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 because they can be obtained as intersection of a plane with a double-napped circular cone. If the plane 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 from a fixed line and a fixed point not on the line.

b b Axis focus Vertex

Directrix

The line is called the directrix of the parabola, and the point is called the focus. A parabola is symmetric about the line that passes through the focus at right angles to the directrix. This line, called the axis or the axis of symmetry of the parabola, intersects the 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 distances from two fixed points is a given positive constant that is greater than the distance between the 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 the midpoint of the line segment joining the foci is called the center.

b b b

b b b Focus Center Focus

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 the line segment across the ellipse, through the center, and perpendicular to the major axis is called the minor axis. The endpoints of the major axis are called vertices.

Minor axis

Vextex b b b b Vextex

Major axis

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

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

Center

b b Focal axis Focus Focus Vertex Vertex MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 3

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

Equations of Parabolas in Standard Position Let p denote the distance between focus and the vertex. The vertex is equidistant from the 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.

pb p b Axis

Directrix

The equation of a parabola is simplest if the vertex is the origin and the axis of symmetry is 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 called the standard equations of a parabola. Orientation Standard Equation y

y2 =4px b x (p, 0) x = p − MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 4

Orientation Standard Equation y

y2 = 4px b x − ( p, 0) − x = p

(0,p) b 2 x x =4py y = p −

y y = p

x 2 b x = 4py (0, p) − −

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

D( p,y) b − P (x,y)

b x F (p, 0)

x = p − Since P is equidistant from the focus and directrix, the distances P F and PD in the above Figure are equal; that is, P F = PD (1.1) From the distance formula, the distance P F and PD are P F = (x p)2 + y2 and PD = (x + p)2 (1.2) − p p MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 5

Substituting in (1.1) and squaring yields

(x p)2 + y2 =(x + p)2 (1.3) − and after simplifying y2 =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 passes through 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, the length of the minor axis by 2b, and the distance between the foci by 2c. c c

b b

a a

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 a point Q at the end of the minor axis.

Q b √b2 + c2 √b2 + c2 b

b b b c c a c P − MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 6

Because P and Q both lie on the ellipse, the sum of the distances from each of them to the foci 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 that all 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 the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure below. These are called the standard positions of an ellipse, and the result equations are called the standard equations of an ellipse. Orientation Standard Equation y b x2 y2 2 + 2 =1 b b x a b a( c, 0) (c, 0) a − −

b − y a b (0,c) x2 y2 2 + 2 =1 x b a b b −

b (0, c) − a − To illustrate how the equations in the above Figure are obtained, we will derive the equation for the ellipse with the foci on the x-axis. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 7

y P (x,y)) b

b b x F 0( c, 0) F (c, 0) −

Let P (x, y) be any point on the ellipse. Since the sum of the distances from P to the foci is 2a, it follows that P F 0 + P F =2a so (x + c)2 + y2 + (x c)2 + y2 =2a − Transposing the second radicalp to the rightp side of the equation and squaring yields

(x + c)2 + y2 =4a2 4a (x c)2 + y2 +(x c)2 + y2 − − − and, on simplifying, p c (x c)2 + y2 = a x − − a Squaring both sides again yieldsp

x2 y2 + =1 a2 a2 c2 − which, by (1.5), can be written as x2 y2 + = 1 (1.7) a2 b2 Conversely it can be shown that any point whose coordinates satisfying (1.7) has 2a as the 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 endpoints (0, 4). ± ± Solution ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 8

Equations of Hyperbolas in Standard Position It is traditional in the study of hyperbola to denote the distance between the vertices by 2a, 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.

a c c b c b b b b a a

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 focus minus the distance to the closer focus is 2a. The equation of a hyperbola is simplest if the center of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure below. These are called the standard positions of a hyperbola, and the result equations are called the standard equations of a hyperbola. Orientation Standard Equation y

b x2 y2 b b x =1 ( c, 0)a (c, 0) a2 − b2 −

b(0,c) a y2 x2 x =1 b a2 − b2

b (0, c) − MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 9

The derivation of these equations are similar to those already given for parabolas and ellipses, 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 hyperbola equation, and then solving for y in terms of x. For example, for the hyperbola x2 y2 =1 a2 − b2 we would write x2 y2 b2 b =0 or y2 = x2 or y = x a2 − b2 a2 ±a 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 ± x. ± 3 Solution ......

