Chapter 1. Topics in Analytic Geometry. 1.1 Conic Sections. Definitions of the Conic Sections

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1 Chapter 1 Topics in Analtic Geometr 1.1 Conic Sections Circles, ellipses, parabolas, and hperbolas are called conic sections or conics because the can be obtained as intersection of a plane with a double-napped circular cone. If the plane passes through the verte 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 fied line and a fied point not on the line. Verte focus Ais Directri The line is called the directri of the parabola, and the point is called the focus. A parabola is smmetric about the line that passes through the focus at right angles to the directri. This line, called the ais or the ais of smmetr of the parabola, intersects the parabola at a point called the verte. Definition 1.2 An ellipse is the set of all points in the plane, the sum of whose distances from two fied points is a given positive constant that is greater than the distance between the fied points. 1

2 MA112 Section : Prepared b Dr. Archara Pacheenburawana 2 The two fied 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. 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 ais, and the line segment across the ellipse, through the center, and perpendicular to the major ais is called the minor ais. The endpoints of the major ais are called vertices. Minor ais Vete Vete Major ais Definition 1.3 A hperbola is the set of all points in the plane, the difference of whose distances from two fied distinct points is a given positive constant that is less than the distance between the fied points. The two fied points are called the foci of the hperbola. The midpoint of the line segment joining the foci is called the center of the hperbola. The line through the foci is called the focal ais, and the line through the center that is perpendicular to the focal ais is called the conjugate ais. The hperbola intersects the focal ais at two points called the vertices. The two separate parts of a hperbola are called the branches. Conjugate ais Center Focus Focus Verte Verte Focal ais

3 MA112 Section : Prepared b Dr. Archara Pacheenburawana 3 Associated with ever hperbola is a pair of lines, called the asmptotes of the hperbola. This line intersect at the center of the hperbola and have the propert that as a point P moves along the hperbola awa from the center, the vertical distance between P and one of the asmptotes approaches zero. Equations of Parabolas in Standard Position Let p denote the distance between focus and the verte. The verte is equidistant from the focus and the directri, so the distance between the verte and the directri is also p; consequentl, the distance between the focus and the directri is 2p. p p 2p 2p Ais Directri Theequationofaparabolaissimplest iftheverte istheoriginandtheaisofsmmetr is along the -ais or -ais. 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 = p (p,0) 2 = 4p

4 MA112 Section : Prepared b Dr. Archara Pacheenburawana 4 Orientation Standard Equation ( p, 0) = p 2 = 4p (0,p) = p 2 = 4p = p (0, p) 2 = 4p To illustrate how the equations in the above Figure are obtained, we will derive the equation for the parabola with focus (p,0) and directri = p. Let P(,) be an point on the parabola. D( p, ) P(,) F(p,0) = p Since P is equidistant from the focus and directri, the distances PF and PD in the above Figure are equal; that is, PF = PD (1.1) From the distance formula, the distance PF and PD are PF = ( p) and PD = (+p) 2 (1.2)

5 MA112 Section : Prepared b Dr. Archara Pacheenburawana 5 Substituting in (3.22) and squaring ields and after simplifing ( p) = (+p) 2 (1.3) 2 = 4p (1.4) The derivations of the other equations in the previous Figure are similar. Eample 1.1 Find the focus and directri of the parabola 2 = 6 and sketch its graph. Eample 1.2 (a) Find an equation of a parabola that has verte at the origin, opens right, and passes through the point P(7, 3). (b) Find the focus. Equations of Ellipses in Standard Position It is traditional in the stud of ellipse to denote the length of the major ais b 2a, the length of the minor ais b 2b, and the distance between the foci b 2c. c c b b a a The number a is called the semimajor ais and the number b the semiminor ais. There is a basic relationship between a, b, and c that can be obtained b eamining the sum of the distances to the foci from a point P at the end of the major ais and from a point Q at the end of the minor ais. b2 +c 2 Q b c c a c b2 +c 2 P

6 MA112 Section : Prepared b 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 from which it follows that or equivalentl 2 b 2 +c 2 = (a c)+(a+c) a = b 2 +c 2 (1.5) c = a 2 b 2 (1.6) From (1.5), the distance from the focus to an end of the minor ais 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 -ais or -ais. 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. a( c, 0) Orientation b (c,0)a Standard Equation 2 a b 2 = 1 b a (0,c) b b 2 b a 2 = 1 (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 -ais.

7 MA112 Section : Prepared b Dr. Archara Pacheenburawana 7 P(,)) F ( c,0) F(c,0) Let P(,) be an point on the ellipse. Since the sum of the distances from P to the foci is 2a, it follows that PF +PF = 2a so (+c) ( c) = 2a Transposing the second radical to the right side of the equation and squaring ields (+c) = 4a 2 4a ( c) ( c) and, on simplifing, ( c)2 + 2 = a c a Squaring both sides again ields which, b (1.5), can be written as 2 a a 2 c 2 = 1 2 a b = 1 (1.7) 2 Conversel it can be shown that an point whose coordinates satisfing (1.7) has 2a as the sum of its distances from the foci, so that such a point is on the ellipse. Eample 1.3 Sketch the graph of the ellipse = 36. Eample 1.4 Find an equation of the ellipse with foci (0,±2) and major ais with endpoints (0, ±4).

