114年升官等薦任-應用數學詳解

 114年公務人員升官等考試

等 級：薦任
類科（別）：天文
科 目：應用數學（包括微積分、微分方程、向量分析）
考試時間： 2 小時

解答:$$\textbf{(一) }\vec r(t)=(t,t^2,t^3) \Rightarrow \vec r'(t)=(1,2t,3t^2) \Rightarrow \vec F(\vec r(t)) =(3t^3,t^2+3t^4,t) \\ \Rightarrow \vec F(\vec r(t)) \cdot \vec r'(t) =(3t^3,t^2+3t^4,t) \cdot (1,2t,3t^2)= 8t^3+6t^5 \Rightarrow 功=\int_0^1 (8t^3+6t^5)\,dt \\= \left. \left[ 2t^4+t^5 \right] \right|_0^1  = \bbox[red, 2pt]3\\\textbf{(二) } \cases{F_1=2xy+z\\ F_2=x^2+3y^2\\ F_3=x} \Rightarrow \nabla \times \vec F = \begin{vmatrix} \vec i& \vec j& \vec k\\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ 2xy+z& x^2+3y^2& x\end{vmatrix} =0\vec i+0\vec j+0\vec j= \vec 0 \\\qquad \Rightarrow 做功與路徑無關 \quad \bbox[red, 2pt]{故得證}$$

解答:$$\begin{bmatrix}\dot x\\ \dot y \end{bmatrix} = \begin{bmatrix}3& 2\\ -2& -1 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} \Rightarrow \cases{x'=3x+2y\\ y'=-2x-y} \Rightarrow \cases{L\{x'\} =3 L\{x\} +2 L\{y\} \\L\{y'\} =-2 L\{x\} - L\{y\}} \\ \Rightarrow \cases{sX(s)-1=3X(s)+2Y(s) \cdots(1 )\\ sY(s)-0=-2X(s)-Y(s) \cdots(2)} \\ 由(1)可得X(s)= {1\over s-3}(2Y(s)+1) 代入(2) \Rightarrow sY(s) ={-2 \over s-3}(2Y(s)+1) -Y(s) \\ \Rightarrow Y(s)=-{2\over (s-1)^2} \Rightarrow y(t)=L^{-1}\{Y(s)\} =-2te^t \Rightarrow y'(t)=-2e^t-2te^t \\ \Rightarrow -2e^t-2te^t=-2x+2te^t \Rightarrow x(t) =e^t+2te^t \Rightarrow \bbox[red, 2pt]{\cases{x(t) =e^t+2te^t \\y(t)=-2te^t}}$$

解答:$$\cases{f(x,y,z)=xyz\\ g(x,y,z)=x^2+y^2+z^2/2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y = \lambda g_y\\ f_z= \lambda g_z \\ g=0} \Rightarrow \cases{yz=\lambda(2x) \\xz= \lambda(2y) \\xy= \lambda (z)\\ x^2+y^2+z^2/2=1} \\ \Rightarrow \cases{y/x=x/y\\ z/x=2x/z} \Rightarrow \cases{x^2=y^2\\ 2x^2=z^2} \Rightarrow x^2+x^2+x^2=1 \Rightarrow x^2={1\over 3} \Rightarrow \cases{y^2=1/3\\ z^2=2/3} \\ \Rightarrow (x,y,z) =(\pm \sqrt{1\over 3},\pm \sqrt{1\over 3}, \pm \sqrt{2\over 3}) \Rightarrow \cases{最大值=f(\sqrt{1/3},\sqrt{1/3},\sqrt{2/3},)=\sqrt{2\over 27} ={\sqrt 6\over 9} \\ 最小值= f(\sqrt{1/3},\sqrt{1/3},-\sqrt{2/3},) =-{\sqrt 6\over 9}}  \\ \Rightarrow \bbox[red, 2pt]{\cases{最大值={\sqrt 6\over 9} \\ 最小值=-{\sqrt 6\over 9}}}$$

解答:$$I=\int_0^1 \int_{\sqrt y}^1 e^{x^3}\,dxdy = \int_0^1 \int_0^{x^2} e^{x^3}\,dydx = \int_0^1 x^2 e^{x^3}\,dx \\ 取u=x^3 \Rightarrow du=3x^2\,dx \Rightarrow I=\int_0^1 {1\over 3}e^u\,du = \bbox[red, 2pt]{{1\over 3}(e-1)}$$

解答:$$二維的散度定理: \oint_C \vec F\cdot \vec n \,ds = \iint_B (\nabla\cdot \vec F)dA = \iint_B \left( {\partial F_1\over \partial x}+{\partial F_2\over \partial y} \right)\,dx \,dy \\ \nabla \cdot F=P_x+ Q_y =3x^2+4y^3 \Rightarrow \cases{P(x,y)=x^3\\ Q(x,y)=y^4} \Rightarrow F=(x^3,y^4) \\ \Rightarrow  \iint_B (3x^2+4y^3)\,dx dy = \oint_C x^3dy-y^4dx =\int_0^{2\pi} \left( (\cos^3 t)(\cos t\,dt)-(\sin^4 t)(-\sin t\,dt) \right) \\=\int_0^{2\pi} (\cos^4 t+\sin^5 t)\,dt = \int_0^{2\pi} \cos^4 t\,dt +0 =\int_0^{2\pi} \left( {3\over 8}+{1\over 2}\cos(2t)+{1\over 8} \cos(4t) \right)\,dt \\= \left. \left[ {3\over 8}t+{1\over 4}\sin(2t)+{1\over 32}\sin(4t) \right] \right|_0^{2\pi} = \bbox[red, 2pt]{{3\over 4}\pi}$$

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