108年升官等-物理-微積分-詳解

108年公務、關務人員升官等考試

等      級：薦任
類科別：物理
科       目：微積分

$$一、試求\lim_{x\to \infty}\frac{x^5}{5^x}。$$

解：
由羅必達法則(L'Hopital's Rule)}求解
$$\lim_{x\to\infty}\frac{x^5}{5^x} = \lim_{x\to\infty} \frac{x^5} {e^{x\ln 5}}= \lim_{x\to\infty} \frac{f(x)}{g(x)} = \lim_{x\to\infty}\frac{f'(x)}{g'(x)}=  \lim_{x\to\infty} \frac{5x^4}{\ln (5) e^{x\ln 5}} \\ = \lim_{x\to\infty}\frac{f''(x)}{g''(x)}=  \lim_{x\to\infty} \frac{20x^3}{(\ln (5))^2 e^{x\ln 5}} = \lim_{x\to\infty} \frac{f'''(x)}{g'''(x)}= \lim_{x\to\infty} \frac{60x^2}{(\ln (5))^3 e^{x\ln 5}} \\= \lim_{x\to\infty}\frac{f^{[4]}(x)}{g^{[4]}(x)}=  \lim_{x\to\infty} \frac{120x}{(\ln (5))^4 e^{x\ln 5}} = \lim_{x\to\infty}\frac{f^{[5]}(x)}{g^{[5]}(x)}=  \lim_{x\to\infty} \frac{120}{(\ln (5))^5 e^{x\ln 5}} \\ = \frac{120}{\infty} = \bbox[red, 2pt]{0}$$

---

二、令$f(x)=2x^3+3x^2-12x-2$，試求$f(x)$在區間[0,10]上的最大值與最小值。

解：

$$f(x)=2x^3+3x^2-12x-2 \Rightarrow f'(x)=0 \Rightarrow 6x^2+6x-12=0 \Rightarrow 6(x+2)(x-1)=0 \\ \Rightarrow x=1,-2 有極值(-2不在[0,10]內) \Rightarrow f''(1)= 12+6=18>0 \\\Rightarrow f在區間內為凹向上 \Rightarrow f(1)=2+3-12-2= -9為區間內之最小值\\ 又\begin{cases}f(0)=-2\\ f(10)=2000+300-120-2= 2178\end{cases} \Rightarrow f(10)為區間內之最大值\\ 答：\bbox[red, 2pt]{最大值為2178，最小值為-9}$$

---

$$三、試求積分\int   (xe^x+e^x\sin   x)dx。$$
解： $$\begin{cases} u=x\\ dv=e^x dx\end{cases} \Rightarrow \begin{cases} du=dx\\ v=e^x \end{cases} \Rightarrow \int xe^x\;dx =xe^x-\int e^xdx =xe^x-e^x \cdots(1)\\ \begin{cases} u=\sin x\\ dv=e^x dx\end{cases} \Rightarrow \begin{cases} du=\cos xdx\\ v=e^x \end{cases} \Rightarrow \int e^x\sin x\;dx =e^x\sin x-\int e^x\cos xdx\\ 又\begin{cases} u=\cos x\\ dv=e^x dx\end{cases} \Rightarrow \begin{cases} du=-\sin xdx\\ v=e^x \end{cases} \Rightarrow \int e^x\cos x\;dx =e^x\cos x+\int e^x\sin xdx \\ 因此\int e^x\sin xdx = e^x\sin x-\left(e^x\cos x+\int e^x\sin x \right) \Rightarrow 2\int e^x\sin x\;dx= e^x\sin x-e^x\cos x \\ \int e^x\sin x\;dx = \frac{1}{2}e^x(\sin x-\cos x)\cdots(2)\\ 由(1)及(2) \Rightarrow \int (xe^x+e^x \sin x)\;dx= \bbox[red, 2pt]{xe^x-e^x+ \frac{1}{2}e^x(\sin x-\cos x)+C}，其中C為常數$$

