113年高考三級-應用數學詳解

 113年公務人員高等考試三級考試試題

類 科：天文
科 目：應用數學（包括微積分、微分方程與向量分析）

解答:$$\textbf{(一)}\;A=\begin{bmatrix} -1&-1 \\ 2&-4\end{bmatrix} \Rightarrow \det(A-\lambda I) =(\lambda+3)(\lambda+2) =0 \Rightarrow \lambda=-3,-2\\ \lambda_1=-3 \Rightarrow (A-\lambda_1 I)v= 0 \Rightarrow \begin{bmatrix} 2&-1 \\ 2&-1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix} =0 \Rightarrow 2x_1=x_2 \Rightarrow v= x_2\begin{bmatrix} 1/2 \\ 1\end{bmatrix},\text{ choose }v_1=\begin{bmatrix} 1/2 \\ 1\end{bmatrix} \\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I)v= 0 \Rightarrow \begin{bmatrix} 1&-1 \\ 2&-2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix} =0 \Rightarrow x_1=x_2 \Rightarrow v= x_1\begin{bmatrix} 1  \\ 1\end{bmatrix},\text{ choose }v_2=\begin{bmatrix} 1  \\ 1\end{bmatrix} \\ \Rightarrow P=[v_1 \; v_2] =\bbox[red, 2pt]{\begin{bmatrix} 1/2& 1 \\ 1&1 \end{bmatrix}}, D= \begin{bmatrix} -3& 0 \\ 0&-2 \end{bmatrix} \\\textbf{(二)}\; \begin{bmatrix} x(t)\\ y(t)\end{bmatrix} =c_1 e^{-3t}\begin{bmatrix} 1/2\\ 1\end{bmatrix}  +c_2 e^{-2t}\begin{bmatrix} 1\\ 1\end{bmatrix} \Rightarrow \bbox[red, 2pt]{\begin{bmatrix} x(t)\\ y(t)\end{bmatrix} =\begin{bmatrix} {1\over 2}c_1e^{-3t}+ c_2e^{-2t}\\ c_1e^{-3t} +c_2e^{-2t}\end{bmatrix} } \\ 又 \lim_{t\to \infty} x(t) =\lim_{t\to \infty} \left({1\over 2}c_1e^{-3t}+ c_2e^{-2t} \right) =0 且\lim_{t\to \infty} y(t)= \lim_{t\to \infty} \left(c_1e^{-3t}+ c_2e^{-2t} \right)=0 \\ \Rightarrow \lim_{t\to \infty} x(t)=\lim_{t\to \infty} y(t)=0. \bbox[red, 2pt]{\text{QED.}}$$

解答:$$\cases{x(t)=\cos t\\ y(t)= \sin t\\ z(t)=3t} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t)=  \cos t\\ z'(t) =3} \Rightarrow 弧長=\int_1^2 \sqrt{x'(t)^2 +y'(t)^2+ z'(t)^2} \,dt =\int_1^2 \sqrt{1+9} \,dt \\= \int_1^2 \sqrt{10}\,dt = \bbox[red, 2pt]{\sqrt{10}}$$

解答:$$算幾不等式:{x^2+2y^4+3z^6 \over 3} \ge \sqrt[3]{6x^2y^4z^6} \Rightarrow 1\ge \sqrt[3]{6x^2y^4z^6} \\ \Rightarrow 6x^2y^4z^6 \le 1 \Rightarrow xy^2z^3 \le {1\over \sqrt 6} \Rightarrow f(x,y,z)最大值 \bbox[red, 2pt]{\sqrt 6\over 6}\\  此時x^2 =2y^4=3z^6 \Rightarrow x^2 =2y^4=3z^6=1 \Rightarrow \cases{x^2=1\\ y^4=1/2\\ z^6=1/3} \Rightarrow \cases{x=\pm 1\\ y=\pm (1/2)^{1/4}\\ z= \pm (1/3)^{1/6}} \\ \Rightarrow (x,y,z)= \bbox[red, 2pt]{\left( \pm 1,\pm ({1\over 2})^{1/4},\pm ({1\over 3})^{1/6} \right)},共八個點$$

