112年中央大學地球科學碩士班-微積分詳解

國立中央大學112學年度碩士班考試入學

所別: 地球科學系地球物理碩士班
科目: 微積分

解答: $$\textbf{(a)}\lim_{x \to a}{a-x\over \ln{x\over a}} =\lim_{x \to a}{(a-x)'\over (\ln{x\over a})'} = \lim_{x \to a}{-1\over {1\over x}} = \bbox[red,2pt]{-1} \\ \textbf{(b)}\lim_{x \to \infty} \frac{3x+4}{\sqrt{2x^2-5}} =\lim_{x \to \infty} \frac{3+4/x}{\sqrt{2-5/x}} ={3\over \sqrt 2} =\bbox[red, 2pt]{3\sqrt 2\over 2}$$

解答: $$\textbf{(a)}\; \frac{d }{dx}(\tanh(x)) =\frac{d }{dx} \left({e^x-e^{-x}\over e^x+e^{-x}}\right)  ={e^x+e^{-x}\over e^x+e^{-x}} -{(e^x-e^{-x})(e^x-e^{-x}) \over (e^x+e^{-x})^2} =1- \left({e^x-e^{-x} \over e^x+e^{-x}} \right) \\\qquad =\bbox[red, 2pt]{1-\tanh^2 (x)}\\ \textbf{(b)}\; y=\sin^{-1}x \Rightarrow \sin y= x \Rightarrow y'\cos y= 1 \Rightarrow y'={1\over \cos y} = \bbox[red, 2pt]{1\over \sqrt {1-x^2}} \;\\\qquad (\sin y=x \Rightarrow \cos y=\sqrt{1-x^2})$$

解答: $$\textbf{(a)}\; \cases{u=e^{ax} \\ dv=\sin(bx)dx} \Rightarrow \cases{du =ae^{ax}dx \\ v=-{1\over b}\cos(bx)} \\\qquad\Rightarrow \int e^{ax}\sin(bx)dx = -{1\over b}e^{ax} \cos(bx)+{a\over b} \int e^{ax} \cos(bx)\,dx \cdots(1) \\\quad \cases{u=e^{ax} \\ dv=\cos(bx)} \Rightarrow \cases{du =ae^{ax} \\ v={1\over b}\sin(bx)} \\\qquad \Rightarrow \int e^{ax} \cos(bx)\,dx = {1\over b}e^{ax} \sin(bx)-{a\over b}\int e^{ax} \sin(bx)\,dx \cdots(2) \\I= \int e^{ax} \sin(bx)\,dx \stackrel{(1)}{= }-{1\over b}e^{ax} \cos(bx)+{a\over b} \int e^{ax} \cos(bx)\,dx \\ \stackrel{(2)}{= }-{1\over b}e^{ax} \cos(bx)+{a\over b}\left( {1\over b}e^{ax} \sin(bx)-{a\over b}\int e^{ax} \sin(bx)\,dx \right) \\= -{1\over b}e^{ax} \cos(bx)+ {a\over b^2}e^{ax} \sin(bx)-{a^2\over b^2}I \Rightarrow \left(1+{a^2 \over b^2} \right) I= -{1\over b}e^{ax} \cos(bx)+ {a\over b^2}e^{ax} \sin(bx) \\ \Rightarrow I={b^2\over a^2+b^2} \left( -{1\over b}e^{ax} \cos(bx)+ {a\over b^2}e^{ax} \sin(bx)\right) =\bbox[red, 2pt]{{a\over a^2+b^2}e^{ax} \sin(bx)-{b\over a^2+b^2} \cos(bx)} \\\textbf{(b)}\; I=\int_{-\infty}^\infty e^{-x^2}\,dx \Rightarrow I^2= \int_{-\infty}^\infty e^{-x^2}\,dx \int_{-\infty}^\infty e^{-y^2}\,dy = \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+ y^2)}\,dy dx \\ \text{Let }\cases{x=r\cos \theta\\ y=r \sin \theta} \Rightarrow I^2 =\int_0^{2\pi} \int_0^\infty re^{-r^2}\,drd\theta =\int_0^{2\pi}  \left. \left[- {1\over 2}e^{-r^2} \right] \right|_0^\infty \,d\theta =\int_0^{2\pi} {1\over 2}\,d\theta =\pi\\ \Rightarrow I= \bbox[red, 2pt]{\sqrt \pi}$$

