- Taylor series $\displaystyle\sum_{n=0}^{\infty}a_nx^n=1+\frac{1}{2}x+\frac{\frac{1}{2}\cdot\left(-\frac{1}{2}\right)}{1\cdot2}x^2+\cdots+\frac{\frac{1}{2}\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{3}{2}\right)\cdots\left(\frac{-2n+3}{2}\right)}{1\cdot2\cdot3\cdots n}x^n+\cdots$ has radius of convergence $r=?$. When $x=r$, is it convergent or divergent? When $x=-r$, is it convergent or divergent? ($20/100$)
- Space curve $\displaystyle\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)=\left(t,t^2,\frac{1}{3}t^3\right)$. When $t=2$, unit tangent vector $\vec{T}=\left(?,?,?\right)$. Principal normal vector $\vec{N}=\left(?,?,?\right)$, bi-normal vector $\vec{B}=\left(?,?,?\right)$, curvature $\kappa=?$, torsion $\tau=?$ ($20/100$)
- $\displaystyle\int_0^1\int_x^1\int_0^{\sqrt{1-y^2}}f\left(x,y,z\right)dzdydx=\int_?^?\int_?^?\int_?^?f\left(x,y,z\right)dxdydz$. ($20/100$)
- A cat starting from $\left(x,y\right)=\left(0,10\right)$ chases a mouse fleeing along the x-axis. If the speed of cat is twice as fast as the speed of mouse, at what point will the mouse be caught? ($20/100$)
- A tiny ball of mass $m=2$ grams is rolling along an inverted cycloid without friction. The equation of cycloid is $x=t-\sin t$, $y=-1+\cos t$. gravity $g=980\mbox{cm}/\mbox{sec}^2$, highest points $\max{y}=-0.3\mbox{cm}$. period $T=?$ seconds. If the highest points are $y=-0.2\mbox{cm}$, will be period $T$ become longer or shorter? ($20/100$)
- 當 $\theta=\theta_{0.2}$ 時有 $u=\cos\theta_{0.2}$；
- 當 $\theta=\pi$ 時有 $u=-1$；
- $\displaystyle d\theta=-\frac{du}{\sqrt{1-u^2}}$。
- 當 $u=-1$ 時有 $s=-1$
- 當 $u=\cos\theta_{0.2}$ 時有 $s=1$
- $\displaystyle ds=\frac{\cos\theta_{0.2}+1}{2}du$
訣竅
利用比值審歛法可求出其收斂半徑；再利用交錯級數審歛法、Raabe 審歛法即可判斷端點的收斂性。解法
設 $\displaystyle a_n=\frac{\left(-1\right)^{n-1}\left(2n-2\right)!}{2^{2n-1}n!\left(n-1\right)!}$，則有$\displaystyle r=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{\displaystyle\frac{\left(-1\right)^n\left(2n\right)!}{2^{2n+1}\left(n+1\right)!n!}}{\displaystyle\frac{\left(-1\right)^{n-1}\left(2n-2\right)!}{2^{2n-1}n!\left(n-1\right)!}}\right|=1$
現在我們檢查 $r=1$ 的情形可得級數為
$\displaystyle1+\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{2n}\frac{1\cdot3\cdot5\cdots\left(2n-3\right)}{2\cdot4\cdot6\cdots\left(2n-2\right)}$
其中容易注意到該項遞減，再由不等式 $\displaystyle0<\frac{1}{2n}\frac{1\cdot3\cdot5\cdots\left(2n-3\right)}{2\cdot4\cdot6\cdots\left(2n-2\right)}<\frac{1}{2n}$，利用夾擠定理可知收斂到 $0$。如此我們由交錯級數審歛法可知在 $r=1$ 時收斂。最後我們檢查 $r=-1$ 的情形可得級數為
