115年中央光電碩士班-工程數學詳解

國立中央大學115學年度碩士班考試入學試題

系所:光電類 科目:工程數學

解答:$$\phi=x+y+z \Rightarrow \nabla \phi = (\phi_x, \phi_y,\phi_z) =(1,1,1) \Rightarrow \nabla \phi(1,2,3) =(1,1,1) \\ F=x^2+{1\over 4}y^2+{1\over 9}z^2-3 \Rightarrow \nabla F=(2x,{1\over 2}y,{2\over 9}z) \Rightarrow \nabla F(1,2,3) = \left( 2,1,{2\over 3} \right) \\ \Rightarrow \Vert \nabla F \Vert= \sqrt{4+1+{4\over 9}} ={7\over 3} \Rightarrow \mathbf n = {\nabla F\over \Vert \nabla F\Vert} = \left( {6\over 7},{3\over 7},{2\over 7} \right) \\ \Rightarrow D_{\mathbf n} \phi=(1,1,1) \cdot  \left( {6\over 7},{3\over 7},{2\over 7} \right) = \bbox[red, 2pt]{11\over 7}$$

解答:$$\vec r=(x,y,z) \Rightarrow |\vec r|= \sqrt{x^2+y^2+z^2} \\ \phi={a\over \vec r} \Rightarrow \nabla \phi= \left( {\partial\over \partial x}({a\over \vec r}), {\partial\over \partial y}({a\over \vec r}), {\partial\over \partial z}({a\over \vec r}) \right) =-a \left( {x\over |\vec r|^3}, {y\over |\vec r|^3}, {z\over |\vec r|^3}  \right) \\ \Rightarrow -\nabla \phi=E \Rightarrow \nabla\times(\nabla \phi)=0 \Rightarrow \nabla\times E=\nabla\times (-\nabla \phi) =0 \Rightarrow \nabla \times E=\bbox[red, 2pt]0$$

解答:$$ A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \Rightarrow A^2= \begin{bmatrix}5&-4\\-4& 5 \end{bmatrix} \Rightarrow \det(A^2-\lambda I) =\lambda^2-10 \lambda+9= (\lambda-1)(\lambda-9)=0 \\ \Rightarrow \bbox[red,2pt]{ \text{eigenvalues: }1,9} \\ \lambda_1=1 \Rightarrow (A^2-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix}4&-4\\-4&4 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_1=x_2 \Rightarrow v=x_1 \begin{bmatrix}1\\1 \end{bmatrix}, \text{choosing }v_1= \begin{bmatrix}1\\1 \end{bmatrix} \\ \lambda_2=9 \Rightarrow (A^2-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-4&-4\\-4&-4 \end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} =0 \Rightarrow x_1=-x_2 \Rightarrow v=x_1 \begin{bmatrix}1\\-1 \end{bmatrix}, \text{ choosing }v_2= \begin{bmatrix}1\\-1 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors: } \begin{bmatrix}1\\1 \end{bmatrix}, \begin{bmatrix}1\\-1 \end{bmatrix}}$$

解答:$$17x^2-30xy+17y^2= \begin{bmatrix}x& y \end{bmatrix} \begin{bmatrix}17& -15\\-15& 17 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} =128 \\ A=\begin{bmatrix}17& -15\\-15& 17 \end{bmatrix}  \Rightarrow \det(A-\lambda I)=0 \Rightarrow \lambda=2,32 \Rightarrow 2u^2+32v^2=128 \Rightarrow {u^2\over 64}+{v^2\over 4}=1\\ \Rightarrow \cases{a=8\\ b=2} \Rightarrow \bbox[red, 2pt]{\cases{\text{major axis: 16} \\\text{minor axis: 4}}}$$

