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Unformatted text preview: 380 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 24. She will have A / n dollars after each withdrawal and 0 dollars just before the next withdrawal. so her average cash balance at any given time is + 0) = A/(Zn). The transaction costs for n withdrawals are nT. The lost interest cost on the average cash balance is [A/(2n)] - R. Thus. the total cost for 71 transactions is AR AR AR AB C : T —. N C’ : a __ I ' __ 2 (n) n + 2n ow (n) T 2n2 and C (n) 0 2n2 T > n 2T > AR . . . . H AR . n : the value of n that minimizes total costs Since C (n) : ——3 < 0. Thus. the optimal average cash 71 A Ax/2T _ x/AT _ AT balance is — 2n 2 2 x/AR _ x/2R 2R" 4.9 Newton's Method 1. (a) The tangent line at m : 1 intersects the arr—axis at as m 2.3. so 382 z 2.3. The tangent line at ac = 2.3 intersects the x—axis at cc 2 3. so x3 x 3.0. (b) :31 : 5 would not be a better first approximation than :61 : 1 since the tangent line is nearly horizontal. In fact. the second approximation for $1 : 5 appears to be to the left of a: = 1. The tangent line at :L' : 9 intersects the m—axis at ac % 6.0. so mg m 6.0. The tangent line at as : 6.0 intersects the ac—axis at cc % 8.0. so x3 x 8.0. 3. Since 2:1 : 3 and y : 5:5 ~ 4 is tangent to y : flan) at cc 2 3. we simply need to find where the tangent line intersects the m—axis. y — 0 5332 4 e 0 -5 x2 ‘ g 4. (a) (b) If :51 : 0. then :02 is negative. and $3 is even If :31 z 1. the tangent line is horizontal and more negative. The sequence of approximations Newton’s method fails. does not converge. that is. Newton‘s method fails. SECTION 4.9 NEWTON'S METHOD 381 (c) (d) If an : 3. then :32 : 1 and we have the same If x1 = 4‘ the tangent line is horizontal and situation as in part (b). Newton‘s method fails Newton‘s method fails. again. (e) If 9:1 : 5. then :32 is greater than 6. x3 gets closer to 6. and the sequence of approximations converges to 6. Newton’s method succeeds! 3 2 n—4 .f(m):z3+2:c—4 é f’(a:)23$2+2.somn+1=$n——mn;§1:2 .Nowxlzl => 1+274 —1 (1.2)3+2(1.2)—4 =1 —1 —1.2 —> —1.2— :3 . . $2 3.12+2 5 5‘3 3(1.2)2+2 11797 _ 3 2 I _ 2 _ f(;cn) _ miimi—l —SE —1 :> 72$.SOSCn+1—.2L’n—fl(mn)iIEn—m. 1V1—1 23722—1 N0 :1:> =1——=2 ,2 ———-—————:. . W131 $2 3—2 : 1’3 2 3.2272‘2 1625 4 / 3 $4 _20 . 23:3: —20 :> m:4a:.son :n— 271* n . 24 —20 (2.125)4 —20 Nowx —2 —> .27 —2 2.125 > 2.125 m . . 1 2 4(2)3 x3 4(2.125)3 21148 5 I n 2 .f(sc)::c5+2 :> f(m)=5m4.s0xn+1:xnim5;; .Noww12—1 => (71)5+2 1 (71.2)5+2 x:—1~—:71——=—1.2 — z 2 5.(_1)4 5 —> $3 1.2 5(_1.2)4 1.1529 f(x):x3+x+3 2 f(.t):3:c2+lso 7 Newton‘s method follows the tangent line at (7]. 1) down to its intersection with the x—axis at (—1.25. 0). giving the second approximation x2 : —1l25l 382 CI CHAPTEFM APPLICATIONS OF DIFFERENTIATION 10. f($)::c4—m~1 => f’(x)=4;c3—1. so 6 xn+1 :xn— M.Nowm1 =1 => 421% *1 A $2 : 1 — fl 2 i _—1 = Newton’s method follows the 4- 13 — 1 3 3 11, 12. 