Math 1105: Calculus II (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 5


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1 Mth 5: Clculus II Mth/Sci mjos) MWF m / pm, Cmpion 35 Witten homewok 5 6.6, p ,33), 6.7, p ,3), 6.875), 7.58,6,6), 7.44,48) Fo pctice not to tun in): 6.6, p. 458,8,,3,4), 6.7, p ,6,8), 7.6,3,38), 7.8,,47,59,7,78) 6.6, p. 458, Poblem 3. When the cicle x + y ) =, whee < <, is evolved bout the xxis the esult is the sufce of tous. Show tht the sufce e of this tous is 4π. Solution. The cicle given is composed of two gphs ove [, ], one given by y = + x nd the othe given by Tking deivtives, we hve nd y = x. y = x ) / x x) = x, y = x x. This mens tht the sufce e fom y is given by π + x ) + x x = nd tht fom y is given by π x ) + x x = This implies tht the totl sufce e is given by the sum: π + x ) x π x ) x. π + x ) x + π x ) x = 4π x = 4π = 4π whee we ve pefomed the substitution u = x/ in the lst line. x du u = 4π sin u) = 4π, 6.6, p. 458, Poblem 33. Suppose sphee of dius is sliced by two hoizontl plnes h units pt. Show tht the sufce e of the esulting zone on the sphee is πh, independent of the loction of the cutting plnes. )
2 Solution. The sphee is the sufce obtined by evolving one hemisphee of the cicle centeed t, ) of dius bout the yxis, i.e. the gph of the function fy) = y bout the xxis. Note tht we hve f y) = y y y) = y. Suppose the given egion is fomed by plnes t heights y = nd y = + h, so tht its sufce e is given by +h π ) y y +h + dy = π y + y y y dy = = +h +h π y π dy = πy +h = π + h ) = πh. y dy 6.7, p. 467, Poblem 8. A cylindicl wte tnk hs height 8m nd dius m. ) If the tnk is full of wte, how much wok is equied to pump the wte to the level of the top of the tnk nd out of the tnk? ) Is it tue tht it tkes hlf s much wok to pump the wte out of the tnk when it is hlf full s when it is full? Explin. Solution ). Let y indicte the height in metes long the cylinde, whee y = is the bse. The wok equied to pump the wte out of the cylinde is given by 8 Ay)ρgDy) dy, whee Ay) is the e of slice t height y, ρ = kg/m 3 is the density of the wte, g = 9.8 m/s is the cceletion due to gvity, nd Dy) is the displcement tht the slice t height y must go though. Since the object is cylinde, Ay) = 4π m. Since wte t height y must be lifted to height y = 8, we hve Dy) = 8 y. We hve: 8 8 ) Ay)ρg8 y)dy = 39π 8 y dy = 39π 8y y 8 = 39π64 3) J. Note tht the units t the end e kg m /s, o joules J ). Solution b). If the tnk is only hlf full, then thee is only wte up to height of 4 m, nd ou clcultion becomes: 4 ) Ay)ρg8 y)dy = 39π 8y y 4 = 39π3 8) 9556 J. Since 3 8 = 4 is moe thn hlf of 64 3 = 3, moe thn hlf the wok is equied when the tnk is hlf full. 6.7, p. 467, Poblem 3. A swimming pool is m long nd m wide, with bottom tht slopes unifomly fom depth of m t one end to depth of m t the othe end. Assuming the pool is full, how much wok is equied to pump the wte to level. m bove the top of the pool?
