CHAPTER 8. CONICS, PARAMETRIC CURVES, AND POLAR CURVES

SECTION 8. PAGE 3 R. A. ADAMS: CALCULUS CHAPTER 8. CONICS, PARAMETRIC CURVES, AND POLAR CURVES Section 8. Conics page 3. The ellipse with foci, ± has major ais along the -ais and c. If a 3, then b 9 5. The ellipse has equation 5 + 9.. The ellipse with foci, and, has c, centre,, and major ais along. If ɛ /, then a c/ɛ and b 6. The ellipse has equation +. 6 8. If +, then + +, or This represents an ellipse with centre at +.,, semi-major ais, semi-minor ais, and foci at 3 ±,. + 3. A parabola with focus, 3 and verte, has a and principal ais. Its equation is 6.. A parabola with focus at, and principal ais along will have verte at a point of the form v,. Its equation will then be of the form + ±v v. The origin lies on this curve if ± v. Onl the sign is possible, and in this case v ±/. The possible equations for the parabola are + ±. 5. The hperbola with semi-transverse ais a and foci, ± has transverse ais along the -ais, c, and b c a 3. The equation is 9. If +, then Fig. 8..8 + + + + This is an ellipse with semi-aes and, centred at,. + 3.,, 6. The hperbola with foci at ±5, and asmptotes ± is rectangular, has centre at, and has transverse ais along the line. Since c 5 and a b because the asmptotes are perpendicular to each other we have a b 5/. The equation of the hperbola is. If, then Fig. 8..9 5. + +, or +. 7. If + +, then + +. This represents the single point,. This represents a hperbola with centre at,, semitransverse ais, semi-conjugate ais, and foci at, ± 5. The asmptotes are ±. 3

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 3 This is a hperbola with centre 3,, and asmptotes the straight lines + 3 ±. 3, 3, +3+ Fig. 8... If + 3, then +. Thus +. This is a parabola with verte,, opening upward. Fig. 8..3. If 9 + 8 + 8 3, then 9 + + + + 9 + +. This represents the single point,.,. If + +, then Fig. 8.. + 3 + + 3 3 +. 5. If 9 + 8 + 8 3, then 9 + + + + 3 + 9 + 36 9 + + 36 + +. 9 This is an ellipse with centre,, and semi-aes and 3., This represents a parabola with verte at 3,, focus at 8, and directri 3 8. ++ 3,,, 9 + 8+83, Fig. 8..5 3, Fig. 8.. 3. If + 3 +, then + 3 9 + 3 9 9 8 6. The equation + simplifies to and hence represents a rectangular hperbola with centre at the origin, asmptotes along the coordinate aes, transverse ais along, conjugate ais along, vertices at, and,, semi-transverse and semi-conjugate aes equal to /, semi-focal separation equal to +, and hence foci at the points, and,. The eccentricit is. 33

SECTION 8. PAGE 3 R. A. ADAMS: CALCULUS, + This is a rectangular hperbola with centre,, semi-aes a b, and eccentricit. The semifocal separation is ; the foci are at ±,. The asmptotes are u ±v +. In terms of the original coordinates, the centre is,, the foci are ± +, ±, and the asmptotes are and. + Fig. 8..6 7. The parabola has focus at 3, and principal ais along. The verte must be at a point of the form v,, in which case a ±3 v and the equation of the parabola must be of the form ±3 v v., Fig. 8..9 This curve passes through the origin if 6 ±v 3v. We have two possible equations for v: v 3v and v 3v +. The first of these has solutions v orv. The second has no real solutions. The two possible equations for the parabola are + or 8 6 or 8 8. The foci of the ellipse are, and 3,, so the centre is 3/, and c 3/. The semi-aes a and b must satisf a b 9/. Thus the possible equations of the ellipse are 3/ 9/ + b + b. 9. For + wehavea C, B. We therefore rotate the coordinate aes see tet pages 7 8 through angle θ π/. Thus cot θ A C/B. The transformation is. We have + + + and A, B, C, D, E and F. We rotate the aes through angle θ satisfing tan θ B/A C θ π. Then A, B, C, D, E and the transformed equation is u + v u v which represents a parabola with verte at u,v, and principal ais along u. The distance a from the focus to the verte is given b a, so a / and the focus is at,. The directri is v. Since u v and u + v, the verte of the parabola in terms of -coordinates is,, and the focus is,. The directri is. The principal ais is. u v, u + v. The given equation becomes u v + u v u + v u v v u v + u v +. /,/ ++ + Fig. 8.. 3

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 3. For 8 + + 7, we have A 8, B, C 7, F. Rotate the aes through angle θ where tan θ B A C 9 3. Thus cos θ 3/5, sin θ /5, and cos θ cos θ 3 5 cos θ 5. We ma therefore take cos θ 5, and sin θ 5. The transformation is therefore u + v 5 5 5 u + 5 v u 5 v 5 + 5 5 The coefficients of the transformed equation are A 8 + + 7 5 5 5 5 B C 8 + 7. 5 5 5 The transformed equation is 5u + v u, or + v. This is an ellipse with centre,, semi-aes a and b, and foci at u ± 3, v. In terms of the original coordinates, the centre is,, 3 3 the foci are ±,. 5 5 Then A, B, C 5, D 5, E and the transformed equation is 5v + 5u v 5 u which represents a parabola with verte at u,v,, focus at 5,. The directri is u and the 5 principal ais is v. Since u v and 5 5 u + v, in terms of the -coordinates, the verte is at,, the focus at, 5 5. The directri is + and the principal ais is. + ++ Fig. 8.. 8 ++7 3. The distance from P to F is +. The distance from P to D is + p. Thus + ɛ + p + ɛ + p + p ɛ + pɛ ɛ p. Fig. 8... We have + + + and A, B, C, D, E and F. We rotate the aes through angle θ satisfing tan θ B/A C 3. Then sec θ + tan θ 5 cos θ 3 3 5 + cos θ cos θ cos θ sin θ 5 ; 5 5. 5 p D F P, Fig. 8..3 35

