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5. (b) (continued) Step 1: (continued) Tumia utofautishaji wa logariti ili kupata mabadiliko ya sehemu katika G. G = 4 R + - L Tofautisha pande zote mbili: (dG)/(G) = 4 (dR)/(R) + (d)/() - (dL)/(L) Mlinganyo huu unaunganisha mabadiliko ya sehemu katika G, R, , na L. Step 2: Badilisha asilimia za mabadiliko zilizotolewa. Imepewa: R imeongezeka kwa 3%, hivyo (dR)/(R) = 0.03. imepungua kwa 5%, hivyo (d)/() = -0.05. L imeongezeka kwa 6%, hivyo (dL)/(L) = 0.06. Badilisha thamani hizi kwenye mlinganyo: (dG)/(G) = 4(0.03) + (-0.05) - (0.06) (dG)/(G) = 0.12 - 0.05 - 0.06 (dG)/(G) = 0.07 - 0.06 (dG)/(G) = 0.01 Step 3: Badilisha mabadiliko ya sehemu kuwa asilimia ya mabadiliko. Asilimia ya mabadiliko katika G = (dG)/(G) × 100\% = 0.01 × 100\% = 1\%. Asilimia ya mabadiliko takriban katika G ni 1\%. 5. (c) Bainisha pointi za kituo za kazi z = x^2 + xy + y^2 - y na kisha bainisha asili yake. Step 1: Tafuta vianzio vya kwanza vya sehemu kwa heshima ya x na y na uviweke sawa na sifuri. ( z)/( x) = 2x + y ( z)/( y) = x + 2y - 1 Weka zote mbili sawa na sifuri ili kupata pointi muhimu: 1) 2x + y = 0 y = -2x 2) x + 2y - 1 = 0 Step 2: Tatua mfumo wa milinganyo kwa x na y. Badilisha y = -2x kutoka (1) hadi (2): x + 2(-2x) - 1 = 0 x - 4x - 1 = 0 -3x - 1 = 0 -3x = 1 x = -(1)/(3) Sasa tafuta y kwa kutumia y = -2x: y = -2(-(1)/(3)) = (2)/(3) Pointi ya kituo ni (-(1)/(3), (2)/(3)). Step 3: Tafuta vianzio vya pili vya sehemu. f_xx = (^2 z)/( x^2) = ()/( x)(2x+y) = 2 f_yy = (^2 z)/( y^2) = ()/( y)(x+2y-1) = 2 f_xy = (^2 z)/( x y) = ()/( x)(x+2y-1) = 1 Step 4: Tumia jaribio la kianzio cha pili (Hessian determinant) kubainisha asili ya pointi ya kituo. Kibaguzi ni D = f_xxf_yy - (f_xy)^2. D = (2)(2) - (1)^2 = 4 - 1 = 3 Kwa kuwa D = 3 > 0 na f_xx = 2 > 0, pointi ya kituo ni kiwango cha chini cha ndani. Step 5: Kokotoa thamani ya z kwenye pointi ya kituo. z = (-(1)/(3))^2 + (-(1)/(3))((2)/(3)) + ((2)/(3))^2 - ((2)/(3)) z = (1)/(9) - (2)/(9) + (4)/(9) - (6)/(9) z = (1 - 2 + 4 - 6)/(9) = -(3)/(9) = -(1)/(3) Pointi ya kituo ni kiwango cha chini cha ndani kwenye (-(1)/(3), (2)/(3), -(1)/(3)). 6. (a) Tathmini kila moja ya viunganishi vifuatavyo: (i) ( 2x - 3x)dx Step 1: Unganisha kila neno kivyake. 