8 MA112 Section : Prepared b Dr. Archara Pacheenburawana 8 Equations of Hperbolas in Standard Position It is traditional in the stud of hperbola to denote the distance between the vertices b 2a, the distance between the foci b 2c, and to define the quantit b as This relationship, which can be epressed as is pictured geometricall in Figure below. b = c 2 a 2 (1.8) c = a 2 +b 2 (1.9) c a a c a b c The number a is called the semifocal ais of the hperbola and the number b the semiconjugate ais. Moreover, for all points on a hperbola, the distance to the farther focus minus the distance to the closer focus is 2a. The equation of a hperbola is simplest if the center of the hperbola is at the origin and the foci are on the -ais or -ais. The two possible such orientations are shown in Figure below. These are called the standard positions of a hperbola, and the result equations are called the standard equations of a hperbola. Orientation b ( c, 0) a (c, 0) Standard Equation 2 a 2 2 b 2 = 1 (0,c) a b (0, c) 2 a 2 2 b 2 = 1

9 MA112 Section : Prepared b Dr. Archara Pacheenburawana 9 The derivation of these equations are similar to those alread given for parabolas and ellipses, so we will leave them as eercises. A Quick Wa to Find Asmptotes The are a trick that can be used find the equations of the asmptotes of a hperbola. The can be obtained, when needed, b replacing 1 b 0 on the right side of the hperbola equation, and then solving for in terms of. For eample, for the hperbola we would write 2 2 a 2 2 b 2 = 1 a 2 2 b = 0 of 2 2 = b2 which are the equations for the asmptotes. a 2 2 or = ± b a Eample 1.5 Sketch the graph of the hperbola = 36, and showing its vertices, foci, and asmptotes. Eample 1.6 Find the equation of the hperbola with vertices (0, ±8) and asmptotes = ± 4 3. Translated Conics Equations of conic that are translated from their standard positions can be obtained b replacing b h and b k in their standard equations. For a parabola, this translates the verte from the origin to the point (h, k); and for ellipses and hperbolas, this translates the center from the origin to the point (h,k). Parabolas with verte (h, k) and ais parallel to -ais ( k) 2 = 4p( h) [Opens right] ( k) 2 = 4p( h) [Opens left] Parabolas with verte (h, k) and ais parallel to -ais ( h) 2 = 4p( k) ( h) 2 = 4p( k) [Opens up] [Opens down]

10 MA112 Section : Prepared b Dr. Archara Pacheenburawana 10 Ellipse with center (h,k) and major ais parallel to -ais ( h) 2 a 2 + ( k)2 b 2 = 1 (a > b) Ellipse with center (h,k) and major ais parallel to -ais ( h) 2 b 2 + ( k)2 a 2 = 1 (a > b) Hperbola with center (h, k) and focal ais parallel to -ais ( h) 2 a 2 ( k)2 b 2 = 1 Hperbola with center (h, k) and focal ais parallel to -ais ( k) 2 a 2 ( h)2 b 2 = 1 Eample 1.7 Find an equation of the parabola with verte V( 4,2) and directri = 5. Sometimes the equations of translated conics occur in epanded form, in which case we are faced with the problem of identifing the graph of a quadratic equation in and : A 2 +C 2 +D+E +F = 0 The basic procedure for determining the nature of such a graph is to complete the squares of the quadratic terms and then tr to match up the resulting equation with one of the forms of a translated conic. Eample 1.8 Show that the curve is a parabola. = Eample 1.9 Discuss and sketch the graph of the equation = 0.

11 MA112 Section : Prepared b Dr. Archara Pacheenburawana 11 Eample 1.10 Show that the curve = 0 is a hperbola. Sketch the hperbola and show the foci, vertices, and asmptotes in the figure. 1.2 Rotation of Aes; Second-Degree Equations Quadratic Equations in and We saw in Eamples 1.8 to 1.10 that equations of the form A 2 +C 2 +D+E +F = 0 (1.10) can represent conic sections. Equation (1.10) is a special case of the more general equation A 2 +B +C 2 +D+E +F = 0 (1.11) which, if A, B, and C are not all zero, is called a quadratic equation in and. If B = 0, then (1.11) reduces to (1.10) and the conic section has its ais or aes parallel to the coordinate aes. However, if B 0, then (1.11) contains a cross-product term B, and the graph of the conic section represented b the equation has its ais or aes tilted relative to the coordinate aes. Rotation of Aes Consider the rotation of aes in Figure below. (0, ) r α θ P(,) P(, ) (,0 ) The aes of an -coordinate sstem have been rotated about the origin through an angle θ to produce a new -coordinate sstem. Each point P in the plane has coordinates (, ) as well as coordinates (,). 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 = rcos(θ+α), = rsin(θ +α) (1.12)

12 MA112 Section : Prepared b Dr. Archara Pacheenburawana 12 and = rcos(α), = rsin(α) (1.13) Appling the addition formulas for the sine and cosine, the relationships in (1.12) can be written as = rcosθcosα rsinθsinα = rsinθcosα+rcosθsinα and on substituting (1.13) in these equations we obtain the following relationships called the rotation equations: = cosθ sinθ (1.14) = sinθ + cosθ Eample 1.11 Suppose that the aes of an -coordinate sstem are rotated through an angle of θ = 45 to obtain an -coordinate sstem. Find the equation of the curve in -coordinates = 0 If the rotation (1.14) are solved for and in terms of and, we obtain: = cosθ +sinθ = sinθ+cosθ (1.15) Eample 1.12 Find the new coordinates of the point (2, 4) if the coordinate aes are rotated through an angle of θ = 30. Eliminating the Cross-Product Term In Eample 1.11, we were able to identif the curve = 0 as an ellipse because the rotation of aes eliminate the -term, thereb reducing the equation to a familiar form. This occurred because the new -aes were aligned with the aes of the ellipses. The follows theorem tells how to determine an appropriate rotation of aes to eliminate the cross-product term of a second-degree equation in and. Theorem 1.1 If the equation A 2 +B +C 2 +D+E +F = 0 (1.16) is such that B 0, and if an -coordinate sstem is obtained b rotating the -aes through an angle θ satisfing cot2θ = A C (1.17) B then, in -coordinates, Equation (1.16) will have the form A 2 +C 2 +D +E +F = 0

