SECTION 7. PAGE 76 R. A. ADAMS: CALCULUS CHAPTER 7. GRATION Section 7. pge 76. B slicing: V = APPLICATIONS OF INTE- Volumes of Solids of Revolution d = 5 cu. units. B shells: V = d = 5/ 5 = cu. units. 5. B slicing: V = = d 5 5 = cu. units. B shells: V = d 5/ = 5 = cu. units. = =, = Fig. 7.. Fig. 7... Slicing:. Slicing: V = d = = cu. units. Shells: V = d = 5 5 = cu. units. Shells: V = d = = cu. units. V = / d = 5 5/ = cu. units. = =, = Fig. 7.. Fig. 7.. 6
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 76 5. About the -is:, V = = = d + d + 5 5 = 6 5 cu. units. = = Fig. 7..6 b About the -is: V = d = = 8 cu. units. = 7. About the -is: V = d = = 7 cu. units. b About the -is: V = [ ] d = 5 8 + d = 5 + 5 5 = 8 5 cu. units. b =, 6. Rotte bout the -is Fig. 7..5 = Fig. 7..7 = b the -is V = = d 5 5 = 5 cu. units. V = d = = cu. units. 6 8. Rotte bout the -is V = [ + sin ] d = sin + sin d = cos + sin = + cu. units. 65
SECTION 7. PAGE 76 R. A. ADAMS: CALCULUS b the -is V = sin d U = dv = sin d du = d V = cos [ ] = cos + cos d = cu. units. /, += =,/ Fig. 7.. 9. About the -is: V = + = = / / sec θ sec θ dθ cos θ dθ d = θ + sin θ cos θ / = 8 = 5 8 Let = tn θ d = sec θ dθ cu. units. b About the -is: V = + d = ln + = ln = ln cu. units.. V = d = + d = + +=. V = [ ] d = 5 5 = = Fig. 7.. cu. units. = = 8 5 cu. units. = + Fig. 7..9. B smmetr, rottion bout the -is gives the sme volume s rottion bout the -is, nmel V = / d 5 = / = 5 cu. units. 8. The volume remining is V = = Fig. 7.. = = d d Let u = du = d udu= = cu. units. u/ 66
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 76 Since the volume of the bll is = cu. units., therefore the volume removed is cu. units. The percentge removed is = 5. 8 About 5% of the volume is removed. = 5. The volume remining is b V = h d b = h b b = hb h b b = h b + cu. units. b h d b + h = = d b. The rdius of the hole is the remining volume is V = L/ L/ Fig. 7.. R L. Thus, b slicing, [R L = L/ R L = 6 L cu. units independent of R. R = R ] d R L Fig. 7..5 6. Let circulr disk with rdius hve centre t point,. Then the disk is rotted bout the -is which is one of its tngent lines. The volume is: V = = = = + d u + u du u u du + = cu. units. Let u = du = d u du Note tht the first integrl is zero becuse the integrnd is odd nd the intervl is smmetric bout zero; the second integrl is the re of semicircle. + = L L Fig. 7.. Fig. 7..6 67
SECTION 7. PAGE 76 R. A. ADAMS: CALCULUS 7. Volume of the smller piece: V = d b = b = b b = b[ + b + b ] = b + b cu. units. 9. The volume of the ellipsoid is V = b = b b d = b cu. units. =b = d b d Fig. 7..9 Fig. 7..7 8. Let the centre of the bowl be t,. Then the volume of the wter in the bowl is [ V = ] d = 6 d = [ ] 9 cm.. The cross-section t height is n nnulus ring hving inner rdius b nd outer rdius b +. Thus the volume of the torus is [ V = b + b ] d = b d = 8b = b cu. units.. We used the re of qurter-circle of rdius to evlute the lst integrl.. Volume of revolution bout the -is is V = e d e R = lim R = cu. units. Fig. 7..8 + = b Volume of revolution bout the -is is V = e d = lim R e e R = cu. units. 68
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 76. The volume is V = =e d Fig. 7.. k d = lim R R k = lim R k + k. k k In order for the solid to hve finite volume we need k <, tht is, k >. R 5. Since ll isosceles right-ngled tringles hving leg length cm re congruent, S does stisf the condition for being prism given in previous editions. It does not stisf the condition in this edition becuse one of the line segments joining vertices of the tringulr cross-sections, nmel the -is, is not prllel to the line joining the vertices of the other end of the hpotenuses of the two bses. The volume os S is still the constnt cross-sectionl re / times the height b, tht is, V = b/ cm. 6. Using heights f estimted from the given grph, we obtin 9 V = f d [ +.8 + 5 + 6.7 + 8 + 8 + 7 + 5. + ] 98 cu. units.. The volume is V = k d. This improper integrl converges if k <, i.e., if k >. The solid hs finite volume onl if k >. 7. Using heights f estimted from the given grph, we obtin = k d 9 V = f d [ +.8 + 5 + 6.7 + 58 ] + 68 + 77 + 85. + 9 57 cu. units. Fig. 7... A solid consisting of points on prllel line segments between prllel plnes will certinl hve congruent cross-sections in plnes prllel to nd ling between the two bse plnes, n solid stisfing the new definition will certinl stisf the old one. But not vice vers; congruent cross-sections does not impl fmil of prllel line segments giving ll the points in solid. For counteremple, see the net eercise. Thus the erlier, incorrect definition defines lrger clss of solids thn does the current definition. However, the formul V = Ah for the volume of such solid is still vlid, s ll congruent cross-sections still hve the sme re, A, s the bse region. 8. Using heights f estimted from the given grph, we obtin 9 V = + f d [ +.8 + 5 + 56.7 + 68 ] + 78 + 87 + 95. + 8 cu. units. 69
SECTION 7. PAGE 76 R. A. ADAMS: CALCULUS 9. The region is smmetric bout = so hs the sme volume of revolution bout the two coordinte es. The volume of revolution bout the -is is V = 8 = 7 = 7 = 7 = 7 = 7 / / d Let = 8 sin u d = sin u cos udu / / sin 5 u cos udu cos u cos u sin udu Let v = cos u dv = sin udu v v dv v v 6 + v 8 dv = 89 5 5 7 + 9 cu. units.. The volume of the bll is R. Epressing this volume s the sum i.e., integrl of volume elements tht re concentric sphericl shells of rdius r nd thickness dr, nd therefore surfce re kr nd volume kr dr, we obtin R R = kr dr = k R. Thus k =. r Fig. 7... Let the bll hve rdius R, nd suppose its centre is units bove the top of the conicl glss, s shown in the figure. Clerl the bll which mimizes wine overflow from the glss must be tngent to the cone long some circle below the top of the cone lrger blls will hve reduced displcement within the cone. Also, the bll will not be completel submerged. R dr h α R Fig. 7.. h+ cos α h sec α R Note tht = sin α, sor = + h sin α. + h Using the result of Eercise #7, the volume of wine displced b the bll is V = R R +. We would like to consider V s function of for R R since V = t ech end of this intervl, nd V > inside the intervl. However, the ctul intervl of vlues of for which the bove formultion mkes phsicl sense is smller: must stisf R h tn α. The left inequlit signifies nonsubmersion of the bll; the right inequlit signifies tht the bll is tngent to the glss somewhere below the rim. We look for criticl point of V, considered s function of. As noted bove, R is function of. We hve = dv d = [ dr R + R dr d + d R + ] dr R + + R = R + R. d Thus 6R sin α = R + = R + R sin α h R sin α = R sin α + R h sin α h sin α R = sin α + sin α = h sin α cos α + sin α. This vlue of R ields positive vlue of V, nd corresponds to = R sin α. Since sin α sin α, h sin α sin α R = + sin α sin α h sin α cos α = h tn α. Therefore it gives the mimum volume of wine displced. 7
