111年高考三級-應用數學詳解

 111年公務人員高等考試三級考試試題

類 科： 天文、 氣象
科 目： 應用數學（ 包括微積分、 微分方程與向量分析）

解答：$$\cases{2x+ y+dz =c\\ x+z= b\\ x+y+2z= a} \Rightarrow \triangle =\begin{vmatrix} 2 & 1 & d\\ 1 & 0 & 1\\ 1 & 1& 2 \end{vmatrix},\triangle_x=\begin{vmatrix}c &  1 & d\\ b& 0 &1  \\ a& 1 & 2 \end{vmatrix}, \triangle_y =\begin{vmatrix} 2 & c & d\\ 1 & b & 1\\ 1 & a & 2\end{vmatrix}, \triangle_z =\begin{vmatrix} 2 & 1 & c\\ 1 & 0 & b\\ 1 & 1 & a\end{vmatrix} \\(一)有唯一解: \triangle \ne 0 \Rightarrow d-3\ne 0 \Rightarrow \bbox[red,2pt]{d\ne 3}\\(二)無窮多解: \triangle =0 \Rightarrow d=3 \Rightarrow \cases{\triangle_x=a+b-c\\ \triangle_y= a+b-c\\ \triangle_z= -a-b+c},因此若 \triangle_x= \triangle_y =\triangle_z=0 \Rightarrow a+b=c\\ \qquad 無窮多解的條件:\bbox[red,2pt]{\cases{a+b=c\\ d=3}}\\(三)無解:由(一)、(二)可知無解的條件:\bbox[red,2pt]{\cases{a+b\ne c\\ d=3}}$$

解答：$$y'=\cfrac{y}{e^{-2y}-2xy} \Rightarrow y\,dx +(2xy-e^{-2y})dy=0，\\取\cases{M(x,y)=2xy-e^{-2y} \Rightarrow M_x=2y\\ N(x,y)=y \Rightarrow N_y=1} \Rightarrow M_x\ne N_y \Rightarrow 非恰當(\text{not exact})\\ 因此我們試著找積分因子\mu=\mu(y) \Rightarrow (y\mu)_y= ((2xy-e^{-2y})\mu)_x \Rightarrow \mu+y\mu_y=2y\mu \\\Rightarrow {1\over u}du = {2y-1\over y}dy \Rightarrow  u=e^{2y}/y \Rightarrow e^{2y}dx+(2xe^{2y}-{1\over y})dy=0\\欲求\Psi(x,y)滿足\cases{\Psi_x=e^{2y} \Rightarrow \Psi = \int e^{2y}dx= xe^{2y}+ p(y)\\ \Psi_y= 2xe^{2y}-{1\over y} \Rightarrow \Psi = \int 2xe^{2y}-{1\over y}dy = xe^{2y} -\ln |y|+q(x)} \\\Rightarrow \Psi = xe^{2y}-\ln |y| \Rightarrow 通解為\bbox[red,2pt]{xe^{2y}-\ln|y|=C,C為常數}$$