13. 14. 74 tangent line at (1. *1) up to its intersection with the ac—axis at (1; . 0). giving the second approximation $2 : 3. To approximate cc 2 {7% (so that x3 = 30). we can take flat) 7 x3 — 30. So f’(a:) : 3x2. and thus. mi-—30 356% until they agree to eight decimal places. 9:1 2 3 => :62 m 3.11111111, :33 z 3.10723734. :04 m 310723251 % 955. So \3/3_ % 3.10723251. to eight decimal places. Here is a quick and easy method for finding the iterations for Newton’s method on a programmable calculator. wn+1 2 xn * . Since \3/27 : 3 and 27 is close to 30. we’ll use m1 = 3. We need to find approximations (The screens shown are from the "TI—83 Plus. but the method is similar on other calculators.) Assign f : m3 A 30 to Y1. and f’(;c) 2 3w2 to Y2. Now store m1 = 3 in X and then enter X — Y1/Y2 —> X to get $2 : 3.1. By successively pressing the ENTER key. you get the approximations $3. :04. . . . . Hui-t1 Hot: Flt-t3 NV1EX3-EB NVEESHE n?3=l KVH= 1V5: HV5= \V?= In Derive. load the utility file SOLVE. Enter NEWTON (XA3 — 3 O , x, 3) and then APPROXIMATE to get [3. 3| l I l I I I I. 3.10723733. 3.10723250. 3.10723250]. You can request a specific iteration by adding a fourth argument. For example. NEWTON (x33 —30,x, 3, 2) gives [3. 311111111. 310723733]. In Maple. make the assignments f z: :1: —> f3 ~ 30;. g 2: .1: —> :c — and x z: 3.;. Repeatedly execute the command I z: g(ac); to generate successive approximations. In Mathematica. make the assignments f[m7] :: x33 — 30. g[z,] :: a: — flx]/f’[$]. and :c = 3. Repeatedly execute the command :3 : g[a:] to generate successive approximations. so; — 1000 7mg agree to eight decimal places. x1 = 3 => :62 % 2.76739173. x3 % 269008741. :04 R: 2.68275645. 535 z 2.68269580 3 m6. So W m 268269580. to eight decimal places. f(;1:) : :07 A 1000 => f’(m) : 7:36. so zen“ : xn — . We need to find approximations until they 2:5: — (Sari —I— 356” +1 633% — 12307; + 3 need to find approximations until they agree to six decimal places. :61 = 2.5 => 362 % 2.285714. mg % 2.228824. x4 % 2.224765. .725 m 2.224745 z 5%. So the root is 2.224745. to six decimal places. ffl):2§i6fi+fiz+1 : fxm:4m?—mm+3 2 mm4:wn— rWe 4 .Tn‘i'xn‘4 . —1.5 —> x z1.323276. 4mi+1 $1 2 $3 % 1.285346, $4 m 1.283784. 335 % 1.283782 m :36. So the root is 1.283782. to six decimal places. f($)=$4+mi4 :> f'($)=4593+1 :> “H’m” 15. since : :32. so : sinx — :62 $n+1 : 3777. 7 sin ac =w2 isnear1. m1 :1 places. 16. 2cosa: : 2:4.so f(:c) = 2cosac —:c :> $n+1 :xn’ of2cosa: : at $3 2 1.013958 % :34. So the positive root is 1.013958. to six decimal places. 17. ~ 2 sm mn i m" cos sun — 23c” 4 2cosxn i m" 4 x2 2 70.725253 :63 m —O.724493 :64 m 70.724492 z 1:5 => f’(ac) = coszr A 2m => . From the figure. the positive root of :> :32 m 0.891396. :63 % 0876985. 