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 8. Let P be the point t, 5 t. The line through P perpendiculr to AB hs eqution = + 5 t, nd meets the curve = t point Q with -coordinte s equl to the positive root of s + 5 ts =. Thus, s = [ 5 t 5 ] + t +. A/, = 5 sec v tn v+ tn / ln sec v + tn v = [ ] 5 5 + ln ln 6 = 57 6 5ln cu. units. = Q P dt += 5 B,/ Section 7. pge 8. V = Other Volumes b Slicing d= = 6m. A horizontl slice of thickness dz t height hs volume dv = zh z dz. Thus the volume of the solid is s t Fig. 7.. V = h hz zh z dz = z h = h 6 units. The volume element t P hs rdius PQ = t s = 5 5 t + nd thickness dt. Hence, the volume of the solid is [ 5 V = / 5 ] t + dt = 5 5 6 5 t + + / [ 5 ] ] t + dt Let u = t 5 du = dt = / / 6 5 u + + u du = 6 u + / u 5 = = / / 5 / u + du tn / tn / tn / Let u = tn v du = sec v dv sec v dv sec v dv. A horizontl slice of thickness dz t height hs volume dv = z z dz. Thus the volume of the solid is. V = 5. V = = V = 6 = z z dz let u z d = udu= = 6 u/ cu. units 6 = units. + z8 z dz = 6 + 6z z dz 6z + z z 6 = ft 6. The re of n equilterl tringle of edge is A = = sq. units. The volume of the solid is V = d= 8 = 5 cu. units. 8 7. The re of cross-section t height is /h A = = sq. units. h 7
SECTION 7. PAGE 8 R. A. ADAMS: CALCULUS The volume of the solid is V = h h d = h cu. units. 8. Since V =, we hve = k d = k = k. + =r r Thus k =. 9. The volume between height nd height z is z. Thus Fig. 7.. z = z At dt, where At is the cross-sectionl re t height t. Differentiting the bove eqution with respect to z, we get z = Az. The cross-sectionl re t height z is z sq. units.. This is similr to Eercise 7. We hve z = At dt, so Az =. Thus the squre cross-section t height z hs side units. r. V = r d r = 8 r d = 8 r r = 6r cu. units. z r Fig. 7.. = r z. The re of n equilterl tringle of bse is =. Hence, the solid hs volume r V = r d = r r = r cu. units.. The cross-section t distnce from the verte of the prtil cone is semicircle of rdius / cm, nd hence re /8cm. The volume of the solid is V = 8 d = = 7 cm. z,, Fig. 7... The volume of solid of given height h nd given crosssectionl re Az t height z bove the bse is given b h V = Az dz. If two solids hve the sme height h nd the sme re function Az, then the must necessril hve the sme volume. 5. Let the -is be long the dimeter shown in the figure, with the origin t the centre of the bse. The crosssection perpendiculr to the -is t is rectngle hving bse r nd height h = + b + b. Thus the volume of the truncted clinder is V = = r r r r + b r + b d r + b r d = r + b cu. units. 7
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 8 7. Cross-sections of the wedge removed perpendiculr to the -is re isosceles, right tringles. The volume of the wedge removed from the log is h V = = d = 6, cm. r = r Fig. 7..5 z 6. The plne z = k meets the ellipsoid in the ellipse tht is, which hs re + b [ = k c k ] + c [ Ak = b b [ k ]. c k ] = c The volume of the ellipsoid is found b summing volume elements of thickness dk: c V = c [ = b [ b k c k k c ] c = bc cu. units. z c k Ak Fig. 7..6 ] dk c + + z b c = one-eighth of the solid is shown b Fig. 7..7 5 = 8. The solution is similr to tht of Eercise 5 ecept tht the legs of the right-tringulr cross-sections re insted of, nd goes from to insted of to. The volume of the notch is V = d = 5 d =,,, 7 cm. 9. The hole hs the shpe of two copies of the truncted clinder of Eercise, plced bse to bse, with + b = inndr = in. Thus the volume of wood removed the volume of the hole is V = / = in.. One eighth of the region ling inside both clinders is shown in the figure. If the region is sliced b horizontl plne t height z, then the intersection is rectngle with re Az = b z z. The volume of the whole region is V = 8 b b z z dz. 7
SECTION 7. PAGE 8 R. A. ADAMS: CALCULUS z Az z b b z z Fig. 7... B the result given in Eercise 8 with = cmnd b = cm, the volume of wood removed is V = 8 z 6 z dz 97.8 cm. We used the numericl integrtion routine in Mple to evlute the integrl. Section 7. pge 87. =, =, ds = L = 5 d = 5 units. Arc Length nd Surfce Are + d. = + b, A B, =. The length is L = B A + d = + B A units.. = /, =, ds = + d 8 8 L = + d= + / = 5 units.. =, = /, = L = + 9 d = 9 5 d = 7 9 5/ = / 8 units. 7 5. = /, = /, ds = + 9 9 / d = / + / d 9 L = / + / d Let u = 9 / + du = 6 / d = / 6 udu= units. 9 7 6. + =, = + + / = + /, ds = + + + 5 d = d L = + 5 d = + 5/ 9 = 5 / / units. 9 7. = +, = ds = + d = + d L = + d = = 6 units. 8. = +, = ds = + L = + d = d = + d 9. = ln, = ds = + d = + d e L = + ln d = + e = + e = e + units.. If = ln 8 then = 8 nd + = +. 8 = 59 units. 7
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 87 Thus the rc length is given b s = + d 8 = + d 8 = + 8 ln = + ln units. 8. s = + sinh d= cosh d = sinh = sinh = e e units.. = ln e e +, = e + e + e e e e e + = e e. The length of the curve is L = = = = ln + e e d e + e d e + e e e d = ln e e e e ln e e e 8 e = ln e e = ln e + e units. /. s = + tn d /6 / = sec d= ln sec + tn /6 = ln + ln + + = ln units.. =,, =. / /6 length = + d Let = tn θ d = sec θ dθ = = sec θ = = = sec θ tn θ + ln sec θ + tn θ = = + + ln + + = 7 + ln + 7 = 7 + ln + 7 units. 5. / + / = /. B smmetr, the curve hs congruent rcs in the four qudrnts. For the first qudrnt rc we hve / / / = = / / / /. Thus the length of the whole curve is 6. The required length is L = L = + / / / = / / d d = / / = 6 units. + d = + 6 6 d. Using clcultor we clculte some Simpson s Rule pproimtions s described in Section 7.: S.599 S 8.65 S.6 S 6.6. To four deciml plces the length is.6 units. 75
SECTION 7. PAGE 87 R. A. ADAMS: CALCULUS 7. = /,, = /. Length = f d, where f = +. We 9/ hve T =.6 M =.6 T 8 =.85 M 8 =.7 T 6 =.78 M 6 =.76. Thus the length is pproimtel.8 units. 8. For the ellipse + =, we hve 6 + =, so = /. Thus ds = + 9 d = The circumference of the ellipse is + 6 + 6 d. d 8.7775 units with little help from Mple s numericl integrtion routine. 9. For the ellipse + =, we hve + =, so = /. Thus ds = + d = d The length of the short rc from, to, / is d.58 units with little help from Mple s numericl integrtion routine.. S = + d Let u = + du = 8d = 7 7 udu= u/ = 6 7 7 sq. units.. =,. ds = + 9 d. The re of the surfce of rottion bout the -is is S = = 8 + 9 d Let u = + 9 du = 6 d udu= 7 / sq. units.. = /,. ds = + 9 d. The re of the surfce of rottion bout the -is is S = / + 9 = 8 = 8 = 8 / u + u du d Let 9 = u 9 d = 8udu Let u = tn v du = sec v dv tn / tn v sec v dv tn / sec 7 v sec 5 v + sec v dv. At this stge it is convenient to use the reduction formul sec n v dv = n secn v tn v + n n sec n v dv see Eercise 6 of Section 7. to reduce the powers of secnt down to, nd then use We hve sec v dv = sec tn + ln sec + tn. I = sec 7 v sec 5 v + sec v dv = sec5 v tn v 5 6 + 6 sec 5 v dv + sec v dv [ = sec5 tn 7 sec v tn v 6 6 + ] sec v dv + sec v dv = sec5 tn 6 = sec5 tn 6 + 7 sec tn + 8 7 sec tn sec tn + ln sec + tn. 6 sec v dv Substituting = rctn/ now gives the following vlue for the surfce re: S = 8 8 + 8 ln + sq. units. 76
