113年中山海下碩士班-工程數學詳解

 國立中山大學113學年度碩士班招生考試

科目名稱:工程數學【海下所碩士班】

解答: $$y(t)=L^{-1} \left\{ {s+1\over s^2+s-6} \right\} =L^{-1} \left\{ {3\over 5(s-2)} +{2\over 5(s+3)}\right\} =\bbox[red, 2pt]{ {3\over 5}e^{2t}+ {2\over 5}e^{-3t}}$$

解答: $$4y''-8y'+3y=0 \Rightarrow 4\lambda^2-8\lambda+3=0 \Rightarrow (2\lambda-1)(2\lambda-3)=0 \Rightarrow \lambda={1\over 2},{3\over 2} \\ \Rightarrow y=c_1e^{x/2} +c_2e^{3x/2} \Rightarrow y'={1\over 2}c_1e^{x/2}+{3\over 2}c_2e^{3x/2} \Rightarrow \cases{y(0)=c_1+c_2=2\\ y'(0)={1\over 2}c_1+ {3\over 2}c_2={1\over 2}} \\ \Rightarrow \cases{c_1=5/2\\ c_2=-1/2} \Rightarrow \bbox[red, 2pt]{y={5\over 2}e^{x/2} -{1\over 2}e^{3x/2}}$$

解答: $$\cases{AB= \left[\begin{matrix}2 & 2 & 0\\0 & -1 & -1\\7 & 0 & -1\end{matrix}\right] \\[1ex]BA= \left[\begin{matrix}0 & -3 & 0\\1 & -4 & 2\\4 & -5 & 4\end{matrix}\right]} \Rightarrow AB\ne BA$$

解答: $$r(t)=t\vec i-t^2\vec j \Rightarrow v={dr\over dt} =\vec i-2t\vec j \Rightarrow |v|=\sqrt{1+4t^2}  \Rightarrow a_T={d\over dt}|v| ={4t\over \sqrt{1+4t^2}}\\ \Rightarrow \mathbf a={d\over dt}v=0\vec i-2\vec j \Rightarrow |\mathbf a|=2 \Rightarrow a_N=\sqrt{|a|^2-a_T^2} =\sqrt{4-{16t^2\over 1+4t^2}} =\sqrt{4 \over 1+4t^2} \\ \Rightarrow \bbox[red, 2pt]{\mathbf a_T={4t\over \sqrt{1+4t^2}},\mathbf a_N=\sqrt{4 \over 1+4t^2}}$$

解答: $$\cases{x(t)=t\\ y(t)=2t\\ z(t)=2t}\Rightarrow \cases{x'(t)=1\\ y'(t)=2\\ z'(t)=2},0\le t\le 3 \Rightarrow \int_0^3 (8t^2, -16t,2)\cdot (1,2,2)\,dt \\=\int_0^3(8t^2-32t+4)\,dt =\left.\left[{8\over 3}t^3-16t^2+4t \right] \right|_0^3= \bbox[red, 2pt]{-60}$$

解答: $$f(x)=f(-x) \Rightarrow f \text{ is even} \Rightarrow b_n=0\\ a_0={1\over 2}\int_{-1}^1 k\,dx =k\\ \Rightarrow a_n= \int_{-1}^1 k \cos({n\pi x \over 2})\, dx =\left. \left[ {2k\over n\pi} \sin{n\pi x\over 2} \right] \right|_{-1}^1 ={4k\over n\pi} \sin{n\pi \over 2} ,n=1,2,\dots\\ \Rightarrow \bbox[red, 2pt]{f(x)= k+ \sum_{n=1}^\infty {4k\over n\pi} \sin{n\pi \over 2} \cos{n\pi x \over 2}}$$

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解題僅供參考,其他歷年試題及詳解