SECTION 8. PAGE 3 R. A. ADAMS: CALCULUS. Let the equation of the parabola be a. The focus F is at a, and verte at,. Then the distance from the verte to the focus is a. At a, aa ±a. Hence, l a, which is twice the distance from the verte to the focus. a a c l a b l a, 7. Fig. 8..6 Fig. 8.. S C 5. We have c a + l. Thus b B l b c a b a b a but c a b b b a. F P F S Therefore l b /a. b a + b C A c l a V Fig. 8..7 Fig. 8..5 6. Suppose the hperbola has equation a. The b vertices are at ±a, and the foci are at ±c, where c a + b. At a + b, Hence, l b a. a + b a b a + b b a a b ± b a. Let the spheres S and S intersect the cone in the circles C and C, and be tangent to the plane of the ellipse at the points F and F, as shown in the figure. Let P be an point on the ellipse, and let the straight line through P and the verte of the cone meet C and C at A and B respectivel. Then PF PA, since both segments are tangents to the sphere S from P. Similarl, PF PB. Thus PF + PF PA+ PB AB constant distance from C to C along all generators of the cone is the same. Thus F and F are the foci of the ellipse. 8. Let F and F be the points where the plane is tangent to the spheres. Let P be an arbitrar point P on the hperbola in which the plane intersects the cone. The spheres are tangent to the cone along two circles as shown in the figure. Let PAVB be a generator of the cone a straight line ling on the cone intersecting these two circles at A and B as shown. V is the verte of the cone. We have PF PA because two tangents to a sphere from 36

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 9 a point outside the sphere have equal lengths. Similarl, PF PB. Therefore PF PF PB PA AB constant, since the distance between the two circles in which the spheres intersect the cone, measured along the generators of the cone, is the same for all generators. Hence, F and F are the foci of the hperbola. Y V X F Q P C P F V A A Fig. 8..9 B Section 8. Parametric Curves page 9 F. If t, t, t then +. This is a straight line segment. Fig. 8..8 t t t Fig. 8.. 9. Let the plane in which the sphere is tangent to the cone meet AV at X. Let the plane through F perpendicular to the ais of the cone meet AV at Y. Then VF VX, and, if C is the centre of the sphere, FC XC. Therefore VC is perpendicular to the ais of the cone. Hence YF is parallel to VC, and we have YV VX VF. If P is on the parabola, FP VF, and the line from P to the verte A of the cone meets the circle of tangenc of the sphere and the cone at Q, then. If t and t + for t <, then + 3 for <, which is a half line. t t+, FP PQ YX VX VF. Since FP VF, FP is the semi-latus rectum of the parabola. See Eercise 8. Therefore F is the focus of the parabola. Fig. 8.. 3. If /t, t, < t <, then. This is part of a hperbola. 37

SECTION 8. PAGE 9 R. A. ADAMS: CALCULUS,3 t ba t a t b a Fig. 8..3 Fig. 8..6. If + t and t for < t <, then + t + + t + t + t +. This curve consists of all points of the circle with centre at, and radius ecept the origin,. 7. If 3 sin πt, cos πt, t, then 9 +. This is an ellipse. 6 t 9 + 6 t t t t t Fig. 8..7 /+t t/+t Fig. 8.. 5. If 3 sin t, 3 cos t, t π/3, then + 9. This is part of a circle. t + 9 8. If cos sin s and sin sin s for < s <, then +. The curve consists of the arc of this circle etending from a, b through, to a, b where a cos and b sin, traversed infinitel often back and forth. cos sin s sin sin s rad t π 3 Fig. 8..5 6. If a sec t and b tan t for π < t < π, then Fig. 8..8 a b sec t tan t. The curve is one arch of this hperbola. 9. If cos 3 t, sin 3 t, t π, then /3 + /3. This is an astroid. 38

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 9 tπ/ /3 + /3 and cos t+sin t cos t+sin t+ cos t sin t. tπ t tπ Hence, + cos t + sin t +. t3π/ Fig. 8..9. If t and + t for t then t. The parametric curve is the left half of the circle of radius centred at,, and is traced in the direction of increasing. t +t t Fig. 8..,. cosh t, sinh t represents the right half branch of the rectangular hperbola.. 3 cosh t, + sinh t represents the left half branch of the hperbola 9 +. 3. t cos t, t sin t, t π represents two revolutions of a spiral curve winding outwards from the origin in a counterclockwise direction. The point on the curve corresponding to parameter value t is t units distant from the origin in a direction making angle t with the positive -ais.. i If cos t and sin t, then cos t sin t [ cos t + sin tcos t sin t cos t sin t cos t + sin t cos t sin t ] ii If sec t and tan t, then and sec t tan t sec t + tan t sec t + tan t + sec t tan t sec t tan t sec t+tan t sec t tan t. Hence, + sec t + tan t +. iii Similarl, if tan t and sec t, then + + sec t tan t + sec t tan t tan t + sec t +. These three parametric curves above correspond to different parts of the parabola + +, as shown in the following diagram. tan t sec t The parabola ++ cos t sin t Fig. 8.. sec t tan t 5. The slope of at is m. Hence the parabola can be parametrized m/, m /, < m <. 6. If, is an point on the circle + R other than R,, then the line from, to R, has slope m. Thus m R, and R + m R R m + Rm + m R [ ] m + m R R m R m + or R. 39

SECTION 8. PAGE 9 R. A. ADAMS: CALCULUS The parametrization of the circle in terms of m is given b m R m + [ m ] R m m R Rm + m + where < m <. This parametrization gives ever point on the circle ecept R,. t b a T Y P, X, slope m R, Fig. 8..8 Fig. 8..6 + R 9. If 3t + t 3, 3t, t, then + t3 7. t a T P, X 3 + 3 7t3 + t 3 3 + t3 7t3 + t 3 3. As t, we see that and,but + 3t + t 3t + t 3 t + t. Thus + is an asmptote of the curve. Fig. 8..7 t Using triangles in the figure, we see that the coordinates of P satisf t t a sec t, a sin t. The Cartesian equation of the curve is a + a. The curve has two branches etending to infinit to the left and right of the circle as shown in the figure. folium of Descartes Fig. 8..9 t 8. The coordinates of P satisf a sec t, b sin t. The Cartesian equation is b + a.. Let C and P be the original positions of the centre of the wheel and a point at the bottom of the flange whose path is to be traced. The wheel is also shown in a subsequent position in which it makes contact with the rail at R. Since the wheel has been rotated b an angle θ, OR arc SR aθ. 3

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 9 Thus, the new position of the centre is C aθ,a. Let P, be the new position of the point; then OR PQ aθ b sinπ θ aθ b sin θ, RC + CQ a + b cosπ θ a b cos θ. C t t T t These are the parametric equations of the prolate ccloid. P Q S b C C θ a t P t, b a θ t A O R P aθ b sin θ a b cos θ Fig. 8.. πa Fig. 8... Let t and θ t be the angles shown in the figure below. Then arc AT t arc T t P t, that is, at bθ t. The centre C t of the rolling circle is C t a b cos t,a b sin t. Thus a b cos t b cosθ t t a b sin t b sinθ t t. Fig. 8.. If a and b, then cos t,. This is a straight line segment. If a and b, then 3 cos t + cos 3t 3 cos t + cos t cos t sin t sin t 3 cos t + cos t cos t sin t cos t cos t + cos 3 t cos t sin t cos 3 t 3 sin t + sin 3t 3 sin t sin t cos t cos t sin t 3 sin t sin t cos t sin t sin t sin t sin t + sin 3 t + sin 3 t sin 3 t This is an astroid, similar to that of Eercise.. a From triangles in the figure, TX OT tan t tan t OY sin π t OY cos t OT cos t cos t cos t. T X Since θ t t a b t t a b b t, therefore a bt a b cos t + b cos b a bt a b sin t b sin. b t O Fig. 8.. Y P, 3

