cee-305_solutions_chapter5_odd_problem-set_homeworks

cee-305_solutions_chapter5_odd_problem-set_homeworks -...

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Unformatted text preview: Problem Set Chapter 05.02 Direct Method of Interpolation COMPLETE SOLUTlON SET l. The following data ofthe velocitywot‘a body is given as a function ot‘time . Time,t(s) 0 its its 22 F4 Velocity,v(m/s) 22 124 l37 25 123 What is the velocity at t 2 14 seconds using linear polynomial interpolation? Solution For first order direct polynomial interpolation (also called linear interpolation), the velocity is given by tit) : (10 + (1,! Since we want to find the velocity at t :14 using a lirst order polynomial, we choose the two data points that are closest to t 2 l4 and that also bracket t : 14 to evaluate it. Those two points are 10 2 O and t] 2 l5. Then t0 = 0, to”): 22 21:15.44): 24 gives rm) 2 ((0 + al (0) = 22 v(l 5) 2 (10 + (“(15): 24 Writing the equations in matrix Form 1 0 1 a0 22 [l 15L, ] “ [24] and solving gives a.) 1: 22 a, :2 0.13333 Hence. to): 22 «t— o.i3333(z), 0 g x s l5 at t 2 14, till): 22 t- 0.l3333(l4) r 23.867 m/s ()filllfi {Hill 3. The following data of the velocity ot‘a bod is given as a function ot‘time. [ Time, t (s) 0 l 15 l8 22 24 { Velocity, v (m/s) 22 L24 37 25 123 What is the velocity at t : 14 seconds using cubic polynomial interpolation? Solution For third order direct polynomial interpolation (also called cubic interpolation), we choose the velocity given by v(t) : (10 + alt + £12!2 + a3t3 Since we want to find the velocity at £214 using a third order polynomial, we need to choose the four points that are closest to t = 14 and that also bracket t = 14 .Then these four points are to 20,1} : l5, 12 =18,andr3 = 22. t0 = 0, i200): 22 t, :15, v(t,)= 24 z, :18, v(:,):37 r, = 22, v(!3)= 25 gives v(0) 2 an + aJ (0)+ a; (0)2 + (13(0)3 2 22 v(15)= no + a. (15)+ (12(15)2 + (1305)"3 2 24 v(l 8) 2 a0 + a, (l 8) + (12(18)2 + (13 (l 8)3 = 37 v(22) 2 a0 + (II (22)+ (12(22)2 + {23(22)3 2 25 Writing the four equations in matrix form 1 0 0 0 7 a0 22 15 225 3375 l a, 24 l l 18 324 5832 a2 37 Ll 22 484 l0648jagj 25 and solving gives (1(1 :2 22 (zI : ~l9.087 a2 2 7.548 a, 2 41058225 Hence v(z):.: 22 ~ worm v 2.54813 ~0.0582252‘. 0 < 22 At t : l4, to) z 22 —19.087(M)+ 2154804): « 0.0532504)“ Hme mm :17.339 mz’s {, "import (litil 5. The following data of the velocity ot'a body is given as a function of time. Time, 2 (s) 7 o 15 is 22 24 Velocity,v m/s) [22 24 37 25 123 Find the value of velocity at t 2: l6 seconds using the following interpolation function for the velocity of the body v(z) :2 a0 + a1 sin(t) + a2 sin(2t) where ama‘, and a2 are the unknown constants. Solution Since we want to find the velocity at t z to usihg a function with three unknown constants, we need to choose the three data points that are closest to t = 16 and that also bracket t = 16 to evaluate it. These three points are IO :15, t, z: 18, t2 = 22. Then to :15, v(:0)=24 1,218,v(t,):37 :2 z 22, v(t2)= 25 gives v(l5) : a0 + a, sin(15) + (12 sin(2>< 