解答:$$\textbf{(一)}\; f(x,y,z)=3xy-2yz \Rightarrow \nabla f=(f_x,f_y,f_z) =(3y,3x-2z,-2y) \Rightarrow \bbox[red, 2pt]{\text{grad }f=(3y,3x-2z,-2y)} \\ \nabla f(1,0,1)=(0,1,0) \Rightarrow  \nabla_{\vec u} f(1,0,1)=\left({1\over \sqrt 2},{1\over 2}, {1\over 2} \right) \cdot (0,1,0)\Rightarrow \bbox[red, 2pt]{\nabla_{\vec u} f(1,0,1) ={1\over 2}} \\ \textbf{(二)}\; \text{curl }(\vec V)= \begin{vmatrix}\vec i & \vec j& \vec k \\\frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 2z& 3y& x-z \end{vmatrix} =(0,1,0) \Rightarrow \bbox[red, 2pt]{\text{curl }(\vec V)=(0,1,0)} \\ \text{div}\left(\text{curl }(\vec V) \right) =\text{div}(0,1,0)= 0+ 0+0 =0\Rightarrow \bbox[red, 2pt]{ \text{div}\left(\text{curl }(\vec V) \right) =0}$$

解答:$$\text{Fourier transform }特性: \cases{\hat{u_t}(\omega,t)={\partial \over \partial t}[\hat{u}(\omega ,t)] \cdots(1)\\ \hat{u_{xx}}(\omega ,t)=(i\omega)^2 \hat{u}(\omega,t) =-\omega^2\hat{u}(\omega,t) \cdots(2)}\\ 將(1)及(2)代入 {\partial \over \partial t}v(x,t)-{\partial^2 \over \partial x^2}v(x,t)=0 \Rightarrow {\partial \over \partial t}\hat {v}(\omega,t)+ \omega^2 \hat{v}(\omega ,t)=0 \\ \Rightarrow {\partial \over \partial t}\hat {v}(\omega,t) \cdot e^{\omega^2 t}+\omega^2 \hat{v}(\omega ,t) \cdot e^{\omega^2 t}=0\Rightarrow {\partial \over \partial t}\left( e^{\omega^2 t} \hat{v}(\omega, t)\right) =0 \Rightarrow   e^{\omega^2 t} \hat{v}(\omega,t) = g(\omega) \\ \Rightarrow \hat{v}(\omega,t)=e^{-\omega^2 t}g(\omega) \Rightarrow \hat v(\omega,0)=g(\omega)\\ 初始值v(x,0)=f(x) \Rightarrow \hat v(\omega,0)=\hat f(\omega) \Rightarrow g(\omega) =\hat f(\omega) \Rightarrow \hat v(\omega,t)= e^{-\omega^2 t} \hat f(\omega)\\ 我們的目的是求v(x,t),也就是e^{-\omega^2 t} \hat f(\omega)的\text{ inverse Fourier transform}\\ 先求\mathcal F^{-1}(e^{-\omega^2 t}) ={1\over 2\pi}\int_{-\infty}^\infty e^{-\omega^2 t} e^{i\omega t}d\omega= {1\over \sqrt{4\pi t}}e^{-x^2/4t} \\ 再由摺積\text{(convolution)}來求v(x,t), 即 v(x,t)= \int_{-\infty}^\infty  {1\over \sqrt{4\pi t}}e^{-(x-y)^2/4t} f(y)\,dy\\ 因此u(x,t)={1\over \sqrt{4\pi t}} \int_{-\infty}^\infty e^{-(x-y)^2/4t}f(y)\,dy 是熱傳導方程式的解. \bbox[red, 2pt]{QED.}$$

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