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 87. If = /,, is rotted bout the -is, the surfce re generted is S = = 8 = 8 = 6 8. We hve S = = / + 9 d Let u = + 9 u udu du = 9 d 5 u5/ / u/ / 5/ // sq. units. 5 = = e + e d Let e = tn θ e d = sec θ dθ + tn θ sec θ dθ = [ = sec θ tn θ + ln sec θ + tn θ Since = tn θ = e, sec θ = ] =. = = tn θ =, sec θ =, = = + e, sec θ dθ therefore [ S = e + e + ln + e + e ln ] + [ = e + e + e + ln ] + e sq. units. + 5. If = sin,, is rotted bout the -is, the surfce re generted is S = sin + cos d Let u = cos du = sin d = + u du Let u = tn θ du = sec θ dθ / = sec θ dθ = / / sec θ dθ / = sec θ tn θ + ln sec θ + tn θ = + ln + sq. units. 6. + = + = + S = d = + 5 8 + + + 6 = 88 + 6 = 75 sq. units. 8 7. For = +,, we hve ds = + d. The surfce generted b rotting the curve bout the -is hs re S = + d = + ln 6 55 = 6 + ln sq. units. 8. The re of the cone obtined b rotting the line = h/r, r, bout the -is is r r S = + h/r d = + h r r = r r + h sq. units. 9. For the circle b + = we hve b + d d = d d = b. Thus b ds = + d = d = b d if >. The surfce re of the torus obtined b rotting the circle bout the line = is S = = = 8b b+ b = 8b sin u d b d u + b u du du u b smmetr = b sq. units. Let u = b du = d 77
SECTION 7. PAGE 87 R. A. ADAMS: CALCULUS. The top hlf of + = is =,so d d =, nd S = + d 6 = 6 d Let = sin θ 6 d = cos θ dθ = / cos θ cos θ dθ = 6 / cos θ dθ = 8 / θ + sin θ cos θ = + sq. units.. For the ellipse + = wehve d d + 8 = d d =. The rc length element on the ellipse is given b ds = = + d d d + 6 d = + d. If the ellipse is rotted bout the -is, the resulting surfce hs re S = = 8 + d + d Let = tn θ d = sec θ dθ = 8 / sec θ dθ = 8 / sec θ tn θ + ln sec θ + tn θ = 8 + ln + = 8 + ln + sq. units.. As in Emple, the rc length element for the ellipse is ds = d b + d = d d. To get the re of the ellipsoid, we must rotte both the upper nd lower semi-ellipses see the figure for Eercise of Section 8.: [ S = c b + ] c + b ds b = 8c d [ ] = 8c of the circumference of the ellipse = 8cEε where ε = b nd Eε = / ε sin tdt s defined in Emple.. From Emple, the length is s = = / / + cos tdt + sin tdt = 5 / + + sin tdt = 5 + E. +. Let the eqution of the sphere be + = R. Then the surfce re between plnes = nd = b R < b R is S = = = R b b b R + d d d R R R d d = Rb sq. units. 78
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 9 Thus, the surfce re depends onl on the rdius R of the sphere, nd the distnce b between the prellel plnes. b The surfce re is S = > d + d =. b c Covering surfce with pint requires ppling ler of pint of constnt thickness to the surfce. Fr to the right, the horn is thinner thn n prescribed constnt, so it cn contin less pint thn would be required to cover its surfce. Fig. 7.. + =R 5. If the curve = k,<, is rotted bout the -is, it genertes surfce of re S = = If k, we hve S k + k k d + k k d. k d, which is infinite. If k, the surfce re S is finite, since k is bounded on, ] in tht cse. Hence we need onl consider the cse < k <. In this cse < k <, nd / S = = + k k d k + k k d < + k k d <. Thus the re is finite if nd onl if k >. 6. S = + d = + d = sec θ dθ Let = tn θ d = sec θ dθ / = sec θ tn θ + ln sec θ + tn θ = [ + ln + ] sq. units. 7. Volume V = d = cu. units. Section 7. Mss, Moments, nd Centre of Mss pge 9. The mss of the wire is L m = δs ds = = L L s cos L L = L. sin s L ds Since δs is smmetric bout s = L/ tht is, δl/ s = δl/ + s, the centre of mss is t the midpoint of the wire: s = L/.. A slice of the wire of width d t hs volume dv = + b d. Therefore the mss of the whole wire is m = L δ + b d L = δ + b + b d = δ L + bl + b L. Its moment bout = is M = = L δ + b d L = δ + b + b d = δ L + bl + b L. Thus, the centre of mss is δ L + bl + b L = δ L + bl + b L L + bl + b L = + bl +. b L 79
SECTION 7. PAGE 9 R. A. ADAMS: CALCULUS. The mss of the plte is m = δ re = δ. The moment bout = is M = = δ d Let u = = δ udu = δ u/ = δ. du = d Thus = M = m = δ δ =. B smmetr, ȳ =. Thus the centre of mss of the plte is,. = d Fig. 7... A verticl strip hs re da = d. Therefore, the mss of the qurter-circulr plte is m = δ d Let u = du = d = δ udu= δ u/ = δ. The moment bout = is M = = δ d Let = sin θ d = cos θ dθ / = δ sin θ cos θ dθ = δ = δ 8 / / The moment bout = is sin θ dθ cos θdθ = δ 6. M = = δ d = δ = 8 δ. Thus, = 6 nd ȳ =. Hence, the centre of mss 8 is locted t 6, 8. 5. The mss of the plte is m = = k = k k d uu / du 8 u/ 5 u5/ B smmetr, M = =, so =. M = = k d Let u = du = d = 56k 5. Let u = du = d = k 6u / 8u / + u 5/ du = k u/ 6 5 u5/ + 7 u7/ = 96k 5. Thus ȳ = 96k 5 5 56k = 6. The centre of mss of the 7 plte is, 6/7. densit k Fig. 7..5 = 6. A verticl strip t h hs re da = h dh. Thus, the mss of the plte is m = h = 5h h dh = h 9 = 5 kg. The moment bout = is M = = h h h = h dh h h dh = 5 kg-m. 8
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 9 The moment bout = is M = = h h h dh = h h + 9 h dh h = h 9 + h = 5 6 kg-m. 5 5 Since the mss is smmetric bout the -is, nd the plte is smmetric bout both the - nd -is, therefore the centre of mss must be locted t the centre of the squre. = dr r Thus, = = 5 nd ȳ = of mss is locted t,. 5 =. The centre Fig. 7..8 dh h Fig. 7..6 7. The mss of the plte is m = B smmetr, ȳ = /. = k d = k. M = = k d = k. Thus = k k =. The centre of mss of the plte is,. 9. m = M = = M = = b b b δ g f d δ g f d δ Centre of mss: g f d M= m, M =. m =g densit ρ = f Fig. 7..9 b densit k. The slice of the brick shown in the figure hs volume dv = 5 d. Thus, the mss of the brick is / Fig. 7..7 8. A verticl strip hs re da = r mss is / [ ] m = kr r dr = k r r dr = dr. Thus, the k g. m = k5 d = 5k = k g. The moment bout =, i.e., the z-plne, is M = = 5k d = 5 k = 5 8k g-cm. 8
