113年台科大機械碩士班甲-工程數學詳解

國立臺灣科技大學113學年度碩士班招生考試

系所組別: 機械工程系碩士班甲組
科目:工程數學

解答: $$\textbf{(1)}\; [A\mid I] =\left[\begin{array}{rrr|rrr}1 & 4 & 3 & 1 & 0 & 0\\2 & 1 & 2 & 0 & 1 & 0\\1 & 2 & 2 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_2-2R_1\to R_2, R_3-R_1\to R_3} \left[\begin{array}{rrr|rrr}1 & 4 & 3 & 1 & 0 & 0\\0 & -7 & -4 & -2 & 1 & 0\\0 & -2 & -1 & -1 & 0 & 1\end{array} \right] \\ \xrightarrow{-(1/7)R_2 \to R_2} \left[\begin{array}{rrr|rrr}1 & 4 & 3 & 1 & 0 & 0\\0 & 1 & \frac{4}{7} & \frac{2}{7} & - \frac{1}{7} & 0\\0 & -2 & -1 & -1 & 0 & 1\end{array} \right] \xrightarrow{R_1-4R_2\to R_1,R_3+2R_2\to R_3} \\ \left[\begin{array}{rrr|rrr}1 & 0 & \frac{5}{7} & - \frac{1}{7} & \frac{4}{7} & 0\\0 & 1 & \frac{4}{7} & \frac{2}{7} & - \frac{1}{7} & 0\\0 & 0 & \frac{1}{7} & - \frac{3}{7} & - \frac{2}{7} & 1\end{array} \right]  \xrightarrow{7R_3\to R_3}\left[\begin{array}{rrr|rrr}1 & 0 & \frac{5}{7} & - \frac{1}{7} & \frac{4}{7} & 0\\0 & 1 & \frac{4}{7} & \frac{2}{7} & - \frac{1}{7} & 0\\0 & 0 & 1 & -3 & -2 & 7b\end{array} \right] \\ \xrightarrow{R_1-(5/7)R_3\to R_1,R_2-(4/7)R_3 \to R_2} \left[\begin{array}{rrr|rrr}1 & 0 & 0 & 2 & 2 & -5\\0 & 1 & 0 & 2 & 1 & -4\\0 & 0 & 1 & -3 & -2 & 7\end{array} \right] \\ \Rightarrow A^{-1}= \bbox[red, 2pt]{\left[ \begin{matrix}2 & 2 & -5\\2 & 1 & -4\\-3 & -2 & 7\end{matrix} \right]} \\\textbf{(2)}\; A= \left[\begin{matrix}1 & 4 & 3\\2 & 1 & 2\\1 & 2 & 2\end{matrix}\right] \xrightarrow{R_2-2R_1\to R_2,R_3-R_1\to R_3} \left[\begin{matrix}1 & 4 & 3\\0 & -7 & -4\\0 & -2 & -1\end{matrix}\right] \xrightarrow{R_3-(2/7)R_2\to R_3} \left[\begin{matrix}1 & 4 & 3\\0 & -7 & -4\\0 & 0 & \frac{1}{7}\end{matrix}\right] \\ \Rightarrow U= \left[\begin{matrix}1 & 4 & 3\\0 & -7 & -4\\0 & 0 & \frac{1}{7}\end{matrix}\right] \Rightarrow L=\left[\begin{matrix}1 & 0 & 0\\a & 1 & 0\\b & c & 1\end{matrix}\right] \Rightarrow A=LU = \left[\begin{matrix}1 & 4 & 3\\a & 4a-7 & 3a-4\\b & 4b-7c & 3b-4c+ \frac{1}{7} \end{matrix} \right] \\ \Rightarrow \cases{a=2\\b =1\\ c=2/7} \Rightarrow A=LU =\bbox[red, 2pt]{\left[\begin{matrix}1 & 0 & 0\\2 & 1 & 0\\1 & 2/7 & 1\end{matrix}\right]  \left[ \begin{matrix}1 & 4 & 3\\0 & -7 & -4\\0 & 0 & \frac{1}{7}\end{matrix}\right] }$$

