Given that z = (x)/(y) y, show that (d z)/(d y) = x (d2 z)/(d y d x).

5. (b) (continued)

• Step 1: (continued) Tumia utofautishaji wa logariti ili kupata mabadiliko ya sehemu katika G.

  $$\ln G = 4 \ln R + \ln \theta - \ln L$$

  Tofautisha pande zote mbili:

  $$\frac{dG}{G} = 4 \frac{dR}{R} + \frac{d\theta}{\theta} - \frac{dL}{L}$$

  Mlinganyo huu unaunganisha mabadiliko ya sehemu katika $G$, $R$, $\theta$, na $L$.

• Step 2: Badilisha asilimia za mabadiliko zilizotolewa. Imepewa: $R$ imeongezeka kwa 3%, hivyo $\frac{dR}{R} = 0.03$. $\theta$ imepungua kwa 5%, hivyo $\frac{d\theta}{\theta} = -0.05$. $L$ imeongezeka kwa 6%, hivyo $\frac{dL}{L} = 0.06$.

  Badilisha thamani hizi kwenye mlinganyo:

  $$\frac{dG}{G} = 4(0.03) + (-0.05) - (0.06)$$ $$\frac{dG}{G} = 0.12 - 0.05 - 0.06$$ $$\frac{dG}{G} = 0.07 - 0.06$$ $$\frac{dG}{G} = 0.01$$

• Step 3: Badilisha mabadiliko ya sehemu kuwa asilimia ya mabadiliko. Asilimia ya mabadiliko katika $G = \frac{dG}{G} \times 100\% = 0.01 \times 100\% = 1\%$. Asilimia ya mabadiliko takriban katika G ni $\boxed{1\%}$.

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.

  $$\frac{\partial z}{\partial x} = 2x + y$$ $$\frac{\partial z}{\partial y} = x + 2y - 1$$

  Weka zote mbili sawa na sifuri ili kupata pointi muhimu:

  1. $2x + y = 0 \implies y = -2x$
  2. $x + 2y - 1 = 0$

• Step 2: Tatua mfumo wa milinganyo kwa $x$ na $y$. Badilisha $y = -2x$ kutoka (1) hadi (2): $x + 2(-2x) - 1 = 0$ $x - 4x - 1 = 0$ $-3x - 1 = 0$ $-3x = 1 \implies x = -\frac{1}{3}$

  Sasa tafuta $y$ kwa kutumia $y = -2x$:
  $y = -2\left(-\frac{1}{3}\right) = \frac{2}{3}$
  Pointi ya kituo ni $\left(-\frac{1}{3}, \frac{2}{3}\right)$.

• Step 3: Tafuta vianzio vya pili vya sehemu.

  $$f_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(2x+y) = 2$$ $$f_{yy} = \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(x+2y-1) = 2$$ $$f_{xy} = \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}(x+2y-1) = 1$$

• Step 4: Tumia jaribio la kianzio cha pili (Hessian determinant) kubainisha asili ya pointi ya kituo. Kibaguzi ni $D = f_{xx}f_{yy} - (f_{xy})^2$.

  $$D = (2)(2) - (1)^2 = 4 - 1 = 3$$

  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.

  $$z = \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)\left(\frac{2}{3}\right) + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)$$ $$z = \frac{1}{9} - \frac{2}{9} + \frac{4}{9} - \frac{6}{9}$$ $$z = \frac{1 - 2 + 4 - 6}{9} = -\frac{3}{9} = -\frac{1}{3}$$

  Pointi ya kituo ni kiwango cha chini cha ndani kwenye $\boxed{\left(-\frac{1}{3}, \frac{2}{3}, -\frac{1}{3}\right)}$.