Translated Conics Equations of conic that are translated from their standard positions can be obtained by replacing x by x h and y by y k in their standard equations. For a parabola, this translates 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 (y k)2 − + − =1 (a > b) a2 b2 MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 10

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

(x h)2 (y k)2 − + − =1 (a > b) b2 a2

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

(x h)2 (y k)2 − − =1 a2 − b2

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

(y k)2 (x h)2 − − =1 a2 − b2

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 we are 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 squares of the quadratic terms and then try to match up the resulting equation with one of the forms 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 the ﬁgure.

Solution ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 11

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. If B = 0, then (1.11) reduces to (1.10) and the conic section has its axis or axes parallel to the coordinate axes. However, if B = 0, then (1.11) contains a cross-product term Bxy, and the graph of the conic section represented6 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. y

0 y P (x,y) b P (x0,y0) x0 b r 0 0 (00,y0) b (x , 0 ) α θ x

The axes of an xy-coordinate system have been rotated about the origin through an angle θ to produce a new x0y0-coordinate system. Each point P in the plane has coordinates (x0,y0) as well as coordinates (x, y). To see how the two are related, let r be the distance from the common 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) and x0 = r cos(α), y0 = r sin(α) (1.13) Applying the addition formulas for the sine and cosine, the relationships in (1.12) can be written as

x = r cos θ cos α r sin θ sin α − y = r sin θ cos α + r cos θ sin α MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 12 and on substituting (1.13) in these equations we obtain the following relationships called the rotation equations: x = x0 cos θ y0 sin θ − (1.14) y = x0 sin θ + y0 cos θ Example 1.11 Suppose that the axes of an xy-coordinate system are rotated through an angle of θ = 45◦ to obtain an x0y0-coordinate system. Find the equation of the curve

x2 xy + y2 6=0 − − in x0y0-coordinates. Solution ......

If the rotation (1.14) are solved for x0 and y0 in terms of x and y, we obtain:

x0 = x cos θ + y sin θ (1.15) y0 = x sin θ + y cos θ − Example 1.12 Find the new coordinates of the point (2, 4) if the coordinate axes are rotated through 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 because the rotation of axes eliminate the xy-term, thereby reducing− the− equation to a familiar form. This occurred because the new x0y0-axes were aligned with the axes of the ellipses. The follows theorem tells how to determine an appropriate rotation of axes to eliminate the cross-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 = 0, and if an x0y0-coordinate system is obtained by rotating the xy-axes 6 through an angle θ satisfying A C cot2θ = − (1.17) B then, in x0y0-coordinates, Equation (1.16) will have the form

A0x02 + C0y02 + D0x0 + E0y0 + F 0 =0

Remark It is always possible to satisfy (1.17) with an angle θ in the range 0 <θ<π/2. We • shall always use such a value of θ. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 13

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

1 cos2θ 1+cos2θ sin θ = − and cos θ = r 2 r 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 ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 14 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- and y-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 the trajectory of the particle or the graph of the equations. The variable t is called the parameter for the equations.

y C

b (x, y)

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

x = t 3 sin t, y =4 3 cos t − −

15 MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 16

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

t x y 0 0.0 1.0 1 -1.5 2.4 y 2 -0.7 5.2

b 3 2.6 7.0 b 4 6.3 6.0 b b 5 7.9 3.1 b 6 6.8 1.1 b b

b 7 5.0 1.7 b

8 5.0 4.4 b 9 7.8 6.7 x 10 11.6 6.5 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 parameter t 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 the circle counterclockwise, starting at (1, 0) when t = 0 and completing one full revolution when t =2π. y (x, y) b

1 y t (0, 1) b x x

z MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 17

Orientation The direction in which the graph of a pair of parametric equations is traces as the parameter increases is called the direction of increasing parameter or sometimes the orientation imposed on the curve by the equation. Thus, we make a distinction between a curve, which is the set of points, and a parametric curve, which is a curve with an orientation. For example, we saw in Example 2.2 that the circle represented parametrically by (2.1) is traced counterclockwise as t increases and hence has counterclockwise orientation. To obtain parametric equation for the unit circle with clockwise orientation, we can replace 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 completes one full revolution when t =2π. y

x (0, 1) bb x t y 1 b (x, y)