13 MA112 Section : Prepared b Dr. Archara Pacheenburawana 13 Remark It is alwas possible to satisf (1.17) with an angle θ in the range 0 < θ < π/2. We shall alwas use such a value of θ. The values of sinθ and cosθ needed for the rotation equations ma be obtained b first calculating cos2θ and then computing sinθ and cosθ from the identities sinθ = 1 cos2θ 2 and cosθ = 1+cos2θ 2 Eample 1.13 Discuss and sketch the graph of the equation = 1. Eample 1.14 Identif and sketch the curve = 0

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 -plane in such a wa that - and -coordinates, as a functions of time, are = f(t), = g(t) We call these the parametric equations of motion for the particle and refer C as the trajector of the particle or the graph of the equations. The variable t is called the parameter for the equations. C (,) Eample 2.1 Sketch the trajector over the time interval 0 t 10 of the particle whose parametric equations of motion are = t 3sint, = 4 3cost 14

15 MA112 Section : Prepared b Dr. Archara Pacheenburawana 15 Solution One wa to sketch the trajector is to plot the (,) coordinate of points on the trajector at those times, and connect the points with a smooth curve. t Eample 2.2 Find the graph of the parametric equations = cost, = sint (0 t 2π) (2.1) Solution One wa to find the graph is to eliminate the parameter t b noting that = sin 2 t+cos 2 t = 1 Thus, the graph is contained in the unit circle = 1. Geometricall, the parameter t can be interpreted as the angle swept out b the radial line from the origin to the point (,) = (cost,sint) 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. 1 t (,) (0,1)

16 MA112 Section : Prepared b Dr. Archara Pacheenburawana 16 Orientation The direction inwhich thegraphof apair of parametric equations is traces asthe parameter increases is called the direction of increasing parameter or sometimes the orientation imposed on the curve b 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 eample, we saw in Eample 3.28 that the circle represented parametricall b (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 b t in (2.1). This ields = cos( t) = cost, = sin( t) = sint (0 t 2π) Here, the circle is traced clockwise b a point that starts at (1,0) when t = 0 and completes one full revolution when t = 2π. 1 t (,) (0,1) Eample 2.3 Graph the parametric curve = 2t 3, = 6t 7 b eliminating the parameter, and indicate the orientation on the graph. Epressing Ordinar Functions Parametricall An equation = f() can be epressed in parametric form b introducing the parameter t = ; this ields the parametric equations = t, = f(t) For eample, the portion of the curve = cos over the interval [ 2π,2π] can be epressed parametricall as = t, = cost ( 2π t 2π)

17 MA112 Section : Prepared b Dr. Archara Pacheenburawana 17 Ifafunctionf isone-to-one,thenithasaninverse functionf 1. Inthiscasetheequation = f 1 () is equivalent to = f(). We can epress the graph of f 1 in parametric form b introducing the parameter t = ; this ields the parametric equations = f(t), = t For eample, the graph of f() = can be represented parametricall as = t, = t 5 +t+1 and the graph of f 1 can be represented parametricall as = t 5 +t+1, = t Tangent Line to Parametric Curves We will be concerned with curves that are given b parametric equations = f(t), = g(t) in which f(t) and g(t) have continuous first derivatives with respect to t. It can be proved that if d/dt 0, then is a differentiable function of, in which case the chain rule implies that d d = d/dt (2.2) d/dt This formula makes it possible to find d/d directl from the parametric equations without eliminating the parameter. Eample 2.4 Find the slope of the tangent line to the curve at the point where t = π/3. = t 3sint, = 4 3cost (t 0) It follows from Formula (2.2) that the tangent line to a parametric curve will be horizontal at those points where d/dt = 0 and d/dt 0, since d/d = 0 at such points. Atpointswhered/dt = 0andd/dt 0, therightsideof (2.2)hasanonzeronumerator and a zero denominator; we will agree that the curve has infinite slope and a vertical tangent line at such point. At points where d/dt and d/dt are both zero, the right side of (2.2) becomes an indeterminate form; we call such points singular points. Eample 2.5 A curve C is defined b the parametric equations = t 2 and = t 3 3t. Find the points on C where the tangent is horizontal or vertical.

