111年台綜大轉學考-工程數學D37詳解

臺灣綜合大學系統111學年度學士班轉學生聯合招生考試

科目名稱：工程數學

類組代碼：D37

解答：$$\mathbf{(a)}\; \text{Green's theorem}只不過是\text{Stokes' theorem }在二維上的特例；\\\qquad 也就是將F_z= dz=0，兩者就相等\\\mathbf{(b)}\;C=\{(x,y,z)\mid x^2+y^2=1,z=0\} \Rightarrow 令\cases{x(t)=\cos t\\ y(t)=\sin t\\ z=0} \Rightarrow r(t)=(\cos t,\sin t, 0),0\le t\lt 2\pi \\ \Rightarrow r'(t)= (-\sin t,\cos t,0) \Rightarrow \oint_C \mathbf F\cdot dr = \int_0^{2\pi } F(r(t))r'(t)\, dt = \int_0^{2\pi } (\cos t,\sin t,0)\cdot (-\sin t,\cos t,0)\,dt\\= \int_0^{2\pi } 0\,dt =\bbox[red,2pt] 0\\ \text{curl }\mathbf F= ({\partial z\over \partial y} -{\partial y\over \partial z}, {\partial x\over \partial z} -{\partial z\over \partial x},{\partial y\over \partial x} -{\partial x\over \partial y}) =0 \Rightarrow \iint_S \text{curl }\mathbf F \cdot \mathbf n\,dS =\bbox[red, 2pt] 0\\ 由這個例子可以證明 \oint_C \mathbf F\cdot dr =\iint_S \text{curl }\mathbf F \cdot \mathbf n\,dS$$

解答：$$A=\left[\begin{matrix}1 & 2 & 1\\2 & 1 & 1\\1 & 3 & 1\end{matrix}\right] \Rightarrow \det(A)=1 \Rightarrow A\text{ is }\bbox[red,2pt]{nonsigular}\\ \det(A-\lambda I)=0 \Rightarrow -(\lambda+1)(\lambda^2-4\lambda-1)=0 \Rightarrow \lambda= -1,2\pm \sqrt{5}\\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I)\mathbf x= \left[\begin{matrix}2 & 2 & 1\\2 & 2 & 1\\1 & 3 & 2 \end{matrix} \right] \left[\begin{matrix} x_1\\x_2 \\x_3\end{matrix}\right] =0 \Rightarrow \cases{4x_1=x_3\\ 4x_2=-3x_3},取v_1=\left[\begin{matrix}1/4 \\-3/4 \\1 \end{matrix} \right] \\ \lambda_2=2-\sqrt 5\Rightarrow (A-\lambda_2 I)\mathbf x= \left[ \begin{matrix}\sqrt{5}-1 & 2 & 1 \\2 & \sqrt{5}-1 & 1 \\1 & 3 & \sqrt{5} -1\end{matrix} \right] \left[\begin{matrix}x_1\\x_2 \\x_3\end{matrix}\right] =0 \\\qquad \Rightarrow \cases{4x_1+(\sqrt 5-1)x_3=0 \\ 4x_2 +(\sqrt 5-1)x_3=0},取v_2= \left[\begin{matrix}(1-\sqrt 5)/4 \\(1-\sqrt 5)/4 \\1\end{matrix}\right] \\\lambda_3=2+\sqrt 5\Rightarrow (A-\lambda_3 I)\mathbf x=  \left[\begin{matrix}-\sqrt{5}-1 & 2 & 1 \\2 & -\sqrt{5}-1 & 1 \\1 & 3 & -\sqrt{5} -1\end{matrix} \right]\left[\begin{matrix}x_1\\x_2 \\x_3\end{matrix}\right] =0 \\\qquad \Rightarrow \cases{4x_1 =(\sqrt 5+1)x_3 \\ 4x_2= (\sqrt 5+1)x_3},取v_3= \left[\begin{matrix}( \sqrt 5+1)/4 \\( \sqrt 5+1)/4 \\1\end{matrix} \right]\\ \Rightarrow 特徵值為\bbox[red, 2pt]{-1,2-\sqrt 5,2+\sqrt 5}，相對應的特徵向量為\bbox[red, 2pt]{\left[\begin{matrix}1/4 \\-3/4 \\1\end{matrix} \right] ,\left[\begin{matrix}(1-\sqrt 5)/4 \\(1-\sqrt 5)/4 \\1\end{matrix}\right], \left[\begin{matrix}( \sqrt 5+1)/4 \\( \sqrt 5+1)/4 \\1\end{matrix} \right] }$$