2x \,dx = (1)/(2) 2x 3x \,dx = - (-(1)/(3) 3x) = (1)/(3) 3x Step 2: Unganisha matokeo na ongeza kigezo cha kuunganisha. ( 2x - 3x)dx = (1)/(2) 2x + (1)/(3) 3x + C (ii) (x^2 + 1)dx Step 1: Unganisha kila neno kivyake kwa kutumia kanuni ya nguvu. x^2 \,dx = x^2+12+1 = (x^3)/(3) 1 \,dx = x Step 2: Unganisha matokeo na ongeza kigezo cha kuunganisha. (x^2 + 1)dx = (x^3)/(3) + x + C (iii) (x(3x^2 + 5x + 7))/(x^2)dx Step 1: Rahisisha kiunganishi. (x(3x^2 + 5x + 7))/(x^2) = (3x^3 + 5x^2 + 7x)/(x^2) = 3x + 5 + (7)/(x) Step 2: Unganisha kila neno kivyake. (3x + 5 + (7)/(x)) dx = 3 x \,dx + 5 \,dx + 7 (1)/(x) \,dx = 3((x^2)/(2)) + 5x + 7 |x| Step 3: Ongeza kigezo cha kuunganisha. (x(3x^2 + 5x + 7))/(x^2)dx = (3x^2)/(2) + 5x + 7 |x| + C 6. (b) (i) Chora mchoro wa eneo lililopakana na mkunjo y = 2x - x^2 na mhimili wa x. Step 1: Tafuta sehemu za kukatiza mhimili wa x kwa kuweka y=0. 2x - x^2 = 0 x(2 - x) = 0 Hivyo, x = 0 na x = 2. Mkunjo unakatiza mhimili wa x kwenye (0,0) na (2,0). Step 2: Tafuta kilele cha parabola. Kuratibu ya x ya kilele ni x = -(b)/(2a) kwa ax^2+bx+c. Hapa y = -x^2 + 2x, hivyo a=-1, b=2. x = -(2)/(2(-1)) = 1. Kuratibu ya y ya kilele ni y = 2(1) - (1)^2 = 2 - 1 = 1. Kilele kipo kwenye (1,1). Step 3: Chora mchoro wa mkunjo. Mchoro utaonyesha parabola inayofunguka chini, ikipita kwenye asili (0,0), ikifikia kiwango cha juu kwenye (1,1), na kisha kuvuka mhimili wa x tena kwenye (2,0). Eneo lililopakana na mkunjo na mhimili wa x ni eneo lililofungwa na mkunjo na sehemu ya mhimili wa x kutoka x=0 hadi x=2. (ii) Tumia ujumuishaji kubainisha eneo la eneo katika b (i). Step 1: Weka kiunganishi dhahiri kwa eneo. Eneo A linatolewa na kiunganishi cha kazi kutoka x=0 hadi x=2. A = _0^2 (2x - x^2) dx Step 2: Tathmini kiunganishi. A = [ x^2 - (x^3)/(3) ]_0^2 A = ( (2)^2 - ((2)^3)/(3) ) - ( (0)^2 - ((0)^3)/(3) ) A = ( 4 - (8)/(3) ) - (0) A = (12)/(3) - (8)/(3) = (4)/(3) Eneo la eneo ni (4)/(3) vitengo vya mraba. 7. (a) (i) Bainisha upanuzi wa mfululizo wa Maclaurin wa (1 + x) katika nguvu zinazopanda za x hadi neno katika x^5. Step 1: Tafuta vianzio vya f(x) = (1+x) na uzitathmini kwenye x=0. f(x) = (1+x) f(0) = (1) = 0 f'(x) = (1)/(1+x) = (1+x)^-1 f'(0) = 1 f''(x) = -(1+x)^-2 f''(0) = -1 f'''(x) = 2(1+x)^-3 f'''(0) = 2 f^(4)(x) = -6(1+x)^-4 f^(4)(0) = -6 f^(5)(x) = 24(1+x)^-5 f^(5)(0) = 24 Step 2: Andika fomula ya mfululizo wa Maclaurin. f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + f^(4)(0)4!x^4 + f^(5)(0)5!x^5 + Step 3: Badilisha vianzio vilivyotathminiwa kwenye fomula. (1+x) = 0 + (1)x + (-1)/(2!)x^2 + (2)/(3!)x^3 + (-6)/(4!)x^4 + (24)/(5!)x^5 + (1+x) = x - (1)/(2)x^2 + (2)/(6)x^3 - (6)/(24)x^4 + (24)/(120)x^5 + (1+x) = x - (x^2)/(2) + (x^3)/(3) - (x^4)/(4) + (x^5)/(5) + (ii) Kisha bainisha _0^0.2 x (1 + x)dx ukitoa jibu sahihi kwa sehemu 3 za desimali. Step 1: Zidisha mfululizo wa Maclaurin kwa (1+x) kwa x. x (1+x) = x ( x - (x^2)/(2) + (x^3)/(3) - (x^4)/(4) + (x^5)/(5) + ) x (1+x) = x^2 - (x^3)/(2) + (x^4)/(3) - (x^5)/(4) + (x^6)/(5) + Step 2: Unganisha mfululizo neno kwa neno kutoka 0 hadi 0.2. _0^0.2 x (1+x)dx = _0^0.2 ( x^2 - (x^3)/(2) + (x^4)/(3) - (x^5)/(4) + (x^6)/(5) + ) dx = [ (x^3)/(3) - (x^4)/(8) + (x^5)/(15) - (x^6)/(24) + (x^7)/(35) + ]_0^0.2 = ( ((0.2)^3)/(3) - ((0.2)^4)/(8) + ((0.2)^5)/(15) - ((0.2)^6)/(24) + ((0.2)^7)/(35) + ) - (0) Step 3: Kokotoa thamani za namba za maneno. Neno la 1: ((0.2)^3)/(3) = (0.008)/(3) ≈ 0.00266667 Neno la 2: -((0.2)^4)/(8) = -(0.0016)/(8) = -0.0002 Neno la 3: ((0.2)^5)/(15) = (0.00032)/(15) ≈ 0.00002133 Neno la 4: -((0.2)^6)/(24) = -(0.000064)/(24) ≈ -0.00000267 Neno la 5: ((0.2)^7)/(35) = (0.0000128)/(35) ≈ 0.00000037 Step 4: Jumlisha maneno na zungusha kwa sehemu 3 za desimali. Jumla ≈ 0.00266667 - 0.0002 + 0.00002133 - 0.00000267 + 0.00000037 Jumla ≈ 0.00248569 Kuzungusha kwa sehemu 3 za desimali, tunapata 0.002. Thamani ya kiunganishi ni 0.002. 7. (b) Tumia theoremu ya Taylor kupanua f(x) = x^4 - 3x^2 + 1 kuhusu x = 1 katika nguvu zinazopanda (x - 1) hadi neno katika (x - 1)^3. Step 1: Tafuta vianzio vya f(x) na uzitathmini kwenye x=1. f(x) = x^4 - 3x^2 + 1 f(1) = (1)^4 - 3(1)^2 + 1 = 1 - 3 + 1 = -1 f'(x) = 4x^3 - 6x f'(1) = 4(1)^3 - 6(1) = 4 - 6 = -2 f''(x) = 12x^2 - 6 f''(1) = 12(1)^2 - 6 = 12 - 6 = 6 f'''(x) = 24x f'''(1) = 24(1) = 24 Step 2: Andika fomula ya mfululizo wa Taylor kuhusu x=a. f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + Hapa a=1. Step 3: Badilisha vianzio vilivyotathminiwa kwenye fomula hadi neno katika (x-1)^3. f(x) = f(1) + f'(1)(x-1) + (f''(1))/(2!)(x-1)^2 + (f'''(1))/(3!)(x-1)^3 + f(x) = -1 + (-2)(x-1) + (6)/(2!)(x-1)^2 + (24)/(3!)(x-1)^3 + f(x) = -1 - 2(x-1) + (6)/(2)(x-1)^2 + (24)/(6)(x-1)^3 + f(x) = -1 - 2(x-1) + 3(x-1)^2 + 4(x-1)^3 + Got more? Send 'em!