SECTION 7. PAGE 9 R. A. ADAMS: CALCULUS Thus, = 5 8k k =. Since the densit is inde- pendent of nd z, ȳ = 5 nd z = 5. Hence, the centre of mss is locted on the cm long centrl is of the brick, two-thirds of the w from the lest dense 5 fce to the most dense such fce. 5 R +R d R z Fig. 7.. Fig. 7... A slice t height z hs volume dv = dz nd densit kz g/cm. Thus, the mss of the cone is. Choose es through the centre of the bll s shown in the following figure. The mss of the bll is R m = + RR d R = R R R = 8 R kg. B smmetr, the centre of mss lies long the -is; we need onl clculte ȳ. M = = R R R = = + RR d R d R 5 R 5 = 5 R5. b m = kz dz b = k z z dz b z = k z b + z b b = k b g. The moment bout z = is b M z= = k z z dz = b k b g-cm. Thus, z = b. Hence, the centre of mss is on the is 5 of the cone t height b/5 cm bove the bse. z b z dz = b z R5 Thus ȳ = 5 8 R = R. The centre of mss is on the line through the centre of the bll perpendiculr to the plne mentioned in the problem, t distnce R/ from the centre of the bll on the side opposite to the plne.. B smmetr, ȳ =. Fig. 7.. 8
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7. PAGE 9 z z Fig. 7.. z A horizontl slice of the solid t height z with thickness dz is hlf-disk of rdius z with centre of mss t = z, b Eercise bove. Its mss is so = δ 5 5 The centre of mss is 8 δ = 6 5. 6 8,, 5 5.. Assume the cone hs its bse in the -plne nd its verte t height b on the z-is. B smmetr, the centre of mss lies on the z-is. A clindricl shell of thickness d nd rdius bout the z-is hs height z = b /. Since it s densit is constnt k, its mss is dm = bk d. Also its centre of mss is t hlf its height, ȳ shell = b Thus its moment bout z = is. dm z= =ȳ shell dm = bk d. dm = δ zdz z, nd its moment bout = is dm = = dm = δ z z z = δ z z /. Thus the mss of the solid is Also, m = δ = δ M z= = δ = δ z z z dz z z z dz z z5 5 nd z = δ 5 8 5 δ = 8 5. Finll, M = = δ = δ = δ 5 u5/ = δ. 8 = δ 5 5, z z / dz Let u = z du = zdz u / du = δ 5 5, 5. Hence m = M z= = bk d = kb 6 bk kb d = nd z = M z= /m = b/5. The centre of mss is on the is of the cone t height b/5 cm bove the bse. + = ds θ dθ Fig. 7..5 Consider the re element which is the thin hlf-ring shown in the figure. We hve dm = ks sds= k s ds. Thus, m = k. Regrd this re element s itself composed of smller elements t positions given b the ngle θ s shown. Then dm = = s sin θsdθ ks ds M = = k = ks ds, s s ds = k. 8
SECTION 7. PAGE 9 R. A. ADAMS: CALCULUS 6. Therefore, ȳ = k k =. B smmetr, =. Thus, the centre of mss of the plte is,. L ds s 7. m = = C = C k / Ce kr r dr r e kr dr u e u du Let u = kr du = kdr U = u dv = ue u du du = du V = e u = C ue u R k / lim R + = C k / + e u du = C k / = C k R / 5.57C k /. e u du θ Fig. 7..6 The rdius of the semicircle is L. Let s mesure the distnce long the wire from the point where it leves the positive -is. Thus, the densit t position s is s δδs = sin g/cm. The mss of the wire is L m = L sin s L ds = L cos s L L = L g. 8. r = rce kr r dr m C = C / k / r e kr dr Let u = kr du = kr dr = k/ k ue u du U = u du = du = k lim R dv = e u du V = e u R ue u + R e u du = + lim k R e e R =. k Since n rc element ds t position s is t height = L sin θ = L s sin, the moment of the wire bout L = is Section 7.5 Centroids pge 99 M = = = L L L s sin L ds sin θ dθ = L θ sin θ cos θ Let θ = s/l dθ = ds/l = L g-cm. Since the wire nd the densit function re both smmetric bout the -is, we hve M = =. Hence, the centre of mss is locted t, L.. A = r M = = = r r d Let u = r du = d r u / du = u/ r = r = r r = r =ȳ b smmetr. r The centroid is, r. 8
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.5 PAGE 99 r = r Thus = ln +, nd ȳ = 8ln +. The centroid is ln +, 8ln +. d = + r Fig. 7.5.. B smmetr, =. A horizontl strip t hs mss dm = 9 d nd moment dm = = 9 d bout =. Thus, m = 9 9 9 d= 9 / = 6 Fig. 7.5. nd 9 M = = 9 d Let u = 9 udu= d = 9u u du = u 5 u5 = 68 5. Thus, ȳ = 68 5 6 = 8. Hence, the centroid is t 5, 8. 5 d 9 =9. The re of the sector is A = 8 r. Its moment bout = is r/ r M = = d + r/ r d = r 6 r r / = r. r/ Fig. 7.5.. The re nd moments of the region re A = M = = M = = / d + = sec θ dθ = ln sec θ + tn θ Let = tn θ d = sec θ dθ / d = + + = ln + d + = tn = = 8. Thus, = r 8 r = 8r. B smmetr, the centroid must lie on the line = tn =. 8 Thus, ȳ = 8r. = Fig. 7.5. r = r r 85
SECTION 7.5 PAGE 99 R. A. ADAMS: CALCULUS 5. B smmetr, =. We hve A = = d / cos θ dθ / = θ + sin θ cos θ = + = M = = d = 5 d = 5 d = + =. Let = sin θ d = cos θ dθ Thus ȳ = 9 = 9. The centroid is, 9. = =b Fig. 7.5.6 d 7. The qudrilterl consists of two tringles, T nd T, s shown in the figure. The re nd centroid of T re given b A = =, = + + = 7, ȳ = + + =. The re nd centroid of T re given b A = =, = + + =, ȳ = + =. It follows tht M,= = 7 = M,= = = M,= = = 8 M,= = = 8. Since res nd moments re dditive, we hve for the whole qudrilterl A = + = 6, M = = + 8 = 8, M = = 8 =. Fig. 7.5.5 6. B smmetr, =. The re is A = b. The moment bout = is Thus = 8 6 = 9, nd ȳ = 9 6 =. The centroid 9 of the qudrilterl is 9,., M = = Thus, ȳ = b = b ] b [ d = b = b. b = b. d T T, Fig. 7.5.7 86
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.5 PAGE 99 8. The region is the union of hlf-disk nd tringle. The centroid of the hlf-disk is known to be t, nd tht of the tringle is t,. The re of the semicircle is nd the tringle is. Hence, M = = + = + 8 ; 6 M = = + =. Since the re of the whole region is +, then = + 8 + nd ȳ = +. = = Fig. 7.5.8 9. A circulr strip of the surfce between heights nd + d hs re ds = d cos θ = r d = r d. The totl surfce re is S = r r d = r. The moment bout = is r M = = r d= r r = r. Thus ȳ = r r = r. B smmetr, the centroid of the hemisphericl surfce is on the is of smmetr of the hemisphere. It is hlfw between the centre of the bse circle nd the verte. ds θ Fig. 7.5.9 r θ,. B smmetr, =ȳ =. The volume is V = r. A thin slice of the solid t height z will hve volume dv = dz = r z dz. Thus, the moment bout z = is r M z= = zr z dz r z = z r = r. Thus, z = r r = r. Hence, the centroid is 8 on the is of the hemisphere t distnce r/8 from the bse. z Fig. 7.5. = r z. The cone hs volume V = r h. See the following figure. The disk-shped slice with verticl width dz hs rdius = r z, nd therefore hs volume h We hve dv = r z h M z= = r h = r h = r h h h dz = r zh z dz h uu du hu u h dz r h h z dz. Let u = h z du = dz = r h. Therefore z = r h r h = h. The centroid of the solid cone is on the is of the cone, t distnce bove the bse equl to one qurter of the height of the cone. 87
SECTION 7.5 PAGE 99 R. A. ADAMS: CALCULUS z h Thus ȳ = 8, nd the centroid is, 8. z dz =r h z =sin r / Fig. 7.5. Fig. 7.5.. A bnd t height z with verticl width dz hs rdius = r z, nd hs ctul slnt width h Its re is ds = + d dz = + r dz h dz. da= r z + r h h dz. Thus the re of the conicl surfce is A = r + r h h The moment bout z = is M z= = r + r h h = r + r z h z dz = r r h + h. z z dz h z h h = r rh + h. Thus, z = rh r + h r r + h = h. B smmetr, =ȳ =. Hence, the centroid is on the is of the conicl surfce, t distnce h/ from the bse.. B smmetr, =. The re nd -moment of the region re given b A = sin d= M = = sin d = sin cos =.. The re of the region is / A = cos d= sin The moment bout = is M = = / cos d / =. U = dv = cos d du = d V = sin / / = sin sin d=. Thus, =. The moment bout = is / M = = cos d = + / sin = 8. Thus, ȳ = 8. The centroid is, 8 =cos d Fig. 7.5. 5. The rc hs length L = r. B smmetr, = ȳ. An element of the rc between nd + d hs length ds = d sin θ = rd. = rd r. 88