334 % 0.876726 z :05. So the positive root is 0.876726. to six decimal .—3. From the figure. the positive root —2 Sin xn — 4:0,. is near 1. m1 : 1 => 132 2 1.014184. SECTION 4.9 NEWTON’S METHOD 383 => : i2sinx — 4:173 From the graph. we see that there appear to be points of intersection near ac : 70.7 and :1: : 1.2. Solving :04 2 1+ 20 is the same as solving f(m):m4ixv1:0.f(22)=$4—m—1 ;> f'(:c)=4w3—1. mfi—zn—l so an : :vn — 425% 71 (£1 : 1.2 232 z 1.221380 25;; 3 1.220745 $4 m 1.220744 % 935 To six decimal places. the roots of the equation are 70.7 24492 and 1.220744. 18. _, 19. *3 3 3 NF From the graph. there appears to be a point of intersection near as : 0.6. Solving em : 3 — 23a is the same as solving f(a:) 2 6I + 2x — 3 : 0. flat) :ex+2x—3 => f’(a:) :ex+2.so 61" + 2.7% — 3 ex" + 2 :03 m 0.594205 z 234. So to six decimal places. the root of the equation is 0.594205. . Now $1 : 0.6 => 372 m 0.594213. mn+1 : 1:71. " From the graph. there appears to be a point of intersection near at = 0.5. 1 Solving tan’ m = l i x is the same as solving f(£v)=tan_1$+m—1:0.f(w):tan’1x+w—1 :> 1 tan—1x +$ —1 I 77. 7L 2 1. : ni—_— f (x) 1+“? + 80%“ x 1/(1+m%)+1 m1 : 0.5 => 952 m 0.520196. :03 2 0.520269 z $4. Soto six decimal places. the root of the equation is 0.520269. 384 C 20. 21. 22. 23. CHAPTER 4 APPLICATIONS OF DIFFERENTIATION From the graph. we see that there appear to be points of intersection near an : —1.2 and SE : 1.5. Solving Va: + 3 = $2 is the same as solving f($):m2ix/a:+ =Olf(m):$2— w—I—3 => 4' I'm-2x ——A —w ——xi’ “+3 — i . n 1 — n — t _3 3 2 m+3 + 295.41/(2 xn+3) ‘1 £131 : 71.2 $1 21.5 $2 R” 71.164526 x2 x 1.453449 £63 3 —1.164035 % .764 {123 Z 1.452627 % 11:4 To six decimal places. the roots of the equation are 71.164035 and 1.452627. From the graph. there appears to be a point of intersection near a: = 0.6. 2 Solving coszc : fl is the same as solving f(x) : cosx — \/_ : 0. f(:c)=cos:c—\/5 :> f’(zc) sinm—l/(2fi).so cos can — Mac” n I n — ,—--—. N = 0.6 => 3: +1 m *SII’IQZn—l/(2fi) owml 9:2 % 0.641928. $3 :5 0.641714 w .724. To six decimal places. the root of the equation is 0.641714. From the graph. there appears to be a point of intersection near :6 = 0.7. 2 Solving tanx = x/ 1 — m2 is the same as solving f(w)=tan:ci\/1—sc2:0.f(w):tan:cg 1—;32 :> 7% 121 tanmnix/l—m2 ’ :se02m+w \/1—$2.soxn =mn*————”—. f (56> / +1 m. + 2:1 : 0.7 :> $2 z 0.652356, :03 % 0.649895. $4 a: 0649889 % $5. 2 To six decimal places, the root of the equation is 0.649889. f($):$5*$4—5$37$2+4$+3 :> f'(sc) : 5304 — 4x3 7 15x2 * 25c + 4 :> 5 _ 4 i 3 i 2 SCn+1 H mu m" w” 5w" m” + 4m” + 3. From the graph off. 556$L i 458% — 159% — 2%; + 4 there appear to be roots near —1.4. 1.1. and 2.7. 