SECTION 8. PAGE 9 R. A. ADAMS: CALCULUS b sec t + tan t +. Thus +. 3. sin t, sint 7. + cos t n n cosnt + sin t n n sinnt represents a ccloid-like curve that is wound around the circle + instead of etending along the - ais. If n is an integer, the curve closes after one revolution and has n cusps. The left figure below shows the curve for n 7. If n is a rational number, the curve will wind around the circle more than once before it closes. Fig. 8..3. sin t, sin3t 5. sint, sin3t Fig. 8.. Fig. 8..7 8. + cos t + cosn t n n + sin t sinn t n n represents a ccloid-like curve that is wound around the inside circle + + /n and is eternall tangent to +. If n is an integer, the curve closes after one revolution and has n cusps. The figure shows the curve for n 7. If n is a rational number but not an integer, the curve will wind around the circle more than once before it closes. Fig. 8..5 6. sint, sin5t Fig. 8..6 Fig. 8..8 Section 8.3 Smooth Parametric Curves and Their Slopes page 53. t + t d d t No horizontal tangents. Vertical tangent at t, i.e., at,. 3

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8.3 PAGE 53. t t t + t d d t t + Horizontal tangent at t, i.e., at 3,. Vertical tangent at t, i.e., at, 3. 3. t t t 3 t d d t 3t Horizontal tangent at t ±, i.e., at, 6 and 8, 6. Vertical tangent at t, i.e., at,.. t 3 3t t 3 + 3t d 3t d 6tt + Horizontal tangent at t, i.e., at,. Vertical tangent at t, i.e., at, 5. At t i.e., at, both d/ and d/ change sign, so the curve is not smooth there. It has a cusp. 5. te t / e t d t e t / d te t Horizontal tangent at t, i.e., at,. Vertical tangent at t ±, i.e. at ±e /, e. 6. sin t sin t t cos t d d cos t t sin t Horizontal tangent at t nπ, i.e., at, n nπ for integers n. Vertical tangent at t n + π, i.e. at, and,. 7. sint sin t d d cost cos t Horizontal tangent at t n + π, i.e., at, ±. Vertical tangent at t n + π, i.e., at ±, / and ±, /. 8. 3t + t 3 d 3 t3 3t + t 3 d + t 3 3t t3 + t 3 Horizontal tangent at t and t /3, i.e., at, and /3, /3. Vertical tangent at t /3, i.e., at /3, /3. The curve also approaches, verticall as t ±. 9. t 3 + t d 3t + At t ; d d t 3 d 3t 3 3 + 3.. t t d t 3 t d t 3 + t 3t + At t ; d d 3 + 3 5.. cost sin t d d sint cos t At t π 6 ; d d cosπ/6 sinπ/3.. e t te t d e t d e t + t At t ; d d e e 3. 3. t 3 t t + t 3 at t d 3t d + 3t at t Tangent line: + t, + t. This line is at, at t. If ou want to be at that point at t instead, use + t t, + t t.. t cos t π d + sin t + sin t at t π d cos t at t π Tangent line: π + + t, t. 5. t 3 t, t is at, at t and t. Since d d t 3t ± ±, the tangents at, at t ± have slopes ±. 6. sin t, sint is at, at t and t π. Since d d cost { if t cos t if t π, the tangents at, at t and t π have slopes and, respectivel. 33

SECTION 8.3 PAGE 53 R. A. ADAMS: CALCULUS 7. t 3 t d 3t d t both vanish at t. d d d has no limit as t. 3t d 3t as t, but d/ changes sign at t. Thus the curve is not smooth at t. In this solution, and in the net five, we are using the Remark following Eample in the tet. 8. t d t 3 t 3 d 3t both vanish at t. Since d t ast, and d/ does not d 3 change sign at t, the curve is smooth at t and therefore everwhere. 9. t sin t t 3 d d sin t + t cos t 3t both vanish at t. d lim t d lim 3t 6t lim, t sin t + t cos t t cos t t sin t but d/ changes sign at t. d/d has no limit at t. Thus the curve is not smooth at t.. t 3 t sin t d lim t 3t d cos t both vanish at t. d d lim 3t t cos t lim 6t 6 and d/ does t sin t not change sign at t. Thus the curve is smooth at t, and hence everwhere.. If t t and t t, then d t, d d d d d d/ t d d d d Directional information is as follows: t t t t 3. The tangent is horizontal at t, i.e.,,, and is vertical at t i.e., at, 3. Observe that d /d >, and the curve is concave up, if t >. Similarl, d /d < and the curve is concave down if t <. t t Fig. 8.3. t t t t. If f t t 3 and gt 3t, then f t 3t, g t 6t, f t 6t; g t 6. Both f t and g t vanish at t. Observe that Thus, d lim t + d, d d 6t 3t t. lim d t d and the curve has a cusp at t, i.e., at,. Since d d 3t 6 6t6t 3t 3 3t < for all t, the curve is concave down everwhere. t d/ + + d/ + curve Fig. 8.3. t 3 3t 3

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8.3 PAGE 53 3. t 3 3t, / + t. Observe that, ± as t ±. d 3t d, t + t d d t 3t + t d 6t, d 3t + t 3 d 3t 3t d + t 3 t6t + t [3t ] 3 Directional information: 6t + 8t + 7t 3 + t 3 t d/ + + d/ + + curve The tangent is horizontal at t, i.e.,,, and vertical at t ±, i.e., ±,. t d d + + curve t t 3 3t t + t t The tangent is horizontal at t, i.e., at 7 8, 9 The tangent is vertical at t ±, i.e.,, and,. Directional information is as follows: t f t + + g t + + curve For concavit, d d 3t t 6t [3t ] 3 t t + 9t 3 which is undefined at t ±, therefore t d d + curve t t 3 3t t t t Fig. 8.3. t, 5. cos t + t sin t, sin t t cos t, t. d d t cos t, t sin t, d cos t t sin t d sin t + t cos t d tan t. Fig. 8.3.3. If f t t 3 3t and gt t t, then f t 3t 3, g t t, f t 6t; g t. d d d d d d 3 t cos 3 t d Tangents are vertical at t n + π, and horizontal at t nπ n,,,... 35