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.5 PAGE 99 Thus M = = r r d r r = r r Hence = r r = r, nd the centroid is r r θ + =r ds +d r Fig. 7.5.5 = r. r, r 6. The solid S in question consists of solid cone C with verte t the origin, height, nd top circulr disk of rdius, nd solid clinder D of rdius nd height sitting on top of the cone. These solids hve volumes V C = /, V D =, nd V S = V C + V D = 6/. B smmetr, the centroid of the solid lies on its verticl is of smmetr; let us continue to cll this the - is. We need onl determine ȳ S. Since D lies between = nd =, its centroid stisfies ȳ D = /. Also, b Eercise, the centroid of the solid cone stisfies ȳ C = /. Thus C nd D hve moments bout = : M C,= = =, M D,= =. = 6. Thus M S,= = + 6 = 7, nd z S = 7/6/ = /6. The centroid of the solid S is on its verticl is of smmetr t height /6 bove the verte of the conicl prt. 7. The region in figure is the union of rectngle of re nd centroid, / nd tringle of re nd centroid /, /. Therefore its re is nd its centroid is, ȳ, where = + ȳ = + = 8 =. Therefore, the centroid is 8/9, /9. 8. The region in figure b is the union of squre of re = nd centroid, nd tringle of re / nd centroid /, /. Therefore its re is 5/ nd its centroid is, ȳ, where 5 = + =. Therefore, = ȳ = /5, nd the centroid is /5, /5. 9. The region in figure c is the union of hlf-disk of re / nd centroid, / b Emple nd tringle of re nd centroid, /. Therefore its re is / + nd its centroid is, ȳ, where = nd + ȳ = + =. Therefore, the centroid is, /[ + ].. The region in figure d is the union of three hlf-disks, one with re / nd centroid, /, nd two with res /8 nd centroids /, / nd /, /. Therefore its re is / nd its centroid is, ȳ, where = + + 8 8 ȳ = + 8 + 8 Therefore, the centroid is, /.. B smmetr the centroid is,.,, Fig. 7.5. = = =. =. The line segment from, to, hs centroid, nd length. B Pppus s Theorem, the surfce re of revolution bout = is A = = sq. units. r Fig. 7.5. 89
SECTION 7.5 PAGE 99 R. A. ADAMS: CALCULUS. The tringle T hs centroid, nd re. B Pppus s Theorem the volume of revolution bout = is V = = 5 cu. units. T = Fig. 7.5.. The ltitude h of the tringle is s. Its centroid is t height h = s bove the bse side. Thus, b Pppus s Theorem, the volume of revolution is s V = s s = s cu. units. The centroid of one side is h = s bove the bse. Thus, the surfce re of revolution is s S = s = s sq. units. 6. The region bounded b = nd = lnsin between = nd = / lies below the -is, so A = = A ȳ = A / / / lnsin d.8879 lnsin d.9 lnsin d.9986. 7. The re nd moments of the region re A = M = = = M = = d + = lim R + d + Let u = + du = d u du u = lim R u + R u The centroid is, 5. d + 6 = lim R = + R = = + 5 = R =. s h Fig. 7.5. 5. For the purpose of evluting the integrls in this problem nd the net, the definite integrl routine in the TI-85 clcultor ws used. For the region bounded b = nd = cos between = nd = /, we hve A = / = A ȳ = A / cos d.78 / / cos d.777 cos d.65. s 8. The surfce re is given b S = Fig. 7.5.7 e + e d. Since lim + ± e =, this epression must be bounded for ll, tht is, + e K for some constnt K. Thus, S K e d = K. The integrl converges nd the surfce re is finite. Since the whole curve = e lies bove the -is, its centroid would hve to stisf ȳ >. However, Pppus s Theorem would then impl tht the surfce of revolution would hve infinite re: S = ȳ length of curve =. The curve cnnot, therefore, hve n centroid. 9
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.6 PAGE 6 9. B nlog with the formuls for the region b, f g, the region c d, f g will hve centroid M = /A, M = /A, where A = M = = M = = d c d c d c g f d [ ] g f d g f d.. Let us tke L to be the -is nd suppose tht plne curve C lies between = nd = b where < < b. Thus, r =, the -coordinte of the centroid of C. Let ds denote n rc length element of C t position. This rc length element genertes, on rottion bout L, circulr bnd of surfce re ds = ds, so the surfce re of the surfce of revolution is =b S = ds= M = = rs. =. Tringle LMN hs re + tn t, nd the -coordinte of its centroid is LMN sec t sec t + + tn t sin t + sec t + tn t sin t = sin t sec t =. Tringle LNP hs re tn t, nd the -coordinte of its centroid is sec t + sec t + sec t + tn t sin t LNP = sec t + tn t sin t =. Therefore, LMNP = [ sin t sec t + tn t 6 ] + sec t + sin t sin t tn t tn t = [ ] sin t sec t tn t + sin t tn t 6 = sin t [ ] 6 cos t + sin t cos t = sin t [ ] 6 cos cos t + sin t t = sin t [ ] 6 cos cos t = sin t [ ] t 6 cos cost t L t / t t / t P N which is positive provided < t </. Thus the bem will rotte counterclockwise until n edge is on top. Section 7.6 pge 6 Other Phsicl Applictions M Fig. 7.5. We need to find the -coordinte LMNP of the centre of buonc, tht is, of the centroid of qudrilterl LMNP. From vrious tringles in the figure we cn determine the -coordintes of the four points: L = sec t, P = sec t, M = sec t + + tn t sin t N = sec t + tn t sin t. The pressure t the bottom is p = 9, 8 6 N/m. The force on the bottom is p = 5, N. b The pressure t depth h metres is 9, 8h N/m. The force on strip between depths h nd h + dh on one wll of the tnk is df = 9, 8h dh = 9, 6 hdh N. Thus, the totl force on one wll is F = 9, 6 6 hdh= 9, 6 8 = 5, 8 N. 9
SECTION 7.6 PAGE 6 R. A. ADAMS: CALCULUS dh h 6m θ h h+dh 6 m Fig. 7.6. m. A verticl slice of wter t position with thickness d is in contct with the botttom over n re 8 sec θ d = 5 d m, which is t depth = + m. The force eerted on this re is then df = ρg + 5 d. Hence, the totl force eerted on the bottom is F = ρg 5 + d = 9.8 5 +.56 6 N. = + d Fig. 7.6.. A strip long the slnt wll of the dm between depths h nd h + dh hs re da= dh cos θ The force on this strip is = 6 dh. df = 9, 8 hda. 6 hdh N. Thus the totl force on the dm is F =. 6 hdh 6. 8 N. θ Fig. 7.6.. The height of ech tringulr fce is m nd the height of the prmid is m. Let the ngle between the tringulr fce nd the bse be θ, then sin θ = nd cos θ =. θ d = + side view of one fce Fig. 7.6. θ front view of one fce dsec θ= d Fig. 7.6. A verticl slice of wter with thickness d t distnce from the verte of the prmid eerts force on the shded strip shown in the front view, which hs re d m nd which is t depth + m. Hence, the force eerted on the tringulr fce is 6 F = ρg + d = [ 98 + 5 ] 6.95 5 N. 9