解答:$$(B)\times: \vec \nabla\cdot (\vec A\times \vec B)= \vec B\cdot (\vec \nabla \times \vec A)-\vec A\cdot (\nabla\times \vec B) \ne \vec B\cdot (\vec \nabla \times \vec A)+\vec A\cdot (\nabla\times \vec B) \\(C)\times: \vec \nabla \cdot \left( {\vec r\over |\vec r|^3} \right) =4\pi \delta^3(\vec r) \ne  4\pi |\vec r| \\(D)\times: \vec\nabla \times (\vec \nabla\times \vec A)= \vec \nabla(\vec \nabla\cdot \vec A)- \vec \nabla^2 \vec A \ne  \vec \nabla(\vec \nabla\cdot \vec A)+ \vec \nabla^2 \vec A \\ \textbf{(a)} \text{ The correct statement: }\bbox[red, 2pt]{(A)} \\ \textbf{(b) } \text{Let }\cases{\phi =\phi(x,y,z)\\ \vec A =(A_x,A_y,A_z)} \Rightarrow \phi\vec A=(\phi A_x, \phi A_y, \phi A_z) \Rightarrow \vec \nabla \times (\phi \vec A) = \begin{vmatrix} \mathbf i& \mathbf j& \mathbf k \\ \partial _x& \partial_y& \partial _z \\ \phi A_x& \phi A_y& \phi A_z\end{vmatrix}\\= ​ \left[ {\partial(\phi A_x) \over \partial y}-{\partial(\phi A_y)\over \partial z} \right] \mathbf i+ \left[ {\partial(\phi A_x)\over \partial z}-{\partial (\phi A_z) \over \partial x} \right] \mathbf j+ \left[ {\partial(\phi A_y) \over \partial x}-{\partial(\phi A_x)\over \partial y} \right] \mathbf k \\= \phi \begin{pmatrix} \displaystyle {\partial A_z\over \partial y}-{\partial A_y\over \partial z}  \\\displaystyle {\partial A_x\over \partial z} -{\partial A_z\over \partial x} \\\displaystyle  {\partial A_y\over x} -{\partial A_x\over \partial y}\end{pmatrix} - \begin{pmatrix} \displaystyle A_y{\partial \phi\over \partial z} -A_z{\partial \phi\over \partial y} \\ \displaystyle A_z{\partial\phi \over \partial x}-A_x{\partial\phi \over \partial z} \\ \displaystyle A_x {\partial\phi \over \partial y} -A_y {\partial \phi\over \partial x}\end{pmatrix} = \phi(\vec \nabla \times \vec A)-\vec A\times \vec \nabla \phi \;\bbox[red, 2pt]{QED.}$$

解答:$$xy'+3y=2x^2 \Rightarrow x^3y'+3x^2y'=2x^4 \Rightarrow \left( x^3y \right)'=2x^4 \Rightarrow x^3 y = \int 2x^4\,dx ={2\over 5}x^5+C \\ \Rightarrow \bbox[red, 2pt] {y={2\over 5}x^2+Cx^{-3}}$$

解答:$$ L \frac{dI}{dt} + RI + \frac{1}{C} Q(t) = E(t) =5\sin(20t) \Rightarrow  L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{1}{C}I = E'(t) =100\cos(20t) \\ \Rightarrow 0.1I''+5I'+{1\over 0.025} I=100\cos(20t) \Rightarrow  I'' + 50 I' + 400 I = 1000\cos(20t) \\ r^2+50r+ 400=0 \Rightarrow (r+10)(r+40)=0 \Rightarrow r_-10,-40 \Rightarrow I_h(t)=c_1 e^{-10t}+c_2 e^{-40t} \\ I_p =A\cos(20t) +B\sin(20t) \Rightarrow I_p'=-20\sin(20t)+20B\cos(20t)\\ \Rightarrow I_p''= -400A \cos(20t) -400B\sin(20t)\\ \Rightarrow I_p''+50I_P'+400I= 1000B\cos(20t) -1000A \sin(20t) =1000\cos(20t) \Rightarrow \cases{A=0\\ B=1} \\ \Rightarrow I_p=\sin(20t) \Rightarrow I=I_h+I_p = c_1e^{-10t}+c_2e^{-40t}+\sin(20t) \\ \Rightarrow I'=-10c_1e^{-10t}-40c_2e^{-40t}+20\cos(20t) \\  L I'(0) + R I(0) + \frac{1}{C} Q(0) = E(0) \Rightarrow 0.1I'(0)+5\cdot 0 +40\cdot 0=5\sin (0) \Rightarrow I'(0)=0 \\ \Rightarrow \cases{I(0)= c_1+c_2=0\\ I'(0)=-10c_1-40c_2+20=0} \Rightarrow \cases{c_1=-2/3\\ c_2=2/3} \\ \Rightarrow \bbox[red, 2pt]{I(t) = -{2\over 3}e^{-10t} +{2\over 3}e^{-40t} +\sin(20t)}$$

