114年中山大學海科碩士班-微積分詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：微積分【海科系碩士班乙組選考】

解答: $$\text{area =} {1\over 2}\int_0^{2\pi} r^2\,d\theta = {1\over 2}\int_0^{2\pi} (3+2\cos \theta)^2\,d\theta  = {1\over 2}\int_0^{2\pi}(9+12\cos \theta+ 4\cos^2 \theta)\,d\theta \\ = {1\over 2}\int_0^{2\pi}(9+0+ 2(\cos 2 \theta+1))\,d\theta ={1\over 2}\int_0^{2\pi}11\,d\theta= \bbox[red, 2pt]{11\pi}$$

解答: $${d\over dx} [2(3^x)+ 5e^x]={d\over dx}[2e^{x\ln 3} +5e^x] = \bbox[red, 2pt]{2 (\ln 3)3^x + 5e^x}$$

解答: $$\text{Change the order of integration, }\int_0^3\int_y^3 e^{x^2}\,dxdy = \int_0^3 \int_0^x e^{x^2}\,dydx = \int_0^3 xe^{x^2}\,dx \\=\left. \left[ {1\over 2}e^{x^2} \right] \right|_0^3 =\bbox[red, 2pt]{{1\over 2}(e^9-1)}$$

解答: $$\cases{u=x^2 \\ dv=e^{-x}\,dx} \Rightarrow \cases{du = 2xdx\\ v=-e^{-x}} \Rightarrow I= \int x^2e^{-x}\,dx = -x^2e^{-x} +2 \int xe^{-x}\,dx \\ \cases{u=x\\ dv= e^{-x}\,dx} \Rightarrow \cases{du =dx \\ v=-e^{-x}} \Rightarrow I= -x^2e^{-x}  +2\left( -xe^{-x} + \int e^{-x}\,dx\right) \\= \bbox[red, 2pt]{-x^2e^{-x}-2xe^{-x}-2e^{-x}+c}$$

解答: $$f(x)=3xe^{(x^2-1)} \Rightarrow f(-x)= -3xe^{(x^2-1)}=-f(x) \Rightarrow f(x) \text{ is odd }\Rightarrow \int_{-1}^1 f(x)=\bbox[red, 2pt]0$$

解答: $$\cases{e^{i\sigma t} = \cos \sigma t+ i\sin \sigma t\\ e^{-\sigma t} =\cos \sigma t -i\sin \sigma t}  \Rightarrow u_+ e^{i\sigma t}+ u_-e^{-i\sigma t}\\ = {1\over 2}[(a_1+b_2)+ i(a_2-b_1)] (\cos \sigma t+ i\sin \sigma t) +{1\over 2}[(a_1-b_2)+ i(a_2+b_1)] (\cos \sigma t -i\sin \sigma t) \\ \Rightarrow Re( u_+ e^{i\sigma t}+ u_-e^{-i\sigma t})  ={1\over 2}\left( (a_1+b_2) \cos \sigma  t +(b_1-a_2) \sin \sigma t + (a_1-b_2) \cos \sigma t+ (a_2+ b_1) \sin \sigma t\right)\\\qquad = a_1\cos \sigma t+ b_1\sin \sigma t=u_1\\ \Rightarrow Im ( u_+ e^{i\sigma t}+ u_-e^{-i\sigma t})   = {1\over 2} \left( (a_2-b_1)\cos \sigma t+ (a_1+ b_2) \sin \sigma t +(a_2+b_1) \cos \sigma t-(a_1-b_2)\sin \sigma t\right) \\\qquad = a_2\cos \sigma t+ b_s \sin \sigma t =u_2\\ \Rightarrow u_1+ iu_2 =u_+e^{i\sigma t}+u_-e^{-i\sigma t} \qquad \bbox[red, 2pt]{QED}$$

解答: $$\nabla \times(\varphi \vec A) =\varphi \nabla \times \vec A+ \nabla \varphi \times \vec A \cdots(1)\\ \nabla \left({\varphi \over \Phi} \right) ={\Phi \nabla \varphi-\varphi \nabla \Phi\over \Phi^2} \cdots(2)\\ -\nabla \times \left({1\over \rho }\nabla p\right) =-{1\over \rho} (\nabla\times\nabla p)-\left( \nabla{1\over \rho} \right)\times \nabla p \qquad \text{by (1)}\\ =-{1\over \rho} (\nabla\times\nabla p)-\left( {\rho \nabla 1-1\nabla \rho\over \rho^2} \right)\times \nabla p \qquad \text{by (2)} \\ =0+\left( { \nabla \rho\over \rho^2} \right)\times \nabla p ={\nabla \rho \times \nabla p \over \rho^2} \Rightarrow   -\nabla \times \left({1\over \rho }\nabla p\right) ={\nabla \rho \times \nabla p \over \rho^2}\qquad \bbox[red, 2pt]{QED}$$

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解題僅供參考，碩士班歷年試題及詳解