3 Solution. Let y indicte the height in metes long the pool, whee y = is the lowest point t the deep end of the pool). As befoe, the wok equied to pump the wte out of the pool is given by Ay)ρgDy) dy, whee Ay) is the e of slice t height y, ρ = kg/m 3, g = 9.8 m/s, nd Dy) is the displcement tht the slice t height y must go though. In this poblem, Dy) = y +. =. y metes, since the wte must be lifted. m bove the sufce of the pool. The e Ay) evidently depends on whethe y is less o gete thn. When y, the e is given by Ay) = m, fixed constnt. As y goes fom to, the length of the slice t height y inceses t constnt te fom to. Thus the length of the slice t height y is given by y fo y between nd. This implies tht Ay) = y m fo y. We my now clculte tht the wok equied is given by: y 9.8. y) dy y) dy ) =.96 6.y y dy +. y dy.y ) =.96 6 y3 ) ) +.y y 3 = ) n+ 6.8, p. 48, Poblem 75. Use left Riemnn sum to ppoximte with unit spcing x between gid points) to show tht > lnn + ). Use this fct to conclude tht 3 n lim + n ) does not exist. n n+ Solution. In ode to build left Riemnn sum fo the integl, we bek up the intevl x [, n + ] into the ptition = x x x... x n x n = n +, whee x k = k + fo ech k =,..., n. On the subintevl [x k, x k ] i.e. [k, k + ]) we let x k = x k = k, the left endpoint. Now the left endpoint Riemnn sum with these choices is given by n n fx k) x k = k, k= k= which is the ptil sum fo the hmonic seies. n Becuse f x) = /x < fo ll x >, the function fx) = /x is stictly decesing fo x >. This implies tht the left endpoint Riemnn sum oveestimtes the ctul integl, so tht n n+ k > = log x n+ = logn + ). x The limit lim logn + ) n doesn t exist becuse the function logn + ) gets bitily lge s n goes to tht is, the limit exists nd equls infinity something we will discuss moe in the coming months). Moeove, the sum k=
4 is gete thn logn + ), so we conclude tht the sum my be bitily lge. Thus 3 n the limit lim + n ) n is lso infinite o doesn t exist, depending on nomencltue). 7., p. 55, Poblem 58. ) Show tht = x x sin x ) + C using eithe u = x o u = x. b) Show tht = x x sin x + C using u = x. c) Pove the identity sin x sin x ) = π. Solution ). Since x x = + x 4 4 x ) = x 4 ), we hve = x x ) =. x x ) 4 Letting u = x so tht du =, we hve = du = x x u sin u) + C = sin x ) + C. Solution b). We hve = x x x x. Letting u = x so tht u = x nd du = / x), we hve = du = x x u sin u) + C = sin x + C. Solution c). diffeent wys. By pt ), x Using pts ) nd b) we my compute the definite integl x dt t t dt t t = sin t ) x = sin x ) sin ) = sin x ) + π. On the othe hnd, by pt b), x dt = t t sin t x = sin x sin ) = sin x. This implies tht nd we hve the identity sin x ) + π = sin x, sin x sin x ) = π. in two 7., p. 55, Poblem 6. Find the e of the entie egion bounded by the cuves y = x 3 /x +) nd y = 8x/x + ). Solution. Let fx) = x 3 /x +) nd gx) = 8x/x +). The two cuves e equl when fx) = gx) which occus when x 3 = 8x. This eqution yields = xx )x + ), 4
5 whose el solutions e x = nd ±. Note tht the egion enclosed by these two cuves occus between nd. In ode to check which cuve is gete on ech intevl [, ] nd [, ], we choose the points in, ) nd in, ). Since f ) = ) 3 / ) + ) = /, g ) = 8 )/ ) + ) = 4, f) = / nd g) = 4, we hve fx) > gx) fo x in, ) nd fx) < gx) fo x in, ). Thus the given e cn be computed vi x 3 8x fx) gx) + gx) fx) = x + + 8x x 3 x + = 8x x 3 x +. The lst equlity holds by doing the substitution u = x nd du = ) in the fist integl, so tht it becomes exctly equl to the second integl. In ode to compute this lst integl, we must find n ntideivtive fo the function x 3 8x)/x + ). We hve: Thus we compute: 8x x 3 x + = xx + ) + 9x x + x 3 8x x + = x + 9x x + = x + 9x x +. = x + 9 x x +. Pefoming the substitution u = x + so tht du = x), we hve: x 3 8x = x x du u = x + 9 log u + C = x + 9 logx + ) + C. Thus the e is given by 8x x 3 x + = x + 9 ) logx + ) = log8 + ) + 9 ) log + ) = 9 log , p. 55, Poblem 6. Conside the egion R bounded by the gph of fx) = /x + ) nd the xxis on the intevl [, 3]. ) Find the volume of the solid fomed when R is evolved bout the xxis. b) Find the volume of the solid fomed when R is evolved bout the yxis. Solution ). The slices pependicul to the xxis e cicles of e π/x + ), so tht the totl volume is given by 3 π x + ) = π ) 3 = π x ) = 3π. Solution b). The slice fomed by the set of points of distnce x fom the y xis is cylinde of dius x nd height /x + ). Using the cylindicl shell method, the volume is given by 3 5 x u π = π du = π x + ) u u u du = πlog u + 5 u ) = π log ) log 5 = π log 5 3 ) 5
6 7., p. 5, Poblem 44. Show x n e x = xn e x n x n e x, fo. Solution. Let u = x n nd dv = e x, so tht du = nx n nd v = e x /. We hve: x n e x = x n ex nx n ex = xn e x n x n e x. 7., p. 5, Poblem 48. Evlute x e 3x. Solution. By pplying the eduction fomul in poblem 44 twice, we hve x e 3x = x e 3x xe 3x 3 3 = x e 3x xe 3x ) e 3x 3 = x e 3x 3 xe3x e3x + C.
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