114年中央大學地球科學碩士班-微積分詳解

 國立中央大學114學年度碩士班考試入學試題

系所: 地球科學學系地球物理 碩士班 不分組(一般生)
地球科學學系地球物理 碩士班 不分組(在職生)
科目: 微積分

解答:$$\textbf{(a) }\lim_{x\to \infty} x\sin{1\over x} = \lim_{x\to \infty} {\sin{1\over x}\over 1/x} = \lim_{t\to 0}{ \sin t\over t} = \lim_{t\to 0}{ (\sin t)'\over (t)'} =\lim_{t\to 0}{ \cos t\over 1} = \bbox[red, 2pt]1 \\\textbf{(b) } L= \left( 1+{1\over x} \right)^x \ln L= x\ln\left( 1+{1\over x} \right) ={\ln\left( 1+{1\over x} \right) \over 1/x} \\ \qquad \Rightarrow \lim_{x\to \infty} {\ln\left( 1+{1\over x} \right) \over 1/x} =\lim_{x\to \infty} {\ln\left( 1+{1\over x} \right) \over 1/x} =\lim_{x\to \infty} { \left( {x\over x+1} \right) (-1/x^2)\over -1/x^2} =\lim_{x\to \infty}{x\over x+1}=1 \\\qquad \Rightarrow \lim_{x\to \infty} L= e^1= \bbox[red, 2pt]e$$

解答:$$\textbf{(a) } {d\over dx} \left(  \int_{A(x)}^{B(x)} e^t\,dt\right) = \bbox[red, 2pt]{e^{B(x)} B'(x)-e^{A(x)}A'(x)} \\\textbf{(b) }{d\over dx} \sqrt{x^2+1} ={1\over 2}(x^2+1)^{-1/2} \cdot (2x) = \bbox[red, 2pt]{x\over \sqrt{x^2+1}}$$

解答:$$\textbf{(a) }  \cases{u=\ln x\\ dv=x\,dx} \Rightarrow \cases{du =dx/x\\ v={1\over 2}x^2} \Rightarrow \int x \ln(x)\,dx ={1\over 2}x^2\ln x-{1\over 2}\int x\,dx = \bbox[red, 2pt]{{1\over 2}x^2\ln x-{1\over 4}x^2+C} \\\textbf{(b) } I= \int_{-\infty}^\infty e^{-x^2}\,dx \Rightarrow I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \,dx dy = \int_0^{2\pi} \int_0^\infty re^{-r^2} \,drd\theta \\\qquad = \int_0^{2\pi } \left. \left[ -{1\over 2}e^{-r^2} \right] \right|_0^\infty\,d\theta = \int_0^{2\pi } {1\over 2}\,d\theta= \pi \Rightarrow I= \bbox[red, 2pt]{\sqrt{\pi}}$$

解答:$$\textbf{(a) } \sum F=ma \Rightarrow f(t)+F_s+F_d=my''(t) \Rightarrow f(t)-ky(t)-cy'(t)=my''(t) \\\qquad \Rightarrow \bbox[red, 2pt]{my''(t)+ cy'(t)+ky(t)=f(t)} \\\textbf{(b) }\cases{c=0\\ f(t)=0} \Rightarrow my''+ky =0 \Rightarrow y''+{k\over m}y=0 \Rightarrow \bbox[red, 2pt]{y(t)=c_1 \cos \sqrt{k\over m}t +c_2 \sin \sqrt{k\over m}t}$$

解答:$$A = \begin{bmatrix}5& 1\\2& 4 \end{bmatrix} \Rightarrow \det(A-\lambda I) = \lambda^2-9\lambda +18=(\lambda-3)(\lambda-6)=0 \Rightarrow \text{eigenvalues: 6,3} \\ \lambda_1=6 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-1& 1\\2& -2 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} =0 \Rightarrow x=y \Rightarrow v= x \begin{bmatrix}1\\ 1 \end{bmatrix} \\ \qquad \text{Choosing }x=1, \text{ we get }v_1= \begin{bmatrix}1\\1 \end{bmatrix} \\ \lambda_2=3 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}2& 1\\2& 1 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} =0 \Rightarrow 2x+y=0 \Rightarrow v=x \begin{bmatrix}1\\ -2 \end{bmatrix} \\ \qquad \text{Choosing }x=-1, \text{ we get }v_2= \begin{bmatrix}-1\\ 2 \end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: }6,3 , \text{ and eigenvectors: }\begin{bmatrix}1\\ 1 \end{bmatrix}, \begin{bmatrix}-1\\ 2 \end{bmatrix}}$$

解答:$$N(t): \text{the remaining quantity} \Rightarrow {d N\over dt}=-\lambda N \Rightarrow \int{dN\over N}=\int-\lambda \,dt \Rightarrow \ln N= -\lambda t+C \\ \Rightarrow N=e^C\cdot e^{-\lambda t} =N_0e^{-\lambda  t}\\ \text{At }t={T_{1/2}}, N(t)={N_0\over 2} \Rightarrow {N_0\over 2}=N_0 e^{-\lambda T_{1/2}} \Rightarrow e^{-\lambda T_{1/2}} ={1\over 2} \Rightarrow -\lambda T_{1/2} =-\ln (2) \Rightarrow \bbox[red, 2pt]{T_{1/2} ={\ln 2\over \lambda}}$$

解答:$$V=\int_0^{2\pi} \int_0^a \int_0^{h-hr/a} r\,dz\,dr\,d\theta  =\int_0^{2\pi} \int_0^a  \left( hr-{h\over a}r^2 \right)\,dr\,d\theta= \int_0^{2\pi} {1\over 6}ha^2\,d\theta =\bbox[red, 2pt]{{1\over 3}a^2\pi h}$$