6. (a) Tathmini kila moja ya viunganishi vifuatavyo:

• (i) $\int (\cos 2x - \sin 3x)dx$

  • Step 1: Unganisha kila neno kivyake. $$\int \cos 2x \,dx = \frac{1}{2} \sin 2x$$ $$\int -\sin 3x \,dx = - \left(-\frac{1}{3} \cos 3x\right) = \frac{1}{3} \cos 3x$$
  • Step 2: Unganisha matokeo na ongeza kigezo cha kuunganisha. $$\int (\cos 2x - \sin 3x)dx = \boxed{\frac{1}{2} \sin 2x + \frac{1}{3} \cos 3x + C}$$

• (ii) $\int (x^2 + 1)dx$

  • Step 1: Unganisha kila neno kivyake kwa kutumia kanuni ya nguvu. $$\int x^2 \,dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$$ $$\int 1 \,dx = x$$
  • Step 2: Unganisha matokeo na ongeza kigezo cha kuunganisha. $$\int (x^2 + 1)dx = \boxed{\frac{x^3}{3} + x + C}$$

• (iii) $\int \frac{x(3x^2 + 5x + 7)}{x^2}dx$

  • Step 1: Rahisisha kiunganishi. $$\frac{x(3x^2 + 5x + 7)}{x^2} = \frac{3x^3 + 5x^2 + 7x}{x^2} = 3x + 5 + \frac{7}{x}$$
  • Step 2: Unganisha kila neno kivyake. $$\int \left(3x + 5 + \frac{7}{x}\right) dx = 3\int x \,dx + \int 5 \,dx + 7\int \frac{1}{x} \,dx$$ $$= 3\left(\frac{x^2}{2}\right) + 5x + 7 \ln|x|$$
  • Step 3: Ongeza kigezo cha kuunganisha. $$\int \frac{x(3x^2 + 5x + 7)}{x^2}dx = \boxed{\frac{3x^2}{2} + 5x + 7 \ln|x| + C}$$

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 $x = -\frac{b}{2a}$ kwa $ax^2+bx+c$. Hapa $y = -x^2 + 2x$, hivyo $a=-1, b=2$. $x = -\frac{2}{2(-1)} = 1$. 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 $x=2$. $$A = \int_0^2 (2x - x^2) dx$$
  • Step 2: Tathmini kiunganishi. $$A = \left[ x^2 - \frac{x^3}{3} \right]_0^2$$ $$A = \left( (2)^2 - \frac{(2)^3}{3} \right) - \left( (0)^2 - \frac{(0)^3}{3} \right)$$ $$A = \left( 4 - \frac{8}{3} \right) - (0)$$ $$A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$$ Eneo la eneo ni $\boxed{\frac{4}{3} \text{ vitengo vya mraba}}$.

7. (a) (i) Bainisha upanuzi wa mfululizo wa Maclaurin wa $\ln(1 + x)$ katika nguvu zinazopanda za $x$ hadi neno katika $x^5$.

• Step 1: Tafuta vianzio vya $f(x) = \ln(1+x)$ na uzitathmini kwenye $x=0$. $f(x) = \ln(1+x) \implies f(0) = \ln(1) = 0$ $f'(x) = \frac{1}{1+x} = (1+x)^{-1} \implies f'(0) = 1$ $f''(x) = -(1+x)^{-2} \implies f''(0) = -1$ $f'''(x) = 2(1+x)^{-3} \implies f'''(0) = 2$ $f^{(4)}(x) = -6(1+x)^{-4} \implies f^{(4)}(0) = -6$ $f^{(5)}(x) = 24(1+x)^{-5} \implies f^{(5)}(0) = 24$

• Step 2: Andika fomula ya mfululizo wa Maclaurin.