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.6 PAGE 6 5. The unblnced force is F = 9, 8 5 6 hdh h = 9, 8 5 8.9 6 N. 6 The work done to empt the pool is W = ρg hah dh [ ] = ρg 6hdh+ h 8h dh = 98 [8h + h 8 ] h =.97 6 N m. 5m 8 m h 6m Ah Fig. 7.6.5 6. The spring force is F = k, where is the mount of compression. The work done to compress the spring cm is N cm = W = k d = k = 9 k. Hence, k = N/cm. The work necessr to compress 9 the spring further cm is W = k d = 9 = 7 9 N cm. 7. A ler of wter in the tnk between depths h nd h +dh hs weight df = ρgdv = ρgdh. The work done to rise the wter in this ler to the top of the tnk is dw = hdf = ρgh dh. Thus the totl work done to pump ll the wter out over the top of the tnk is Fig. 7.6.8 9. A ler of wter between depths nd + d hs volume dv = d nd weight df = 9, 8 d N. The work done to rise this wter to height h m bove the top of the bowl is dw = h + df = 9, 8h + d N m. Thus the totl work done to pump ll the wter in the bowl to tht height is W = 9, 8 h + h d = 9, 8 [h + ] h [ ] h = 9, 8 + + 8h = 9, 8 = 5 + 8h N m. W = ρg 6 hdh= 9, 8 8 7.56 5 N m. d 8. The horizontl cross-sectionl re of the pool t depth h is { 6, if h ; Ah = 8h, if < h. Fig. 7.6.9 9
SECTION 7.6 PAGE 6 R. A. ADAMS: CALCULUS. Let the time required to rise the bucket to height h m be t minutes. Given tht the velocit is m/min, then t = h. The weight of the bucket t time t is 6 kg kg/mint min = 6 h kg. Therefore, the work done required to move the bucket to height of mis W = g 6 h dh = 9.8 6h h = N m. 6. The present vlue of continuous pments of $, per er for ers t discount rte of 5% is V =,e.5t dt =, = $7,869.9..5 e.5t 7. The present vlue of continuous pments of $, per er for ers beginning ers from now t discount rte of 8% is V =,e.8t dt =,.8 e.8t = $5,865.6. Section 7.7 Applictions in Business, Finnce, nd Ecolog pge 9. Cost = $, + = $,., 6 + 6 6 d. The number of chips sold in the first er ws, 5 tht is, bout 96,58.. The monthl chrge is te t/ dt =, 6, e 6/5 + dt let t = u t u =8 + u du = 8 du + u =$8 ln +. 8. The present vlue of continuous pments of $, per er for 5 ers beginning ers from now t discount rte of 5% is V = 5,e.5t dt =, 5.5 e.5t = $8,655.. 9. The present vlue of continuous pments of $, per er for ll future time t discount rte of % is V =,e.t dt =, = $5,... The present vlue of continuous pments of $, per er beginning ers from now nd continuing for ll future time t discount rte of 5% is V =,e.5t dt =,.5 e.5 = $,.6.. The price per kg t time t ers is $ + 5t. Thus the revenue per er t time t is + 5t/ +.t $/er. The totl revenue over the er is + 5t +.t dt $, 75.7. 5. The present vlue of continuous pments of $, per er for ers t discount rte of % is V =,e.t dt =,. e.t = $9,6.6.. After t ers, mone is flowing t $, + t per er. The present vlue of ers of pments discounted t 5% is V = + te.5t dt U = + t dv = e.5t dt du = dt V = e.5t.5 = + t e.5t.5 + e.5t dt.5 = 6. + = $, 77.5..5 e.5t 9
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.7 PAGE 9. After t ers, mone is flowing t $,. t per er. The present vlue of ers of pments discounted t 5% is V =, e t ln. e.5t dt, = ln..5 etln..5. The mount fter ers is A = 5, e.5t dt = 5,.5 e.5t = $, 65.. = $6,87... Let T be the time required for the ccount blnce to rech $,,. The $5,. t dt deposited in the time intervl [t, t + dt] grows for T t ers, so the blnce fter T ers is T 5,. t.6 T t dt =,, T. t.6 T,, dt = =.6 5, [.6 T. T ] = ln./.6.6. T.6 T = ln..6. This eqution cn be solved b Newton s method or using clcultor solve routine. The solution is T 6.5 ers. 5. Let Pτ be the vlue t time τ < t tht will grow to $P = Pt t time t. If the discount rte t time τ is δτ, then d Pτ = δτpτ, dτ or, equivlentl, dpτ Pτ = δτdτ. Integrting this from to t, we get t ln Pt ln P = δτdτ = λt, nd, tking eponentils of both sides nd solving for P, we get P = Pte λt = Pe λt. The present vlue of strem of pments due t rte Pt t time t from t = tot = T is T t Pte λt dt, where λt = δτdτ. 6. The nlsis crried out in the tet for the logistic growth model showed tht the totl present vlue of future hrvests could be mimized b holding the popultion size t vlue tht mimizes the qudrtic epression Q = k δ. L If the logistic model d/dt = k /L is replced with more generl growth model d/dt = F, ectl the sme nlsis leds us to mimize Q = F δ. For relistic growth functions, the mimum will occur where Q =, tht is, where F = δ. 7. We re given L = 8,, k =., nd δ =.5. According to the nlsis in the tet, the present vlue of future hrvests will be mimized if the popultion level is mintined t = k δ L k =.7 8, =,.. The nnul revenue from hrvesting to keep the popultion t this level given price of $6 per fish is 6.,.,. = $, 9. 8, 8. We re given tht k =., L = 5,, p = $,. The growth rte t popultion level is d dt =.. 5, The mimum sustinble nnul hrvest is d dt =.75,.5 = 75 whles. =L/ b The resulting nnul revenue is $75p = $7, 5,. c If the whole popultion of 75, is hrvested nd the proceeds invested t %, the nnul interest will be 75, $,. = $5,,. 95
SECTION 7.7 PAGE 9 R. A. ADAMS: CALCULUS d At 5%, the interest would be 5/$5, = $7, 5,. e The totl present vlue of ll future hrvesting revenue if the popultion level is mintined t 75, nd δ =.5 is e.5t 7, 5, dt = 7, 5,.5 = $5,,. 9. If we ssume tht the cost of hrvesting unit of popultion is $C when the popultion size is, then the effective income from unit hrvested is $p C. Using this epression in plce of the constnt p in the nlsis given in the tet, we re led to choose to mimize [ Q = p C k ] δ. L A resonble cost function C will increse s decreses the whles re hrder to find, nd will eceed p if, for some positive popultion level. The vlue of tht mimizes Q must eceed,sothe model no longer predicts etinction, even for lrge discount rtes δ. However, the optimizing popultion m be so low tht other fctors not ccounted for in the simple logistic growth model m still bring bout etinction whether it is economicll indicted or not. Section 7.8 Probbilit pge. The epected winnings on toss of the coin re $.9 + $.9 + $5. = $.7. If ou p this much to pl one gme, in the long term ou cn epect to brek even.. We need 6 n= Kn =. Thus K =, nd K = /. b PrX = / + + = /7.. From the second previous Eercise, the men winings is µ = $.7. Now σ =.9 +.9 +,5. µ 5.5 6. = 6.5. The stndrd devition is thus σ $6.8.. Since PrX = n = n/, we hve µ = σ = 6 n= nprx = n = + + +6 6 6 n PrX = n µ = + + +6 µ n= = 69 9 = 9. σ =.9. 5. The men of X is µ = 9 6 + + + + 5 + 6 6.58. 6 The epecttion of X is =. EX = 9 6 + + + +5 6 +6 6 5.75. Hence the stndrd devition of X is 5.75.58.759. Also PrX = 9 6 + 6 = 9 6.8. 6. Clculting s we did to construct the probbilit function in Emple, but using the different vlues for the probbilities of nd 6, we obtin f = 9 6 9 6.5 f = 9 6 =.5 6 f = 9 6 6 + 6 =.778 f 5 = 9 6 6 + 6 =.56 f 6 = 9 6 6 + 6 =. f 7 = 9 6 6 + 6 =.66 f 8 = 6 6 + 6 =. f 9 = 6 6 + 6 =.67 f = 6 6 + 6 =.889 f = 6 =.6 6 f = 6 =.6. 6 96