解答: $$\textbf{(1)}\; L\{y''\} +L\{y\} =L\{\delta(t-2\pi)\} \Rightarrow s^2Y(s)-s +Y(s)=e^{-2\pi s} \Rightarrow Y(s)={e^{-2\pi s}+s \over s^2+1}\\ \qquad \Rightarrow y(t)= L^{-1}\{Y(s)\} =L^{-1}\left\{ e^{-2\pi s} \over s^2+1\right\} +L^{-1}\left\{{s\over s^2+1} \right\}\\ \Rightarrow \bbox[red, 2pt]{y(t)=u(t-2\pi)\sin(t-2\pi)+ \cos(t)}\\ \textbf{(2)}\; y''-2y'+y=0 \Rightarrow (\lambda-1)^2=0 \Rightarrow \lambda=1 \Rightarrow y_h=c_1e^x+ c_2xe^x\\ \qquad y_p=Ax^2e^x  \Rightarrow y_p'=2Axe^x+Ax^2e^x \Rightarrow y_p''=2Ae^x+4Axe^x+  Ax^2e^x \\ \qquad \Rightarrow y_p''-2y_p'+y_p= 2Ae^x =e^x \Rightarrow A={1\over 2} \Rightarrow y_p={1\over 2}x^2e^x\\ \qquad \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^x +c_2xe^x+{1\over 2}x^2e^x} \\\textbf{(3)}\; (xy)\frac{\text{d}y}{\text{d}x} =2y^2+3x^2 \Rightarrow (2y^2+3x^2)dx+(-xy)dy=0 \Rightarrow \cases{P(x,y) =2y^2 +3x^2\\ Q(x,y)=-xy} \\ \Rightarrow \cases{P_y=4y\\ Q_x=-y} \Rightarrow {P_y-Q_x\over Q}=-{5\over x} \Rightarrow u'=-{5\over x}u \Rightarrow \text{integration facotr }u={1\over x^5} \\ \Rightarrow \cases{uP={2y^2\over x^5}+{3\over x^3}\\ uQ=-{y\over x^4}} \Rightarrow (uP)_y={4y\over x^5} =(uQ)_x \Rightarrow \text{Exact} \\ \Rightarrow \Phi(x,y)= \int \left( {2y^2\over x^5}+{3\over x^3}\right)\,dx =\int -{y\over x^4}dy \Rightarrow {y^2\over 2x^4}+{3\over 2x^2}+c_1=0 \\ \Rightarrow {y^2\over x^4}=c_2-{3\over x^2} \Rightarrow \bbox[red, 2pt]{ y= \pm x\sqrt{c_2x^2-3}}$$

解答: $$\cases{A(1,1,1)\\ B(5,-7,3)\\ C(7,4,8) \\D(10,7,4)} \Rightarrow \cases{\overrightarrow{AB} =(4,-8,2)\\ \overrightarrow{AC}=(6,3,7) \\ \overrightarrow{AD} =(9,6,3)} \Rightarrow V={1\over 6} \begin{Vmatrix}4 & -8& 2 \\6 & 3& 7\\ 9 & 6 & 3 \end{Vmatrix} ={474\over 6} =\bbox[red, 2pt]{79}$$

解答: $$\textbf{(1)}\;S_1: \vec r(u,v)=(u\cos v,u\sin v,b),0\le v\le 2\pi, 0\le u\le a \Rightarrow \cases{\vec t_u=(\cos v,\sin v,0) \\ \vec t_v=(-u\sin v,u\cos v,0)} \\\quad \Rightarrow \vec t_u\times \vec t_v=(0,0,u) \Rightarrow \iint_{S_1} \vec F\cdot d\vec S =\int_0^{2\pi} \int_0^a (u^3\cos^3v,u^3\cos^2 v\sin v, bu^2\cos^2 v) \cdot (0,0,u)dudv \\=\int_0^{2\pi} \int_0^a bu^3\cos^2 v\,dudv =\int_0^{2\pi} {a^4b\over 4} \cos^2 v\,dv =\int_0^{2\pi} {a^4b\over 4} \cdot {1\over 2}(\cos 2v+1)  \,dv ={a^4b\over 8} \cdot 2\pi= \bbox[cyan,2pt]{{a^4b\over 4}\pi} \\S_2: \vec r(u,v)=(a\cos v,a\sin v,u),0\le v\le 2\pi, 0\le u\le b \Rightarrow \cases{\vec t_u=(0,0,1) \\ \vec t_v=(-a\sin v,a\cos v,0)} \\\quad \Rightarrow \vec t_u\times \vec t_v=(-a\cos v,-a\sin v,0) \Rightarrow \text{ we choose }\vec t_u\times \vec t_v=(a\cos v,a\sin v,0) \\\Rightarrow \iint_{S_2} \vec F\cdot d\vec S =\int_0^{2\pi} \int_0^b (a^3\cos^3v,a^3\cos^2 v\sin v, a^2u\cos^2 v) \cdot (a\cos v,a\sin v,0)dudv \\=\int_0^{2\pi} \int_0^b a^4\cos^2 v\,dudv =\int_0^{2\pi}a^4b\cos^2 v\,dv =\int_0^{2\pi}{a^4b\over 2}(1+\cos 2v)\,dv = \bbox[cyan,2pt]{a^4b\pi} \\S_3: \vec r(u,v)=(u\cos v,u\sin v,0),0\le v\le 2\pi, 0\le u\le a \Rightarrow \cases{\vec t_u=(\cos v,\sin v,0) \\ \vec t_v=(-u\sin v,u\cos v,0)} \\\quad \Rightarrow \vec t_u\times \vec t_v=(0,0,u) \Rightarrow  \iint_{S_3} \vec F\cdot d\vec S =\int_0^{2\pi} \int_0^a 0\;dudv = \bbox[cyan,2pt]{0} \\ \Rightarrow \iint_{S} \vec F\cdot d\vec S =\iint_{S_1} \vec F\cdot d\vec S +\iint_{S_2} \vec F\cdot d\vec S +\iint_{S_3} \vec F\cdot d\vec S ={a^4b\over 4}\pi+a^4b\pi+0= \bbox[red, 2pt]{{5\over 4}a^4b\pi}\\ \textbf{(2)}\; \vec F=(x^3,x^2y,x^2z) \Rightarrow \nabla\cdot \vec F=3x^2+x^2+x^2= 5x^2 \\ \Rightarrow I= \iiint_R \nabla\cdot \vec F\,dV=\iiint_R 5x^2\,dV,  \text{ where }R=\{(r,\theta, z) \mid 0\le r\le a, 0\le \theta\le 2\pi, 0\le z\le b\} \\ \cases{x=r\cos \theta\\ y=r\sin \theta\\ z=z} \Rightarrow I=  \int_0^{2\pi}\int_0^b \int_0^a 5r^3\cos^2 \theta\,dr\,dzd\theta  = \int_0^{2\pi} \int_0^b{5\over 4}a^4\cos^2\theta\,dz\, d\theta  \\=\int_0^{2\pi} {5\over 4}a^4b \cos^2 \theta\,d\theta = {5\over 8}a^4b \int_0^{2\pi}  (\cos(2\theta)+1)\,d\theta = \bbox[red, 2pt]{{5\over 4}a^4b\pi}$$