解答：
$$令\cases{\vec F=(F_1,F_2,F_3)\\ \vec G=(G_1,G_2,G_3)}及\cases{D_1={\partial \over \partial x} \\ D_2={\partial \over \partial y} \\ D_3={\partial \over \partial z} } \Rightarrow \nabla(\vec F\cdot \vec G) =\nabla (\sum_{i=1}^3F_iG_i) \\
\nabla (\vec{F} \cdot \vec{G}) \ =\ \nabla ( F_{1} G_{1} +F_{2} G_{2} +F_{3} G_{3})\\
=( D_{1}( F_{1} G_{1} +F_{2} G_{2} +F_{3} G_{3}) ,\ D_{2}( F_{1} G_{1} +F_{2} G_{2} +F_{3} G_{3}) ,\ D_{3}( F_{1} G_{1} +F_{2} G_{2} +F_{3} G_{3}))\\
=( D_{1} F_{1} G_{1} +F_{1} D_{1} G_{1} \ +D_{1} F_{2} G_{2} +F_{2} D_{1} G_{2} +D_{1} F_{3} G_{3} +F_{3} D_{1} G_{3} \ ,\\
D_{2} F_{1} G_{1} +F_{2} D_{2} G_{1} +D_{2} F_{2} G_{2} +F_{2} D_{2} G_{2} +D_{2} F_{3} G_{3} +F_{3} D_{2} G_{3} ,\\
D_{3} F_{1} G_{1} +F_{1} D_{3} G_{1} +D_{3} F_{2} G_{2} +F_{2} D_{3} G_{2} +D_{3} F_{3} G_{3} +F_{3} D_{3} G_{3})\\
=(( F_{1} D_{1} G_{1} +F_{2} D_{1} G_{2} +F_{3} D_{1} G_{3}) +( G_{1} D_{1} F_{1} +G_{2} D_{1} F_{2} +G_{3} D_{1} F_{3}) ,\\
( F_{1} D_{2} G_{1} +F_{2} D_{2} G_{2} +F_{3} D_{2} G_{3}) +( G_{1} D_{2} F_{1} +G_{2} D_{2} F_{2} +G_{3} D_{2} F_{3}) ,\\
( F_{1} D_{3} G_{1} +F_{2} D_{3} G_{2} +F_{3} D_{3} G_{3}) +( G_{1} D_{3} F_{1} +G_{2} D_{3} F_{2} +G_{3} D_{3} F_{3}))\\
=(( F_{1} D_{1} G_{1} +{F_{2} D_{2} G_{1}} +{F_{3} D_{3} G_{1}}) +( F_{2} D_{1} G_{2} -{F_{2} D_{2} G_{1}}) \ +( F_{3} D_{1} G_{3} -{F_{3} D_{3} G_{1}})\\
+( G_{1} D_{1} F_{1} +G_{2} D_{2} F_{1} +G_{3} D_{3} F_{1}) \ +( G_{2} D_{1} F_{2} -G_{2} D_{2} F_{1}) \ +( G_{3} D_{1} F_{3} -G_{3} D_{3} F_{1}) ,\\
( F_{1} D_{1} G_{2} +F_{2} D_{2} G_{2} +F_{3} D_{3} G_{2}) +( F_{3} D_{2} G_{3} -F_{3} D_{3} G_{2}) +( F_{1} D_{2} G_{1} -F_{1} D_{1} G_{2})\\
+( G_{1} D_{1} F_{2} +G_{2} D_{2} F_{2} +G_{3} D_{3} F_{2}) +( G_{3} D_{2} F_{3} -G_{3} D_{3} F_{2}) +( G_{1} D_{2} F_{1} -G_{1} D_{1} F_{2}) ,\\
( F_{1} D_{1} G_{3} +F_{2} D_{2} G_{3} +F_{3} D_{3} G_{3}) +( F_{1} D_{3} G_{1} -F_{1} D_{1} G_{3}) +( F_{2} D_{3} G_{2} -F_{2} D_{2} G_{3})\\
+( G_{1} D_{1} F_{3} +G_{2} D_{2} F_{3} +G_{3} D_{3} F_{3}) +( G_{1} D_{3} F_{1} -G_{1} D_{1} F_{3}) +( G_{2} D_{3} F_{2} -G_{2} D_{2} F_{3}))\\
=(\vec{F} \cdot \nabla ) G_{1} +F_{2}( D_{1} G_{2} -D_{2} G_{1}) +F_{3}( D_{1} G_{3} -D_{3} G_{1})
+(\vec{G} \cdot \nabla ) F_{1} +G_{2}( D_{1} F_{2} -D_{2} F_{1}) +G_{3}( D_{1} F_{3} -D_{3} F_{1}) ,\\
(\vec{F} \cdot \nabla ) G_{2} +F_{3}( D_{2} G_{3} -D_{3} G_{2}) +F_{1}( D_{2} G_{1} -D_{1} G_{2})
+(\vec{G} \cdot \nabla ) F_{2} +G_{3}( D_{2} F_{3} -D_{3} F_{2}) +G_{1}( D_{2} F_{1} -D_{1} F_{2}) ,\\
(\vec{F} \cdot \nabla ) G_{3} +F_{1}( D_{3} G_{1} -D_{1} G_{3}) +F_{2}( D_{3} G_{2} -D_{2} G_{3})
+(\vec{G} \cdot \nabla ) F_{3} +G_{1}( D_{3} F_{1} -D_{1} F_{3}) +G_{2}( D_{3} F_{2} -D_{2} F_{3}))\\=(\vec{F} \cdot \nabla )\vec{G} +\vec{F}\times ( \nabla\times \vec{G}) +(\vec{G} \cdot \nabla )\vec{F} +\vec{G}\times ( \nabla\times \vec{F})\\ \Rightarrow \nabla (\vec{F} \cdot \vec{G})=(\vec{F} \cdot \nabla )\vec{G} +\vec{F} \times( \nabla\times \vec{G}) +(\vec{G} \cdot \nabla )\vec{F} +\vec{G}\times ( \nabla\times \vec{F})\\ \Rightarrow \nabla(\vec F\cdot \vec G)-\vec F\times(\nabla \times \vec G) -\vec G\times (\nabla \times \vec G)= \bbox[red,2pt]{(\vec F\cdot \nabla)\vec G +(\vec G\cdot \nabla)\vec F}$$