SECTION 8.3 PAGE 53 R. A. ADAMS: CALCULUS tπ t3π/ t tπ Fig. 8.3.5 Section 8. Arc Lengths and Areas for Parametric Curves page 58. 3t d 6t Length 6 3 t 3 t d 6t 6t + 6t t + t Let u + t du t udu u 3/ units. If + t 3 and t for t, then the arc length is s 3t + t t 9t + + t 9t + Let u 9t + du 8t 3 udu + 8 3 3 + 6 units. 7 3. a cos 3 t, a sin 3 t, t π. The length is 3a π a π 9a cos t sin t + 9a sin t cos t π/ 6a sin t cos t sin t π/ cos t 6a units... If ln + t and tan t for t, then d The arc length is t + s + t π/ t + t ; d + t. + t sec θ dθ ln sec θ + tan θ Let t tan θ sec θ dθ π/ 5. t sin t, t cos t, t π. ln + units. d t sin t + t cos t d t cos t t sin t ds [ t sin t + t sin t cos t + t cos t ] + cos t t sin t cos t + t sin t t + t. The length of the curve is π 6. cos t + t sin t d t cos t Length t + t Let u + t du t +π u / du +π 3 u3/ 8 + π 3/ units. 3 π π 7. t + sin t d + cos t sin t t cos t t π d t sin t t cos t + t sin t t t π π units. cos t t π d sin t 36

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 58 Length π π sin t 8. sin t d sin t cos t Length π/ π/ π/ + cos t + cos t + sin t π cos t/ cos t π units. cos t t π/ d sin t sin t cos t + sin t sin t + cos t Let cos t tan u sin t sec udu sec 3 udu π/ sec u tan u + lnsec u + tan u + ln + units. 9. at sin t d a cos t Length a π a π π a cos t t π d a sin t a cos t + cos t + sin t π cos t a sin t sin t a cos t π a units.. If at a sin t and a a cos t for t π, then d d a a cos t, a sin t; ds a a cos t + a sin t a cos t a t sin t a sin. a The surface area generated b rotating the arch about the -ais is S π π π ds π a a cos ta sin π t 6πa sin 3 π [ t 6πa cos t Let u cos du t sin 3πa u du 3πa [ u 3 u3 ] 6 3 πa3 sq. units. t ] t sin b The surface area generated b rotating the arch about the -ais is π S π ds π t π at a sin ta sin π [ t πa t sin π t πa t sin π 8πa πa [ t cos cos ] t t sin t cos t sin t π π + πa [π + ] 6π a sq. units.. e t cos t d e t cos t sin t Arc length element: ] t cos e t sin t t π/ d e t sin t + cos t ds e t cos t sin t + e t sin t + cos t e t. 37

SECTION 8. PAGE 58 R. A. ADAMS: CALCULUS The area of revolution about the -ais is tπ/ t π ds π π/ π et 5 e t sin t sin t cos t π e π + sq. units. 5 π/ 5. t 3 t, t, t. Area t 3t 3t t 5 t3 3 3t 5 56 5 sq. units.. The area of revolution of the curve in Eercise about the -ais is t 3 t t tπ/ t π ds π π/ π et 5 e t cos t cos t + sin t π e π sq. units. 5 3. 3t t 3 t d d 6t 6t Arc length element: ds 36t + t 6t + t. The area of revolution about the -ais is t t π ds 36π 8π 8π π/ t 3 + t Let u + t du t u udu 5 u5/ 3 u3/ 7π 5 + sq. units.. The area of revolution of the curve of Eercise 3 about the -ais is t t π ds π π π π π/ π/ t + t tan u sec 3 udu Let t tan u sec udu sec 7 u sec 5 u + sec 3 u du sq. units. 7 + 3ln + We have omitted the details of evaluation of the final integral. See Eercise of Section 8.3 for a similar evaluation. A Fig. 8..5 6. Area of R a sin 3 t 3a sin t cos t π/ a sin t cos t π/ [ t a 6 sint ] sin3 t π/ 6 8 See Eercise 3 of Section 6.. 3 8 πa sq. units. a 7. sin t, cos t, Area π/ π/ a R a Fig. 8..6 a cos 3 t a sin 3 t t π. cos t sin 3 t cos t cos 5 t cos t sin t u 5 u 7 du 6 a Let u cos t du sin t 6 sq. units. 8 6 38

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8. PAGE 58 sin t cos t t π/ A A +sin t cos t +sin t sin t t π Fig. 8..7 8. If cos s sin s sin s and sin s cos s for s π, then + sin s + cos s which is the right half of the circle with radius and centre at,. Hence, the area of R is [ π ] π 8 sq. units. cos s sin s sin s Fig. 8..9. To find the shaded area we subtract the area under the upper half of the hperbola from that of a right triangle: Shaded area Area ABC Area sector ABC sec t tan t sec t tan t sec t tan t t t tan tsec t tan t sec 3 t sec t [ sec t tan t+ ln sec t + tan t ln sec t + tan t ln sec t + tan t sq. units. ] t tt Fig. 8..8 9. + sin t cos t, + sin t sin t, t π. This is just the polar curve r + sin θ. Area π π π + sin t sin t d + sin t cos t sin t + sin tcos t sin t sin t [ sin t + sin 3 t + sin t sin t cos t sin t cos t π + [ cos t + π π + π + 9π [ sin t 6 cos t cos t ] sq. units. ] ] cos t R t Fig. 8.. sec t tan t. See the figure below. The area is the area of a triangle less the area under the hperbola: A t cosh t sinh t sinh t t sinh t sinh t cosh t sinh t sinh t + t t sq. units. 39

SECTION 8. PAGE 58 R. A. ADAMS: CALCULUS Section 8.5 Polar Coordinates and Polar Curves page 6 A Fig. 8.. cosh t,sinh t. If f t at a sin t and gt a a cos t, then the volume of the solid obtained b rotating about the -ais is V π tπ t π π d π tπ t [gt] f t a a cos t a a cos t π πa 3 cos t 3 π πa 3 3 cos t + 3 cos t cos 3 t [ πa 3 π + 3 π ] + cos t [ πa 3 π + 3 ] π 5π a 3 cu. units. t d at a sin t a a cos t Fig. 8.. tπ 3. Half of the volume corresponds to rotating a cos 3 t, a sin 3 t t π/ about the -ais. The whole volume is V π π/ π/ π d a sin 6 t3a cos t sin t π/ 6πa 3 cos t 3 cos t sin t Let u cos t du sin t 6πa 3 3u + 3u u 6 u du 6πa 3 3 3 5 + 3 7 3πa3 cu. units. 9 5. r 3 sec θ r cos θ 3 3 vertical straight line.. r csc θ r sin θ a horizontal line. 3. r 5/3 sin θ cos θ 3r sin θ r cos θ 5 3 5 straight line.. r sin θ + cos θ r r sin θ + r cos θ + + + a circle with centre, and radius 5. r csc θ r sin θ r sin θ cos θ a rectangular hperbola. 6. r sec θ tan θ r cos θ r sin θ r cos θ a parabola. 7. r sec θ + tan θ r cos θ + tan θ + a parabola. 8. r cos θ + sin θ r cos θ + r sin θ + an ellipse. 9. r cos θ r r + + + + + a parabola.. 33