解答: $$u(x,t) =X(x)T(t) \Rightarrow \frac{\partial u}{\partial t}=c^2 \frac{\partial^2u }{\partial x^2} \Rightarrow XT'=c^2X''T \Rightarrow {T'\over c^2 T}={X''\over X} \\ BC:\cases{u(0,t)=0\\ u(L,t)=0} \Rightarrow \cases{X(0)T(t)=0 \\ X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\ \text{Let }{T'\over c^2 T}={X''\over X}=\lambda\\ \textbf{Case I }\lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+ c_2 \Rightarrow BC: \cases{X(0)=c_2=0\\ X(L)=c_1L+c_2=0} \Rightarrow \cases{c_1=0\\c_2=0 } \Rightarrow X=0 \\ \textbf{Case II }\lambda =\rho^2\gt 0  \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x} +c_2e^{-\rho x} \Rightarrow BC: \cases{X(0)=c_1+c_2=0\\ X(L)=c_1e^{\rho L}+ c_2e^{-\rho L}=0} \\ \qquad \Rightarrow c_1e^{\rho L}- c_1e^{-\rho L}= 0 \Rightarrow c_1(e^{2\rho L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case III }\lambda=-\rho^2 \lt 0 \Rightarrow X^2+\rho^2X=0 \Rightarrow X=c_1\cos (\rho x)+c_2 \sin (\rho x) \\\qquad \Rightarrow BC:\cases{X(0)=c_1=0\\ X(L)=c_2\sin(\rho L)=0} \Rightarrow \rho L=n\pi \Rightarrow \rho ={n\pi \over L} \Rightarrow X_n=\sin {n\pi x\over L},n\in \mathbb N \\ \Rightarrow T'=\lambda c^2 T=-{n^2 c^2\pi^2\over L^2} T \Rightarrow T_n =e^{-n^2c^2 \pi^2\ t/L^2},n \in \mathbb N \\ \Rightarrow u(x,t)= \sum_{n=1}^\infty A_n \sin{n\pi x\over L} e^{-n^2c^2 \pi^2\ t/L^2} \Rightarrow u(x,0)= f(x)=\sum_{n=1}^\infty A_n \sin{n\pi x\over L} \\ \Rightarrow A_n= {2\over L} \int_0^L f(x) \sin {n\pi x \over L}\,dx ={2\over L}\left( \int_0^{L/2} x \sin{n\pi x\over L}\,dx +\int_{L/2}^L (L-x)\sin {n\pi x\over L}\,dx \right) \\={2\over L} \cdot {L^2\over n^2\pi^2} \sin {n\pi\over 2} ={2L\over n^2\pi^2} \sin{n\pi \over 2}\\ \Rightarrow \bbox[red, 2pt]{ u(x,t)=\sum_{n=1}^\infty A_n \sin{n\pi x\over L} e^{-n^2c^2 \pi^2\ t/L^2}, A_n= {2L\over n^2\pi^2} \sin{n\pi \over 2}}$$

 

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解題僅供參考,其他歷年試題及詳解