114年台師大機電碩士班-工程數學詳解

 國立臺灣師範大學 114學年度碩士班招生考試試題

科目:工程數學
適用系所:機電工程學系

解答:$$\textbf{(a) } {d^2y\over dx^2}+4{dy\over dx}+4y=0 \Rightarrow r^2+4r+4=0 \Rightarrow(r+2)^2=0 \Rightarrow r=-2 \\\quad \Rightarrow \bbox[red, 2pt]{y(x)= c_1e^{-2x} +c_2xe^{-2x}} \\ \textbf{(b) }y'=(c_2-2c_1)e^{-2x} -2c_2xe^{-2x} \Rightarrow \cases{y(0)=c_1=2\\ y'(0) =c_2-2c_1=0} \Rightarrow \cases{c_1=2\\ c_2=4} \\\quad \Rightarrow \bbox[red, 2pt]{y(x)=2e^{-2x}+4xe^{-2x}}$$

解答:$$\textbf{(a) }  \int{x^4+1\over x(x^2+1)^2} \,dx = \int \left( {1\over x}-{2x\over (x^2+1)^2} \right)\,dx = \bbox[red, 2pt]{\ln |x| +{1\over x^2+1}+C} \\\textbf{(b) }y''-2y'+2y=0 \Rightarrow r^2-2r+2=0 \Rightarrow y=1\pm i \Rightarrow \bbox[red, 2pt]{y=e^x(c_1 \cos x+ c_2\sin x)} \\\textbf{(c) }yy''+(y')^2+1=0 \Rightarrow (yy')'+1=0 \Rightarrow yy'= \int -1\,dx =-x+c_1 \\ \quad \Rightarrow \int y\,dy =\int(-x+c_1)\,dx \Rightarrow {1\over 2}y^2 =-{1\over 2}x^2+c_1x +c_2\Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{-x^2+2c_1x+ 2c_2}}$$

解答:$$my''+cy'+ky=0 \Rightarrow mr^2+cr+k=0 \Rightarrow r={-c\pm \sqrt{c^2-4mk} \over 2m} \\\textbf{Case I: }\text{Overdamped System }(c^2\gt 4mk) \Rightarrow y(t)=C_1e^{r_1t} +C_2e^{r_2t}, \\ \qquad \text{ where } r_1= {-c+ \sqrt{c^2-4mk} \over 2m}, r_2 = {-c-\sqrt{c^2-4mk} \over 2m} \\\textbf{Case II: }\text{Critically Damped System }(c^2=4mk) \Rightarrow y(t)=(C_1+C_2 t)e^{rt},r= -{c\over 2m}\\ \textbf{Case III: }\text{Underdamped System }(c^2 \lt 4mk) \Rightarrow y(t)=e^{\alpha t} (C_1\cos(\beta t)+ C_2 \sin (\beta t)), \\\qquad \text{ where }\alpha=-{c\over 2m}, \beta ={\sqrt{4mk-c^2} \over 2m} \\ \textbf{Case IV: }\text{Undamped System }(c=0) \Rightarrow y(t)=C_1 \cos (\omega t)+ C_2 \sin (\omega t), \omega =\sqrt{k\over m}$$

解答:$$\textbf{(a) }L^{-1} \left\{ {s-1\over (s-1)^2+4}\right\} =e^tL^{-1} \left\{ {s\over s^2+4}\right\} = \bbox[red, 2pt]{e^t\cos 2t} \\\textbf{(b) }L^{-1} \left\{ {s+1\over (s^2+1)(s^2+4s+13)}\right\} =L^{-1} \left\{ {s+2\over 20(s^2+1) } +{-s-6\over 20(s^2+4s+13)}\right\} \\\qquad =L^{-1} \left\{ {s\over 20(s^2+1) }+ {2\over 20(s^2+1) }-{1\over 20}\cdot {s+2\over (s+2)^2+9} -{1\over 5} \cdot {1 \over ((s+2)^2 + 9)}\right\} \\\qquad = \bbox[red, 2pt]{{1\over 20}\cos t+{1\over 10} \sin t-{1\over 20}e^{-2t} \cos(3t) -{1\over 15}e^{-2t} \sin(3t)}$$

解答:$$\textbf{(a) } \cases{f_0(x )=1\\ f_1(x)=x\\ f_2(x)=3x^2-1} \Rightarrow \cases{\langle f_0,f_1\rangle = \int_{-1}^1 x\,dx =0 \\ \langle f_0,f_2\rangle = \int_{-1}^1 (3x^2-1)\,dx = \left. \left[ x^3-x \right] \right|_{-1}^1=0 \\ \langle f_1,f_2\rangle = \int_{-1}^1 x(3x^2-1)\,dx \int_{-1}^1 (3x^3-x)\,dx =0}  \\\qquad \Rightarrow \{1,x,3x^2-1\} \text{ is an orthogonal set} \; \bbox[red, 2pt]{Q.E.D.} \\\textbf{(b) } f(x) =\sin{\pi x\over 2} \approx P(x) =c_0f_0+ c_1f_1+ c_2f_2 \Rightarrow c_k= {\langle f,f_k \rangle\over \langle f_k,f_k \rangle} \\ \cases{f_0 \text{ is an even function} \\f_2 \text{ is an even function}\\f \text{ is an odd function} } \Rightarrow \cases{\langle f_0,f \rangle = \displaystyle \int_{-1}^1 f_0\cdot f\,dx =0 \\ \langle f_2,f\rangle = \displaystyle \int_{-1}^1 f_2\cdot f\,dx=0} \\ \cases{\langle f_1, f\rangle = \displaystyle \int_{-1}^1 x \sin(\pi x/2)\,dx = {8\over \pi^2} \\ \langle f_1,f_1\rangle = \displaystyle \int_{-1}^1 x^2\,dx ={2\over 3}} \Rightarrow c_1= {\langle f,f_1\rangle \over \langle f_1,f_1 \rangle} ={8/\pi^2 \over 2/3} ={12\over \pi^2} \Rightarrow P(x) = \bbox[red, 2pt]{{12\over \pi^2}x^2}$$

