108年普考微積分詳解

108年公務人員普通考試

類 科 ：天文、氣象
科 目：微積分

解：$$利用柯西不等式\Rightarrow \left( x^2+y^2+(\sqrt{2}z)^2 \right) \left( 2^2+1^2+(-1/\sqrt{2})^2 \right) \ge \left( 2x+y-z \right)^2\\ \Rightarrow \left( x^2+y^2+2z^2 \right) \left( 4+1+1/2 \right) \ge \left( 2x+y-z \right)^2 \Rightarrow {11\over 2}\left( x^2+y^2+2z^2 \right)  \ge \left( 2x+y-z \right)^2\\ \Rightarrow \sqrt{{11\over 2}\left( x^2+y^2+2z^2 \right) } \ge 2x+y-z \ge -\sqrt{{11\over 2}\left( x^2+y^2+2z^2 \right) }\\ \Rightarrow \sqrt{{11\over 2} } \ge 2x+y-z \ge -\sqrt{{11\over 2} } \Rightarrow \bbox[red,2pt]{\begin{cases}最大值為\sqrt{{11\over 2} }\\最小值為-\sqrt{{11\over 2} }\end{cases}}$$

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解：

上圖為積分區域，我們將積分順序對調，即$$\int_{y=0}^{y=1} \int_{x=\sqrt{y}}^{x=1}{{1\over \sqrt{1+x^3}}\,dx\,dy} =\int_{x=0}^{x=1}\int_{y=0}^{y=x^2}{{1\over \sqrt{1+x^3}}\,dy\,dx} =\int_{x=0}^{x=1}{{x^2\over \sqrt{1+x^3}}\,dx}\\ =\left.\left[ \frac{2}{3}\sqrt{1+x^3} \right]\right|_0^1 = \bbox[red,2pt]{\frac{2}{3}\left(\sqrt{2}-1 \right)}$$

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解：
$$S=\int_a^b{2\pi f(x)\sqrt{1+(f'(x))^2}\,dx}= 2\pi\int_0^{\pi/2}{\sin{x}\sqrt{1+\cos^2{x}}\,dx}\\
令u=\cos{x}\Rightarrow du=-\sin{x}dx\Rightarrow \int{\sin{x}\sqrt{1+\cos^2{x}}\,dx}=-\int{\sqrt{1+u^2}\,du}\\
=-{1\over 2}\left( \sqrt{1+u^2}\cdot u+\ln{\left( \sqrt{1+u^2}+u \right)} \right)
\\ \Rightarrow S= 2\pi\left. \left[-{1\over 2}\sqrt{\cos^2{x}+1}\cos{x}-{1\over 2}\ln{(\sqrt{1+\cos^2{x}} +\cos{x})} \right]\right|_0^{\pi/2}\\ =2\pi \left( \frac{1}{2}\ln{(\sqrt{2}+1)}+\frac{1}{2}\sqrt{2} \right) = \bbox[red,2pt]{\left( \ln{(\sqrt{2}+1)+\sqrt{2}} \right)\pi}$$

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解：
(一)$${dP\over dt}=\alpha P(\gamma-P) \Rightarrow {d^2P\over dt^2}=\alpha {dP\over dt}(\gamma-P)+\alpha P(-{dP\over dt}) = \alpha \gamma{dP\over dt}-\alpha {dP\over dt}P- \alpha {dP\over dt}P\\=\alpha{dP\over dt}(\gamma-2P)不一定大於0\Rightarrow   {dP\over dt}\bbox[red,2pt]{不是}一直遞增,當\gamma-2P=0,即當\bbox[red,2pt]{P={\gamma\over 2}}時遞增或遞減會改變$$(二)$${dP\over dt}=\alpha P(\gamma-P) \Rightarrow \int{{1\over P(\gamma-P)}\,dP}=\int{\alpha\,dt} \Rightarrow {1\over \gamma }\int{\left({1\over P}+{1\over(\gamma-P)}\right)\,dP}=\int{\alpha\,dt}\\ \Rightarrow {1\over \gamma }\left(\ln{P}-\ln(P- \gamma) \right)=\alpha t+C_1 \Rightarrow \ln{{P\over P-\gamma}}=\alpha\gamma t+C_2 \Rightarrow {P\over P-\gamma} = C_3e^{ \alpha\gamma t} \\ \Rightarrow 1+{\gamma\over P-\gamma}= C_3e^{ \alpha\gamma t} \Rightarrow P={\gamma\over C_3e^{ \alpha\gamma t}-1}+\gamma \Rightarrow \lim_{t\to\infty}{P}=\lim_{t\to\infty}{{\gamma\over C_3e^{ \alpha\gamma t}-1}+\gamma}\\ =0+\gamma =\gamma\\ \Rightarrow \bbox[red,2pt]{\lim_{t\to\infty}{P(t)}=\gamma,與P_0無關}$$

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