解答：$$\mathbf{(a)}\; \int_0^4\int_0^\sqrt y xe^{y^2}\,dxdy =\int_0^4 \left.\left[{1\over 2}x^2 e^{y^2} \right]\right|_0^\sqrt y \, dy =\int_0^4 {1\over 2}ye^{y^2}\,dy = \left.\left[ {1\over 4}e^{y^2} \right]\right|_0^4 = \bbox[red, 2pt]{{1\over 4}(e^{16}-1)} \\\mathbf{(b)}\; \int_0^4\int_0^1 \int_0^{x^2} {1\over \sqrt{x^2-y^2}}\,dydxdz =\int_0^4\int_0^1 \left.\left[ \sin^{-1}{y\over x}\right]\right|_0^{x^2}\,dxdz =\int_0^4\int_0^1 \sin^{-1}x\,dxdz \\ =\int_0^4 \left.\left[ x\sin^{-1}x +\sqrt{1-x^2} \right]\right|_0^1 dz =\int_0^4 \left({\pi\over 2}-1\right)\,dz =\bbox[red, 2pt]{ 2\pi-4}$$

解答：$$\cases{x_1-2x_2+x_3=2 \cdots(1)\\ 3x_1-x_2+2x_3=5\cdots(2)\\ 2x_1+x_2+ x_3=1 \cdots(3)} \Rightarrow \begin{bmatrix} 1 & -2 & 1\\ 3 & -1 & 2\\ 2 & 1& 1\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =\begin{bmatrix} 2\\5\\ 1\end{bmatrix}  \equiv A\mathbf  x=\mathbf b\\ A=\begin{bmatrix} 1 & -2 & 1\\ 3 & -1 & 2\\ 2 & 1& 1\end{bmatrix}  \Rightarrow   rref(A)=\left[\begin{matrix}1 & 0 & \frac{3}{5}\\0 & 1 & - \frac{1}{5}\\0 & 0 & 0\end{matrix}\right] \Rightarrow rank=\bbox[red,2pt]2;\\  \left[\begin{array}{rrr|r} 1 & -2 & 1 &2\\ 3 & -1 & 2 & 5\\ 2 & 1& 1 & 1\end{array}\right] \xrightarrow{\matrix{-2R_1+R_2\to R_2\\ -R_1+R_3\to R_3}} \left[\begin{array}{rrr|r} 1 & -2 & 1 &2\\ 1 & 3 & 0 & 1\\ 1 & 3& 0 & -1\end{array}\right] \Rightarrow \cases{x_1-2x_2+x_3=1\\ x_1+3x_2=1 \\x_1+3x_2= -1} \Rightarrow \bbox[red,2pt]{無解}$$

解答：$$\cases{x_1- x_2-2 x_3=0  \\ 6x_1 +3x_3=0 \\ 2x_1+ 4x_2+ 5x_3=0  } \Rightarrow \begin{bmatrix} 1 & -1& -2\\ 6 & 0 & 3\\ 2 & 4& 5\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =\begin{bmatrix} 0\\0\\ 0\end{bmatrix}  \equiv A\mathbf  x=\mathbf b\\ A=\begin{bmatrix} 1 & -1& -2\\ 6 & 0 & 3\\ 2 & 4& 5\end{bmatrix}   \Rightarrow    \det(A)=-36\ne 0\Rightarrow 只有一明顯解: \bbox[red, 2pt]{x_1=x_2=x_3=0}$$

解答：$$y'''-2y''-y'+2y=0 \Rightarrow \lambda^3-2\lambda^2-\lambda+ 2 =0 \Rightarrow (\lambda-2)(\lambda-1)(\lambda+1)=0 \\ \Rightarrow \lambda=2,1,-1 \Rightarrow y=C_1e^{2x}+ C_2e^x+ C_3e^{-x} \Rightarrow y'=2C_1e^{2x} +C_2e^x-C_3e^{-x}\\ \Rightarrow y''=4C_2e^{2x} +C_2e^x+C_3e^{-x} \Rightarrow \cases{y(0)=C_1+C_2 +C_3= 3\\ y'(0)= 2C_1+C_2-C_3= -1\\ y''(0)= 4C_1 +C_2+C_3 = 3} \Rightarrow \cases{C_1=0\\ C_2=1\\ C_3=2} \\ \Rightarrow \bbox[red, 2pt]{ y=e^2+ 2e^{-x}}$$

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