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 parameter t = x; this yields the parametric equations

x = t, y = f(t)

For example, the portion of the curve y = cos x over the interval [ 2π, 2π] can be expressed parametrically as − x = t, y = cos t ( 2π t 2π) − ≤ ≤ MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 18

If a function f is one-to-one, then it has an inverse function f −1. In this case the equation y = f −1(x) is equivalent to x = f(y). We can express the graph of f −1 in parametric form by 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 proved that if dx/dt = 0, then y is a differentiable function of x, in which case the chain rule implies 6 that dy dy/dt = (2.2) dx dx/dt This formula makes it possible to find dy/dx directly from the parametric equations without eliminating 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 horizontal at those points where dy/dt = 0 and dx/dt = 0, since dy/dx = 0 at such points. 6 At points where dx/dt = 0 and dy/dt = 0, the right side of (2.2) has a nonzero numerator and a zero denominator; we will agree that6 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. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 19

Solution ...... Example 2.6 The curve represented by the parametric equations x = t2 y = t3 (

x = t2, y = t3 (

Arc Length of Parametric Curve The following result provides a formula for ﬁnding the arc length of a curve from parametric equations 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 continuous functions for a t b, then the arclength L of the curve is given by ≤ ≤ 2 2 b dx dy L = + dt (2.3) a s dt dt Z 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 ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 20

2.2 Polar Coordinates

Polar Coordinate Systems A polar coordinate system in a plane consists of a ﬁxed point O, called the pole (or origin), and a ray emanating from a pole, called the polar axis. In such a coordinate system 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 ray OP . P (r, θ)

θ O Origin Polar axis The number r is called the radial distance of P and the number θ the angular coordinate (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 11π/3

π/3 −

(1, 5π ) (1, π ) (1, 11π ) 3 − 3 3 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 negative values of r as well. To motivate an appropriate deﬁnition, consider the point P with polar coordinates (3, 5π/4). We can reach this point by MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 21

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

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

5π/4 Polar axis Terminalπ/ side4 Polar axis

Terminal side b b P (3, 5π/4) P (3,π/4)

This suggest that the point (3, 5π/4) might also be denoted by ( 3,π/4), with minus sign serving to indicate that the point is on the extension of the angle’s− terminal side rather than on the terminal side itself. In general, the terminal side of the angle θ + π is the extension of the terminal side of θ, 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 of a polar coordinate system, making the positive x-axis coincide with the polar axis. Then every point P will have both rectangular coordinates (x, y) and polar coordinates (r, θ). y

P (r, θ)= P (x, y)

r y

θ x x

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

x = r cos θ, y = r sin θ (2.4) MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 22

These equation are well suited for ﬁnding x and y when r and θ are known. However, to ﬁnd r and θ when x and y are known, we use equations y r2 = x2 + y2, tan θ = (2.5) x

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 to be measured in radians. Some examples of such equations are

r =1, θ = π/4, r = θ, r = sin θ, r = cos2θ

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 = cos2θ in polar coordinates.

Solution Instead of plotting points, we will use the graph of r = cos2θ in rectangular coordinates to visualize how the polar graph of this equation is generated. This curve is called a four-petal rose. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 23

r 1

θ π 3π 2 π 2 2π

1 − r varies from r varies from y = cos2θ 1 to 0 as θ 0 to 1 as θ varies from varies− from 0 to π/4. π/4 to π/2.

r varies from r varies from r varies from 1 to 0 as θ 0 to 1 as θ 1 to 0 as θ varies− from varies from varies from π/2 to 3π/4. 3π/4 to π. π to 5π/4.

r varies from r varies from r varies from 0 to 1 as θ 1 to 0 as θ 1 to 0 as θ − − − varies from varies from varies from 5π/4to 3π/2. 3π/2 to 7π/4. 7π/4 to 2π. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 24

Symmetry Tests Observe that the polar graph r = cos2θ in above Figure is symmetric about the x-axis and the 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 its equation produces an equivalent equation. −

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

(c) A curve in polar coordinates is symmetric about the origin if replacing θ by θ + π, or replacing r by r in its equation produces an equivalent equation. − π/2 π/2 π/2 (r, π θ) (r, θ) (r, θ) − (r, θ)

0 0 0 (r, θ + π) (r, θ) or − ( r, θ) − (a) (b) (c)

Example 2.15 Show that the graph of r = cos2θ is symmetric about the x-axis and y-axis.

