112年台北科大光電系碩士班-工程數學詳解

 國立臺北科技大學 l12學 年度碩 士班招生考試

系所組別 :2401、 2402、 2403光 電工程系碩士班
第一節 工程數學 試題

解答： $$y'={dy\over dx}={y-2xy^2-2x^3 \over x+2x^2y+2y^3} \Rightarrow (2x^3+2xy^2-y)dx +(x+ 2x^2y+ 2y^3)dy=0 \\ \Rightarrow (2x^3+2xy^2)dx +(2x^2y+2y^3)dy= ydx-xdy \Rightarrow 2x(x^2+y^2)dx+ 2y(x^2+y^2)dy=ydx-xdy \\ \Rightarrow 2(xdx +ydy) = {ydx-xdy\over x^2+y^2} \Rightarrow \bbox[red, 2pt]{x^2+y^2 = \tan^{-1}{y\over x}+C}$$

解答： $$先求齊次解：y''-5y'+6y=0 \Rightarrow \lambda^2-5\lambda+6=0 \Rightarrow (\lambda-3)(\lambda-2)=0\\ \Rightarrow \lambda=2,3 \Rightarrow y_h=c_1e^{2x}+ c_2e^{3x}\\ 利用參數變換法求特定解y_p, 令\cases{y_1=e^{3x}\\ y_2=e^{2x} \\ r(x)=xe^{2x}} \Rightarrow \cases{y_1'=3e^{3x}\\ y_2'=2e^{2x}} \Rightarrow W=\begin{vmatrix} y_1& y_2\\ y_1'& y_2'\end{vmatrix} =-e^{5x} \\ \Rightarrow y_p= -y_1\int {y_2 r\over W}\,dx +y_2\int {y_1r\over W}\,dx = e^{3x} \int xe^{-x}\,dx - e^{2x} \int x\,dx \\ =e^{3x}\left(-xe^{-x}-e^{-x} \right)-e^{2x}\cdot {1\over 2}x^2 =-e^{2x}-xe^{2x}-{1\over 2}x^2e^{2x} \\ \Rightarrow y=y_h+y_p  \Rightarrow \bbox[red, 2pt]{y=C_1e^{2x} +C_2e^{3x}-xe^{2x}-{1\over 2}x^2e^{2x}}$$

解答： $$令\cases{A(1,1,2) \\ B(3,1,5)\\ C(1,7,4)\\ D(4,5,2)} \Rightarrow \cases{\overrightarrow{AB} =(2,0,3) \\\overrightarrow{AC} =(0,6,2) \\\overrightarrow{AD} =(3,4,0) } \Rightarrow 四面體體積= {1\over 6}\begin{Vmatrix} 2& 0 & 3\\ 0 & 6&2\\ 3& 4& 0\end{Vmatrix} ={70\over 6} =\bbox[red, 2pt]{35\over 3}$$

解答：

$$\mathbf{(a)}\;A=\left[\begin{matrix} \frac{2}{3} & a & \frac{2}{3} \\\frac{-2}{3} & b & \frac{1}{3} \\ \frac{1}{3} & c & \frac{-2}{3}\end{matrix}\right] \Rightarrow A^T=\left[\begin{matrix}\frac{2}{3} & \frac{-2}{3} & \frac{1}{3} \\ a & b & c \\\frac{2}{3} & \frac{1}{3} & \frac{-2}{3}\end{matrix} \right] \Rightarrow AA^T=I(\because A \text{ is orthogonal})\\ \Rightarrow \left[\begin{matrix} \frac{9a^2+8}{9} & \frac{9ab-2}{9} & \frac{9ac-2}{9} \\\frac{9ab-2}{9} & \frac{9b^2+5}{9} & \frac{9bc-4}{9} \\\frac{9ac-2}{9} & \frac{9bc-4}{9} & \frac{9c^2+5}{9}\end{matrix}\right] =\begin{bmatrix}1 & 0 & 0 \\0 & 1& 0 \\0 & 0& 1\end{bmatrix} \Rightarrow \cases{a^2=1/9\\ b^2=c^2=4/9\\ ab=ac=2/9\\ bc=4/9} \\\Rightarrow \bbox[red, 2pt]{\cases{a=1/3\\ b=c=2/3}或\cases{a=-1/3\\ b=c=-2/3}} \\ \mathbf{(b)}\; A^{-1}=A^T=\left[\begin{matrix}\frac{2}{3} & \frac{-2}{3} & \frac{1}{3} \\ a & b & c \\\frac{2}{3} & \frac{1}{3} & \frac{-2}{3}\end{matrix} \right] =\bbox[red, 2pt]{ \left[\begin{matrix}\frac{2}{3} & \frac{-2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\\frac{2}{3} & \frac{1}{3} & \frac{-2}{3}\end{matrix} \right] 或\left[\begin{matrix}\frac{2}{3} & \frac{-2}{3} & \frac{1}{3} \\ -\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} \\\frac{2}{3} & \frac{1}{3} & \frac{-2}{3}\end{matrix} \right]}$$

解答： $$\text{Stokes' Theorem }\oint_C \vec F\cdot d\vec r =\iint_S \text{curl }\vec F\cdot d\vec S\\曲面S的底面邊界為一圓:x^2+y^2=4,其逆時鐘路徑可表示成\vec r(t)=(2\cos t,2\sin t ,2),0\le t\le 2\pi\\ \Rightarrow d\vec r=(-2\sin t,2\cos t,0)dt;\;同理\vec F=(y,-2xz,yz^2)= (2\sin t,-8\cos t, 8\sin t)\\ 因此左式\oint_C \vec F\cdot d\vec r= \int_0^{2\pi} (2\sin t,-8\cos t, 8\sin t)\cdot (-2\sin t,2\cos t,0)dt\\= \int_0^{2\pi} (-4\sin^2 t-16\cos^2 t)\,dt= \int_0^{2\pi} (-4 -12\cos^2 t)\,dt=  \int_0^{2\pi} (-10 -6\cos  2t)\,dt= \bbox[cyan,2pt]{-20\pi}\\接著計算右式,\text{curl }\vec F=\begin{vmatrix}\vec i &\vec j  & \vec k \\\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\y & -2xz &yz^2 \end{vmatrix} =(2x+z^2, 0,-2z-1)\\ 令\vec r=(x,y,{1\over 2}(x^2+y^2)) \Rightarrow \cases{\vec r_x=(1,0,x)\\ \vec r_y=(0,1,y)} \Rightarrow \vec r_x\times \vec r_y=(-x,-y,1)\\ 因此右式\iint_S \text{curl }\vec F\cdot d\vec S =\iint_S (2x+z^2 ,0,-2z-1) \cdot (-x,-y,1)\,dxdy \\=\iint_S -2x^2-xz^2-2z-1\,dxdy=\iint_S -2x^2-{1\over 4}x(x^2+y^2)^2-(x^2+y^2)-1\,dxdy\\ =\int_0^{2\pi }\int_0^2 (-2r^2\cos^2\theta-{1\over 4}r^5\cos \theta-r^2-1)r\,drd\theta =\int_0^{2\pi} (-8\cos^2\theta -{32\over 7}\cos\theta -6)\,d\theta \\= \int_0^{2\pi} (-4\cos(2\theta)-{32\over 7}\cos\theta-10)\,d\theta = \bbox[cyan,2pt]{-20\pi}\\\Rightarrow 左式=右式=-20\pi，\bbox[red,2pt]{故得證}$$

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解題僅供參考，其他歷年試題及詳解