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8.5 PAGE 6. r cos θ r r cos θ r + + + + 3 + an ellipse. π/3 3. r sin θ r + r + + 8 + 3 8. r + sin θ r + r sin θ r + + a parabola. a hperbola. π/3 Fig. 8.5.5 6. If r sin θ, then r atθ π 6 and 5π 6. θ 5π 6 θ π 6 3. r + sin θ cardioid Fig. 8.5.6 r sin θ 7. r + cos θ 3 Fig. 8.5.3. If r cos θ + π, then r atθ π and 7π. This is a cardioid. Fig. 8.5.7 8. If r sin θ, then r atθ, ± π and π., θ π 5. r + cos θ r ifθ ±π/3. r cosθ+ π Fig. 8.5. r sin θ Fig. 8.5.8 9. r cos 3θ three leaf rosette r atθ ±π/6, ±π/, ±5π/6. 33

SECTION 8.5 PAGE 6 R. A. ADAMS: CALCULUS π/6 θ π 6 r cos 3θ Fig. 8.5.9. If r cos θ, then r atθ ± π 8, ± 3π 8, ± 5π 8 and Fig. 8.5. 3. r sin 3θ. Thus r ± sin 3θ. This is a lemniscate. r atθ, ±π/3, ±π/3, π. π/3 ± 7π 8. an eight leaf rosette θ 3π 8 θ π 8 Fig. 8.5. r cos θ. r sin θ. Thus r ± sin θ. This is a lemniscate. r atθ, θ ±π/, and θ π. Fig. 8.5.3. If r ln θ, then r atθ. Note that sin θ r sin θ ln θ sin θ θ ln θ θ as θ +. Therefore, the negative -ais is an asmptote of the curve. rln θ Fig. 8.5. Fig. 8.5.. If r cos 3θ, then r atθ ± π 6, ± π and ± 5π. This equation defines two functions of r, namel 6 r ± cos 3θ. Each contributes 3 leaves to the graph. 5. r 3 cos θ, and r sin θ both pass through the origin, and so intersect there. Also sin θ 3 cos θ tan θ 3 θ π/3, π/3. Both of these give the same point [ 3/,π/3]. Intersections: the origin and [ 3/,π/3]. 6. r cosθ, r. cosθ / θ ±π/6 orθ ±5π/6. Intersections: [, ±π/6] and [, ±5π/6]. 33

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8.5 PAGE 6 7. r + cos θ, r 3 cos θ. Both curves pass through the origin, so intersect there. Also 3 cos θ +cos θ cos θ / θ ±π/3. Intersections: the origin and [3/, ±π/3]. 8. Let r θ θ and r θ θ + π. Although the equation r θ r θ has no solutions, the curves r r θ and r r θ can still intersect if r θ r θ for two angles θ and θ having the opposite directions in the polar plane. Observe that θ nπ and θ n π are two such angles provided n is an integer. Since r θ nπ r n π, the curves intersect at an point of the form [nπ, ] or [nπ,π]. 9. If r /θ for θ>, then lim lim sin θ θ + θ + θ Thus is a horizontal asmptote. r/θ. 3. r sinθsinmθ For odd m there are m + petals, each of m + / different sizes. For even m there are alwas m petals. The are of n different sizes if m n orm n. 35. r + sinθsinmθ These are similar to the ones in Eercise 3, but the petals are joined, and less distinct. The smaller ones cannot be distinguished. There appear to be m + petals in both the even and odd cases. 36. r C + cos θ cosθ The curve alwas has 3 bulges, one larger than the other two. For C these are 3 distinct petals. For < C < there is a fourth supplementar petal inside the large one. For C the curve has a cusp at the origin. For C > the curve does not approach the origin, and the petals become less distinct as C increases. 37. r C + cos θ sin3θ For C < there appear to be 6 petals of 3 different sizes. For C there are onl of sizes, and these coalesce as C increases. Fig. 8.5.9 3. The graph of r cos nθ has n leaves if n is an even integer and n leaves if n is an odd integer. The situation for r cos nθ is reversed. The graph has n leaves if n is an odd integer provided negative values of r are allowed, and it has n leaves if n is even. 3. If r f θ, then 38. r lnθ r cos θ f θ cos θ r sin θ f θ sin θ. 3. r cos θ cosmθ For odd m this flower has m petals, large ones and each of m / smaller sizes. For even m the flower has m + petals, one large and each of m/ smaller sizes. 33. r + cos θ cosmθ These are similar to the ones in Eercise 3, but the curve does not approach the origin ecept for θ π in the case of even m. The petals are joined, and less distinct. The smaller ones cannot be distinguished. Fig. 8.5.38 We will have [ln θ,θ ] [ln θ,θ ]if θ θ + π and ln θ ln θ, that is, if ln θ + lnθ + π. This equation has solution θ.99956. The corresponding intersection point has Cartesian coordinates ln θ cos θ, ln θ sin θ.8,.353. 333

SECTION 8.5 PAGE 6 R. A. ADAMS: CALCULUS 39. r /θ A rθ Fig. 8.5.39 r lnθ The two intersections of r ln θ and r /θ for <θ π correspond to solutions θ and θ of ln θ θ, ln θ θ + π. π/ 3. Area sin θ a π/ Fig. 8.6. a cos θ dθ a sq. units. r a cos θ The first equation has solution θ.7638, giving the point.86,.556676, and the second equation has solution θ.77677, giving the point.888,.7866. Fig. 8.6.3 Section 8.6 Slopes, Areas, and Arc Lengths for Polar Curves page 68. Area π θ dθ π π. π/3 π/3. Area sin 3θ dθ cos 6θdθ θ 6 π/3 sin 6θ π sq. units. θ π 3 A r θ rsin 3θ θ θπ Fig. 8.6.. Area π θ dθ θ 3 6 Fig. 8.6. π 3 π 3 sq. units. 5. Total area 6 π/8 θ + π/8 cos θ dθ + cos 8θdθ sin 8θ π/8 π 8 sq. units. 33

