115年中央地科碩士班-微積分詳解

國立中央大學115學年度碩士班考試入學試題

系所:地球科學 科目:微積分

解答:$$\textbf{(a) }y=\cos^{-1}  \theta\Rightarrow \cos y=\theta \Rightarrow  -\sin y\cdot y'=1 \Rightarrow -\cos y\cdot (y')^2-\sin y\cdot y''=0 \\ \qquad \Rightarrow y''=-{\cos y\cdot (y')^2\over \sin y} =-{\theta\cdot {1\over 1-\theta^2} \over \sqrt{1-\theta^2}} = \bbox[red, 2pt]{-{\theta\over (1-\theta^2)^{{3/2}}}} \\\textbf{(b) }\lim_{x\to \pi/2} {\tan x-5\over \sec x+4} =\lim_{x\to \pi/2} {(\sin \theta/\cos \theta)-5\over (1/\cos \theta)+4} =\lim_{x\to \pi/2} {\sin x-5\cos x\over 1+4\cos x} ={1-0\over 1+0}= \bbox[red, 2pt]1$$

解答:$$\textbf{(a) } {d\over dx} \left(  \int_{A(x)}^{B(x)} e^t\,dt  \right) = \bbox[red, 2pt]{e^{B(x)}\cdot B'(x)-e^{A(x)} \cdot A'(x)} \\ \textbf{(b)} f(x)={1\over x} \Rightarrow f'(x)=-{1\over x^2} \Rightarrow f''(x)={2\over x^3} \Rightarrow f'''(x)=-{6\over x^4} \Rightarrow \cdots \\\quad \Rightarrow f^{(n)} (x)= \bbox[red, 2pt]{(-1)^n\cdot {n!\over x^{n+1}}}$$

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

解答:$$\vec v = \left[ \ln(x^2+y^2), 2\tan^{-1} \left( {y\over x} \right),0 \right] \Rightarrow \text{curl }\vec v = \begin{vmatrix} \mathbf i& \mathbf j& \mathbf k\\ {\partial \over \partial x} & {\partial \over \partial y}& {\partial \over \partial z} \\\ln(x^2+y^2)& 2\tan^{-1} \left( {y\over x} \right)& 0\end{vmatrix} \\= 0\mathbf i+0\mathbf j-{4y\over x^2+y^2}\mathbf k = \bbox[red, 2pt]{\left[ 0,0, -{4y\over x^2+y^2} \right]}$$

解答:$$ A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{bmatrix} \Rightarrow  \det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 & 0 \\ 1 & 3-\lambda & 1 \\ 0 & 1 & 2-\lambda \end{vmatrix} = 0 \Rightarrow  (2-\lambda)(\lambda - 1)(\lambda - 4) = 0 \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvalues are: }1,2,4} \\ \lambda=1 \Rightarrow (A-\lambda I)v=0 \Rightarrow \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{x=-y\\ z=-y} \Rightarrow v=y \begin{bmatrix}-1\\1\\-1 \end{bmatrix} \\ \qquad \text{choosing }v_1= \begin{bmatrix}1\\-1\\1 \end{bmatrix} \\ \lambda=2 \Rightarrow (A-\lambda I)v=0 \Rightarrow   \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{x=-z\\ y=0} \Rightarrow v= x \begin{bmatrix}1\\0\\-1 \end{bmatrix} \\ \qquad \text{choosing }v_2= \begin{bmatrix}1\\0\\-1 \end{bmatrix} \\ \lambda =4 \Rightarrow (A-\lambda I)v=0 \Rightarrow    \begin{bmatrix} -2 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{y=2x\\ x=z} \Rightarrow v=x \begin{bmatrix}1\\2\\1 \end{bmatrix} \\ \qquad \text{choosing }v_3 = \begin{bmatrix}1\\2\\1 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors are:}\begin{bmatrix}1\\-1\\1 \end{bmatrix}, \begin{bmatrix}1\\0\\-1 \end{bmatrix}, \begin{bmatrix}1\\1\\1 \end{bmatrix}}$$