031 : —1.4 $1 :1.1 {131 : 2.7 :82 z —1.39210970 232 m 1.07780402 932 m 272046250 903 m 7139194698 mg m 107739442 353 m 271987870 x4 m ~1.39194691 m 275 :04 % 107739428 Ft: $5 :64 % 2.71987822 :3 935 To eight decimal places. the roots of the equation are —1.39194691. 1.07739428. and 271987822. SECTION 4.9 NEWTON'S METHOD l: 385 4 i . 24. 5 Solving x2 (4 — :52) 2 $2 + 1 IS the same as solvmg 4 8.7: = 2— 4- :0.’ 28 —4$3+—— => A A W e x W) m (M2 22.5 2.5 4 ,2 — 4 — 4 ,2; 1 " xn+1 2 mn — From the graph of 8:0" — 4.1% + 8xn/(scfi + 1) ‘5 there appear to be roots near as 2 :l:1.9 and a: 2 ::0.8. Since f is even. we only need to find the positive roots. (131 = 0.8 $1 = 1.9 x2 2 084287645 mg m 194689103 :03 z 084310820 $3 a 194383891 204 z 0.84310821 z 335 3:4 % 194382538 2 $5 To eight decimal places. the roots of the equation are 1084310821 and ::1.94382538. 25. 3 From the graph. y 2 $2\/ 2 — z — m2 and y 2 1 intersect twice. at ‘ xw—Zandatccmil.f(w)=$2\/2—m—$2—1 => f’(m) 2 2:2 - %(2 — m — $2)’1/2(—1 — 256) + (2 — x 2 $2)1/2 - 2a: _3-“'A- : gm 2 m _ 15—1/2 [35H _ 2:6) +49 _ x _ 952)] _- _ “849349) " 2 (2+:c)(1—:c)‘ mix/2—mn—le—1 ‘ , , t . so can“ 2 xn — —. Trying 9:1 2 72 won‘t work because f (—2) is undefined. so we ll mn(8 2 5% — 6.2721) 2 (2 + mn)(1— xn) try 561 2 —1.95. 231 2 —1.95 x1 — —0.8 322 % 7198580357 x2 x —0.82674444 x3 x —1.97899778 x3 2 7082646236 x4 % —1.97807848 m4 5: —0.82646233 % x5 x5 7: —1.97806682 336 m —1.97806681 2 237 To eight decimal places. the roots of the equation are 71197806681 and 2082646233. 26. From the equations y 2 3 sin(:c2) and y 2 2m and the graph, we deduce .231 I 0.7 232 m 969303689 $3 :3 069299996 2: 504 that one root of the equation 3 sin(w2) 2 2a: is :1: 2 0. We also see that the graphs intersect at approximately at 2 0.7 and a: 2 1.4. 2 3sin(m2) — 2x => 2 3COS(1’2) - 2:0 7 2. so 33in(xi) — 2:13n :c 21: — . "+1 n 69:71 003(3021) — 2 {131 Z $2 2 139530295 333 z 139525078 :64 z 139525077 % 935 To eight decimal places‘ the roots of the equation are 069299996 and 1.39525077. 386 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 27. 3 2 From the graph. we see that y : e’xz and y = a: i x intersect twice. Good first approximations are a: = —0.5 and 5c 2 1.1. ,2 A- 2 -— 70.5 —0.51036446 $3 8 7051031156 ,8 964 can“ : zn — —2—. —2:cne’2n i 23:" —I— 1 1131 = 1.1 x2 x 120139754 x3 x 119844118 324 m 1.19843871 m :65 ~ ~ To eight decimal places. the roots of the equation are —0.51031156 and 119843871. 28. 2 mac wmi2andatatm1.f(sc)=ln(4—ac2) gm :> From the graph. y = ln(4 — 9:2) and y : ac intersect twice. 2 = e‘z — $2 + ac => f’(w) : 7212(i—I2 * 2m + 1. so —22 2 I k 6 IL — {En +xn at 71.99. *3 r I _2$ 1n(4 —.7:ZL) —wn : — 1H : — ———-————. ' f (as) 4 — 9:2 9° “+1 9”" [Jinn/(4 v x%)] — 1 Trying :01 — —2 won‘t work because it’s not in the domain of T4 y : 111(4 7 x2). Trying x1 : 71.9 also fails after one iteration because the approximation m2 is less than 72. We try 931 : 932 m 7197753026 203 % —1.96741777 :04 a —1.96475281 125 m —1.96463580 :06 m —1.96463560 % m7 m2 AN. 105864851 333 z 105800655 334 % 1.05800640 3 :65 To eight decimal places. the roots of the equation are —1.96463560 and 105800640. 