112年中央大學環境工程碩士班-工程數學詳解

 國立中央大學112學年度碩士班考試入學試題

所別: 環境工程研究所
科目:工程數學

解答:  $$\left( \begin{array} {rrr|rrr} 3 & 2 & 1 & 1 & 0 & 0 \\0 & -1 & 4 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 & 0 & 1 \end{array} \right) \xrightarrow {R_1/3\to R_1,-R_2 \to R_2, R_3/2\to R_3} \left( \begin{array} {rrr|rrr} 1 & \frac{2}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & -4 & 0 & -1 & 0 \\0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right) \\ \xrightarrow{4R_3+R_2\to R_2, -R_3/3+R_1 \to R_1} \left( \begin{array} {rrr|rrr}1 & \frac{2}{3} & 0 & \frac{1}{3} & 0 & \frac{-1}{6} \\0 & 1 & 0 & 0 & -1 & 2 \\0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right) \xrightarrow{-2R_2/3+R_1\to R_2}\\ \left( \begin{array} {rrr|rrr} 1 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & \frac{-3}{2} \\0 & 1 & 0 & 0 & -1 & 2 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right) \Rightarrow \bbox[red, 2pt]{A^{-1}= \begin{bmatrix}1/3 & 2/3 & -3/2 \\0 & -1& 2\\ 0 & 0 & 1/2 \end{bmatrix}}$$

解答: $$\text{div }\vec V =\frac{\partial }{\partial x}(4x) +\frac{\partial  }{\partial y}(-8y) +\frac{\partial  }{\partial z}(4z)=4-8+4=\bbox[red, 2pt] 0$$

解答: $$r(t)=\langle 2t, 2t^2,2t^3 \rangle \Rightarrow r'(t)= \langle 2, 4t,6t^2 \rangle \Rightarrow r''(t)= \langle 0, 4,12t \rangle \\ \Rightarrow r'(t)\times r''(t)=\langle 24t^2,-24t,8 \rangle \Rightarrow 曲率\kappa={|r'(t)\times r''(t)| \over |r'(t)|^3} ={|\langle 24t^2,-24t,8 \rangle| \over |\langle 2, 4t,6t^2 \rangle|^3} \\= \bbox[red, 2pt]{\sqrt{576t^4+576t^2+64} \over (36t^4+16t^2+4)^{3/2}}$$

解答: $$A=\begin{bmatrix}5 & 7&-5 \\0 & 4& -1\\ 2&8& -3 \end{bmatrix} \Rightarrow \cases{ A^2=\left[ \begin{matrix}15 & 23 & -17 \\-2 & 8 & -1 \\4 & 22 & -9 \end{matrix} \right] \\[1ex]2A= \left[ \begin{matrix} 10 & 14 & -10 \\0 & 8 & -2 \\4 & 16 & -6\end{matrix} \right]} \Rightarrow A^2+2A = \bbox[red, 2pt]{\left[ \begin{matrix} 25 & 37 & -27 \\ -2 & 16 & -3 \\ 8 & 38 & -15 \end{matrix} \right]}$$

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