114年高考三級-應用數學詳解

 114年公務人員高等考試三級考試試題

類 科：天文
科 目：應用數學（包括微積分、微分方程與向量分析）

解答:$$\begin{bmatrix} 1 & 1 & 0 \\2 & 1 & -1 \\3 & -1 & -1\end{bmatrix} \underrightarrow{R_2-2R_1\to R_2, R_3-3R_1\to R_3}\begin{bmatrix} 1 & 1 & 0 \\0 & -1 & -1 \\0 & -4 & -1\end{bmatrix}  \underrightarrow{R_3-4R_2 \to R_3} \\\begin{bmatrix} 1 & 1 & 0 \\0 & -1 & -1 \\0 & 0 & 3 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{U= \begin{bmatrix} 1 & 1 & 0 \\0 & -1 & -1 \\0 & 0 & 3 \end{bmatrix} , L=\begin{bmatrix} 1 & 0 & 0 \\2 & 1 & 0 \\3 & 4 & 1\end{bmatrix}}$$

解答:$$y'''+y''+y'+y=0 \Rightarrow \lambda^3+ \lambda^2+ \lambda+1=0 \Rightarrow (\lambda+1)(\lambda^2+1) =0 \Rightarrow \lambda=-1, \pm i\\ \Rightarrow y_h=c_1e^{-x} +c_2\cos x+c_3\sin x\\ y_p=Axe^{-x}+ Bx+C \Rightarrow y_p'=A(1-x)e^{-x} +B \Rightarrow y_p''=A(x-2)e^{-x} \Rightarrow y_p'''=A(3-x)e^{-x} \\ \Rightarrow y_p'''+y_p'' +y_p'+y_p =2Ae^{-x}+Bx+B+C=e^{-x}+4x \Rightarrow \cases{A=1/2\\ B=4\\ C=-4} \\ \Rightarrow y_p={1\over 2}xe^{-x}+4x-4 \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{ y=c_1e^{-x} +c_2\cos x+c_3\sin x +{1\over 2}xe^{-x}+4x-4}$$

解答:$$\cases{f(x,y,z) =x^2+ 2y^2+2z^2-20\\ g(x,y,z) = x^2+y^2+z-4} \Rightarrow \cases{\nabla f=(2x,4y,4z) \\ \nabla g=(2x,2y,1)} \Rightarrow \cases{\nabla f(0,1,3)=(0,4,12) \\ \nabla g(0,1,3)=(0,2,1)} \\ \Rightarrow 切線向量\vec v=(0,4,12) \times (0,2,1) =(-20,0,0) \Rightarrow 切線方程式: \bbox[red, 2pt]{\{(t,1,3) \mid t\in \mathbb R\}}$$

解答:$$球坐標\cases{x=  R\sin\phi\cos\theta \\y= R\sin\phi\sin\theta \\z=R\cos\phi} \Rightarrow \vec r=(R\sin\phi\cos\theta ,  R\sin\phi\sin\theta ,R\cos\phi) \\ \Rightarrow \cases{\displaystyle {\partial \vec r\over \partial \phi} =(R\cos \rho \cos \theta, R\cos \phi \sin \theta, -R\sin \rho) \\ \displaystyle {\partial \vec r\over \partial \theta} =(-R\sin \phi \sin \theta, R\sin \phi \cos \theta, 0)}  \\\Rightarrow \frac{\partial\mathbf{r}}{\partial\phi} \times \frac{ \partial \mathbf{r}}{\partial\theta} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ R\cos\phi\cos\theta & R\cos\phi\sin\theta & -R\sin\phi \\ -R\sin\phi\sin\theta & R\sin\phi\cos\theta & 0 \end{vmatrix} \\= (R^2\sin^2 \phi \cos \theta, R^2 \sin^2\phi\sin \theta, R^2 \sin \phi \cos \phi)\\ \Rightarrow  \left\| \frac{\partial\mathbf{r}}{\partial\phi} \times \frac{\partial\mathbf{r}}{\partial \theta}\right\| = \sqrt{(R^2\sin^2\phi \cos\theta)^2 + (R^2\sin^2\phi\sin\theta)^2 + (R^2\sin \phi \cos\phi)^2} =R^2\sin \phi\\\Rightarrow 表面積S=\iint_D \begin{Vmatrix} {\partial r\over \partial \phi} \times{\partial r\over \partial \theta} \end{Vmatrix} \,d\phi d\theta  = \int_0^{2\pi} \int_0^\pi R^2\sin\phi \, d\phi\,d\theta = R^2 \int_0^{2\pi} (-\cos\pi - (-\cos0)) \, d\theta \\= R^2 \int_0^{2\pi} (-(-1) - (-1)) \, d\theta = R^2 \int_0^{2\pi} (1+1) \, d\theta =4\pi R^2 = \bbox[red, 2pt]{4\pi} (單位球半徑R=1)$$