29. (a) f(x) : 3:2 i a :> f'(.z) : 21'. so Newton’s method gives a 1 a v 1 m | 2x” 7 2 " ' ‘ $.21 — a 1 2% _ '77" 2 2:0", _ 2 (b) Using (a) with a : 1000 and x1 V800 : 30. we get 932 m 31.666667. :53 % 31.622807. and 034 z 31.622777 z 2:... So m z 31.622777. 1 l/scn — a ——.SO£L'n+1 :IKn (b) Using (a) with a : 1.6894 and x1 : l : a $n+1 : 2'!” xn + 1:1; I mm 2 2 axon — : $71 I $71 2$n 7 amvr 30. (a) m) = i —a 4 my) : m2 2 0.5. we get :02 : 0.5754163 z 0.588485. and 734 m 0.588789 m 25. So 1/16984 z 0.588789. 31. : $3 7 32: + 6 : f'(a:) : 3:62 — 3. If an : 1. then f'(a:1) : 0 and the tangent line used for approximating :02 is horizontal. Attempting to find 302 results in trying to divide by zero. 32.x3—m:1<=> m3—$71:0.f(m):m3—:c~1 2 f’($):3$271.somn+1:xni (a) 11:1 : 1. :62 : 1.5. 2:3 % 1.347826. 56.; % 1.325200. m5 3 1.324718 % m6 SECTION4.9 NEWTON'S METHOD I 387 (b) $1 : 0.6. $2 = 17.9. $3 $6 11.946802. $4 % 7.985520. $5 % 5.356909. $6 % 3.624996. $7 % 2.505589. $8 % 1.820129. $9 z 1.461044. $10 % 1.339323. $11 % 1.324913. $12 % 1.324718 % $13 (C) $1 : 0.57. $2 % 754165455. $3 % —36.114293. $4 z —24.082094. $5 % —16.063387. $6 % 710721483. $7 8 77.165534. $8 % 74.801704. $9 $3 A3.233425. $10 % ~2.193674. $11 % —1.496867. $12 8 70.997546. $13 % —0.496305. $14 % 72.894162. $15 a 71.967962. $15 m —1.341355. $17 8 70.870187. $18 % 70.249949. $19 % —1.192219.$20 % —0.731952. $21 z 0.355213. —1.753322. $23 % —1.189420. $24 % 70.729123. $25 % 0.377844. $26 % —1.937872. 22 $22 $27 ’3 —1.320350. $28 % 70.851919. $29 % —0.200959. $30 % ~1.119386. $31 % v0.654291. $32 % 1.547010, $33 % 1.360051. $34 2 1.325828. $35 % 1.324719. $35 z 1.324718 8 $37. (d) From the figure. we see that the tangent line corresponding to $1 : 1 results in a sequence of approximations that converges quite quickly ($5 2: $6). The tangent line corresponding to $1 : 0.6 is close to being horizontal. so $2 is quite far from the root. But the sequence still converges —just a little more slowly ($12 x $13). Lastly. the tangent line corresponding to $1 : 0.57 is very nearly horizontal. $2 is farther away from the root. and the sequence takes more iterations to converge ($36 % $37)- 33. For f($) : $1/3. f'($) = %$_2/3 and fan) wag _ $71 fxmn> Axews 3 n $n+1 2 $0,; 7 : 9:7z 3$n = 2$n. Therefore. each successive approximation becomes twice as large as the previous one in absolute value. so the sequence of approximations fails to converge to the root. which is 0. In the figure, we have $1 : 0.5. $2 : ~2(0.5) : *1. and $3 = —2(71)= 2. 34. According to Newton’s Method. for $7. > 0. W7 $n+1 : wn — — : $n v 2$n : 7:23,. and for$n < 0. 