解答:$$\textbf{(a) }\det(A-\lambda I) =\lambda^2+6\lambda+8 =(\lambda+2)(\lambda+4)=0 \Rightarrow \lambda_1=-2, \lambda_2=-4\\ \lambda_1=-2 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}-1&1\\1&-1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=x_2 \\\qquad\Rightarrow v= x_1 \begin{bmatrix}1\\1 \end{bmatrix} \Rightarrow \text{choosing }v_1= \begin{bmatrix}1\\ 1 \end{bmatrix} \\ \lambda_2 =-4\Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}1& 1\\1& 1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1+x_2=0 \\ \qquad \Rightarrow v= x_1 \begin{bmatrix} 1\\ -1 \end{bmatrix} \Rightarrow \text{ choosing }v_2 = \begin{bmatrix}1\\ -1 \end{bmatrix} \\ \Rightarrow V=[v_1\; v_2] \Rightarrow \bbox[red, 2pt]{V= \begin{bmatrix}1& 1\\1& -1 \end{bmatrix}} \\\textbf{(b) } \lambda_i \lt 0 \Rightarrow  x^TAx \text{ is }\bbox[red, 2pt]{\text{negative definite}} \\\textbf{(c) }A=VDV^{-1} = \begin{bmatrix}1& 1\\ 1&-1 \end{bmatrix} \begin{bmatrix}-2&0\\0&-4 \end{bmatrix} \begin{bmatrix}1/2& 1/2\\ 1/2&-1/2 \end{bmatrix} \Rightarrow e^A = \begin{bmatrix}1& 1\\ 1&-1 \end{bmatrix} \begin{bmatrix}e^{-2}&0\\0& e^{-4} \end{bmatrix} \begin{bmatrix}1/2& 1/2\\ 1/2&-1/2 \end{bmatrix} \\ \qquad = \begin{bmatrix}e^{-2}& e^{-4}\\ e^{-2}& -e^{-4} \end{bmatrix} \begin{bmatrix}1/2& 1/2\\ 1/2&-1/2 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}{1\over 2}(e^{-2}+e^{-4} ) &{1\over 2}(e^{-2}-e^{-4} ) \\{1\over 2}(e^{-2}-e^{-4} ) &{1\over 2}(e^{-2}+e^{-4} )  \end{bmatrix}} \\\textbf{(d) } x(t)= \xi e^{\lambda t} \Rightarrow \dot x(t)= \lambda \xi e^{\lambda t}=A\xi e^{\lambda t} \Rightarrow A \xi=\lambda \xi \Rightarrow \lambda \text{ is an eigenvalue and }\xi \text{ the eigenvector.} \\\qquad \Rightarrow x(t)= c_1e^{\lambda_1 t}v_1+ c_2 e^{\lambda_2 t}v_2 = c_1e^{-2t} \begin{bmatrix}1\\ 1 \end{bmatrix} + c_2 e^{-4t} \begin{bmatrix}1\\ -1 \end{bmatrix} \Rightarrow x(0) = \begin{bmatrix}c_1+c_2\\ c_1-c_2 \end{bmatrix}= \begin{bmatrix}1\\ 0 \end{bmatrix} \\ \qquad \Rightarrow c_1=c_2={1\over 2} \Rightarrow x(t) = {1\over 2}e^{-2t} \begin{bmatrix}1\\ 1 \end{bmatrix} + {1\over 2} e^{-4t} \begin{bmatrix}1\\ -1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}\displaystyle {1\over 2}(e^{-2t}+e^{-4t}) \\ \displaystyle {1\over 2} (e^{-2t}-e^{-4t})\end{bmatrix}}$$

解答:$$\textbf{(a) }T(x,y,z)=2x^2-2xy+2y^2+z^2 \Rightarrow \nabla T(x,y,z) =(Tx,T_y,T_z) =(4x-2y,-2x+4y,2z) \\ \qquad \nabla T(2,1,1) =(6,0,2) \Rightarrow \text{ the direction of most rapid decrease is: } -(6,0,2)= \bbox[red, 2pt]{(-6,0,-2)} \\\textbf{(b) }\nabla T\cdot v=0 \Rightarrow (6,0,2)\cdot (v_1,v_2,v_3) =6v_1+2v_3=0 \Rightarrow v_3=-3v_1 \\ \qquad \Rightarrow v= \bbox[red, 2pt]{(v_1,v_2,-3v_1), \text{ where }v_1 \text{ and }v_2\text{ are not both zero}} \\\textbf{(c) }\cases{x(t)= \cos t\\ y(t)= \sin t\\ z(t)=1} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t)=\cos t\\ z'(t)=0} \Rightarrow ds =\sqrt{(x')^2+ (y')^2+ (z')^2} \,dt =\sqrt{1}\,dt =dt \\ \qquad \Rightarrow T(t)=2\cos^2 t-2\cos t\sin t+2\sin^2 t+1=3-\sin(2t) \\\qquad \Rightarrow \int_C T\,ds = \int_0^{\pi/2}(3-\sin(2t))\,dt = \left. \left[ 3t+{1\over 2}\cos(2t) \right] \right|_0^{\pi /2}=\bbox[red, 2pt]{{3\over 2}\pi-1}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解