18 MA112 Section : Prepared b Dr. Archara Pacheenburawana 18 Eample 2.6 The curve represented b the parametric equations = t 2 = t 3 ( < t < + ) is called a semicubical parabola. The parameter t can be eliminated b cubing and squaring, from which it follows that 2 = 3. = t 2, = t 3 ( < t < + ) The graph of this equation consists of two branches: an upper branch obtained b graphing = 3/2 and a lower branch obtained b graphing = 3/2. The two branches meet at the origin, which corresponds to t = 0 in the parametric equations. This is a singular point because the derivatives d/dt = 2t and d/dt = 3t 2 are both zero there. Eample 2.7 Without eliminating the parameter, find d/d and d 2 /d 2 at (1,1) and (1, 1) on the semicubical parabola given b the parametric equations in Eample 2.6 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 b the parametric equations = (t), = (t) (a t b) is traced more than once as t increase from a to b, and if d/dt and d/dt are continuous functions for a t b, then the arclength L of the curve is given b b (d ) 2 ( ) 2 d L = + dt (2.3) dt dt a Eample 2.8 Find the circumference of a circle of radius a form the parametric equations = acost, = asint (0 t 2π)

19 MA112 Section : Prepared b Dr. Archara Pacheenburawana Polar Coordinates Polar Coordinate Sstems A polar coordinate sstem in a plane consists of a fied point O, called the pole (or origin), and a ra emanating from a pole, called the polar ais. In such a coordinate sstem 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 ais to the ra OP. P(r,θ) r O Origin θ Polar ais The number r is called the radial distance of P and the number θ the angular coordinate (or polar angle) of P. Eample 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) The polar coordinates of a point are not unique. For eample, the polar coordinates all represent the same point. 5π/3 (1, 5π/3), (1, π/3), (1, 11π/3) 11π/3 π/3 (1, 5π 3 ) (1, π 3 ) (1, 11π 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 an 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 b

20 MA112 Section : Prepared b Dr. Archara Pacheenburawana 20 rotating the polar ais 5π/4 and then moving forward from the pole 3 units along the terminal side of the angle, or rotating the polar ais π/4 and then moving backward from the pole 3 units along the etension of the terminal side. Terminal side P(3,5π/4) 5π/4 Polar ais P(3,π/4) π/4 Terminal side Polar ais This suggest that the point (3,5π/4) might also be denoted b ( 3,π/4), with minus sign serving to indicate that the point is on the etension 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 etension of the terminal side of θ, we define negative radial coordinates b agreeing that to be polar coordinates for the same point. ( r,θ) and (r,θ+π) Relationship Between Polar and Rectangular Coordinates Frequentl, it will be useful to superimpose a rectangular -coordinate sstem on top of a polar coordinate sstem, making the positive -ais coincide with the polar ais. Then ever point P will have both rectangular coordinates (, ) and polar coordinates (r, θ). P(r,θ) = P(,) r θ From the above Figure, these coordinates are related b the equations = rcosθ, = rsinθ (2.4)

21 MA112 Section : Prepared b Dr. Archara Pacheenburawana 21 These equation are well suited for finding and when r and θ are known. However, to find r and θ when and are known, we use equations r 2 = 2 + 2, tanθ = (2.5) Eample 2.10 Find the rectangular coordinates of the point P whose polar coordinates are (4,2π/3). Eample 2.11 Find polar coordinates of the point P whose rectangular coordinates are (1, 1). 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 eamples of such equations are Eample 2.12 Sketch the graph of r = 1, θ = π/4, r = θ, r = sinθ, r = cos2θ (a) r = 1 (b) θ = π/4 Eample 2.13 Sketch the graph of r = 2cosθ in polar coordinates b plotting points. Eample 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.

22 MA112 Section : Prepared b Dr. Archara Pacheenburawana 22 1 r 1 π 2 π 3π 2 2π = cos2θ θ r varies from 1 to 0 as θ varies from 0 to π/4. r varies from 0 to 1 as θ varies from π/4 to π/2. r varies from 1 to 0 as θ varies from π/2 to 3π/4. r varies from 0 to 1 as θ varies from 3π/4 to π. r varies from 1 to 0 as θ varies from π to 5π/4. r varies from 0 to 1 as θ varies from 5π/4 to 3π/2. r varies from 1 to 0 as θ varies from 3π/2 to 7π/4. r varies from 1 to 0 as θ varies from 7π/4 to 2π.

23 MA112 Section : Prepared b Dr. Archara Pacheenburawana 23 Smmetr Tests Observe that the polar graph r = cos2θ in above Figure is smmetric about the -ais and the -ais. This smmetr could have been predicted from the following theorem. Theorem 2.1 (Smmetr Tests). (a) A curve in polar coordinates is smmetric about the -ais if replacing θ b θ in its equation produces an equivalent equation. (b) A curve in polar coordinates is smmetric about the -ais if replacing θ b π θ in its equation produces an equivalent equation. (c) A curve in polar coordinates is smmetric about the origin if replacing θ b θ+π, or replacing r b r in its equation produces an equivalent equation. π/2 (r,θ) (r,π θ) π/2 (r,θ) π/2 (r,θ) (r, θ) (r,θ+π) or ( r, θ) (a) (b) (c) Eample 2.15 Show that the graph of r = cos2θ is smmetric about the -ais and -ais. Families of Lines and Ras Through the Pole For an constant θ 0, the equation θ = θ 0 (2.6) is satisfied b the coordinates of the form P(r,θ 0 ), regardless of the value of r. Thus, the equation represents the line that passes through the pole and makes an angle of θ 0 with the polar ais. π/2 θ 0 θ = θ 0

24 MA112 Section : Prepared b Dr. Archara Pacheenburawana 24 Families of Circles We will consider three families of circles in which a is assumed to be a positive constant: r = a r = 2acosθ r = 2asin2θ The equation r = a represents a circle of radius a, centered at the pole The equation r = 2acosθ represents a circle of radius a, centered on the -ais and tangent to the -ais at the origin. The equation r = a represents a circle of radius a, centered on the -ais and tangent to the -ais at the origin. π/2 π/2 π/2 r = 2asinθ a r = a 0 r = 2acosθ r = 2acosθ (a,π) (a,0) 0 (a, π 2 ) 0 (a, π 2 ) r = 2asinθ Eample 2.16 Sketch the graphs of the following equations in polar coordinates. (a) r = 4cosθ (b) r = 5sinθ (c) r = 3 Families of Rose Curves In polar coordinates, equations of the form r = asinnθ or r = acosnθ in which a > 0 and n is a positive integer represent families of flower-shaped curves called roses. The rose consists of n equall spaced petals of radius a if n is odd and 2n equall spaced petals of radius a if n is even.