解答:$$\cases{u=t^v\\ dw = e^{-t}\,dt} \Rightarrow \cases{du=vt^{v-1}\,dt\\ w=-e^{-t}}\Rightarrow \Gamma(v+1)= \int_0^\infty e^{-t}t^v \,dt = \left. \left[ -e^{-t}t^v \right] \right|_0^\infty+ \int_0^\infty e^{-t}vt^{v-1}\,dt\\ =v \int_0^\infty e^{-t} t^{v-1}\,dt =v\Gamma(v) \Rightarrow \Gamma(v+1)=v\Gamma(v) \\ \Gamma(1) =\int_0^\infty e^{-t}t^0\,dt = \left. \left[ -e^{-t} \right] \right|_0^\infty=0-(-1)=1 \quad \bbox[red, 2pt]{QED.}$$

解答:$$y''+2y'+y=0 \Rightarrow \alpha^2+2\alpha+1=0 \Rightarrow (\alpha+1)^2= 0 \Rightarrow \alpha=-1 \Rightarrow y_h=c_1e^{-t} +c_2te^{-t}\\ y_p= A\cos(2t)+ B\sin(2t) \Rightarrow y_p'=-2A\sin(2t)+ 2B\cos(2t) \Rightarrow y_p''=-4A\cos(2t)-4B\sin(2t) \\ \Rightarrow y_p''+2y_p'+y_p=(-3A+4B)\cos(2t)+(-4A-3B)\sin(2t) =\sin(2t) \\ \Rightarrow \cases{-3A+4B=0\\ -4A-3B=1} \Rightarrow \cases{A=-4/25\\ B=-3/25} \Rightarrow y_p=-{4\over 25}\cos(2t)-{3\over 25}\sin(2t) \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{-t} +c_2te^{-t}-{4\over 25}\cos(2t)-{3\over 25}\sin(2t)}$$

解答:$$u(x,t) =X(x)T(t) \Rightarrow XT''=c^2X''T \Rightarrow {X''\over X}={T''\over c^2T}=-k^2 \Rightarrow \cases{X''+k^2 X =0\\T''+c^2k^2 T=0} \\ \text{boundary conditions: }\cases{u(0,t)=X(0)T(t)=0\\ u(L,t)=X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0}\\ X''+k^2X=0 \Rightarrow X=A\cos(kx)+B\sin(kx) \Rightarrow X(0)=A=0 \Rightarrow X(x)=B\sin(kx) \\ \Rightarrow X(L)= B\sin(kL)=0 \Rightarrow kL=n\pi \Rightarrow k_n={n\pi\over L} \Rightarrow X_n(x)=\sin{n\pi k\over L},n=1,2,\dots \\ \Rightarrow T_n''+c^2 \left( {cn\pi\over L} \right)^2T_n=0 \Rightarrow T_n= C_n \cos\left( {cn\pi t\over L} \right)+D_n \sin\left( {cn\pi t\over L} \right) \\ \Rightarrow u(x,t)= \sum_{n=1}^\infty X_nT_n = \sum_{n=1}^\infty \left[  C_n \cos\left( {cn\pi t\over L} \right)+D_n \sin\left( {cn\pi t\over L} \right)  \right] \sin{n\pi x\over L} \\ \Rightarrow  u_t(x,t) = \sum_{n=1}^{\infty} \left[ -C_n \left(\frac{cn\pi}{L}\right) \sin\left(\frac{cn\pi t}{L}\right) + D_n \left(\frac{cn\pi}{L}\right) \cos\left(\frac{cn\pi t}{L}\right) \right] \sin\left(\frac{n\pi x}{L}\right) \\ \Rightarrow u_t(x,0)=  \sum_{n=1}^{\infty} D_n \left(\frac{cn\pi}{L}\right) \sin\left(\frac{n\pi x}{L}\right) = 0 \Rightarrow D_n=0 \\ \Rightarrow  u(x,t) = \sum_{n=1}^{\infty} C_n \cos\left(\frac{cn\pi t}{L}\right) \sin\left(\frac{n\pi x}{L}\right) \Rightarrow u(x,0)  = \sum_{n=1}^{\infty} C_n \sin\left(\frac{n\pi x}{L}\right) = f(x) \\ \Rightarrow  C_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \\ \Rightarrow \bbox[red, 2pt] { u(x,t) = \sum_{n=1}^{\infty} C_n \cos\left(\frac{cn\pi t}{L}\right) \sin\left(\frac{n\pi x}{L}\right),C_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx}$$