解答:$$\textbf{(a) }\bbox[red, 2pt]{u(t-a)= \begin{cases} 0& \text{if }t\lt a\\ 1& \text{if }t\ge a\end{cases} }\Rightarrow L\{u(t-a)\} = \int_0^\infty u(t-a)e^{-st}\,dt = \int_a^\infty e^{-st}\,dt\\ = \left. \left[ -{1\over s}e^{-st} \right] \right|_a^\infty ={1\over s}e^{-as} \Rightarrow \bbox[red, 2pt]{ L\{u(t-a)\} ={1\over s}e^{-as}} \\\textbf{(b) } L^{-1} \{F(s)\} =L^{-1}\left\{ {e^{-2s} \over s^2+\pi^2} \right\} +L^{-1}\left\{{e^{-2s} \over (s+2)^2} \right\} \\=u(t-2)L^{-1}\left\{ {1 \over s^2+\pi^2} \right\} (t-2) +u(t-2) L^{-1}\left\{ {1 \over (s+2)^2} \right\}(t-2) \\=u(t-2){1\over \pi}\sin(\pi (t-2)) +u(t-2)e^{-2(t-2)}(t-2) = \bbox[red, 2pt]{u(t-2) \left( {1\over \pi}\sin(\pi(t-2)) +(t-2)e^{-2(t-2)}\right)}$$

解答:$$u(x,t)=X(x)T(t) \Rightarrow XT'=c^2X''T \Rightarrow {T'\over c^2T} ={X''\over X} = -\lambda \Rightarrow \cases{X''+\lambda X=0\\ T'+c^2\lambda T=0} \\ \text{B.C.:} \cases{u(0,t)=0\\ u(L,t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\ \textbf{Case I }\lambda =0 \Rightarrow \text{ Trivial solution }X=0\\ \textbf{Case II }\lambda \lt  \Rightarrow \text{ Trivial solution }X=0 \\ \textbf{Case III }\lambda \gt 0 \Rightarrow \lambda=\beta^2 \Rightarrow X= A\cos \beta x+ B\sin \beta x \Rightarrow \cases{X(0)=A=0 \\ X(L)=A\cos \beta L+ B\sin \beta L=0} \\\qquad \Rightarrow \sin \beta L=0 \Rightarrow \beta L= n\pi \Rightarrow \beta_n={n\pi\over L} \Rightarrow \lambda_n={n^2\pi^2 \over L^2} \Rightarrow X_n= \sin{n\pi x\over L}, n\in \mathbb N \\ \Rightarrow T'+c^2\cdot {n^2\pi^2\over L^2}T=0 \Rightarrow T_n= C_n e^{-(c n\pi/L)^2t} \Rightarrow u_n=X_nT_n \\ \Rightarrow u(x,t)=  \sum_{n=1}^\infty d_n \sin{n\pi x\over L} e^{-(c n \pi/L)^2 t} \Rightarrow u(x,t)= f(x)=  \sum_{n=1}^\infty d_n \sin{n\pi x\over L} \\ \Rightarrow d_n={2\over L} \int_0^L f(x)\sin{n\pi x\over L} \\ \Rightarrow \bbox[red, 2pt]{ u(x,t)= \sum_{n=1}^\infty d_n \sin{n\pi x\over L} e^{-(c n \pi/L)^2 t}, \text{ where } d_n={2\over L} \int_0^L f(x)\sin{n\pi x\over L}}$$

解答:$$f(x)= \sum_{n=1}^\infty b_n \sin {n\pi x\over L} \Rightarrow b_n= {2\over L} \int_0^L f(x) \sin{n \pi x\over L} \,dx \\= {2\over L} \int_0^{L/2} {2k\over L}x \sin{n\pi x\over L}\,dx + \int_{L/2}^L {2k\over L}(L-x)\sin {n\pi x\over L}\,dx ={4k\over L^2} (I_1+ I_2) \\ \cases{u=x\\ dv= \sin(n\pi x/L)dx} \Rightarrow \cases{du =dx\\ v=-L/(n\pi) \cos(n\pi x/L)} \Rightarrow I_1= \int_0^{L/2} x\sin {n\pi x\over L}\,dx \\= \left. \left[ -{Lx\over n\pi}\cos {n \pi x\over L} \right] \right|_0^{L/2}+{L\over n\pi} \int_0^{L/2} \cos {n\pi x\over L}\,dx=-{L^2\over 2n\pi}\cos{n\pi\over 2}+{L^2\over n^2\pi^2} \sin{n\pi\over 2} \\I_2= \int_{L/2}^L (L-x)\sin{n\pi x\over L}\,dx = \left. \left[ -{L(L-x) \over n\pi}\cos {n\pi x\over L} \right] \right|_{L/2}^L -{L\over n\pi}\int_{L/2}^L \cos {n\pi x\over L}\,dx \\\quad ={L^2\over 2n\pi} \cos{n\pi\over 2}+{L^2\over n^2\pi^2} \sin{n\pi\over 2} \\ \Rightarrow I_1+ I_2 ={2L^2\over n^2\pi^2} \sin{n\pi\over 2} \Rightarrow b_n ={8k\over n^2\pi^2} \sin{n\pi\over 2} \Rightarrow \bbox[red, 2pt]{f(x)={8k\over \pi^2} \sum_{n=1}^\infty {1\over n^2}\sin{n\pi \over 2} \sin{n\pi x\over L}}$$

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解題僅供參考，碩士班歷年試題及詳解