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8.6 PAGE 68 rcos θ Fig. 8.6.5 π/8 π/ 8. Area πa + a sin θ dθ πa π/ + cos θ a sin θ + πa 3 + a θ + cos θ π/ sin θ 5π a sq. units. ra dθ 6. The circles r a and r a cos θ intersect at θ ±π/3. B smmetr, the common area is area of sector area of right triangle see the figure, i.e., [ 6 πa a ] 3a π 3 3 a sq. units. 6 Fig. 8.6.8 A ra sin θ ra ra cos θ 9. For intersections: + cos θ 3 cos θ. Thus cos θ and θ ±π/3. The shaded area is given b π A Fig. 8.6.6 7. Area cos θ π/ dθ π π + cos θ cos θ + dθ π π/ 3 π π sin θ π sin θ π π/ π + sq. units. [ π + cos θ dθ 9 π/3 π π/3 + cos θ + π/ π/ π/3 + cos θ 9 + cos θdθ π/3 3 π sin θ π + sin θ + 3 9 sin θ π/ θ + π/3 π 3 3 8 + 9 3 π π/3 ] cos θ dθ dθ π/3 sq. units. r cos θ r r+cos θ r3 cos θ Fig. 8.6.7 Fig. 8.6.9 335

SECTION 8.6 PAGE 68 R. A. ADAMS: CALCULUS. Since r cos θ meets r atθ ± π 6 and ± 5π 6, the area inside the lemniscate and outside the circle is π/6 sin θ [ cos θ ] dθ π/6 π 3 3 π sq. units. 3 r A A r cos θ Fig. 8.6.. r atθ ±π/3. The shaded area is π π/3 π + cos θ dθ π/3 + cos θ + + cos θ dθ π π π 3 + sin θ 3 + sin θ π/3 π/3 π 3 3 + π 3 3 sq. units. π/3 r+ cos θ 3. r e aθ dr, π θ π. dθ aeaθ. ds e aθ + a e aθ dθ + a e aθ dθ. The length of the curve is. s π a a π π + a + a e aθ dθ e aπ e aπ units. a π a a [ θ a a + a θ dθ θπ θ + θ dθ Let θ tan u sec 3 udu dθ sec udθ sec u tan u + ln sec u + tan u + θ + ln + θ + θ θπ θ ] θπ θ [ π + π + lnπ + + π 5. r cos θ r dr dr θ sin θ sin dθ dθ r ds cos θ + sin θ cos θ dθ sec θ dθ Length π/ sec θ dθ. ] units. 3 r cos θ π/3 Fig. 8.6. π dr π. s + r dθ dθ θ + θ dθ π θ + θ dθ Let u + θ du θ dθ +π +π udu 3 u3/ [ ] + π 3/ 8 units. 3 6. If r cos θ, then and r dr dθ ds Fig. 8.6.5 sin θ dr dθ sin θ cos θ cos θ + sin θ cos θ dθ dθ. cos θ 336

INSTRUCTOR S SOLUTIONS MANUAL SECTION 8.6 PAGE 68 a Area of the surface generated b rotation about the - ais is S π π π/ π/ π cos θ r sin θ ds cos θ sin θ π/ dθ cos θ π sq. units. b Area of the surface generated b rotation about the - ais is S π 7. For r + sin θ, π π/ π/ π/ π sin θ r cos θ ds cos θ cos θ π/ dθ cos θ π sq. units. tan ψ r dr/dθ + sin θ. cos θ If θ π/, then tan ψ + and ψ 3π/8. If θ 5π/, then tan ψ and ψ π/8. The line meets the cardioid r + sin θ at the origin at an angle of 5, and also at first and third quadrant points at angles of 67.5 and.5 as shown in the figure. r+sin θ [ ], π i.e., at P and P,. For r sin θ we have r dr dθ cos θ. At P we have r and dr/dθ. Thus the angle ψ between the curve and the radial line θ π/ isψ π/. For r cos θ we have dr/dθ sin θ, so the angle between this curve and the radial line θ π/ satisfies tan ψ r dr/dθ, and ψ 3π/. The two θπ/ curves intersect at P at angle 3π π π. The Figure shows that at the origin, P, the circle meets the lemniscate twice, at angles and π/. r sin θ Fig. 8.6.8 r cos θ 9. The curves r cos θ and r sin θ intersect on the ras θ π/ and θ 5π/, as well as at the origin. At the origin their cusps clearl intersect at right angles. For r cos θ, tan ψ cos θ/sin θ. At θ π/, tan ψ, so ψ π/8. At θ 5π/, tan ψ +, soψ 3π/8. For r sin θ, tan ψ sin θ/ cos θ. At θ π/, tan ψ, so ψ π/8. At θ 5π/, tan ψ +, so ψ 3π/8. At π/ the curves intersect at angle π/8 π/8 π/. At 5π/ the curves intersect at angle 3π/8 3π/8 3π/ or π/ if ou use the supplementar angle. r cos θ ψ θ π/ r sin θ Fig. 8.6.7 8. The two curves r sin θ and r cos θ intersect where sin θ cos θ sin θ cos θ cos θ sin θ cos θcos θ sin θ cos θ or cos θ, Fig. 8.6.9. We have r cos θ + sin θ. For horizontal tangents: d dθ d cos θ sin θ + sin θ dθ cos θ sin θ + sin θ cos θ cos θ sin θ tan θ. 337

SECTION 8.6 PAGE 68 R. A. ADAMS: CALCULUS Thus θ π 8 or 3π. The tangents are horizontal at [ 8 π π cos sin, π ] and 8 8 8 [ 3π 3π cos + sin, 3π ]. 8 8 8 For vertical tangent: d dθ d dθ cos θ + cos θ sin θ cos θ sin θ + cos θ sin θ sin θ cos θ tan θ. [ Thus θ π/8 of5π/8. There are vertical tangents at π π cos + sin, π ] and 8 8 8 [ 5π 5π cos + sin, 5π ]. 8 8 8 rcos θ+sin θ. We have r cos θ, and r dr sin θ. For horizontal tangents: dθ d r sin θ r cos θ + sin θ dθ cos θ cos θ sin θ sin θ cos θ sin θcos θ sin θ cos θ cos θ or cos θ 3 sin θ. sin θ r There are no points on the curve where cos θ. Therefore, horizontal tangents occur onl where [ tan θ /3. There are horizontal tangents at, ± π ] [ and 6, ± 5π ]. 6 For vertical tangents: d r cos θ r sin θ + cos θ dθ cos θ sin θ sin θ cos θ cos θ sin θsin θ sin θ cos θ sin θ or 3cos θ sin θ. sin θ r There are no points on the curve where tan θ 3, so the onl vertical tangents occur where sin θ, that is, at the points with polar coordinates [, ] and [, π]. r cos θ Fig. 8.6.. r cos θ. tan ψ r cot θ. dr/dθ For horizontal tangents we want tan ψ tan θ. Thus we want tan θ cot θ, and so θ ±π/ or±3π/. The tangents are horizontal at [, ±π/]. For vertical tangents we want tan ψ cot θ. Thus we want cot θ cot θ, and so θ, ±π/, or π. There are vertical tangents at the origin and at [, ]. Fig. 8.6. θπ/ r cos θ θ π/ Fig. 8.6. sin θ 3. r sin θ. tan ψ cos θ tan θ. For horizontal tangents: tan θ tan θ tan θ tan tan θ θ tan θ + tan θ tan θ tan θ. Thus θ, π, ± tan, π ± tan. There are horizontal tangents at the origin and the points [ 3, ± ] [ tan and 3,π ± ] tan. Since the rosette r sin θ is smmetric about, there must be vertical tangents at the origin and at the points [ ] 3, ± tan and [ ] 3,π ± tan. 338