---

$$四、令f(x,y)=(5+xy^2+yx^2)+e^{5+xy^2+yx^2}，求\frac{\partial f}{\partial x}，\frac{\partial f}{\partial y}，\frac{\partial f}{\partial x^2}。$$
解：
$$f(x,y)=(5+xy^2+yx^2)+e^{5+xy^2+yx^2}\\ \Rightarrow \begin{cases}\frac{\partial f}{\partial x}= \frac{\partial }{\partial x}(5+xy^2+yx^2)+ \frac{\partial }{\partial x}e^{5+xy^2+yx^2} =y^2+2xy+ (y^2+2xy)e^{5+xy^2+yx^2}\\ \frac{\partial f}{\partial y}= \frac{\partial }{\partial y}(5+xy^2+yx^2)+ \frac{\partial }{\partial y} e^{5+xy^2+yx^2} =2xy+x^2 +(2xy+x^2)e^{5+xy^2+yx^2}\end{cases} \\ \Rightarrow \frac{\partial f}{\partial x^2}=\frac{\partial }{\partial x}\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}(y^2+2xy+ (y^2+2xy)e^{5+xy^2+yx^2}) = 2y+ \left(2y+(y^2+2xy)^2 \right)e^{5+xy^2+yx^2} \\ \Rightarrow \bbox[red, 2pt]{\begin{cases} \frac{\partial f}{\partial x}= y^2+2xy+ (y^2+2xy)e^{5+xy^2+yx^2}\\\frac{\partial f}{\partial y}= 2xy+x^2 +(2xy+x^2)e^{5+xy^2+yx^2} \\\frac{\partial f}{\partial x^2}= 2y+ \left(2y+(y^2+2xy)^2 \right)e^{5+xy^2+yx^2} \end{cases} }$$

---

五、試求曲面$4x^2+3y^2 +z^2=20$在點$P(1,2,2)$的切平面方程式

解：
$$4x^2+3y^2 +z^2=20 \Rightarrow \nabla F= \left. ( 8x,6y,2z) \right|_{(1,2,2)} = (8, 12, 4) \Rightarrow (x-1,y-2,z-2) \cdot (8,12,4)=0 \\ \Rightarrow  2(x-1)+3(y-2)+(z-2)=0 \Rightarrow 2x+3y+z=10 \\\Rightarrow 切平面方程式為  \bbox[red, 2pt]{2x+3y+z=10}$$

---

六、試求下列積分：$\iint_D\sin{\sqrt{x^2+y^2}}\;dA$，此處$D=\{(x,y)\mid \pi\le   \sqrt{x^2+y^2} \le 2\pi \}$。

解：
$$\begin{cases} x=r\cos \theta \\ y=r\sin \theta \end{cases} \Rightarrow D=\{(r\cos \theta,r\sin \theta)\mid \pi\le r\le 2\pi\} \Rightarrow \iint_D \sin \sqrt{x^2+y^2}\;dA\\ = \int_\pi^{2\pi} \int_0^{2\pi}r\sin r\;d\theta dr =2\pi \int_\pi^{2\pi}r\sin r\;dr = 2\pi\left .\left[ -r\cos r+\sin r\right] \right|_\pi^{2\pi} =2\pi (-2\pi-\pi) = \bbox[red, 2pt]{-6\pi^2}$$

---

七、今$f(x,y,z)=y^2-x^2$及$G=\{(x,y,z)\mid 2x^2+y^2+3z^2=1\}$，試求$f(x,y,z)$在$G$上的最大值與最小值。

解：
$$\begin{cases} f(x,y,z)=y^2-x^2\\ g(x,y,z)=2x^2+y^2 +3z^2-1 \end{cases} \Rightarrow \begin{cases} f_x=\lambda g_x\\ f_y=\lambda g_y\\ f_z=\lambda g_z\\ g=0 \end{cases} \Rightarrow \begin{cases} -2x=4\lambda x\\ 2y=2\lambda y\\ 0=6\lambda z\\ 2x^2+y^2 +3z^2=1 \end{cases} \Rightarrow \begin{cases} \lambda =-1/2\\ \lambda =1\\ \lambda =0\\ 2x^2+y^2 +3z^2=1 \end{cases} \\ \lambda=-\frac{1}{2} \Rightarrow \begin{cases} 2y=-y\\ 0=-3z\\ 2x^2+y^2 +3z^2=1 \end{cases} \Rightarrow \begin{cases} y=0\\ z=0\\ x= \pm{1\over \sqrt{2}} \end{cases} \Rightarrow f(\pm{1\over \sqrt{2}}, 0,0)= -{1\over 2}\\ \lambda=1 \Rightarrow \begin{cases} -2x=4x\\ 0=6z\\ 2x^2+y^2 +3z^2=1 \end{cases} \Rightarrow \begin{cases} x=0\\ z=0\\ y= \pm 1 \end{cases} \Rightarrow f(0, \pm 1,0)= 1\\ \lambda=0 \Rightarrow \begin{cases} -2x=0\\ 2y=0\\ 2x^2+y^2 +3z^2=1 \end{cases} \Rightarrow \begin{cases} x=0\\ y=0\\ z= \pm{1\over \sqrt{3}} \end{cases} \Rightarrow f(0, 0,\pm{1\over \sqrt{3}})= 0\\ \Rightarrow f在G上之\bbox[red, 2pt]{最大值為1,最小值為-{1\over 2}}$$

---

考選部未公布答案，解題僅供參考