解答: $$y''+3y'+2y=0 \Rightarrow \lambda^2+3\lambda+2=0 \Rightarrow (\lambda+2) (\lambda+1)=0 \Rightarrow \lambda=-1,-2 \Rightarrow y_h=c_1e^{-x} +c_2e^{-2x} \\ y_p= ax^2 +bx+c \Rightarrow y_p'=2ax+b \Rightarrow y_p'=2a \\\Rightarrow y_p''+3y_p'+2y_p =2ax^2+(6a+2b)x +2a+3b+2c =12x^2 \Rightarrow \cases{2a=12\\ 6a+2b=0\\ 2a+3b+2c=0} \Rightarrow \cases{a=6\\ b=-18\\ c=21} \\ \Rightarrow y_p=6x^2-18x+21 \Rightarrow y=y_h+y_p \Rightarrow \bbox[red,2pt]{y= c_1e^{-x} +c_2e^{-2x}+6x^2-18x+21}$$

解答: $$u(x,t) =X(x)T(t) \Rightarrow XT''=c^2X''T \Rightarrow {T''\over c^2T} ={X''\over X} \\ BC\; \cases{u(0,t)=0\\ u(L,t)=0} \Rightarrow \cases{X(0)T(t)=0\\ X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\ \text{Let }{T''\over c^2T} ={X''\over X} =k\\ \textbf{Case 1. }\mathbf{k=0}\\ \qquad X''=0 \Rightarrow X=c_1x+ c_2 \Rightarrow BC\; \cases{X(0)=c_2=0\\ X(L)= c_1L+c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0} \Rightarrow X=0\\\textbf{Case 2. } \mathbf{k \gt 0} \\\qquad k=\rho^2 (\rho \gt 0) \Rightarrow X''-\rho^2X=0 \Rightarrow X=c_1 e^{\rho x} +c_2 e^{-\rho x} \Rightarrow BC\; \cases{X(0)= c_1+c_2=0 \\X(L)=c_1e^{\rho L}+ c_2e^{-\rho L}=0} \\\qquad \Rightarrow c_2=-c_1 \Rightarrow c_1e^{\rho L}-c_1e^{-\rho L} =0 \Rightarrow c_1(e^{2\rho L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case 3. }\mathbf{k\lt 0} \\\qquad k=-\rho^2 (\rho\gt 0) \Rightarrow X''+\rho^2 X=0 \Rightarrow X=c_1 \cos(\rho x)+ c_2\sin(\rho x) \Rightarrow BC\; \cases{X(0)=c_1=0 \\ X(L)=c_2 \sin(\rho L)=0} \\\qquad \Rightarrow \sin(\rho L)=0 \Rightarrow \rho ={n\pi \over L} \Rightarrow X_n= \sin{n\pi x\over L},n\in \mathbb N\\ \Rightarrow {T''\over c^2T}=-\rho^2 \Rightarrow T''+\rho^2c^2 T=0 \Rightarrow T= c_3\cos(\rho c t) +c_4\sin(\rho c t) \Rightarrow T'=c_4 \rho c\cos(\rho c t)-c_3 \rho c\sin( \rho c t)\\ u_t(x,t) |_{t=0} =X(x)T'(0)=0 \Rightarrow T'(0)=0 \Rightarrow c_4 \rho c=0 \Rightarrow c_4=0 \Rightarrow T=c_3\cos(\rho c t) \\ \Rightarrow T_n= \cos({n\pi c t \over L}) \Rightarrow u_n(x,t) =X_n(x)T_n(t) = \sin{n\pi x\over L} \cos {n\pi c t\over L} \\ \Rightarrow u(x,t)=\sum_{n=1}^\infty a_n \sin{n\pi x\over L} \cos {n\pi c t\over L} \Rightarrow u(x,0)=f(x) =\sum_{n=1}^\infty a_n \sin{n\pi x\over L}  \\ \Rightarrow a_n={2\over L} \int_0^L f(x)\sin{n\pi x\over L}\,dx\\ \Rightarrow \bbox[red, 2pt] {u(x,t)=\sum_{n=1}^\infty a_n \sin{n\pi x\over L} \cos {n\pi c t\over L}, a_n={2\over L} \int_0^L f(x)\sin{n\pi x\over L}\,dx}$$