INSTRUCTOR S SOLUTIONS MANUAL REVIEW EXERCISES 8 PAGE 69 r sin θ rsin θ Fig. 8.6.3. We have r e θ and dr dθ eθ. For horizontal tangents: d dθ r sin θ eθ cos θ + e θ sin θ tan θ θ π + kπ, where k, ±, ±,... At the points [e kπ π/, kπ π/] the tangents are horizontal. For vertical tangents: d dθ r cos θ eθ cos θ e θ sin θ tan θ θ π + kπ. At the points [e kπ+π/, kπ + π/] the tangents are vertical. 5. r sin θ, tan ψ sin θ. cos θ For horizontal tangents tan ψ cot θ, so sin θ sin θ cos θ cos θ cos θ, or sin θ. The solutions are θ ±π/, ±π/6, and ±5π/6. θ π/ corresponds to the origin where the cardioid has a cusp, and therefore no tangent. There are horizontal tangents at [, π/], [,π/6], and [, 5π/6]. For vertical tangents tan ψ cot θ, so sin θ cos θ cos θ sin θ sin θ sin θ cos θ sin θ sin θ sin θ sin θ sin θ + The solutions here are θ π/ the origin again, θ π/6 and θ 5π/6. There are vertical tangents at [3, π/6] and [3, 5π/6]. Fig. 8.6.5 6. r cos θ f θ cos θ, r sin θ f θ sin θ. d dθ f d θ cos θ f θ sin θ, dθ f θ sin θ + f θ cos θ ds f θ cos θ f θ sin θ + f θ sin θ + f θ cos θ dθ [ f θ cos θ f θ f θ cos θ sin θ + f θ sin θ / + f θ sin θ + f θ f θ sin θ cos θ + f θ cos θ] dθ f θ + f θ dθ. Review Eercises 8 page 69. + + Ellipse, semi-major ais a, along the -ais. Semiminor ais b. c a b. Foci: ±,.. 9 36 9 Hperbola, transverse ais along the -ais. Semi-transverse ais a, semi-conjugate ais b 3. c a + b 3. Foci: ± 3,. Asmptotes: 3 ±. 3. + + 3 Parabola, verte,, opening to the left, principal ais. a /. Focus: 5/,.. + 8 8 + + 8 + 6 + 9 7 + 37 + 3 37/. Ellipse, centre, 3, major ais along 3. a 37, b 37/, c a b /. Foci: ± /, 3. 5. t, t, t. Straight line segment from, to,. 339

REVIEW EXERCISES 8 PAGE 69 R. A. ADAMS: CALCULUS 6. sin3t, cos3t, t Part of a circle of radius centred at the origin from the point, clockwise to sin 6, cos 6. 7. cosh t, sinh t. Parabola, or, traversed left to right. 8. e t, e t, t. Part of the curve from /e, e to e, /e. 9. cost/, sint/, t π. The first quadrant part of the ellipse 6 + 6, traversed counterclockwise.. cos t + sin t, cos t sin t, t π The circle +, traversed clockwise, starting and ending at,.. + t t 3 3t d 8t d + t 3t Horizontal tangent at t ±, i.e., at, ±. Vertical tangent at t, i.e., at,. Self-intersection at t ± 3, i.e., at,. t t± 3 t Fig. R-8. t 3. t 3 3t t 3 d 3t d 3t Horizontal tangent at t, i.e., at,. Vertical tangent at t ±, i.e., at, and,. Slope d { d t > if t > t < if t < Slope ast ±. t Fig. R-8.3 t. t 3 3t t 3 t d 3t d 3t Horizontal tangent at t ±, i.e., at, 6 and, 6. Vertical tangent at t ±, i.e., at, and,. Slope d { d t > if t > or t < t < if < t < Slope ast ±.,6. t 3 3t t 3 + 3t d 3t d 3t + Horizontal tangent: none. Vertical tangent at t ±, i.e., at, and,. Slope d { d t + > if t > t < if t < Slope ast ±.,, t 3 3t t 3 t, 6 Fig. R-8., t 3 3t t 3 +3t 5. The curve t 3 t, t 3 is smmetric about since is an odd function and is an even function. Its self-intersection occurs at a nonzero value of t that makes, namel, t ±. The area of the loop is Fig. R-8., A t t d t 6 + 3 t sq. units. t 3 t3t 3

INSTRUCTOR S SOLUTIONS MANUAL REVIEW EXERCISES 8 PAGE 69 r θ t 3 t t 3 t± t Fig. R-8.5 6. The volume of revolution about the -ais is. r θ, π θ π Fig. R-8.9 r θ V π π 3π 3π t t d t 6 t + t 3t t 8 t 6 + t 9 7 + 5 8π cu. units. 5 Fig. R-8.. r + cosθ 7. e t t, e t/, t. Length is L e t + e t e t + e t + e t + t e + units. Fig. R-8. r+cos θ 8. Area of revolution about the -ais is S π e t/ e t +. r + cosθ r+cosθ 8π 3 e3t/ + e t/ 6π 3 e3 + 3e sq. units. Fig. R-8. 9. r θ, 3π θ 3π 3. r + cosθ 3

REVIEW EXERCISES 8 PAGE 69 R. A. ADAMS: CALCULUS. r sin3θ r+ cos θ Fig. R-8.3 r sin3θ 7. r + sin θ approaches the origin in the directions for which sin θ /, that is, θ 3π/ and θ π/. The smaller loop corresponds to values of θ between these two values. B smmetr, the area of the loop is A π/ π/ π/ π/ + sin θ + sin θdθ + sin θ cosθdθ θ cos θ π/ sinθ π + π 3 sq. units. π/ π/6 r+ sin θ 3π/ π/ 5. Area of a large loop: A π/3 π/3 Fig. R-8. + cosθ dθ [ + cosθ+ + cosθ] dθ 3θ + sinθ+ π/3 sinθ π + 3 3 6. Area of a small loop: A π/ π/3 π/ π/3 sq. units. + cosθ dθ [ + cosθ+ + cosθ] dθ 3θ + sinθ+ π/ sinθ π 3 3 sq. units. π/3 Fig. R-8.7 8. r cos θ / and r + cos θ intersect where + cos θ cos θ cos θ + cos θ cos θ ± 6 + 6 8 ±. Onl / is between and, so is a possible value of cos θ. Let θ cos. Then sin θ +. B smmetr, the area inside r + cos θ to the left of the line / is A π + cos θ + + cosθ θ 3 π θ + sin θ + π sinθ θ + + 3 π cos dθ + cos θ sin θ + + 9 sq. units. 8 3