解答:$$ y(x) = \sum_{n=0}^{\infty} c_n x^n = c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + c_5x^5 + \dots \\ \Rightarrow  y'(x) = \sum_{n=1}^{\infty} n c_n x^{n-1} = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + 5c_5x^4 + \dots \\ \Rightarrow  y''(x) = \sum_{n=2}^{\infty} n(n-1) c_n x^{n-2} = 2c_2 + 6c_3x + 12c_4x^2 + 20c_5x^3 + \dots \\ \text{initial conditions:}\cases{y(0)=c_0=1\\ y'(0)=c_1=0} \Rightarrow y''-3xy'+4x^2y=2x\\ \Rightarrow (2c_2 + 6c_3x + 12c_4x^2 + 20c_5x^3 + \dots) - 3x(2c_2x + 3c_3x^2 + 4c_4x^3 + \dots) + 4x^2(1 + c_2x^2 + c_3x^3 + \dots)\\ =2c_2+ 6c_3x +(12c_4 -6c_2+4)x^2+ (20c_5-9c_3)x^3 +\cdots= 2x \\ \Rightarrow \cases{2c_2 =0\\ 6c_3=2\\ 12c_4-6c_2+4=0\\ 20c_5-9c_3=0} \Rightarrow \cases{c_2=0\\ c_3=1/3\\ c_4=-1/3\\ c_5=3/20} \Rightarrow \bbox[red, 2pt]{y(x)=  1 + \frac{1}{3}x^3 - \frac{1}{3}x^4 + \frac{3}{20}x^5+\cdots}$$

解答:$$L\{\delta(t-2\pi)\} = \int_0^\infty e^{-st} \delta(t-2\pi)\,dt =e^{-s(2\pi)} = \bbox[red, 2pt]{e^{-2\pi s}}$$

 

解答:$$\textbf{(a) }f_0(x) = \begin{cases}0,& x\le -2\\ 1, & -2\le x\lt -1\\ 3, & -1\le x\le 1\\ 1,& 1\lt x\le 2 \\ 0,& x\gt 2\end{cases} \Rightarrow f_0(x) \text{ is even} \Rightarrow \mathcal{F}\{f_0(x)\} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f_0(x)e^{-i\omega x} dx \\  = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f_0(x)(\cos (\omega x)-i \sin(\omega x)) dx  = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f_0(x) \cos (\omega x)  dx  \\= \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty} f_0(x) \cos (\omega x)  dx  = \sqrt{\frac{2}{\pi}} \left[ \int_{0}^{1} 3\cos(\omega x) dx + \int_{1}^{2} 1\cos(\omega x) dx \right] \\  = \sqrt{\frac{2}{\pi}} \left( \frac{3\sin(\omega)}{\omega} + \frac{\sin(2\omega) - \sin(\omega)}{\omega} \right)   = \bbox[red, 2pt]{\sqrt{\frac{2}{\pi}} \frac{2\sin(\omega) + \sin(2\omega)}{\omega}}\\ \textbf{(b) }\text{We are given that $f(x) = f_0(x)$ for $\vert{}x\vert{} \leq 2$ and $f(x) = f(x+8)$} \\\quad \Rightarrow T=8 \Rightarrow \text{ the half-period }L={T\over 2}=4 \\ a_0={1\over L} \int_{-L}^L f(x)\,dx = {2\over L} \int_0^L f(x)\,dx   = \frac{1}{2} \left[ \int_{0}^{1} 3 dx + \int_{1}^{2} 1 dx \right] =  2 \\ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx = \frac{1}{4} \int_{-4}^{4} f(x) \cos\left(\frac{n\pi x}{4}\right) dx   = \frac{2}{4} \int_{0}^{2} f_0(x) \cos\left(\frac{n\pi x}{4}\right) dx \\=  \frac{1}{2} \left[ \int_{0}^{1} 3\cos \left(\frac{n \pi x}{4}\right) dx + \int_{1}^{2} 1\cos\left(\frac{n\pi x}{4}\right) dx \right]  \\ = \frac{1}{2} \left[ \frac{12}{n\pi} \sin\left(\frac{n\pi}{4}\right) + \frac{4}{n\pi} \sin\left(\frac{n\pi}{2}\right) - \frac{4}{n\pi} \sin \left( \frac{n \pi}{4}\right) \right]   = \frac{2}{n\pi} \left[ 2\sin \left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{2}\right) \right] \\ \Rightarrow  \bbox[red, 2pt]{f(x) = 1 + \sum_{n=1}^{\infty} \frac{2}{n\pi} \left[ 2\sin \left(\frac{n \pi}{4}\right) + \sin\left(\frac{n\pi}{2}\right) \right] \cos\left(\frac{n\pi x}{4}\right)}$$

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解題僅供參考，碩士班歷年試題及詳解