Solution ......

Families of Lines and Rays Through the Pole

For any constant θ0, the equation θ = θ0 (2.6) is satisﬁed by the coordinates of the form P (r, θ0), regardless of the value of r. Thus, the equation represents the line that passes through the pole and makes an angle of θ0 with the polar axis. π/2

θ 0

θ = θ0 MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 25

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 and • tangent to the y-axis at the origin.

The equation r = a represents a circle of radius a, centered on the y-axis and tangent • to the x-axis at the origin. π/2 π/2 π/2 r = 2a sin θ

r = 2a cos θ r = 2a cos θ π r = a − b (a, 2 ) a 0 b b 0 0 (a, π) (a, 0) π b (a, ) − 2

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 sin nθ or r = a cos nθ in which a> 0 and n is a positive integer represent families of ﬂower-shaped curves called roses. The rose consists of n equally spaced petals of radius a if n is odd and 2n equally spaced petals of radius a if n is even. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 26

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

r = a sin 5θ r = a sin 6θ

r = a cos2θ r = a cos3θ r = a cos4θ

r = a cos5θ r = a cos6θ

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 possible shapes 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. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 27

a/b < 1 a/b =1 1 < a/b < 2 a/b 2 ≥ Limacon with Cardioid Dimpled limacon Convex limacon inner loop

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 the spiral 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 lines to polar curves of the form r = f(θ) in which r is a differentiable function of θ. A curve of this form can be expressed parametrically in terms of parameter θ by substituting f(θ) for r in the equations x = r cos θ and y = r sin θ. This yields

x = f(θ) cos θ, y = f(θ) sin θ MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 28 from which we obtain dx dr = f(θ) sin θ + f 0(θ) cos θ = r sin θ + cos θ dθ − − dθ dy dr = f(θ) cos θ + f 0(θ) cos θ = r cos θ + sin θ dθ dθ Thus, if dx/dθ and dy/dθ are continuous and if dx/dθ = 0, then y is a diﬀerentiable function of x, and Formula (2.2) with θ in place of t yields 6

dr dy dy/dθ r cos θ + sin θ = = dθ (2.7) dx dx/dθ dr r sin θ + cos θ − dθ

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

Solution ......

Example 2.20 Find the points on the cardioid r =1 cos θ at which there is a horizontal tangent 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 if dr/dθ =0 at θ = θ0, then the line θ = θ0 is tangent to the curve at the origin. 6 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 to β, and if dr/dθ is continuous for α θ β, then the arc length L from θ = α to θ = β is ≤ ≤

2 β β dr L = [f(θ)]2 + [f 0(θ)]2 dθ = r2 + dθ (2.8) dθ α α s Z p Z 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 ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 29

Area in Polar Coordinates r = f(θ)

R θ = β θ = α 0

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 the region 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 the area A of the region R enclosed by the polar curve r = f(θ) (α θ ≤ β)≤ and the lines θ = α and θ = β is ≤ ≤

β β 1 2 1 2 A = 2 [f(θ)] dθ = 2 r dθ (2.9) Zα Zα The hardest part of applying (2.9) is determining the limits of integration. This can be done 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 the region R?”

Step 4 The answer in Step 3 will determine the lower and upper limits of integration. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 30

Example 2.23 Find the area of the region in the ﬁrst quadrant that is within the cardioid r =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 = cos2θ.

Solution ......