1/(2x/33—n) a _$n $n+1: In — ——— : In — ~2 —$n : far”. So we can see that after choosing any value $1 the subsequent values will alternate between 7931 and $1 and never approach the root. 35. (a) f($) : 3$4 ~ 28$3 + 6$2 + 24$ :> f’($) = 12a:3 i 84x2 + 12$ + 24 => f’(:r1) “ 2 f”($1) 3 $3 m 0.6455 => $4 % 0.6452 => $5 2 0.6452. Now try $1 = 6 :> $2 : 7.12 => $3 % 6.8353 :> $4 % 68102 :> $5 ’r‘v' 6.8100. Finally try $1 : —0.5 2 $2 % —0.4571 :> $3 2 —0.4552 => $4 a —0.4552. Therefore. at : —0.455. 6.810 and 0.645 are all critical numbers correct to three decimal places. f"($) : 36$2 — 168$ + 12. Now to solve f’($) : 0. try $1 — -> $2 — $1 —> [Oh—1 388 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (b) f(—1) : 13. f(7) : 71939. f(6.810) w. 41949.07. f(—O.455) m 76.912. f(0.645) x 10.982. Therefore. f(6.810) % 71949.07 is the absolute minimum correct to two decimal places. 36. flat) : :02 + sinzc :> f’(x) : 22: + cos x. f’(:c) exists for all :5. so to ‘0 find the minimum of f. we can examine the zeros of f'. From the graph of V f’. we see that a good choice for $1 is 0:1 : —0.5. Use 9(73) 2 250 + cosx _10 ‘ m and g’(m) : 2 — sinsc to obtain :02 2 —0.450627. 333 z 70.450184 m :04. Since f”(;z:) = 2 — sinx > 0 for all :0. “0 f(~0.450184) % 70.232466 is the absolute minimum. 37. From the figure. we see that y : f(a:) : 6”” is periodic with period 27r. To find the m-coordinates of the IP. we only need to approximate the zeros ofy” on [0.7r]. f’(:c) = —eC°”sinw => f”(w) = e“°””(sin2 ac — cos Since em” 7f 0. we will use Newton‘s method with g(.~c) 2 sim2 av — cos x. g'(:zr) : 2sin$ cosm + sin :0. and :01 = 1. :02 % 0904173. 223 m 0.904557 % 264. Thus. (0.9045577 1.855277) is the IP. 38. flat) 2 —sinw => f’($) : i cosac. At :r, = a. the slope of the tangent line is f’(a) = — cos a. The line through the origin and (a. f(a)) . — sin a — 0 . . . . IS y : If this line 18 to be tangent to f at a: : a. then its a , —sina slope must equal f'(a). Thus. : e cosa : tana = a. a To solve this equation using Newton‘s method. let g(:t) = tanx * :5. tan xn — xn g'(z) 2 sec2 :1: ~ Land mn+1 : JCn i with 931 = 4.5 (estimated from the figure). :32 % 4.493614. sec2 1'" — 1 003 3 4493410. 904 z 4.493409 2 905. Thus. the slope of the line that has the largest slope is f'(m5) z 0.217234. 39. 56000 The volume of the silo. in terms of its radius, is V(’F) : 7rr2(30) —I— %(§7rr3) : 3071'?“2 + $073. From a graph of V. we see that V('r) : 15.000 at r z 11 ft. Now we use 7‘ Newton‘s method to solve the equation V(r) — 15.000 : 0. 20 dV 2 # 30wri + gm: — 15.000 ___. 7 607??" + 2777" . so n+1 _ Tn 607T” + 2m% 0 . Takin d7" g n : 11. we get m w 11.2853. 