25 MA112 Section : Prepared b Dr. Archara Pacheenburawana 25 r = asin2θ r = asin3θ r = asin4θ r = asin5θ r = asin6θ r = acos2θ r = acos3θ r = acos4θ r = acos5θ r = acos6θ Families of Cardioids and Limacons Equations with an of the four forms r = a±bsinθ r = a±bcosθ 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.

26 MA112 Section : Prepared b Dr. Archara Pacheenburawana 26 a/b < 1 a/b = 1 1 < a/b < 2 a/b 2 Limacon with inner loop Cardioid Dimpled limacon Conve limacon Eample 2.17 Sketch the graph of the equation in polar coordinates. Families of Spirals r = 2(1 cosθ) A spiral is a curve that coils around a central points. The most common eample 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. Eample 2.18 Sketch the graph of r = θ (θ 0) in polar coordinates b plotting points. 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 epressed parametricall in terms of parameter θ b substituting f(θ) for r in the equations = rcosθ and = rsinθ. This ields = f(θ)cosθ, = f(θ)sinθ

27 MA112 Section : Prepared b Dr. Archara Pacheenburawana 27 from which we obtain d dθ d dθ = f(θ)sinθ+f (θ)cosθ = rsinθ+ dr dθ cosθ = f(θ)cosθ+f (θ)cosθ = rcosθ+ dr dθ sinθ Thus, if d/dθ and d/dθ arecontinuous and if d/dθ 0, then is a differentiable function of, and Formula (2.2) with θ in place of t ields dr d d = d/dθ rcosθ +sinθ d/dθ = dθ rsinθ +cosθ dr dθ (2.7) Eample 2.19 Find the slope of the tangent line to the cardioid r = 1+sinθ at the point where θ = π/3. Eample 2.20 Find the points on the cardioid r = 1 cosθ at which there is a horizontal tangent line, a vertical tangent line, or a singular point. Theorem 2.2 If the polar curve r = f(θ) passes through the origin at θ = θ 0, and if dr/dθ 0 at θ = θ 0, then the line θ = θ 0 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 to β, 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 r 2 + ( ) 2 dr dθ (2.8) dθ Eample 2.21 Find the arc length of the spiral r = e θ between θ = 0 and θ = π. Eample 2.22 Find the total arc length of the cardioid r = 1+cosθ.

28 MA112 Section : Prepared b Dr. Archara Pacheenburawana 28 Area in Polar Coordinates r = f(θ) θ = β 0 R θ = α Area Problem in Polar Coordinates Suppose that α and β are angles that satisf 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 b the polar curve r = f(θ) and the ras θ = α and θ = β. Area in Polar Coordinates If α and β are angles that satisf 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 b the polar curve r = f(θ) (α θ β) and the lines θ = α and θ = β is A = β α β 1 2 [f(θ)]2 1 dθ = 2 r2 dθ (2.9) α The hardest part of appling (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 arbitrar radial line from the pole to the boundar curve r = f(θ). Step 3 Ark, Over what intervals must θ var 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.

29 MA112 Section : Prepared b Dr. Archara Pacheenburawana 29 Eample 2.23 Find the area of the region in the first quadrant that is within the cardioid r = 1 cosθ. Eample 2.24 Find the entire area within the cardioid of Eample Eample 2.25 Find the area enclosed b the rose curve r = cos2θ. Eample 2.26 Find the area of the region that is inside of the cardioid r = 4 4cosθ and outside of the circle r = 6.

30 Chapter 3 Three-Dimensional Space; Vectors 3.1 Rectangular Coordinates in 3-Space; Spheres; Clindrical Surfaces Rectangular Coordinate Sstems To begin, consider three mutuall perpendicular coordinate lines, called the -ais, the -ais, and the z-ais, positioned so that their origin coincide. z O The three coordinate aes form a three-dimensional rectangular coordinate sstem (or Cartesian coordinate sstem) The point of intersection of the coordinate aes is called the origin of the coordinate sstem. The coordinate aes, taken in pairs, determine three coordinate planes: the plane, the z-plane, and the z-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 -coordinate, - coordinate, and z-coordinate of P, respectivel, and we denote the point P b (a,b,c) or b P(a,b,c). The following facts about three-dimensional rectangular coordinate sstems: 30