解答: $$f_{odd}(x)=\begin{cases} f(x)& 0\lt x\lt L\\ -f(-x)& -L\lt x\lt 0\end{cases} \\ \Rightarrow b_n ={2\over L} \int_0^L f(x)\sin{n\pi x\over L}\,dx ={2\over L}\left(\int_0^{L/2} {2k\over L}x \sin{n\pi x\over L} \,dx +\int_{L/2}^L {2k\over L}(L-x) \sin {n\pi x\over L} \,dx\right) \\={2\over L}\left( -{kL\over n\pi}\cos{n\pi \over 2}  + {2kL\over n^2\pi^2} \sin{n\pi \over 2}+ {2kL\over n^2\pi^2} \sin{n\pi\over 2}+ {kL\over n\pi} \cos{n\pi \over 2}\right) ={2\over L}\cdot {4kL\over n^2\pi^2} \sin{n\pi \over 2} \\={8k\over n^2\pi^2} \sin{n\pi \over 2}  \Rightarrow \bbox[red, 2pt]{f_{odd}(x) =\sum_{n=1}^\infty {8k\over n^2\pi^2} \sin{n\pi \over 2} \sin{n\pi x\over L}}$$

解答: $$[A\mid I] = \left[ \begin{array}{rrr|rrr} -1 & 1 & 2 & 1 & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-1 & 3 & 4 & 0 & 0 & 1\end{array} \right] \xrightarrow{-R_1+R_3\to R_3, 3R_1+R_2\to R_2}    \left[ \begin{array}{rrr|rrr} -1 & 1 & 2 & 1 & 0 & 0\\0 & 2 & 7 & 3 & 1 & 0\\0 & 2 & 2 & -1 & 0 & 1 \end{array} \right] \\ \xrightarrow{-R_1\to R_1, R_2-R_3 \to R_3} \left[ \begin{array}{rrr|rrr} 1 & -1 & -2 & -1 & 0 & 0\\0 & 2 & 7 & 3 & 1 & 0\\0 & 0 & -5 & -4 & -1 & 1 \end{array} \right] \xrightarrow{(1/2)R_2 \to R_2, -(1/5)R_3 \to R_3} \\\left[ \begin{array}{rrr|rrr} 1 & -1 & -2 & -1 & 0 & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{array} \right] \xrightarrow{R_1+R_2 \to R_1, R_2-(7/2)R_3 \to R_2}\left[ \begin{array}{rrr|rrr}  1 & 0 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0\\0 & 1 & 0 & - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right] \\ \xrightarrow{R_1-(3/2)R_3 \to R_1}\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & - \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\0 & 1 & 0 & - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{array} \right] \Rightarrow A^{-1}= \bbox[red, 2pt]{ \left[ \begin{array}{rrr|rrr} - \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\- \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\\frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{array} \right] }$$

解答: $$y'(t)=-\lambda y(t) \Rightarrow {1\over y}dy =-\lambda \,dt \Rightarrow \ln y=-\lambda t+  y_0 \Rightarrow y(t)=y(0)e^{-\lambda t} \\ \text{half-life }\Rightarrow e^{-\lambda T}={1\over 2} \Rightarrow -\lambda T=-\ln 2 \Rightarrow T_{1/2}= \ln 2/\lambda \Rightarrow \lambda T_{1/2}=\ln 2, \bbox[red, 2pt]{Q.E.D.}$$

解答: $$\textbf{(b)}  \qquad \qquad \iint_S \vec F\cdot d\vec S = \iiint_E \text{div }\vec F\,dV, \\ \text{where }\vec F\text{ is a vector field and its compoenets are continuous and first partial derivative. }\\ \text{E is closed bounded region and }S \text{ is the boundary surface of }E. \\ \textbf{(c)}\; \href{https://byjus.com/physics/derivation-of-heat-equation/}{\text{derivation of heat equation}}$$

解答: $$\cases{x(t)=a\cos t\\ y(t)=a\sin t},0\le t\le 2\pi \Rightarrow \cases{x'(t)=-a\sin t\\ y'(t)=a\cos t}\Rightarrow \text{length of a circle:}\int_0^{2\pi} \sqrt{(x'(t))^2+ (y'(t))^2}\,dt \\=\int_0^{2\pi} \sqrt{a^2(\cos^2t+ \sin^2 t)}\,dt =\int_0^{2\pi} a\,dt =2\pi a,\bbox[red, 2pt]{Q.E.D.}$$
 

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解題僅供參考,其他歷年試題及詳解