13 R: 11.2807 z m. So in order for the silo to hold 15.000 ft3 of grain. its radius must be about 11.2807 ft. SECTION 4.9 NEWTON'S METHOD 389 40. Let the radius of the circle be 1“. Using 5 : r19. we have 5 : TB and so r : 5/0. From the Law of Cosines we gel 42 = 7‘2 +7"2 7 2 ~ 7" - r - c0sl9 <=> 16 : 2r2(1— c036) : 2(5/6)2 (1 i 0030). Multiplying by 92 gives 1662 2 50(1 7 cos 6). so we take f(6) : 1602 + 50 cos0 * 50 and f’(9) : 320 — 50 sin 9. The formula 10 1662 + cos 6” — ’ ' : A —."———_ F [h for Newton 5 method is t9n+1 6” 326" ‘ 5081116" rom e 0 3 graph of f. we can use 01 : 2.2. giving us 62 2 2.2662. 63 x 2.2622 x (94. So correct to four decimal places. the angle is 2.2622 radians % 1300. ‘15 R . ,n 41. In this case. A : 18000. R = 375. and n : 5(12) : 60. So the formula A = [1 — (1 —l— 2) ] becomes 18.000 2 E [1 — (1+ @450] <4 4823 : 1 — (1+ my“ [multiply each term by (1 + @601 <:> :c 48$(1+ x)“ e (1 + (5)60 + 1 : 0. Let the LHS be called f(m). so that f/(ac) : 48:0(60)(1+ x)59 + 48(1+ 3:)60 — 60(1+ cc)59 212(1 + at)59 [450(60) + 4(1+ m) e 5] : 12(1+ m)59(244x — 1) 48:87.0 —l— $11)“) — (1+ mn)60 +1 12(1+xn)59(244a:n — 1) estimate for w = i. So let :21 : 1% : 0.01. and we get 202 % 0.0082202. 333 3 0.0076802. $4 % 0.0076291, 3:5 R: 0.0076286 % we. Thus. the dealer is charging a monthly interest rate of 0.76286% (or 9.55% per year. compounded monthly). mn+1 i an — . An interest rate of 1% per month seems like a reasonable 42. (a) p(:c) = x5 i (2 —l—7")w4 + (1 + 2r)m3 — (1 — 79202 + 2(17 r)x+ r — 1 => p’(a:) : 5x4 — 4(2 + r):c3 + 3(1+ 2r)x2 7 2(1— r)m + 2(17 r). So we use sci~(2+r)wi+(1+2r)x§l—(1~r)zi+2(1—r)mn+r—1 5xfi — 4(2 +r)x% + 3(1+ 2r):c% — 2(1— 109:” + 2(1— 7') mn+1 = 36.. i . We substitute in the value 7‘ z 3.04042 X 10—6 in order to evaluate the approximations numerically. The libration point L1 is slightly less than 1 AU from the Sun. so we take x1 = 0.95 as our first approximation. and get $2 2 0.96682. x3 "NV 0.97770. :64 % 0.98451. :05 m 0.98830. .16 2 0.98976. :37 ,8 0.98998. 258 % 0.98999 z 209. So. to five decimal places. L1 is located 0.98999 AU from the Sun (or 0.01001 AU from Earth). (b) In this case we use Newton‘s method with the function i 27222 :1‘5 — (2+r)a:4+ (1-l-27")x3 — (1 +r)$2 +2(1 7r):c+r *1 => [p(x) — 2112], = 5:114 — 4(2 + r)w3 + 3(1—l— 270302 v 2(1-l— 79x + 2(1 — 7"). So :13: — (2+r)x4n +(1+2r)a:i — (1+7")aci+2(1 —7")acn+r— 1 5.23% 7 4(2 + 79:82 —l— 3(1+ 2r)m% — 2(1+ 7‘).’En —l— 2(1~ r) mn+1 ‘ atn . Again. we substitute 7" m 3.04042 X 10’6. L2 is slightly more than 1 AU from the Sun and. judging from the result of part (:1). probably less than 0.02 AU from Earth. So we take :01 = 1.02 and get x2 % 1.01422. x3 8 1.01118. 35.; m 1.01018. $5 a 1.01008 m :06. So. to five decimal places. L2 is located 1.01008 AU from the Sun (or 0.01008 AU from Earth). ...
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9 - 380 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 24. She...

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