31 MA112 Section : Prepared b Dr. Archara Pacheenburawana 31 Region Description -plane Consists of all points of the form (,,0) z-plane Consists of all points of the form (,0,z) z-plane Consists of all points of the form (0,,z) -ais Consists of all points of the form (,0,0) -ais Consists of all points of the form (0,,0) z-ais 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 P 1 ( 1, 1 ) and P 2 ( 2, 2 ) is d = ( 2 1 ) 2 +( 2 1 ) 2 The distance formula in 3-space has the same form, but it has a third term to account for the added dimension. The distance between the points P 1 ( 1, 1,z 1 ) and P 2 ( 2, 2,z 2 ) is d = ( 2 1 ) 2 +( 2 1 ) 2 +(z 2 z 1 ) 2 Eample 3.1 Find the distance d between the points (2, 3,4) and ( 3,2, 5). Recall that the standard equation of a circle in 2-space that has center ( 0, 0 ) and radius r is ( 0 ) 2 +( 0 ) 2 = r 2 Analogousl, standard equation of the sphere in 3-space that has center ( 0, 0,z 0 ) and radius r is ( 0 ) 2 +( 0 ) 2 +(z z 0 ) 2 = r 2 (3.1) If the terms in(3.1) are epanded and like terms are collected, then the resulting equation has the form z 2 +G+H +Iz +J = 0 (3.2) Eample 3.2 Find the center and radius of the sphere z z +68 = 0 In general, completing the squares in (3.2) produces an equation of the form ( 0 ) 2 +( 0 ) 2 +(z z 0 ) 2 = k 2 If k > 0, then the graph of this equation is a sphere with center ( 0, 0,z 0 ) and radius k. If k = 0, then the sphere has radius zero, so the graph is the single point ( 0, 0,z 0 ).

32 MA112 Section : Prepared b Dr. Archara Pacheenburawana 32 If k < 0, the equation is not satisfied b an values of,, and z, so it has no graph. Theorem 3.1 An equation of the form z 2 +G+H +Iz +J = 0 represents a sphere, a point, or has no graph. Clindrical Surfaces Theorem 3.2 An equation that contains onl two of the variables,, and z represents a clindrical surface in an z-coordinate sstem. The surface can be obtained b graphing the equation in the coordinate plane of the two variables that appear in the equation and then translating that graph parallel to the ais of the missing variable. Eample 3.3 Sketch the graph of 2 +z 2 = 1 in 3-space. Eample 3.4 Sketch the graph of z = sin in 3-space. 3.2 Vectors Man phsical quantities such as area, length, mass, and temperature are completel described once the magnitude of the quantit is given. Such quantities are called scalar. Other phsical quantities, called vectors are not completel determined until both magnitude and a direction are specified. Vectors can be represented geometricall b 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 tpe such as a, k, v, w, and. When discussing vectors, we will refer to real numbers as scalars. Scalar will be denoted b lowercase italic tpe such as a, k, w, and. Two vectors, v and w, are considered to be equal (also called equivalent) if the have the same length and same direction, in which case we write v = w. Geometricall, two vectors are equal if the are translations of one another; thus, the three vectors in the following figure are equal, even though the are in different positions.

33 MA112 Section : Prepared b Dr. Archara Pacheenburawana 33 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 A 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 b 0. The zero vector does not have a specific direction, so we will agree that it can be assigned an convenient direction in a specific problem. 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 2v 1 v 2 ( 1)v ( 3 2 )v 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 the do not. Thus, we sa that v and kv are parallel vectors. Observe also that the vector ( 1)v has the same length as v but is oppositel directed. We call ( 1)v the negative of v and denote it b v. In particular, 0 = ( 1)0 = 0. Vector subtraction is defined in terms of addition and scalar multiplication b v w = v+( w) In the special case where v = w, their difference is 0; that is, v+( v) = v v = 0

34 MA112 Section : Prepared b Dr. Archara Pacheenburawana 34 Vectors in Coordinate Sstems If a vector v is positioned with its initial point at the origin of the rectangular coordinate sstem, then the terminal point will have coordinates of the form (v 1,v 2 ) or (v 1,v 2,v 3 ), 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 = v 1,v 2 or v = v 1,v 2,v 3 In particular, the zero vectors in 2-space and 3-space are respectivel. v = 0,0 and v = 0,0,0 z (v 1,v 2 ) (v 1,v 2,v 3 ) v v Theorem 3.3 Two vectors are equivalent if and onl if their corresponding components are equal. For eample, a,b,c = 2,3,5 if and onl if a = 2, b = 3, and c = 5. Arithmetic Operations on Vectors Theorem 3.4 If v = v 1,v 2 and w = w 1,w 2 are vectors in 2-space and k is an scalar, then v+w = v 1 +w 1,v 2 +w 2 v w = v 1 w 1,v 2 w 2 kv = kv 1,kv 2 Similarl, if v = v 1,v 2,v 3 and w = w 1,w 2,w 3 are vectors in 3-space and k is an scalar, then v+w = v 1 +w 1,v 2 +w 2,v 3 +w 3 v w = v 1 w 1,v 2 w 2,v 3 w 3 kv = kv 1,kv 2,kv 3

35 MA112 Section : Prepared b Dr. Archara Pacheenburawana 35 Eample 3.5 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 P 1 ( 1, 1 ) and P 2 ( 2, 2 ) are points in 2-space and we interested in finding the components of the vector P 1 P 2. As illustrated in the following figure, we can write this vector as P 1 P 2 = OP 2 OP 1 = 2, 2 1, 1 = 2 1, 2 1 P 1 ( 1, 1 ) OP 1 P 1 P 2 OP2 P 2 ( 2, 2 ) O Thus, wehaveshownthatthecomponentsofthevector P 1 P 2 canbeobtainedbsubtracting 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 3.5 If P 1 P 2 is a vector in 2-space with initial point P 1 ( 1, 1 ) and terminal point P 2 ( 2, 2 ), then P 1 P 2 = 2 1, 2 1 Similarl, if P 1 P 2 is a vector in 3-space with initial point P 1 ( 1, 1,z 1 ) and terminal point P 2 ( 2, 2,z 2 ), then P 1 P 2 = 2 1, 2 1,z 2 z 1 Eample 3.6 In 2-space the vector from P 1 (3,2) to P 2 ( 1,4) is P 1 P 2 = 1 3,4 2 = 4,2 and in 3-space the vector from A(1, 2,0) to B( 3,1,2) is AB = 1 ( 3), 2 1,0 2 = 4, 3, 2