解答:$$ f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) \Rightarrow  b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \\ \Rightarrow b_n = \frac{2}{L} \left[ \int_{0}^{\frac{L}{2}} \frac{2k}{L}x \sin\left(\frac{n\pi x}{L}\right) dx + \int_{\frac{L}{2}}^{L} \frac{2k}{L}(L-x) \sin\left(\frac{n\pi x}{L}\right) dx \right] \\ \qquad =  \frac{4k}{L^2} \left[ \underbrace{\int_{0}^{\frac{L}{2}} x \sin\left(\frac{n\pi x}{L}\right) dx}_{I_1} + \underbrace{\int_{\frac{L}{2}}^{L} (L-x) \sin\left(\frac{n\pi x}{L}\right) dx}_{I_2} \right] \\ \cases{u=x \\ dv= \sin(n\pi x/L)\,dx } \Rightarrow \cases{du=dx\\ v=-(L/n\pi) \cos(n\pi x/L)} \\ \qquad\Rightarrow I_1=  \left[ -x \frac{L}{n\pi} \cos\left(\frac{n\pi x}{L}\right) \right]_{0}^{\frac{L}{2}} - \int_{0}^{\frac{L}{2}} \left( -\frac{L}{n\pi} \cos\left(\frac{n\pi x}{L}\right) \right) dx  \\\qquad = -\frac{L^2}{2n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{L^2}{n^2\pi^2} \sin\left(\frac{n\pi}{2}\right) \\ \cases{u=L-x\\ dv= \sin(n\pi x/L)dx} \Rightarrow \cases{du=-dx\\ v=-L/(n\pi) \cos(n\pi x/L)} \\ \qquad \Rightarrow I_2=  \left[ -(L-x) \frac{L}{n\pi} \cos\left(\frac{n\pi x}{L}\right) \right]_{\frac{L}{2}}^{L} - \int_{\frac{L}{2}}^{L} \left( \frac{L}{n\pi} \cos\left(\frac{n\pi x}{L}\right) \right) dx \\ \qquad =  \frac{L^2}{2n\pi} \cos\left(\frac{n\pi}{2}\right) - \frac{L^2}{n^2\pi^2} \left( \sin(n\pi) - \sin\left(\frac{n\pi}{2}\right) \right)\\ \Rightarrow b_n= {4k\over L^2} (I_1+I_2) =  \frac{4k}{L^2} \left[ \frac{2L^2}{n^2\pi^2} \sin\left(\frac{n\pi}{2}\right) \right]   = \frac{8k}{n^2\pi^2} \sin\left(\frac{n\pi}{2}\right) \\ \Rightarrow \bbox[red, 2pt]{ f(x) = \sum_{n=1}^{\infty} \frac{8k}{n^2\pi^2} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi x}{L}\right) }$$

解答:$$ 假設N(t) 為在時間 t 的放射性元素數量 \Rightarrow  \frac{dN}{dt} = -\lambda N , 其中衰減常數\lambda \gt 0 \\ \Rightarrow \int{1\over N}dN = \int-\lambda\,dt \Rightarrow \ln N=-\lambda t+C  \Rightarrow N= N_0e^{-\lambda t}\\ 半衰期T \Rightarrow N(T)={1\over 2}N_0 \Rightarrow {1\over 2}N_0=N_0e^{-\lambda t} \Rightarrow \bbox[red, 2pt]{T={\ln 2\over \lambda}}$$

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解題僅供參考，碩士班歷年試題及詳解