108年專技高考_電子工程技師-工程數學詳解

108年專門職業及技術人員高等考試

等        別：高等考試
類        科：電子工程技師
科        目：工程數學

解：

$$\vec{r}=(x,y,z) \Rightarrow r=|\vec{r}|= \sqrt{x^2+y^2+z^2}\\
(一)\nabla r= \left( {\partial r\over \partial x}, {\partial r\over \partial y}, {\partial r\over \partial z}\right) =\bbox[red, 2pt]{ \left( {x\over \sqrt{x^2+y^2+z^2}} ,{y\over \sqrt{x^2+y^2+z^2}} ,{z\over \sqrt{x^2+y^2+z^2}} \right)}\\
(二) \nabla \cdot \vec{r}=\left( {\partial \over \partial x}, {\partial \over \partial y}, {\partial \over \partial z}\right) \cdot (x,y,z) ={\partial x\over \partial x}+ {\partial y\over \partial y}+ {\partial z\over \partial z}= 1+1+1= \bbox[red, 2pt]{3}\\
(三)\nabla \times \vec{r}=  \begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\{\partial \over \partial x}& {\partial \over \partial y}& {\partial \over \partial z}\\ x & y & z \end{vmatrix} = \left( {\partial z \over \partial y}- {\partial y \over \partial z}\right)\vec{i} + \left( {\partial x \over \partial z}- {\partial z \over \partial x}\right)\vec{j} +\left( {\partial y \over \partial x}- {\partial x \over \partial y}\right)\vec{k} = \bbox[red, 2pt]{(0,0,0)}$$

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解：
$$T={1\over 10y}e^{-(x^2+z^2)} \Rightarrow \nabla T=( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}) =\left( -{x\over 5y} e^{-(x^2+z^2)}, -{1\over 10y^2}e^{-(x^2+z^2)}, -{z\over 5y} e^{-(x^2+z^2)}\right) \\ \Rightarrow \nabla T(0,10,0)=\left(0,-{1\over 1000},0\right)\\
又\vec{a}=(1,1,1) \Rightarrow {\vec{a}\over |\vec{a}|}=(1/\sqrt{3}, 1/\sqrt{3},1/\sqrt{3})\\  \Rightarrow T(0,10,0)\cdot {\vec{a}\over |\vec{a}|}= \left(0,-{1\over 1000},0\right) \cdot (1/\sqrt{3}, 1/\sqrt{3},1/\sqrt{3}) =-{1\over 1000\sqrt{3}} =\bbox[red, 2pt]{- {\sqrt{3}\over 3000}}$$

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解：
(一)$$Res_{z=0} \cot z =\lim_{z\to 0} {z\cos z\over \sin z} =\lim_{z\to 0} {\cos z -z\sin z\over \cos z}=1  \Rightarrow \oint_C \cot z\;dz= -2\pi i \times Res_{z=0} \cot z=  \bbox[red, 2pt]{-2\pi i}$$(二)$$Res_{z=i/2}{z^3-6\over 2z-i} =\lim_{z\to i/2} {(z-i/2)(z^3-6)\over 2z-i} =\lim_{z\to i/2} {z^3 -6 \over 2}={-i/8 -6 \over 2}= -3-{i\over 16}\\  \Rightarrow \oint_C {z^3-6\over 2z-i}\;dz= -2\pi i \times Res_{z=i/2} {z^3-6\over 2z-i}= -2\pi i\times \left(-3-{i\over 16} \right) = \bbox[red, 2pt]{-{\pi \over 8}+6\pi i}$$

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解：
(一)一階二次
(二)一階三次
(三)二階一次

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解：
$$e^{2x}\left( 2\cos y\;dx-\sin y\;dy\right)=0 \Rightarrow 2\cos y\;dx=\sin y\;dy \Rightarrow 2\;dx = {\sin y\over \cos y}dy  \Rightarrow 2x = -\ln |\cos y|+C\\ y(0)=0 \Rightarrow 0=0+C \Rightarrow C=0 \Rightarrow 2x = -\ln |\cos y| \Rightarrow \cos y=e^{-2x} \Rightarrow \bbox[red, 2pt]{y=\pm \cos^{-1}\left( e^{-2x}\right)}$$

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解：
$$L(y(t))=F(s) \Rightarrow  \begin{cases}L(y'(t)) = sF(s)-y(0) =sF(s)-0.16\\L(y''(t))= s^2F(s)-sy(0)-y'(0) =s^2F(s)-0.16s\end{cases} \\  \Rightarrow L(y''+y'+9y)= L(y'')+ L(y')+9L(y) = s^2F(s)-0.16s +sF(s)-0.16+ 9F(s)=0\\ \Rightarrow F(s)= {0.16(s+1)\over s^2+s+9} ={0.16(s+1) \over (s+1/2)^2+(\sqrt{35}/2)^2}\\ =0.16\cdot{s+1/2 \over (s+1/2)^2+(\sqrt{35}/2)^2} +{0.16 \over \sqrt{35}}\cdot{\sqrt{35}/2 \over (s+1/2)^2+(\sqrt{35}/2)^2} \\ \Rightarrow y(t)=0.16L^{-1}\left( {s+1/2 \over (s+1/2)^2+(\sqrt{35}/2)^2}\right)+ {0.16 \over \sqrt{35}}L^{-1}\left({\sqrt{35}/2 \over (s+1/2)^2+(\sqrt{35}/2)^2} \right)\\ =0.16e^{-1/2t}\cos{\sqrt{35}t\over 2}+ {0.16 \over \sqrt{35}} e^{-1/2t}\sin{\sqrt{35}t\over 2}\\  \Rightarrow \bbox[red, 2pt]{y(t)=0.16e^{-1/2t} \left(\cos{\sqrt{35}t\over 2}+ {1 \over \sqrt{35}} \sin{\sqrt{35}t\over 2}\right)}$$

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解：


$$f(x) =\begin{cases}0, & -2<x<-1\\ 6, & -1<x<1 \\ 0, & 1<x<2\end{cases} \Rightarrow f(x)=f(-x) \Rightarrow f(x)為偶函數 \Rightarrow b_n=0\\

a_0= \frac{1}{4} \int_{-2}^2 f(x)\;dx= \frac{1}{4} \int_{-1}^1 6\;dx= 3\\

a_n={1\over 2}\int_{-2}^2 f(x)\cos {n\pi x\over 2}dx ={1\over 2}\int_{-1}^1 6\times\cos {n\pi x\over 2}dx =3\int_{-1}^1 \cos {n\pi x\over 2}dx = 3\left .\left[ {2\over n\pi}\sin {n\pi x\over 2}\right] \right|_{-1}^1\\ ={6\over n\pi}\left( \sin{n\pi \over 2}-\sin{-n\pi \over 2}\right) ={12\over n\pi}\sin{n\pi \over 2}\\ \Rightarrow f(x)=a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over 2}=3 +\sum_{n=1}^\infty {12\over n\pi}\sin{n\pi \over 2}\cos{n\pi x\over 2}=3+{12\over \pi}\sum_{n=1}^\infty {1\over n}\sin{n\pi \over 2}\cos{n\pi x\over 2}\\ \Rightarrow \bbox[red, 2pt]{f(x)=3+{12\over \pi}\sum_{n=1}^\infty {1\over n}\sin{n\pi \over 2}\cos{n\pi x\over 2}}$$

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考選部未公布申論題答案，解題僅供參考