108年高考三級-微積分與微分方程詳解

108年公務人員高等考試三級考試

類 科 ：核子工程
科 目：微積分與微分方程

解： $$f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\\ 取n=2\Rightarrow f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \\ f(x)=\cos{x}\Rightarrow f'(x)=-\sin{x}\Rightarrow f''(x)=-\cos{x}\\ \Rightarrow \cos{x}\approx \cos{a}-\sin{a}\cdot(x-a)-\frac{\cos{a}}{2}(x-a)^2\\ x=0.1,a=0代入上式\Rightarrow \cos{0.1}\approx \cos{0}-\sin{0}\cdot 0.1-\frac{\cos{0}}{2}(0.1)^2\\=1-0.1^2/2=0.995\Rightarrow \cos{0.1}\approx 0.995\\ 同理f(x)=\ln{x}\Rightarrow f'(x)=1/x\Rightarrow f''(x)=-1/x^2\\ \Rightarrow \ln{x}=\ln{a}+\frac{1}{a}(x-a)-\frac{1}{2a^2}(x-a)^2\\ x=0.995, a=1代入上式\;\Rightarrow \ln{0.995}=\ln{1}+{1\over 1}(0.995-1)-{1\over 2}(0.995-1)^2\\=0-0.005-0.0000125\approx \bbox[red,2pt]{-0.005}$$

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解：
$$\begin{cases}T(t):在時間t的溫度\\V(t):在時間t的水量\end{cases} \Rightarrow \begin{cases}T(0)=25\\V(0)=400 \end{cases} \Rightarrow \begin{cases}T'(t)=\sqrt{T}/V(t)\\V(t)=400+10t \end{cases} \Rightarrow T'(t)={\sqrt{T}\over 400+10t}\\ \Rightarrow {dT\over dt} ={\sqrt{T}\over 400+10t} \Rightarrow \int{1\over \sqrt{T}}dT =\int{1\over 400+10t}dt \Rightarrow 2\sqrt{T}={1\over 10}\ln{(400+10t)}+C \\ T(0)=25\Rightarrow 2\sqrt{25}={1\over 10}\ln{(400)}+C \Rightarrow C=10-\frac{1}{5}\ln{20}\\ \Rightarrow 2\sqrt{T}={1\over 10}\ln{(400+10t)}+10-\frac{1}{5}\ln{20}\\ t=10 \Rightarrow 2\sqrt{T}={1\over 10}\ln{(500)}+10-\frac{1}{5}\ln{20} = {1\over 10}\ln{(500)}+10-\frac{1}{10}\ln{400} \\ ={1\over 10}\ln{(5/4)}+10\Rightarrow \sqrt{T}={1\over 20}\ln{(5/4)}+5 \Rightarrow T=\bbox[red,2pt]{\left( 5+{1\over 20}\ln{(5/4)} \right)^2}$$

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解：
$$y=x^2/4\Rightarrow \begin{cases}x=t\\y=t^2/4\end{cases}\Rightarrow \begin{cases}dx=dt\\dy=tdt/2\end{cases} \Rightarrow \int_1^2{\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt}=\int_1^2{\sqrt{1+(t^2/4)}\,dt}\\=\left.\left[ \ln{ \left(\sqrt{1+t^2/4}+t/2 \right)} \right]\right|_1^2 =\ln{\left(\sqrt{2}+1\right)}-\ln{\left(\sqrt{5/4}+1/2 \right)}=\bbox[red,2pt]{\ln{\left(\left(\sqrt{2}+1 \right)\left(\sqrt{5}-1 \right)/2\right)}}$$

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解：
$$\left( x^2+(\sqrt{2}y)^2\right)\left(2^2+(1/\sqrt{2})^2 \right)\ge \left(2x+y \right)^2 \Rightarrow \left( x^2+2y^2\right)\left(4+1/2 \right)\ge \left(2x+y \right)^2 \Rightarrow 10\cdot \frac{9}{2}\ge \left(2x+y \right)^2\\ \Rightarrow 45\ge (2x+y)^2 \Rightarrow \sqrt{45}\ge 2x+y\ge -\sqrt{45}\Rightarrow 3\sqrt{5}+10\ge 2x+y+10\ge 10-3\sqrt{5}\\ \Rightarrow 2x+y+10的最小值為\bbox[red,2pt]{10-3\sqrt{5}}$$

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解：

區域$S$如上圖，可以拆成兩上下兩塊$S_1$與$S_2$，其中$S_2$是一個矩形，該矩形繞Y軸旋轉可得一個圓柱體。該圓柱體的半徑為$\overline{FC}=\overline{AB}=1$，高為$\overline{BC}=\sqrt{3}$，因此體積為$1^2\pi\times \sqrt{3}=\sqrt{3}\pi$；

$$S_1繞Y軸旋轉所得體積為\int_\sqrt{3}^2{(4-y^2)^2\pi\,dy} = \pi\int_\sqrt{3}^2{\left( y^4-8y^2+16\right) \,dy}\\ = \pi\left. \left[ 16y-\frac{8}{3}y^3+\frac{1}{5}y^5 \right] \right|_\sqrt{3}^2 = \left( \frac{256}{15}-\frac{49}{5}\sqrt{3}\right)\pi\\ \Rightarrow S繞Y軸旋轉所得體積為\sqrt{3}\pi+ \left( \frac{256}{15}-\frac{49}{5}\sqrt{3}\right)\pi= \bbox[red,2pt]{\left( \frac{256}{15}-\frac{44}{5}\sqrt{3}\right)\pi}$$

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