36 MA112 Section : Prepared b Dr. Archara Pacheenburawana 36 Rules of Vector Arithmetic Theorem 3.6 For an vectors u, v, and w and an scalars k and l, the following relationships hold: (a) u+v = v+u (b) (u+v)+w = u+(v+w) (c) u+0 = 0+u = u (d) u+( u) = 0 (e) k(lu) = (kl)u (f) k(u+v) = ku+kv (g) (k +l)u = ku+lu (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 b v. This distance does not change if the vector is translated. The norm of a vector v = v 1,v 2 in 2-space is given b v = v 2 1 +v 2 2 and the norm of a vector v = v 1,v 2,v 3 in 3-space is given b v = v 2 1 +v2 2 +v2 3 Eample 3.7 Find the norm of v = 4, 2, and w = 1,3,5. is, For an vector v and scalar k, the length of kv must be k times the length of v; that kv = k v Thus, for eample, 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 -coordinate sstem the unit vectors along the -ais and -ais are denoted b i and j, respectivel; and in z-coordinate

37 MA112 Section : Prepared b Dr. Archara Pacheenburawana 37 sstem the unit vectors along the -ais, -ais and z-ais are denoted b i, j, and k, respectivel. z (0,1) (0,0,1) j i (1, 0) i k j (0,1,0) Thus, (1,0,0) i = 1,0, j = 0,1 In 2-space i = 1,0,0, j = 0,1,0, k = 0,0,1 In 3-space Ever vector in 2-space is epressible uniquel in terms of i and j, and ever vector in 3-space is epressible uniquel in terms of i, j, and k as follows: Eample 3.8 v = v 1,v 2 = v 1,0 + 0,v 2 = v 1 1,0 +v 2 0,1 = v 1 i+v 2 j v = v 1,v 2,v 3 = v 1 1,0,0 +v 2 0,1,0 +v 3 0,0,1 = v 1 i+v 2 j+v 3 k 2-space 3, 4 = 3i 4j 3-space 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 3(2i 4j) = 6i 12j (2i j+3k)+(3i+2j k) = i+j+2k 2(3i+4j k) = 6i+8j 2k 3i+4j = = 5 2i j+3k = 2 2 +( 1) = 14 v 1 i+v 2 j = v 2 1 +v2 2 v 1,v 2,v 3 = v 2 1 +v2 2 +v2 3 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 b multipling v b the reciprocal of its length; that is, u = 1 v v = v v

38 MA112 Section : Prepared b Dr. Archara Pacheenburawana 38 is a unit vector with the same direction as v. The process of multipling v b the reciprocal of its length to obtain a unit vector with the same direction is called normalizing v. Eample 3.9 Find the unit vector that has the same direction as v = 2i j 2k. Vectors Determined b Length and Angle If v is a nonzero vector with its initial point at the origin of an -coordinate sstem, and if θ is the angle from the positive -ais to the radial line through v, then the -component of v can be written as v cosθ and the -component as v sinθ; v θ v cosθ v sinθ and hence v can be epressed in trigonometric form as v = v cosθ,sinθ or v = v cosθi+ v sinθj (3.3) In the special case of a unit vector u this simplifies to u = cosθ,sinθ or u = cosθi+sinθj (3.4) Eample 3.10 (a) Find the vector of length 3 that makes an angle of π/3 with the positive -ais. (b) Find the angle that the vector v = i 3j makes with the positive -ais.

39 MA112 Section : Prepared b Dr. Archara Pacheenburawana 39 Vectors determined b length and a vector in the same direction It is a common problem in man applications that a direction in 2-space or 3-space is determined b 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 b epressing v as v = v u and then reading off the components of v u. Eample 3.11 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). 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 F 1 and F 2 are applied at the same point on an object, then the two forces have the same effect on the object as the single force F 1 +F 2 applied at the point. F 2 F 1 +F 2 F 1 Phsicists and engineers call F 1 +F 2 the resultant of F 1 and F 2, and the sa that the forces F 1 and F 2 are concurrent to indicate that the are applied at the same point. Eample 3.12 Suppose that two forces are applied to an ee bracket, as show in Figure below. Find the magnitude of the resultant and the angle θ that it makes with the positive -ais. F 2 = 300N 40 F 1 = 200N 30

40 MA112 Section : Prepared b Dr. Archara Pacheenburawana Dot Product; Projection Definition of the Dot Product Definition 3.3 If u = u 1,u 2 and v = v 1,v 2 are vectors in 2-space, then the dot product of u and v is written as u v and is defined as u v = u 1 v 1 +u 2 v 2 Similarl, if u = u 1,u 2,u 3 and v = v 1,v 2,v 3 are vectors in 3-space, then their dot product is defined as u v = u 1 v 1 +u 2 v 2 +u 3 v 3 Note that the dot product of two vectors is a scalar. For eample, 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 3.7 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 their initial pointscoincide. Wedefine the angle between u and v tobe theangleθ determined b the vectors that satisfies the condition 0 θ π. u θ v u θ v u θ v u θ v Theorem 3.8 If u and v are nonzero vectors in 2-space or 3-space, and if θ is the angle between them, then cosθ = u v (3.5) u v