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.8 PAGE b Multipling ech vlue f n b n nd summing, we get µ = nfn 7.665. Similrl, n= EX = n f n 57.78, n= so the stndrd devition of X is σ = EX µ.. The men is somewht lrger thn the vlue 7 obtined for the unweighted dice, becuse the weighting fvours more 6s thn s showing if the roll is repeted mn times. The stndrd devition is just tin bit smller thn tht found for the unweighted dice.5; the distribution of probbilit is just slightl more concentrted round the men here. 7. The smple spce consists of the eight triples H, H, H, H, H, T, H, T, H, T, H, H, H, T, T, T, H, T, T, T, H, nd T, T, T. b We hve b The smple spce for the three bll selection consists of ll eight triples of the form,, z, where ech of,, z is either Red or Blue. Let X be the number of red blls mong the three blls pulled out. Arguing in the sme w s in, we clculte PrX = = PrB, B, B = 8 7 9 6 8 = 85.9 PrX = = PrR, B, B + PrB, R, B + PrB, B, R = 8 9 7 8 = 8 95.97 PrX = = PrR, R, B + PrR, B, R + PrB, R, R = 9 8 8 = 95.6 PrX = = PrR, R, R = 9 8 = 57.9 Thus the epected vlue of X is EX = 85 + 8 95 + 95 + 57 = 9 5 =.8. PrH, H, H =.55 =.6675 PrH, H, T = PrH, T, H = PrT, H, H =.55.5 9. We hve f = C on [, ]. C is given b =.65 PrH, T, T = PrT, H, T = PrT, T, H =.55.5 = C d = C =.75 = 9 C. PrT, T, T =.5 =.95. c The probbilit function f for X is given b Hence, C = 9. b The men is f =.5 =.95 f =.55.5 =.5 f =.55.5 =.875 f =.55 =.6675. d PrX = PrX = =.98875. e EX = f + f + f + f =.65. 8. The number of red blls in the sck must be.6 =. Thus there re 8 blue blls. The probbilit of pulling out one blue bll is 8/. If ou got blue bll, then there would be onl 7 blue blls left mong the 9 blls remining in the sck, so the probbilit of pulling out second blue bll is 7/9. Thus the probbilit of pulling out two blue blls is 8 7 9 = 95. µ = EX = 9 Since EX = 9 vrince is d = 7 =. d = 6 σ = EX µ = 9 =, nd the stndrd devition is σ = /. c We hve Prµ σ X µ + σ = 9 = µ + σ µ σ 9 = µσ 9 µ+σ µ σ = 9, the d.685. 97
SECTION 7.8 PAGE R. A. ADAMS: CALCULUS c We hve. We hve f = C on [, ]. To find C, wehve = C d = C = C. µ+σ Prµ σ X µ + σ = d µ σ = µ + σ µ σ = + 8.668. 8 Hence, C =. b The men is µ = EX = Since EX = vrince is d = 9 d = 6 = 9.556. = 5, the σ = EX µ = 5 96 8 = 6 nd the stndrd devition is σ = 6.8. c We hve Prµ σ X µ + σ = = µ + σ µ σ. We hve f = C on [, ]. C is given b = = µσ µ+σ µ σ C d = C = C. d.5875. Hence, C =. b The men, vrince, nd stndrd devition re µ = EX = σ = EX µ = σ = /8. d = d 9 6 = 5 9 6 = 8. We hve f = C sin on [,]. To find C, we clculte = C sin d= Ccos Hence, C =. b The men is Since = C. µ = EX = sin d U = dv = sin d du = d V = cos = [ ] cos + cos d = =.57. EX = sin d U = dv = sin d du = d V = cos = [ ] cos + cos d U = dv = cos d du = d V = sin = [ ] + sin sin d =. Hence, the vrince is σ = EX µ = = 8.67 nd the stndrd devition is 8 σ =.68. 98
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.8 PAGE c Then µ+σ Prµ σ X µ + σ = µ σ = [ ] cosµ + σ cosµ σ = sin µ sin σ = sin σ.6.. We hve f = C on [, ]. C is given b = sin d C d = C = C 6. Hence, C = 6. b The men, vrince, nd stndrd devition re c We hve µ = EX = 6 σ = EX µ = 6 = = σ = /. d = d /+σ Prµ σ X µ + σ = 6 d / σ [ /+σ = 6 / σ ] d Let u = du = d σ [ ] [ σ = u du = σ ] = [ ].66. 6. It ws shown in Section 6. p. 9 tht n e d = n e + n n e d. If I n = n e d, then I n = lim R Rn e R + ni n = ni n if n. Since I = e d =, therefore I n = n! for n. Let u = k; then n e k d = k n+ Now let f = Ce k on [,. To find C, observe tht = C u n e u du = k n+ I n = n! k n+. e k d = C k. Hence, C = k. b The men is µ = EX = k e k d = k k = k. 6 Since EX = k e k d = k k = 6 k, then the vrince is σ = EX µ = 6 k k = k nd the stndrd devition is σ = k. c Finll, 5. We hve Prµ σ X µ + σ µ+σ = k e k d Let u = k µ σ du = kd kµ+σ = ue u du kµ σ kµ+σ = ue u + kµ σ kµ+σ kµ σ e u du = + e + + e e + + e.78. = C Thus C = /. e d = C e d = C. 99
SECTION 7.8 PAGE R. A. ADAMS: CALCULUS b The men, vrince, nd stndrd devition re µ = e d = e σ = + e d σ = c We hve U = du = d = + dv = e d V = e = e + e d = + + =.6. Prµ σ X µ + σ = µ+σ = Let = z/ d = dz/ µ+σ µ σ e z / dz. µ σ e d But µ σ.95 nd µ + σ.. Thus, if Z is stndrd norml rndom vrible, we obtin b interpoltion in the tble on pge 86 in the tet, Prµ σ X µ + σ = Pr.95 Z..99.577.68. 6. No. The identit constnt C. 7. f µ,σ = σ e µ /σ men = σ Cd = is not stisfied for n e µ /σ d Let z = µ σ dz = σ d = µ + σ ze z / dz = µ e z / dz = µ vrince = E µ = σ µ e µ /σ d = σ σ z e z / dz = σ VrZ = σ 8. Since f = + > on[, nd d + = lim R tn R = =, therefore f is probbilit densit function on [,. The epecttion of X is µ = EX = d + = lim R ln + R =. No mtter wht the cost per gme, ou should be willing to pl if ou hve n dequte bnkroll. Your epected winnings per gme in the long term is infinite. 9. The densit function for the uniform distribution on [, b] isgivenb f = /b, for b. B Emple 5, the men nd stndrd devition re given b µ = b +, σ = b. Since µ + σ = b + + b > b, nd similrl, µ σ <, therefore Pr X µ σ =. b For f = ke k on [,, we know tht µ = σ = Emple 6. Thus µ σ < nd k µ + σ =. We hve k Pr X µ σ = Pr X k = k /k = e k e k d /k = e.5. c For f µ,σ = σ e µ /σ, which hs men µ nd stndrd devition σ,we hve Pr X µ σ = PrX µ σ µ σ = σ e µ /σ d Let z = µ σ dz = σ d = e z dz = PrZ. =.6
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.9 PAGE 9 from the tble in this section.. The densit function for T is f t = ke kt on [,, where k = µ = PrT = see Emple 6. Then e t/ dt = = + e t/ = e /.59. e t/ dt The probbilit tht the sstem will lst t lest hours is bout.59.. If X is distributed normll, with men µ = 5,, nd stndrd devition σ =, then PrX 55 = e 5 / d 55 Let z = 5 dz = d = e z / dz 5/ = PrZ 5/ = PrZ 5/.6 from the tble in this section.. If X is the rndom vrible giving the spinner s vlue, then PrX = / = / nd the densit function for the other vlues of X is f = /. Thus the men of X is. µ = EX = Pr X = + Also, EX = 6 Pr X = + σ = EX µ = 9 96 9 6 = 9. Thus σ = /9. Section 7.9 First-Order Differentil Equtions pge 9 d d = d = d ln = ln + C = C f d = 8 + = 8. f d = + 6 = 9 96... 5. 6. 7. 8. d d = d = d ln =ln + ln C = C = + C. d d = d = d = + C, or = C d d = d = d = + C = + C. dy dt = ty dy Y = tdt ln Y = t + C, or Y = Ce t / d = e sin t dt e d = sin tdt e = cos t C = lncos t + C. d d = d = d + + d = d ln + = + C + = Ce or = Ce Ce + d d = + d + = d tn = + C = tn + C.