Example 2.26 Find the area of the region that is inside of the cardioid r = 4+4cos θ and outside 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, the y-axis, and the z-axis, positioned so that their origin coincide. z

O y

The three coordinate axes form a three-dimensional rectangular coordinate system (or Cartesian coordinate system) The point of intersection of the coordinate axes is called the 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 each point P in 3-space corresponds to ordered triple of real numbers (a, b, c) which measure its 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 MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 32

Region Description xy-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

2 2 d = (x2 x1) +(y2 y1) − − The distance formula in 3-space hasp the same form, but it has a third term to account for the added dimension. The distance between the points P1(x1,y1, z1) and P2(x2,y2, z2) is

2 2 2 d = (x2 x1) +(y2 y1) +(z2 z1) − − − Example 3.1 Find the distancep 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) and radius r is 2 2 2 (x x0) +(y y0) = r − − Analogously, standard equation of the sphere in 3-space that has center (x0,y0, z0) and radius r is 2 2 2 2 (x x0) +(y y0) +(z z0) = r (3.1) − − − If the terms in (3.1) are expanded and like terms are collected, then the resulting equation has 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

2 2 2 2 (x x0) +(y y0) +(z z0) = k − − − 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). • MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 33

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 magnitude and a direction are specified. Vectors can be represented geometrically by arrows in 2-space or 3-space: the direction of the arrow specifies the direction of the vector and the length of the arrow describes its magnitude. The tail of the arrow is called the the initial point of the vector, and the tip of the arrow the terminal point. We will denote vectors with lowercase boldface type such as a, k, v, w, and x. When discussing vectors, we will refer to real numbers as scalars. Scalar will be denoted by lowercase italic type such as a, k, w, and x. Two vectors, v and w, are considered to be equal (also called equivalent) if they have 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 the following 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. b B

A b

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 speciﬁc direction, so we will agree that it can be assigned any convenient direction in a speciﬁc problem. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 34

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

v v + w

Definition 3.2 If v is a nonzero vector and k is a nonzero real number (a scalar), then the scalar multiple kv is defined to be the vector whose length is k times the length of | | v 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 1 2 v

( 1)v − ( 3 )v − 2 Observe that if k and v are nonzero, then the vectors v and kv lie on the same line if their initial points coincide and lies on parallel or coincident lines if they do not. Thus, we say that v and kv are parallel vectors. Observe also that the vector ( 1)v has the same length 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 coordinate system, 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 the components of v, and we write v in component form using the bracket notation

v = v1, v2 or v = v1, v2, v3 h i h i In particular, the zero vectors in 2-space and 3-space are v = 0, 0 and v = 0, 0, 0 h i h i respectively. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 35 y z

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

x y

Theorem 3.2 Two vectors are equivalent if and only if their corresponding components are equal. For example, a, b, c = 2, 3, 5 h i h− i 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 h i h i

v + w = v1 + w1, v2 + w2 h i v w = v1 w1, v2 w2 − h − − i kv = kv1, kv2 h i Similarly, if v = v1, v2, v3 and w = w1,w2,w3 are vectors in 3-space and k is any scalar, then h i h i

v + w = v1 + w1, v2 + w2, v3 + w3 h i v w = v1 w1, v2 w2, v3 w3 − h − − − i kv = kv1, kv2, kv3 h i Example 3.3 If v = 1, 2, 3 and w = 2, 0, 4 , then h− − i h − i 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 − h i−h i h − − i MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 36

P2(x2,y2) −−−→P1P2 b P1(x1,y1) b

−−→OP1 −−→OP2

x O

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

−−→P1P2 = 1 3, 4 2 = 4, 2 h− − − i h− i 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 h− − − − − i h− i Rules of Vector Arithmetic Theorem 3.5 For any vectors u, v, and w and any scalars k and `, the following relationships 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, the norm, or the magnitude of v and is denoted by v . This distance does not change if k k MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 37

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

2 2 v = v1 + v2 k k q and the norm of a vector v = v1, v2, v3 in 3-space is given by h i

2 2 2 v = v1 + v2 + v3 k k q Example 3.5 Find the norm of v = 4, 2 , and w = 1, 3, 5 . h − i h− i Solution ......