  $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \dots$$

• Step 3: Badilisha vianzio vilivyotathminiwa kwenye fomula.

  $$\ln(1+x) = 0 + (1)x + \frac{-1}{2!}x^2 + \frac{2}{3!}x^3 + \frac{-6}{4!}x^4 + \frac{24}{5!}x^5 + \dots$$ $$\ln(1+x) = x - \frac{1}{2}x^2 + \frac{2}{6}x^3 - \frac{6}{24}x^4 + \frac{24}{120}x^5 + \dots$$ $$\ln(1+x) = \boxed{x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} + \dots}$$

• (ii) Kisha bainisha $\int_0^{0.2} x \ln(1 + x)dx$ ukitoa jibu sahihi kwa sehemu 3 za desimali.

  • Step 1: Zidisha mfululizo wa Maclaurin kwa $\ln(1+x)$ kwa $x$.

    $$x \ln(1+x) = x \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} + \dots \right)$$ $$x \ln(1+x) = x^2 - \frac{x^3}{2} + \frac{x^4}{3} - \frac{x^5}{4} + \frac{x^6}{5} + \dots$$

  • Step 2: Unganisha mfululizo neno kwa neno kutoka 0 hadi 0.2.

    $$\int_0^{0.2} x \ln(1+x)dx = \int_0^{0.2} \left( x^2 - \frac{x^3}{2} + \frac{x^4}{3} - \frac{x^5}{4} + \frac{x^6}{5} + \dots \right) dx$$ $$= \left[ \frac{x^3}{3} - \frac{x^4}{8} + \frac{x^5}{15} - \frac{x^6}{24} + \frac{x^7}{35} + \dots \right]_0^{0.2}$$ $$= \left( \frac{(0.2)^3}{3} - \frac{(0.2)^4}{8} + \frac{(0.2)^5}{15} - \frac{(0.2)^6}{24} + \frac{(0.2)^7}{35} + \dots \right) - (0)$$

  • Step 3: Kokotoa thamani za namba za maneno. Neno la 1: $\frac{(0.2)^3}{3} = \frac{0.008}{3} \approx 0.00266667$ Neno la 2: $-\frac{(0.2)^4}{8} = -\frac{0.0016}{8} = -0.0002$ Neno la 3: $\frac{(0.2)^5}{15} = \frac{0.00032}{15} \approx 0.00002133$ Neno la 4: $-\frac{(0.2)^6}{24} = -\frac{0.000064}{24} \approx -0.00000267$ Neno la 5: $\frac{(0.2)^7}{35} = \frac{0.0000128}{35} \approx 0.00000037$

  • Step 4: Jumlisha maneno na zungusha kwa sehemu 3 za desimali. Jumla $\approx 0.00266667 - 0.0002 + 0.00002133 - 0.00000267 + 0.00000037$ Jumla $\approx 0.00248569$ Kuzungusha kwa sehemu 3 za desimali, tunapata 0.002. Thamani ya kiunganishi ni $\boxed{0.002}$.

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 $x=1$. $f(x) = x^4 - 3x^2 + 1 \implies f(1) = (1)^4 - 3(1)^2 + 1 = 1 - 3 + 1 = -1$ $f'(x) = 4x^3 - 6x \implies f'(1) = 4(1)^3 - 6(1) = 4 - 6 = -2$ $f''(x) = 12x^2 - 6 \implies f''(1) = 12(1)^2 - 6 = 12 - 6 = 6$ $f'''(x) = 24x \implies f'''(1) = 24(1) = 24$

• Step 2: Andika fomula ya mfululizo wa Taylor kuhusu $x=a$.

  $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots$$

  Hapa $a=1$.

• Step 3: Badilisha vianzio vilivyotathminiwa kwenye fomula hadi neno katika $(x-1)^3$.

  $$f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \dots$$ $$f(x) = -1 + (-2)(x-1) + \frac{6}{2!}(x-1)^2 + \frac{24}{3!}(x-1)^3 + \dots$$ $$f(x) = -1 - 2(x-1) + \frac{6}{2}(x-1)^2 + \frac{24}{6}(x-1)^3 + \dots$$ $$f(x) = \boxed{-1 - 2(x-1) + 3(x-1)^2 + 4(x-1)^3 + \dots}$$