41 MA112 Section : Prepared b Dr. Archara Pacheenburawana 41 Eample 3.13 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 Eample 3.14 Find the angle ABC if A = (1, 2,3), B = (2,4, 6), and C = (5, 3,2). A(1, 2,3) θ B(2,4, 6) C(5, 3,2) Interpreting the Sign of the Dot Product It will often be convenient to epress Formula (3.5) as u v = u v cosθ (3.6) which epresses 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 cosθ. 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 θ v u θ v u θ 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 eample, if v = v 1 i+v 2 j+v 3 k, then cosα = v i v i = v 1 v j, cosβ = v v j = v 2 v k, cosγ = v v k = v 3 v

42 MA112 Section : Prepared b Dr. Archara Pacheenburawana 42 z j v k v β α i γ α β j i Theorem 3.9 The direction cosines of a nonzero vector v = v 1 i+v 2 j+v 3 k are cosα = v 1 v, cosβ = v 2 v, cosγ = v 3 v The direction cosines of a vector v = v 1 i+v 2 j+v 3 k can be computed b normalizing v and reading off the components of v/ v, since v v = v 1 v i+ v 2 v j+ v k v k = (cosα)i+(cosβ)j+(cosγ)k Moreover, we can show that the direction cosines of a vector satisf the equation cos 2 α+cos 2 β +cos 2 γ = 1 Eample 3.15 Find the direction angles of the vector v = 4i 5j+3k. Decomposing Vectors into Orthogonal Components Our net objective is to develop a computational procedure for decomposing a vector into sum of orthogonal vectors. For this purpose, suppose that e 1 and e 2 are two orthogonal unit vectors in 2-space, and suppose that we want to epress a given vector v as a sum v = w 1 +w 2 so that w 1 is a scalar multiple of e 1 and w 2 is a scalar multiple of e 2. w 2 e 2 v e 1 w 1

43 MA112 Section : Prepared b Dr. Archara Pacheenburawana 43 That is, we want to find scalars k 1 and k 2 such that v = k 1 e 1 +k 2 e 2 (3.7) We can find k 1 b taking the dot product of v with e 1. This ields Similarl, v e 1 = (k 1 e 1 +k 2 e 2 ) e 1 = k 1 (e 1 e 1 )+k 2 (e 2 e 1 ) = k 1 e = k 1 v e 2 = (k 1 e 1 +k 2 e 2 ) e 2 = k 1 (e 1 e 2 )+k 2 (e 2 e 2 ) = 0+k 2 e 2 2 = k 2 Substituting these epressions for k 1 and k 2 in (3.7) ields v = (v e 1 )e 1 +(v e 2 )e 2 (3.8) In this formula we call (v e 1 )e 1 and (v e 2 )e 2 the vector components of v along e 1 and e 2, respectivel; and we call v e 1 and v e 2 the scalar components of v along e 1 and e 2, respectivel. If θ denote the angle between v and e 1, then v e 1 = v cosθ and v e 2 = v sinθ and the decomposition (3.7) can be epressed as Eample 3.16 Let v = 2,3, e 1 = v = ( v cosθ)e 1 +( v sinθ)e 2 (3.9) 1 2, 1, and e 2 = 1, Find the scalar components of v along e 1 and e 2 and the vector components of v along e 1 and e 2. Eample 3.17 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 eert 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?

44 MA112 Section : Prepared b Dr. Archara Pacheenburawana 44 Orthogonal Projections Thevectorcomponentsofv alonge 1 ande 2 in(3.8)arealsocalledtheorthogonal projections of v on e 1 and e 2 and are denoted b proj e1 v = (v e 1 )e 1 and proj e2 v = (v e 2 )e 2 In general, if e is a unit vector, then we define the orthogonal projection of v on e to be proj e v = (v e)e (3.10) The orthogonal projection of v on an arbitrar nonzero vector b can be obtained b normalizing b and then appling Formula (3.10); that is, ( )( ) b b proj b v = v b b which can be rewritten as proj b v = v b b 2 b (3.11) Moreover, if we subtract proj b v from v, then the resulting vector v proj b v will be orthogonal to b; we call this the vector component of v orthogonal to b. v proj b v v v v proj b v b proj b v proj b v b Acute angle between v and b Obtuse angle between v and b Eample 3.18 Find the orthogonal projection of v = i +j+k on b = 2i+2j, and then find the vector component of v orthogonal to b. Work Recall that we define the work W done on the object b 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 W = F d

45 MA112 Section : Prepared b Dr. Archara Pacheenburawana 45 Moreover, 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 epressed entirel 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 makes an angle θ with the displacement vector, then we define the work W done b F to be W = ( F cosθ) PQ = F PQ (3.13) P F F Work = F PQ Q P F F θ F cosθ Work = ( F cosθ) PQ Q Eample 3.19 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? Eample 3.20 A wagon is pulled horizontall b eerting a constant force of 10lb on the handle at an angle of 60 with the horizontal. How much work is done in moving the wagon 50 ft? 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 matri of real number is defined b a 1 a 2 b 1 b 2 = a 1b 2 a 2 b 1. Definition 3.5 The determinant of a 3 3 matri of real number is defined as a combination of three 2 2 determinants, as follows: a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 +a 3 b 1 b 2 c 1 c 2 (3.14) Note that Equation (3.14) is referred to as an epansion of the determinant along the first row.