For any vector v and scalar k, the length of kv must be k times the length of v; that is, | | kv = k v k k | |k k Thus, for example,

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

y z

(0, 1) (0, 0, 1)

j k y x j (0, 1, 0) i (1, 0) i (1, 0, 0) x Thus,

i = 1, 0 , j = 0, 1 In 2-space h i h i i = 1, 0, 0 , j = 0, 1, 0 , k = 0, 0, 1 In 3-space h i h i h i MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 38

Every vector in 2-space is expressible uniquely in terms of i and j, and every vector in 3-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 h i h i h i h i h i

v = v1, v2, v3 = v1 1, 0, 0 + v2 0, 1, 0 + v3 0, 0, 1 = v1i + v2j + v3k h i h i h i h i Example 3.6

2-space 3-space 3, 4 =3i 4j 2, 3, 5 =2i +3j 5k h − i − h − i − 5, 0 =5i +0j =5i 0, 0, 3 =3k h i h i 0, 0 =0i +0j = 0 0, 0, 0 =0i +0j +0k = 0 h i h i (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 k k k − k − 2 2 2 2 2 v1i + v2j = v1 + v2 v1, v2, v3 = pv1 + v2 + v3 k k kh ik p p Normalizing a Vector A common problem in applications is to ﬁnd a unit vector u that has the same direction as some given nonzero vector v. This can be done by multiplying v by the reciprocal of its length; that is, 1 v u = v = v v k k k k 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 with the 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, and if θ is the angle from the positive x-axis to the radial line through v, then the x-component of v can be written as v cos θ and the y-component as v sin θ; k k k k MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 39

v k k v sin θ θ k k x v cos θ k k

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) k kh i k k k k 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) h i 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 is determined by some known unit vector u, and it is interest to find the components of a vector v that has the same direction as u and some specified length v. This can be done by expressing v as v = v u k k and then reading off the components of v u. k k 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 the forces and the point at which it is applied. If two forces F1 and F2 are applied at the same point on an object, then the two forces have the same eﬀect on the object as the single force F1 + F2 applied at the point. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 40

F1 + F2

Physicists and engineers call F1 + F2 the resultant of F1 and F2, and they say that the forces 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 Figure below. Find the magnitude of the resultant and the angle θ that it makes with the positive x-axis. y

F2 = 300N k k

◦ F1 = 200N 40 k k 30◦ b x

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 h i h i 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 dot product is defined ash i h i 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 h − i·h i − 1, 2, 3 4, 1, 2 = (1)(4)+(2)( 1)+( 3)(2) = 4 h − i·h − i − − − MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 41

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 · k k (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 their initial points coincide. We deﬁne the angle between u and v to be the angle θ determined by the vectors that satisﬁes the condition 0 θ π. ≤ ≤

u θ v

u b θ θ u v u θ v v

Theorem 3.7 If u and v are nonzero vectors in 2-space or 3-space, and if θ is the angle between them, then u v cos θ = · (3.5) u v k kk k Example 3.11 Find the angle between (a) u = 4, 3, 1 and v = 2, 3, 5 h − − i h− − i (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). − − − A(1, 2, 3) −

B(2, 4, 6) C(5, 3, 2) − −

Solution ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 42

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

u u u θ θ θ b v v v

u v > 0 u v < 0 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 the direction cosines of v. Formulas for the direction cosines of a vector can be obtained form Formula (3.5). For example, if v = v1i + v2j + v3k, then

v i v1 v j v2 v k v3 cos α = · = , cos β = · = , cos γ = · = v i v v j v v k v k kk k k k k kk k k k k kk k k k y z

j k v v

β γ α β x y i j α i x

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

v1 v2 v3 cos α = , cos β = , cos γ = v v v k k k k k k MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 43

The direction cosines of a vector v = v1i + v2j + v3k can be computed by normalizing v and reading oﬀ the components of v/ v , since k k v v1 v2 v = i + j + k k = (cos α)i + (cos β)j + (cos γ)k v v v v k k k k k k k 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 into sum of orthogonal vectors. For this purpose, suppose that e1 and e2 are two orthogonal unit 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.

e2 v

e1 w1

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) · 2 · = k1 e1 +0 = k1 k k Similarly,

2 v e2 =(k1e1 + k2e2) e2 = k1(e1 e2)+ k2(e2 e2)=0+ k2 e2 = k2 · · · · k k Substituting these expressions for k1 and k2 in (3.7) yields

v =(v e1)e1 +(v e2)e2 (3.8) · · MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 44

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

If θ denote the angle between v and e1, then

v e1 = v cos θ and v e2 = v sin θ · k k · k k and the decomposition (3.7) can be expressed as

v =( v cos θ)e1 +( v sin θ)e2 (3.9) k k k k Example 3.14 Let

1 1 1 1 v = 2, 3 , e1 = , , and e2 = , h i √2 √2 −√2 √2 Find the scalar components of v along e1 and e2 and the vector components of v along e1 and e2.

