w1, w2, w3〉 are vectors in 3-space, then thenumber

u · (v ×w)

is called the scalar triple product of u, v, and w. This value can be obtained directlyfrom the formula

u · (v ×w) =

∣

∣

∣

∣

∣

∣

u1 u2 u3

v1 v2 v3w1 w2 w3

∣

∣

∣

∣

∣

∣

(3.20)

Example 3.23 Calculate the scalar triple product u · (v×w) of the vectors

u = 3i− 2j− 5k, v = i + 4j− 4k, w = 3j+ 2k

Solution . . . . . . . . .

Geometric Properties of the Scalar Triple Product

Theorem 3.12 Let u, v and w be nonzero vectors in 3-space.

(a) The volume V of the parallelepiped that has u, v and w as adjacent edges is

V = |u · (v ×w)| (3.21)

(b) u · (v×w) = 0 if and only if u, v and w lie in the same plane.

Example 3.24 Find the volume of the parallelepiped with three adjacent edges formed bythe vectors u = 〈7, 8, 0〉, v = 〈1, 2, 3〉 and w = 〈4, 5, 6〉.

Solution . . . . . . . . .

Algebraic Properties of the Scalar Triple Product

u · (v ×w) = w · (u× v) = v · (w × u)

u · v ×w = u× v ·w

chapter 1 topics in analytic geometry

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