解答：$$先求兩圖形\cases{x^2+y^2+z=1\\ y+z=1}的交集\Rightarrow z=1-x^2-y^2=1-y \Rightarrow x^2+y^2-y=0\\ \Rightarrow x^2+(y-{1\over 2})^2= {1\over 4} 為一圓，其\cases{圓心(0,1/2)\\ 半徑=1/2}\\ 利用圓柱坐標: \cases{x= r\cos\theta \\ y= r\sin \theta +{1\over 2}\\ z=z} \Rightarrow \cases{z=1-x^2-y^2 ={3\over 4}-r^2-r\sin \theta\\ z=1-y = {1\over 2}-r\sin\theta }\\ \Rightarrow 欲求之體積= \int_0^{2\pi} \int_0^{1/2} \int_{1/2-r\sin\theta}^{3/4-r^2-r\sin\theta} r\,dzdrd\theta = \int_0^{2\pi} \int_0^{1/2} {1\over 4}r-r^3\, drd\theta =\int_0^{2\pi}{1\over 64}\,d\theta\\ =\bbox[red, 2pt]{\pi\over 32}$$

解答：$$\triangle u(r,\theta)={1\over \sqrt{r^2+c^2}}=f(r) \Rightarrow {\partial^2 \over \partial r^2}u+ {1\over r}{\partial \over \partial r}u +{1\over r^2}{\partial^2 \over \partial \theta^2}u ={1\over \sqrt{r^2+c^2}}\\ 由於是所有徑向，即與\theta 無關，因此上式變為{\partial^2 \over \partial r^2}u+ {1\over r}{\partial \over \partial r}u   ={1\over \sqrt{r^2+c^2}} \Rightarrow u''+{1\over r}u'={1\over \sqrt{r^2+c^2}} \\先求齊次解:u''+{1\over r}u'=0，令u'=r^m \Rightarrow u''=mr^{m-1} \Rightarrow mr^{m-1}+r^{m-1}=0 \Rightarrow m=-1\\ \Rightarrow u'=C_1r^{-1} \Rightarrow u_h=C_1\ln x+C_2\\ 接下來用參數變換法來求u_p: \cases{u_1=\ln r\\ u_2=1} \Rightarrow W(u_1,u_2)= \begin{vmatrix} \ln r& 1\\ 1/x & 0 \end{vmatrix} =-{1\over r}\\ \Rightarrow u_p =-y_1\int { y_2f(r)\over W(y_1,y_2)}dr +y_2\int { y_1f(r)\over W(y_1,y_2)}dr =-\ln r \int {1/\sqrt{r^2+c^2}\over -1/r}dr +\int {\ln r/\sqrt{r^2+c^2}\over -1/r}dr \\=\ln r\int{r\over \sqrt{r^2+c^2}}\,dr- \int{r\ln r\over \sqrt{r^2+c^2}}dr =\ln r\left(\sqrt{r^2+c^2} \right) -\left(\ln r\sqrt{r^2 +c^2}- \int{\sqrt{r^2+c^2}\over r}dr \right)\\ =\int{\sqrt{r^2+c^2}\over r}dr =\sqrt{r^2+c^2} -c\tanh^{-1}{\sqrt{r^2+c^2}\over c} (取r=c\tan(v) \Rightarrow dr=c\sec^2 vdv)\\\Rightarrow u=u_h+u_p \Rightarrow \bbox[red, 2pt]{u =\sqrt{x^2+c^2}-c\tanh^{-1}\left({\sqrt{x^2+c^2} \over c} \right) +C_1\ln x+C_2}$$

=================== END ========================

考選部未公告答案，解題僅供參考，其他國考試題及詳解