15) 2 24 v(l8) = (to + a, sin(18)+ (:2 sin(2 x18): 37 v(22) = a0 + (II sin(22) + (12 sin(2 x 22) = 25 Writing the three equations in matrix form 1 0.6502 - 0.9880 (:0 24 1 —0.7509 ~0.99l7 a, 2 37 l — 0.00885l3 0.017702 :22 25 and solving gives an 2 25.008 cu :2 412637 (12 = —5.0770 Hence v(t) : 25.008 — 9.2637 sin(t) — 5.0770 sin(2(), l 5 S t S 22 At I :16 v(16)= 25.008 ~ 9.2637 sin(l 6) — 5.07703in(2 x l6) 2 24.875 m/s sé’iti‘; it 7. You are given data for the upward velocity of a rocket as a function of time in the table below. 1 1,(s) 0 10 l 15 lzo {22.5 30 l } 12(1), (m/s) 0 227.04 l 362.78 517.35 J 602.97 901.67 21) Use a linear interpolant approximation of velocity to find the acceleration at t : 16 seconds. b) Use a cubic interpolant approximation of velocity to find the acceleration at t 2 16 seconds. c) Find the distance covered by the rocket from t =5 to 16 seconds? Use any method. Solution a) For first order direct polynomial interpolation (also called linear interpolation), the velocity is given by v(t) : a<1 + alt Since we want to find the acceleration at t 2 l6 using a first order polynomial, we choose the two data points that are closest to t 2 l6 and that also bracket 2‘ 2 l6 to evaluate it. Those two points are t2 :15 and t3 2 20. Then 12 :15, v(t2): 362.78 :3 .—. 20, veg): 517.35 gives v(l 5): a0 + a, (15) 2 362.78 v(20) = a0 + a,(20)= 517.35 Writing the equations in matrix form 1 15 a0 fl 362.78 [1 20L} _ [517.35] and solving gives a0 2 —100.93 a1 2 30.9l4 Hence v(t): 400.93 + 30.9141, 15 s 1 _<_ 20 The acceleration is a(z) = iI—(~ 100.93 + 30.9141) dt 230.914 m/sz, 15sts20 a(16):30.9l4 m/s2 05.01.13 b) For third order direct polynomial interpolation (also called cubic interpolation), we choose the velocity given by 13(1): a0 + a1t+ (12:2 + a3t3 Since we want to find the acceleration at I = 16 , and we are using a third order polynomial, we need to choose the four points closest to t : 16 and that also bracket 1‘ z 16 to evaluate it. The four points are 1‘, = 10,12 2 15, t3 = 20 and 1‘4 2 22.5. Then 1, :10, 12(5): 227.04 :2 :15, 10.): 362.78 13 = 20, 145): 517.35 4 : 22.5, 12(14): 602.97 Chapter 051.) 1 gives 11(10): an + a, (10)+ (12(10)2 + (13(10)3 2 227.04 v(15) = a0 + a, (15)+ (12(15)2 + (73(15)3 2 362.78 12(20): 210 + a. (20)+ a2 (20)2 + 1523(20)3 = 517.35 v(22.5) = a0 + a, (22.5)+ (12(225)2 + (13022.5)3 : 602.97 Writing the four equations in matrix form, we have 1 10 100 1000 1.70 227.04 1 15 225 3375 al 362.78 1 20 400 8000 a2 2 517.35 1 22.5 506.25 11391 a3 602.97 Solving the above four equations gives 110 = —4.33 a, 2 21.282 02 = 0.13085 (13 = 0.0054612 Hence 12(1) 2 a0 + alt + 1:1th + 613$} = 4.33 + 21 .282t + 0.1308512 + 000546121”, 10 s t _<_ 22.5 The acceleration is 21(1) 2 {EL 4.33 + 21.232: +0.13085z2 + 0.0054612t3) = 0.0163841‘2 + 0.2617t+21.289, 10 St S 22.5 At t:16 a(16) = 0016384106)2 + 0.2617(16)+ 21.289 = 29.784 m/s2 e) We need to choose the polynomia1 interpolant that will find the distance covered from t 2 5 to t = 16. A third order polynomial is chosen for interpolation