115年中山大學海下所碩士班-工程數學詳解

國立中山大學 115學年度碩士班考試入學招生考試試題

科目名稱:工程數學【海下所碩士班】

解答:$$\textbf{(1) }x^2-2x-3=(x-3)(x+1)=0 \Rightarrow x=3,-1 \Rightarrow \bbox[red, 2pt]{(3,0),(-1,0)} \\ \textbf{(2) }f(x)=x^2-2x-3 \Rightarrow f'(x)=2x-2 \Rightarrow f'(3)=6-2= \bbox[red, 2pt]4 \\ \textbf{(3)} y=f(x)=(x-1)^2-4 \Rightarrow f(x) \le 0, x\in[-1,3] \Rightarrow 面積= \left|\int_{-1}^3 f(x)\,dx \right| \\= \left|\int_{-1}^3 (x^2-2x-3)\,dx \right| = \left| \left. \left[ {1\over 3}x^3-x^2-3x \right] \right|_{-1}^3 \right|= \bbox[red, 2pt]{32\over 3}$$

解答:$$\cases{P(1,0,-1) \\Q(-3,2,-4) \\R(1,-6,2)} \Rightarrow \cases{\overrightarrow{PQ} =(-4,2,-3) \\ \overrightarrow{PR} =(0,-6,3)} \Rightarrow \vec u= \overrightarrow{PQ} \times \overrightarrow{PR} =(-12,12,24) \\ \Rightarrow {\vec u\over |\vec u|}= \bbox[red, 2pt]{{\sqrt 6\over 6}(-1,1,2)}$$

解答:$$假設長方形\cases{長:a\\ 寬:b} \Rightarrow 2(a+b)=12 \Rightarrow a+b=6 \Rightarrow b=6-a \Rightarrow 面積=f(a)=a(6-a) \\ \Rightarrow f(a)=6a-a^2 \Rightarrow f'(a)=0 \Rightarrow 6-2a=0 \Rightarrow a=3 \Rightarrow b=3 \\ \Rightarrow \bbox[red, 2pt]{最大面積的長方形為一正方形，長與寬皆為3}$$

解答:$$\textbf{(1) }{dy\over dx}={2x-3\over 2+3y^2} \Rightarrow \int (2+3y^2)\,dy= \int(2x-3)\,dx \Rightarrow  \bbox[red, 2pt]{2y+y^3=x^2-3x+C} \\\textbf{(2) }y(1)=2 \Rightarrow 2\cdot 2+2^3=1-3+C \Rightarrow C=14 \Rightarrow \bbox[red, 2pt]{2y+y^3=x^2-3x+14}$$

解答:$$\textbf{(1) }f(x)= \sin x \Rightarrow f'(x)=\cos x\Rightarrow f''(x)=-\sin x \Rightarrow f'''(x)=-\cos x \Rightarrow f^{(4)}(x)=\sin x =f(x)\\ \quad \Rightarrow \cases{f^{(2k)}(0)=0\\ f^{(4k+1)}(0)=1\\ f^{(4k+3)}(0)=-1} ,k=0,1,2,\dots \Rightarrow f(x)= \bbox[red, 2pt]{\sin x=x-{1\over 3!}x^3+{1\over 5!}x^5-{1\over 7!}x^7+\cdots} \\ g(x)=\cos \Rightarrow g'(x)= -\sin x \Rightarrow g''(x) =-\cos x \Rightarrow g'''(x)=\sin x \Rightarrow g^{(4)}(x)=\cos(x) =g(x) \\ \Rightarrow \cases{g^{(2k-1)}(0)=0 \\ g^{(4k)}(0)=1\\ g^{(4k+2)}(0)=-1} \Rightarrow g(x)= \bbox[red, 2pt]{\cos x=1-{1\over 2!}x^2+{1\over 4!}x^4-{1\over 6!}x^6+\cdots} \\ \textbf{(2) }將\alpha=\beta=x代入合角公式 \Rightarrow \sin(x+x) =\sin x\cos x+\cos x\sin x \Rightarrow \sin (2x)=2\sin x\cos x\\ \quad \Rightarrow \sin x\cos x={1\over 2}\sin(2x) \\ 再依泰勒展開式: \sin(2x) =2x-{1\over 3!}(2x)^3+{1\over 5!}(2x)^5-\cdots \Rightarrow {1\over 2}\sin 2x=x-{2^2\over 3!}x^3+{2^4\over 5!}x^5-\cdots \\ 當x很小時, x^3,x^5..可以忽略不計,即{1\over 2}\sin 2x \approx x, 由以上可得 \sin x\cos x={1\over 2}\sin 2x \approx x \; \bbox[red, 2pt]{故得證} \\\textbf{(3) }A_3=2\times \left({1\over 2}(R\sin \theta_3) \times (R\cos\theta_3) \right) \times 3=R^2\sin\theta_3\cos \theta_3 \times 3, \theta_3={1\over 2}{2\pi\over 3} ={\pi\over 3} \\ A_4= 2\times \left({1\over 2}(R\sin \theta_4) \times (R\cos\theta_4) \right) \times 4=R^2\sin\theta_4 \cos \theta_4 \times 4, \theta_4={1\over 2}{2\pi\over 4} ={\pi\over 4} \\ \Rightarrow \bbox[red, 2pt]{A_n= R^2\sin \theta_n \cos \theta_n \times n, \theta_n={\pi\over n}} \\\textbf{(4) }\lim_{n\to \infty} A_n = \lim_{n\to \infty} R^2\sin {\pi\over n} \cos {\pi\over n} \times n \approx R^2\cdot {\pi\over n} \cdot n= \bbox[red, 2pt]{R^2\pi}$$

解答:$$\cases{|V|= \sqrt{3^2+4^2} =5 \\ |V'|= \sqrt{(-5.362)^2+4.499^2} \approx 7} \Rightarrow m={7\over 5} \Rightarrow \begin{bmatrix}-5.362\\ 4.499 \end{bmatrix} ={7\over 5} \begin{bmatrix}\cos \theta&-\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} \begin{bmatrix}3\\4 \end{bmatrix} \\ \Rightarrow \begin{bmatrix}-5.362\\ 4.499 \end{bmatrix} = \begin{bmatrix}4.2\cos \theta-5.6\sin \theta\\ 4.2\sin \theta+5.6\cos \theta \end{bmatrix} \Rightarrow \cases{4.2\cos \theta-5.6\sin \theta=-5.362\\ 4.2\sin \theta+5.6\cos \theta=4.499} \Rightarrow \cases{\cos\theta \approx 0.05457\\ \sin \theta \approx 0.9984} \\ \Rightarrow \theta= \cos^{-1}(0.05457) \approx 87^\circ \Rightarrow \bbox[red, 2pt]{\cases{m=1.2\\ \theta=\cos^{-1}(0.05457) \approx 87^\circ}}$$

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