SECTION 7.9 PAGE 9 R. A. ADAMS: CALCULUS 9. d = + e dt e d e + = dt lne + = t + C d + e = dt e + = C e t, or = ln. We hve. d d = d = d = + K. Epnd the left side in prtil frctions: Hence, Therefore, = A + B + C = A + B + C { A + C = ; A B = ; B =. Ce t A = B = C =. d = + + d = ln ln. ln = + K. d d = liner µ = ep d = d d = d d = = + C, so = + C. We hve d d + µ =. Let = d = ln = ln, then e µ =, nd d d = d d + d = d + = = + C. = d = + C =. µ = ep d d + = d d e = e + = e d = e e = e + C = + Ce. We hve d d + = e. Let µ = d =, then e µ = e, nd 5. d d e = e d d d + e = e d + = e e = e d = e + C. Hence, = e + Ce. d d + = µ = ep d = e d d e = e + = e e = e d = e e + C = + Ce 6. We hve d d + e = e. Let µ = e d = e, then Therefore, d e e e d = e d d + e e e = e e d d + e e e = e e e d = Hence, = + Ce e. = e e e. Let u = e du = e d e u du = ee + C.
INSTRUCTOR S SOLUTIONS MANUAL SECTION 7.9 PAGE 9 7. 8. d dt + =, = µ = dt = t d dt et t d = e dt + et = e t e t t = et + C = e = e + C C = e = + e t. d d + =, = µ = d = d d e d = e d + e = e e = e d = e + C = = + C C = = + e. 9. + = e /, = e + = e/ µ = d = d e / = e / + d = e / = d = + C = e = + C C = = + e /.. + cos = e sin, = µ = cos d= sin d d esin = e sin + cos = e sin = d= + C = = + C C = = e sin.. = + d d =, = + C t dt t = i.e. d= d = + C C = = +. t. = + dt = + t d d = +, i.e. d/ = d/ + = tn + C = + C C = = / tn.. = + t dt tt + = d d = +, for > d = d + = d d + ln = ln + + ln C = C, + = C/ = +.. = + e dt = d d = e, i.e. e d = d e = + C = ln + C = = ln C C = e = ln + e. 5. Since > b > nd k >, b e b kt lim t = lim t t be b kt b = = b. 6. Since b > > nd k >, b e b kt lim t = lim t t be b kt e bkt b = lim t b e bkt b = =. b
SECTION 7.9 PAGE 9 R. A. ADAMS: CALCULUS 7. The solution given, nmel = b e b kt be b kt, is indeterminte / if = b. If = b the originl differentil eqution becomes d dt = k, which is seprble nd ields the solution = d = k dt = kt + C. Since =, we hve C =,so = kt +. Solving for, we obtin = kt + kt. This solution lso results from evluting the limit of solution obtined for the cse = b s b pproches using l Hôpitl s Rule, s. 8. Given tht m dv dt = mg kv, then dv g k = m v dt m k ln g k m v = t + C. Since v =, therefore C = m k ln g. Also, g k m v remins positive for ll t >, so m k ln g g k = t m v g k m v = e kt/m g v = vt = mg e kt/m. k mg Note tht lim vt =. This limiting velocit cn be t k obtined directl from the differentil eqution b setting dv =. dt 9. We proceed b seprtion of vribles: m dv = mg kv dt dv = g k dt m v dv g k = dt m v dv mg = k k v m dt = kt m + C. Let = mg/k, where >. Thus, we hve dv v = kt m + C ln + v v = kt m + C ln + v v = kt kg m + C = m t + C + v v = C e tkg/m. Assuming v =, we get C =. Thus + v = e tkg/m v v + e t kg/m = e t kg/m mg = e t kg/m k mg e tkg/m v = k e tkg/m + mg Clerl v s t. This lso follows from k setting dv = in the given differentil eqution. dt. The blnce in the ccount fter t ers is t nd =. The blnce must stisf d =. dt,, d = 5 dt 6 d dt 5 = 6 5 + 5 d = ln ln 5 = t C 5 = = e C t/ 5 e C t/ +. t 6 C 5
INSTRUCTOR S SOLUTIONS MANUAL REVIEW EXERCISES 7 PAGE Since =, we hve nd = = 5 e C + = The blnce fter er is = 5 99e t/ +. C = ln 99, 5 99e / $,.. + As t, the blnce cn grow to 5 lim t = lim t t e.6.t + = 5 = $,. + For the ccount to grow to $5,, t must stisf, 5, = t = 99e t/ + 99e t/ + = t = ln 99 6 ers.. The hperbols = C stisf the differentil eqution + d d =, or d d =. Curves tht intersect these hperbols t right ngles must therefore stisf d d =,ord = d, seprted eqution with solutions = C, which is lso fmil of rectngulr hperbols. Both fmilies re degenerte t the origin for C =.. Let t be the number of kg of slt in the solution in the tnk fter t minutes. Thus, = 5. Slt is coming into the tnk t rte of g/l L/min =. kg/min. Since the contents flow out t rte of L/min, the volume of the solution is incresing t L/min nd thus, t n time t, the volume of the solution is + t L. Therefore the concentrtion of slt is L. Hence, slt is being t + t removed t rte t + t Therefore, kg/l L/min = 5t 5 + t d =. 5 dt 5 + t d dt + 5 5 + t =.. kg/min.. Let µ = 5 5 + t dt = 5ln 5 + t =ln5 + t5 for t >. Then e µ = 5 + t 5, nd d [ ] 5 + t 5 = 5 + t 5 d dt d + 55 + t d = 5 + t 5 d + 5 5 + t =.5 + t 5. Hence, 5 + t 5 =. 5 + t 5 dt =.5 + t 6 + C =.5 + t + C5 + t 5. Since = 5, we hve C =.5 5 nd =.5 + t +.5 5 5 + t 5. After min, there will be =.5 +.5 5 5 5 = 8. kg of slt in the tnk. Review Eercises 7 pge cm 5cm cm cm Fig. R-7. cm cm cm5cm The volume of thred tht cn be wound on the left spool is 5 = cm. The height of the winding region of the right spool t distnce r from the centrl is of the spool is of the form h = A + Br. Since h = ifr =, nd h = 5if r =, we hve A = nd B =, so h = + r. The volume of thred tht cn be wound on the right spool is r + r dr = r + r The right spool will hold thred. = cm., = 8. m of 5