as is given by 1:33.131: .1 0201.1, g 5 J 1”“) ,1. w I" . WW w W v0) = a“) + alt + agz‘2 + (131‘3 113113, we need to cheese the four points Closest to t = 5 and t 2 16 that also bracket them to evaluate it. The four points are t0 2 0, t1 = 10, t2 = 15, andz‘3 = 20. :0 z 0, we): 0 , :10, 17(11): 227.04 :2 =15, veg) = 362.78 {3 = 20, 17(73): 517.35 gives v(0) 2 a0 + a1 (0)+ (12(0)Z + 1573(0)3 2 0 12(10): a0 + a, (10)+ 612(10)2 + a3(10)3 = 227.04 11(15): a0 + a! (15)+ c1205)2 + (13(15)3 : 362.78 12(20): a0 + (11(20)+ 612(20)2 + (13(20)3 2 517.35 Writing the four equations in matrix form 1 0 0 0 a0 0 10 100 1000 al 22704 l 1 15 225 3375 at2 362.78 1 20 400 8000 ar3 517.35 and solving gives (:0 = 0 al : 20.344 a2 = 0.19585 (13 = 0.0040167 Hence v(t) = a0 + alt + azt2 + a3t3 = 0 + 20.344t + 0.19585t2 + 0.00401671‘3 , 0 S t s 20 integrate the velocity interpolant from t = 5 to t: 16 to find the distance covered by the rocket. 05.01 .14 {Lhapter 95.0} 16 6(16)—~ 3(5) 2 (20.344: + 0.19585:2 + 0.0040167t3dt 5 = [10.17272 + (106528313 + 0001004274]:6 = (10.177206)2 + 0.065283(l6)3 + 0.0010042(16)4) m (10.172(5)2 + 0.065283(5)3 + 0.0010042(5)4) = 2937.2 — 263.09 = 2674.1 m Alternatively the Newton’s divided difference polynomial method and the trapezoidal rule may be used. ) (227.04 —0 5 = 0 W 5 mo v1 > + 10 __0 ( > = 113.52 12(16): 362.78 + WM —15) 20 ~ 15 = 393.69 Thus so 6)“ 5(5) z (1 13.52 : 227.04)(10 5)+ (227.04 .5 362.78) (1 5 1 O) + (362.78 : 393.63) (16 _15) 2 2704.2 951.913 igflmm Problem Set Chapter 05.03 Newton’s Divided Difference Polynomial COMPLETE SOLUTlON SET 1. The following data of the velocity ofa body is given as a function oftime. ITime, t (s) 0 15 18 122 24 l | Velocity, 12 (m/s) 22 24 37 25 123 1 Find the coefficients of the Newton’s divided difference interpolating first order polynomial to find the velocity at t = 14 seconds. Solution For linear interpolation, the velocity is given by 11(1): b1) +bi(t”“t0) Since we want to find the velocity at t = 14 using a first order polynomial, we choose the two data points that are closest to t = 14 and that also bracket t = 14 to evaluate it. Those two points are t = 0 and t 2 15 . Then to : 0, 1200): 22 :1 =15, v(tl):24 gives be :"(to) = 22 b1 : V01) “V00 ‘1 “to _ 24—22 15—0 2 0.1333 Hence v(t):bo+b,(z—to) 222+0.1333(t~0), Ogtgls At t: 14 v04) 2 22 + 0.1333(14) = 23.867 m/s {35132.4 m (Liliapter 05,93 3. The following data ofthe velocity ofa body is given as a function oftime. I Time, t (s) l 0 15 1 18 22 24 lVelocity,v(m/s) l 22 24 I 37 25 123 Find the coefficients of the Newton’s divided difference interpolating third order polynomial to find the velocity at t 2 14 seconds. Solution The velocity profile is chosen as v(t) :bo +bl(t 40) +b2(t~t0)(l~t,)+b3(t—t0)(t~tl)(t~t2) Since we want to find the velocity at t z 14 , using a third order polynomial, we need to choose the four points closest to t 2 l4 and that also bracket t = 14 to evaluate it. These four points are t0 = 0, t1 = 15, t2 = 18, and t3 = 22. 