Solution ......

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

30◦

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

Solution ......

Orthogonal Projections

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

proje v =(v e1)e1 and proje v =(v e2)e2 1 · 2 · In general, if e is a unit vector, then we deﬁne the orthogonal projection of v on e to be projev =(v e) e (3.10) · MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 45

The orthogonal projection of v on an arbitrary nonzero vector b can be obtained by normalizing b and then applying Formula (3.10); that is, b b projbv = v · b b k k k k which can be rewritten as v b projbv = · b (3.11) b 2 k k 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 v projbv v v projbv − −

b projbv projbv b

Acute angle between v and b Obtuse angle between v and b

Example 3.16 Find the orthogonal projection of v = i + j + k on b = 2i +2j, and then ﬁnd 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 F acting 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 k k W = F d k k Moreover, if we assume that the object moves along a line from point P to point Q, then d = −→P Q , so that the work can be expressed entirely in vector form as k k W = F −→P Q k kk k The vector −→P Q 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 makes an angle θ with the displacement vector, then we deﬁne the work W done by F to be

W =( F cos θ) −→P Q = F −→P Q (3.13) k k k k · MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 46

F F k k F θ k k F b b b F cos θ b k k P Q P Q

Work = F −→P Q Work = ( F cos θ) −→P Q k kk k k k k k

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 the handle at an angle of 60◦ with the horizontal. How much work is done in moving the wagon 50 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 a2 = a1b2 a2b1. b1 b2 −

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 a3 b2 b3 b1 b3 b1 b2 b1 b2 b3 = a1 a2 + a3 (3.14) c2 c3 − c1 c3 c1 c2 c1 c2 c3

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 the h i h i cross product u v is the vector deﬁned by × u2 u3 u1 u3 u1 u2 u v = i j + k (3.15) × v2 v3 − v1 v3 v1 v2

MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 47 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

i j k u v = u1 u2 u3 (3.17) × v1 v2 v3

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

Algebraic Properties of the Cross Product The cross product is deﬁned only for vectors in 3-space, whereas the dot product is • deﬁned for vectors in 2-space and 3-space. The cross product of two vectors is a vector, whereas the dot product of two vectors • is 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 with them: i j = k j k = i k i = j (3.18) j × i = k k× j = i i ×k = j × − × − × − These results are easy to obtain; for example, i j k 0 0 1 0 1 0 i j = 1 0 0 = i j + k × 1 0 − 0 0 0 1 0 1 0

However, rather than computing these cross products each time you need them, you can use the diagram in Figure below. MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 48

i j

Geometric Properties of the Cross Product The following theorem shows that the cross product of two vectors is orthogonal to both factors.

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 orthogonal to both u and v. ×

Solution ......

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

(a) u v = u v sin θ k × k k kk k (b) The area A of the parallelogram that has u and v as adjacent sides is

A = u v (3.19) k × k

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

Example 3.21 Find the area of the parallelogram with two adjacent sides formed by the vectors u = 1, 2, 2 and v = 3, 0, 1 . h − i h i 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 ...... MA112 Section 750001: Prepared by Dr. Archara Pacheenburawana 49

Scalar Triple Products

If u = u1,u2,u3 , v = v1, v2, v3 , and w = w1,w2,w3 are vectors in 3-space, then the numberh i h i h i u (v w) · × is called the scalar triple product of u, v, and w. This value can be obtained directly from the formula u1 u2 u3 u (v w)= v1 v2 v3 (3.20) · × w1 w2 w3

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 by the vectors u = 7, 8, 0 , v = 1, 2, 3 and w = 4, 5, 6 . h i h i h i Solution ......

Algebraic Properties of the Scalar Triple Product

u (v w)= w (u v)= v (w u) · × · × · × u v w = u v w · × × ·