INSTRUCTOR S SOLUTIONS MANUAL CHALLENGING PROBLEMS 8 PAGE 69 / r+cos θ θ A C S Fig. R-8.8 C P F Challenging Problems 8 page 69 F C. The surface of the water is elliptical see Problem below whose semi-minor ais is cm, the radius of the clinder, and whose semi-major ais is sec θ cm because of the tilt of the glass. The surface area is that of the ellipse This area is A sec θ cos t, sin t, tπ/ t π/ 3 sec θ d sec θ cos t cos t π/ t π. + cost 6π sec θ cm. cm θ sec θ cm S A Fig. C-8. Let P be an point on C. Let A A be the line through P that lies on the clinder, with A on C and A on C. Then PF PA because both lengths are of tangents drawn to the sphere S from the same eterior point P. Similarl, PF PA. Hence PF + PF PA + PA A A, which is constant, the distance between the centres of the two spheres. Thus C must be an ellipse, with foci at F and F. 3. Given the foci F and F, and the point P on the ellipse, construct N PN, the bisector of the angle F PF. Then construct T PT perpendicular to N N at P. B the reflection propert of the ellipse, N N is normal to the ellipse at P. Therefore T T is tangent there. T θ θ N P N T F F Fig. C-8.. Let S and S be two spheres inscribed in the clinder, one on each side of the plane that intersects the clinder in the curve C that we are tring to show is an ellipse. Let the spheres be tangent to the clinder around the circles C and C, and suppose the are also tangent to the plane at the points F and F, respectivel, as shown in the figure. Fig. C-8.3. Without loss of generalit, choose the aes and ais scales so that the parabola has equation. If P is the point, on it, then the tangent to the parabola at P has equation +, 33

CHALLENGING PROBLEMS 8 PAGE 69 R. A. ADAMS: CALCULUS which intersects the principal ais at,. Thus R, and Q,. Evidentl the verte V, bisects RQ. 6. P [r,θ] Q P, r a θ [a,θ ] L V θ 5. R Fig. C-8. To construct the tangent at a given point P on a parabola with given verte V and principal ais L, drop a perpendicular from P to L, meeting L at Q. Then find R on L on the side of V opposite Q and such that QV VR. Then PR is the desired tangent. b c ft Fig. C-8.5 ft Let the ellipse be a +, with a and foci at b, ± so that c and b a + c 8. The volume of the barrel is V π d π d 8 8π 3 π ft 3. 3 a Fig. C-8.6 a Let L be a line not passing through the origin, and let [a,θ ] be the polar coordinates of the point on L that is closest to the origin. If P [r,θ]isan point on the line, then, from the triangle in the figure, a r cosθ θ a, or r cosθ θ. b As shown in part a, an line not passing through the origin has equation of the form r gθ a cosθ θ a secθ θ, for some constants a and θ. We have g θ a secθ θ tanθ θ g θ a secθ θ tan θ θ + a sec 3 θ θ gθ + g θ gθg θ a sec θ θ + a sec θ θ tan θ θ a sec θ θ tan θ θ a sec θ θ a [ ] sec θ θ + tan θ θ sec θ θ. c If r gθ is the polar equation of the tangent to r f θ at θ α, then gα f α and g α f α. Suppose that f α + f α f α f α >. B part b we have gα + g α gαg α. 3

INSTRUCTOR S SOLUTIONS MANUAL CHALLENGING PROBLEMS 8 PAGE 69 7. Subtracting, and using gα f α and g α f α, we get f α < g α. It follows that f θ < gθ for values of θ near α; that is, the graph of r f θ is curving to the origin side of its tangent at α. Similarl, if f α + f α f α f α <, then the graph is curving to the opposite side of the tangent, awa from the origin. B R t Fig. C-8.7 When the vehicle is at position, as shown in the figure, the component of the gravitational force on it in the direction of the tunnel is mar cos θ mgr R θ r cos θ mg R. B Newton s Law of Motion, this force produces an acceleration d / along the tunnel given b that is m d mg R, d + ω, where ω g R. This is the equation of simple harmonic motion, with period T π/ω π R/g. For R 396 mi.9 7 ft, and g 3 ft/s,we have T 579 s 8.6 minutes. This is a rather short time for a round trip between Atlanta and Baghdad, or an other two points on the surface of the earth. A 8. Take the origin at station O as shown in the figure. Both of the lines L and L pass at distance cos ɛ from the origin. Therefore, b Problem 6a, their equations are L : cos ɛ r cos [ θ cos ɛ ] π ɛ sinθ + ɛ L : cos ɛ r cos [ θ cos ɛ ] π + ɛ sinθ ɛ. The search area Aɛ is, therefore, Aɛ π +ɛ cos ɛ π ɛ sin θ ɛ cos ɛ sin dθ θ + ɛ π 5, cos +ɛ ɛ csc θ ɛ csc θ + ɛ dθ π ɛ 5, cos ɛ [ cot π + ɛ cot π + cot π ɛ ] [ cos π 5, cos + ɛ ɛ sin π + ɛ + sin π + ɛ ] cos π + ɛ, cos ɛ [ csc π + ɛ ], cos ɛsecɛ mi. For ɛ 3 π/6, we have Aɛ.8 square miles. Also A ɛ, cos ɛ sin ɛsecɛ +, cos ɛ secɛ tanɛ A π/6 865. When ɛ 3, the search area increases at about 865π/8 5 square miles per degree increase in ɛ. mi O π/ ɛ ɛ Fig. C-8.8 Area Aɛ L L 9. The easiest wa to determine which curve is which is to calculate both their areas; the outer curve bounds the larger area. The curve C with parametric equations sin t, sint, t π 35

CHALLENGING PROBLEMS 8 PAGE 69 R. A. ADAMS: CALCULUS has area Let u cos t du sin t tπ/ A t π/ π/ d sint cos t sin t cos t u du sq. units. 3 The curve C with polar equation r cosθ has area A π/ π/ cosθdθ sinθ sq. units. C is the outer curve, and the area between the curves is /3 sq. units. Fig. C-8.9 36