0 =0, v(t0):22 r, :15, v(t,):24 :2 =18, v(t2)=37 t3 : 22, v(z3)= 25 then b0 :Vitoi :V(t0) :22 bl :V[[1?t()] : V(t1)—V(t0) [1"‘0 24—22 2 l5~0 =0.l333 b2 : vitz’tl’tll] V(t2)“v(ti) V(ti)—V(to) 37 m 24 24 — 22 Wig—TS" “ is — o _ 4.333 — o. 1333 _ 18 : 0.2333 Ncwmn‘s {filiévédcd Diffcrcmm 05.025 3 3"}[13’5353‘0] _ v[t3,tz,t,1—»{zpzpt0} ‘ t3~10 Eggvtz]“v[t2atz] t3 «t1 V(t3)“v(t2) 2‘3 —t2 ” 25—37 ” 22—18 2—3 v(t2)~v(ti) [2 “[1 fl 37‘24 18*15 24.333 V[t3at2]—V[t29t1] t3 —tl ” —3fi4.333 ” 2245 24.0476 v[t2,z,,zo]:0.2333 b3 ZVUB’tDtlJO] M v[t3,t2,tl]~v[t2,tl,to] ” t3 —t0 fl —1.0476—0.2333 ” 224) :~0.058225 3113,54,] : V[t3st2]: VUZ’II]: V[t3’tzatn]: Hence V0) Zbe +1710“to)+b2([—to)(t"t1)+b3(t“t0)(t_t1)(t“tz) z 22 + 0.1333(t~ 0) + 0.23330 —0)(t~15) ,09322 * 0.058225(t — 0)(t —15)(t ~18) At {214, v(14): 22 + 0.1333(14)+ 0.2333(14)(14 —15)— 0.058225(14)(14 -15)(14 ~18) £35.:32fi (fjhaptcr {35.03 217.339 m/s fi/T 05.02.10 (fhagite1‘0502 5. The aceeleratiomtime data for a small rocket is given in the table below. Time, ((5) 10 12 114 16 l 18 20 22 24 Acceleration, 6: _ (m/SZ) 106.6 94.1 80.9 68.0 56.2 45.8 37.1 30.1 21) Use Newton's divided differenCe quadratic polynomial interpolation to find the acceleration at t 2 15.5 seconds. b) Use the quadratic interpolant of part (a) to find the change in the velocity ofthe rocket between 1:14.! and [215.8 seconds. Solution a) The acceleration profile is chosen as a(t) : bO +bl(t —I0)+b2(t—t0)(t—tl) Since we want to find the acceleration at t 2 15.5 using a second order polynomial, we need to choose the three data points that are closest to t 2 l5.5 and that also bracket t 2 15.5 to evaluate it. These three points are 1‘2 214, [3 216, and I4 218. t2 2 14, (1(5) 2 80.9 {3 216, a(t3) 2 68.0 t4 2 l8, £104) 2 56.2 then b0 2 (1(0)) 2 80.9 bl : a(’1)"a(to) £1 ‘to _ 68.0 —80.9 16 * 14 2 «6.45 ffiffljflgfl , EEQLEIEQ b2 2 [2 ‘11 ‘1 “to [2 “to €931§§3 _ £918.89. 2 18 — 16 16 214 18 ~14 _ 25.92(— 6.45) 4 0.1375 11 Nan/111116 Divided 011101121100 .4” N84 then 11(1) 2 [10 +0.0 —IO)+b2(t ~to)(t-t,) : 80.9 a 6.450 “14)49 0.13750 — 1410 ~ 16), At 1215.5, 605.5) : 80.9«~6.45(15.5 414) + 0.1375055 414x155 ~16), 271.122 m/s2 [431318 b) Ifwe expand a0): 80.9 a 6.450 — 14) + 0.13750 41410 ~16), 14 g z :18 we get a(t) : 202 410.5751 + 0.1375:2 15.8 v(15.8)—v(l4.1): [202410.5751+0.137512dt 14.} 15.8 2 [2021 ~5.2875t2 + 0.04583313],4V1 z (202(15.8)— 5.28750 5.8)2 + 0.04583 30 5.8)3) ~(202(14.1)~5.2875(14.l)2 +0.045833(14.1)3) : 2052.4 4 1925.5 2126.94 m/s 05.02.! I Problem Set Chapter 05.05 Spline Method of Interpolation COMPLETE SOLUTION SET l. The Following y vs. x data is given. )6 n y a) Set up the equations to solve for the quadratic spline interpolants that go through the data. b) Use a program such as MATLAB tO solve the equations and then write down the spline interpolants. c) Estimate the value Of y(3.6). Solution a) Since there are 3 data points, 2 quadratic splines pass through them. ZSxS3 33x36 y(x) 2 (1,):2 + blx + c, , 2 (tax? + bzx + c2 , The equations are setup as follows. 