解答:$$u(r,\theta) =R(r) \Theta(\theta) \Rightarrow R''\Theta+{1\over r}R'\Theta+{1\over r^2}R\Theta'' =0 \Rightarrow {r^2R''+rR'\over R} =-{\Theta''\over \Theta} =\lambda \\ \Rightarrow \cases{\Theta''+ \lambda \Theta=0\\ r^2R''+rR'-\lambda R=0}; 又邊界條件\cases{u(r,0)=R(r)\Theta(0) =0\\ u(r,\pi/2)=R(r)\Theta(\pi/2) =0} \Rightarrow \cases{\Theta(0)=0\\ \Theta(\pi/2)=0} \\  \textbf{Case I }\lambda=0 \Rightarrow \Theta''=0 \Rightarrow \Theta=c_1\theta+c_2 \Rightarrow \cases{\Theta(0)=c_2=0\\ \Theta(\pi/2)=c_1\cdot (\pi/2)+c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0} \Rightarrow \Theta=0\\\qquad \Rightarrow u(r,\theta)=0, \text{ 明顯解,不討論} \\\textbf{Case II }\lambda \lt 0 \Rightarrow \lambda= -k^2 (k\gt 0) \Rightarrow \Theta''-k^2\Theta=0 \Rightarrow \alpha^2-k^2=0 \Rightarrow \alpha=\pm k \\\qquad \Rightarrow \Theta=c_1e^{kx} +c_2e^{-kx} \Rightarrow \cases{\Theta(0)= c_1+c_2=0\\ \Theta(\pi/2)=c_1e^{k\pi/2} +c_2e^{-k\pi/2}=0} \Rightarrow c_1e^{k\pi/2} -c_1e^{-k\pi/2}=0 \\\qquad \Rightarrow c_1(e^{k\pi}-1) =0 \Rightarrow c_1=0\Rightarrow c_2=0 \Rightarrow \Theta=0 \Rightarrow u(r,\theta)=0 , \text{ 明顯解,不討論} \\\textbf{Case III }\lambda\gt 0 \Rightarrow \lambda=k^2(k\gt 0) \Rightarrow \Theta''+k^2\Theta=0 \Rightarrow \alpha^2+k^2=0 \Rightarrow \alpha=\pm ki \\\qquad \Rightarrow \Theta= c_1\cos k\theta+ c_2 \sin k\theta \Rightarrow \Theta(0)=c_1=0 \Rightarrow \Theta(\pi/2)=c_2\sin{k\pi\over 2} =0 \Rightarrow {k\pi\over 2}=n\pi\\ \qquad \Rightarrow k=2n \Rightarrow \Theta_n=c_n \sin(2n\theta),n=1,2,\dots \\ \lambda=k^2 \Rightarrow r^2R''+rR'-k^2R=0 \Rightarrow R=r^\alpha \Rightarrow R'=\alpha r^{\alpha-1} \Rightarrow R''=\alpha (\alpha-1)r^{\alpha-2} \\ \Rightarrow \alpha(\alpha-1)r^\alpha+\alpha r^\alpha-k^2 r^\alpha=0 \Rightarrow (\alpha^2-k^2)r^\alpha=0 \Rightarrow \alpha =\pm k \Rightarrow R=Ar^k+ Br^{-k} \\ \Rightarrow R_n= a_nr^{2n} +b_nr^{-2n} \Rightarrow u_n(r,\theta) =R_n\Theta_n =(a_nr^{2n} +b_nr^{-2n} )c_n \sin(2n\theta) \\ \Rightarrow \bbox[red, 2pt]{u(r,\theta) =\sum_{n=1}^\infty \left( A_nr^{2n} \sin(2n\theta) +B_n r^{-2n}\sin (2n\theta) \right)}$$

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