1. Each quadratic spline passes through two consecutive data points. {131% i All alxZ + blx + 61 passes through x 2 2 and x 2 3 , my.)2 +1),(2)+cl 24.75 (“(3)2 +1;l (3) +c, : 5.25 azx2 +£72x+c2 passes through x 2 3 and x 2 6, (22(3)2 + [12(3) + 02 2 5.25 02(6)2 +b2(6)+cz :45 Quadratic splines have continuous derivatives at the interior data points. At x 2 3 , 2a, (3) +2), 2 2a2(3)—b2 : 0 . Assuming the first spline avle +1),x+cl is linear, (11:0 (1) (2) (3) (4) (5) (6) 0‘? (H 2 2 g k 4 2 1 0 0 0 a, 4.75 9 3 I 0 0 0 b, 5.25 0 0 0 9 3 1 c] 5.25 o 0 0 36 6 1 a, 2 45 6 I 0 ~ 6 —1 0 b2 0 1 0 0 O 0 0 e, 0 b) Solving the above 6 equations gives the 6 unknowns as ‘ i a, b, c, ’ 1 0 0.5 3.75 2 4.25 ~ 25 42 Therefore, the splines are given by y(x) : 22.704x + 3.75, 2 S x _<_ 3 : 4.25x2 ~ 25x + 42, 3 g x s 6 C) At x z 3.6, y(3.6) = 4.25(3.6)2 * 25(3.6) + 42 = 7.08 {,‘hziptcr 05,04 0.5.04.6 Chapter {15.04 3. The following incomplete y vs. x data is given. 3: 1 2.2 3.7 5.1 6 1 1’ 4.25 6 5.25 15.1 7???? The data is fit by quadratic spline interpolants given by f(x):1.4583x + 2.7917, 13 x S 2.2 *4 ~1.3056x2 + 7.2028x«3.5272 , 2.2 g x s 3.7 =cx2 +gx+h, 3.7335351 :jx3+kx+1,5.leS6 _ where c, g, h, j, k, and! are constants. What is the value of 3? Show all your steps clearly. Solution We know cx2 + gx + h passes through x = 3.7 and x : 5.1 giving c(3.7)2 + g(3.7)+ h = 5.25 (1) c(5.1)Z +g(5.l)+h215.1 (2) The first derivatives of consecutive quadratic splines are continuous at the common interior point. At 2: z 3.7 , 3191305628 +7.2028x~3.52721 : 31(ch + gx+ 71‘ dx x=3.7 t x=3.7 (" 2.61 1,2x + 72028sz37 = (2cx + glt fl 2.6112(3.7)+ 7.2028 2 2c(3.7)+ g 7.4a + g : 4.4586 (3) This gives three equations and three unknowns. 1n matrix form :37 13.69 3.7 l c 5.25 26.01 5.1 1 g 2 15.1 7.4 1 0 h —2.4586 Solving gives 6' : 6.7817 g = $2.643 h 2 107.19 The value of g is g = —52.643 s £2 05,048 a gig (,‘hugtcr 85.04 5. The thllowing incomplete y vs. .1: data is given. )6 1 72.2 13.7 l5.1 T6 l y 4.25 6 l 5.25 j ‘2??? 1 2???? The data is lit by quadratic Spline interpolants given by f(x) 2 1.45833: + 2.7917, l S. x S 2.2 2 *l 3056.162 +7.2028x #35272 , 2.2 S x S 3.7 :cx2+gx+h, 3.7Sx35.l :jx2+/c1c+l,5.leS6 2.5 where c, g, h, j, k, and! are constants. What is the value of I f (x)ch ‘2 l 5 Solution To integrate the function from x: 1.5 to x225 , we need to use the splines valid in the domain (1.5.2.5), that is, f(x) = l.4583x+2.79l7, 1g x g 2.2 = *1 .3056x2 + 7.2028x 93.5272, 2.2 s x s 3.7 to give 2.5 2.2 2.5 jfmdx = “1.45833: + 2.791 7).?1: + j(— 1 3056.12 + 7.2028x a 352.72%: l5 ii 22 2 [0.72915112 + 2.7917xf'52 0.4352963 + 3.6Ol4x2 —3.5272xl§j§ z [(0.72915(2.2)2 + 2.7917(2.2))—(0.72915(1.5)2 + 2.7917(1.5))] (— 0.4352(2.5)3 +3.6014(2.5)2 —3.5272(2.5)) + ~(~0.4352(2.2)3 + 3.6014(22)2 93.5272(2.2)) z [(9.6708)~(5.8281)]+ [(6.8908)fl(5.0369)] : [3.8427]+[1.8538] : 5.6965 ...
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This note was uploaded on 10/28/2011 for the course CEE 305 taught by Professor Nguyen,d during the Fall '08 term at Old Dominion.

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