$\displaystyle1-\sum_{n=1}^{\infty}\frac{1}{2n}\frac{1\cdot3\cdot5\cdots\left(2n-3\right)}{2\cdot4\cdot6\cdots\left(2n-2\right)}$
計算極限式 $\displaystyle\lim_{n\to\infty}n\left(1-\frac{\displaystyle\frac{1}{2n+2}\frac{1\cdot3\cdot5\cdots\left(2n-3\right)\left(2n-1\right)}{2\cdot4\cdot6\cdots\left(2n-2\right)\left(2n\right)}}{\displaystyle\frac{1}{2n}\frac{1\cdot3\cdot5\cdots\left(2n-3\right)}{2\cdot4\cdot6\cdots\left(2n-2\right)}}\right)=\frac{3}{2}>1$，因此由 Raabe 審歛法可知在 $r=-1$ 時收斂。訣竅
根據定義計算即可。解法
首先根據定義$\displaystyle\vec{T}\left(t\right)=\frac{\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}=\frac{\left(1,2t,t^2\right)}{\left|\left(1,2t,t^2\right)\right|}=\left(\frac{1}{\sqrt{1+4t^2+t^4}},\frac{2t}{\sqrt{1+4t^2+t^4}},\frac{t^2}{\sqrt{1+4t^2+t^4}}\right)$
由此可知$\displaystyle\vec{T}\left(2\right)=\left(\frac{1}{\sqrt{33}},\frac{4}{\sqrt{33}},\frac{4}{\sqrt{33}}\right)$
繼續按照定義，我們有$\displaystyle\vec{N}\left(2\right)=\frac{\vec{T}'\left(2\right)}{\left|\vec{T}'\left(2\right)\right|}$
為此我們先計算 $\vec{T}'\left(t\right)$ 如下：$\displaystyle\vec{T}'\left(t\right)=\left(-\frac{4t+2t^3}{\left(1+4t^2+t^4\right)^{3/2}},\frac{2-2t^4}{\left(1+4t^2+t^4\right)^{3/2}},\frac{2t+4t^3}{\left(1+4t^2+t^4\right)^{3/2}}\right)$
從而有$\displaystyle\vec{T}'\left(2\right)=\left(-\frac{24}{33^{3/2}},-\frac{30}{33^{3/2}},\frac{36}{33^{3/2}}\right)$ 、 $\displaystyle\lvert\vec{T}'\left(2\right)\rvert=\frac{2\sqrt{21}}{33}$
由此可得$\displaystyle\vec{N}\left(2\right)=\left(-\frac{4}{\sqrt{77}},-\frac{5}{\sqrt{77}},\frac{6}{\sqrt{77}}\right)$
再按照定義則有$\displaystyle\begin{aligned}\vec{B}\left(2\right)=\vec{T}\left(2\right)\times\vec{N}\left(2\right)=&\left(\frac{1}{\sqrt{33}},\frac{4}{\sqrt{33}},\frac{4}{\sqrt{33}}\right)\times\left(-\frac{4}{\sqrt{77}},-\frac{5}{\sqrt{77}},\frac{6}{\sqrt{77}}\right)\\=&\left(\frac{4}{\sqrt{21}},-\frac{2}{\sqrt{21}},\frac{1}{\sqrt{21}}\right)\end{aligned}$
另一方面曲率可以利用公式計算如下：$\displaystyle\begin{aligned}\kappa\left(t\right)=&\frac{\lvert\left(x'\left(t\right),y'\left(t\right),z'\left(t\right)\right)\times\left(x''\left(t\right),y''\left(t\right),z''\left(t\right)\right)\rvert}{\lvert\left(x'\left(t\right),y'\left(t\right),z'\left(t\right)\right)\rvert^3}\\=&\frac{\lvert\left(1,2t,t^2\right)\times\left(0,2,2t\right)\rvert}{\lvert\left(1,2t,t^2\right)\rvert^3}\\=&\frac{2\sqrt{t^4+t^2+1}}{\left(1+4t^2+t^4\right)^{3/2}}\end{aligned}$
故 $\displaystyle\kappa\left(2\right)=\frac{2\sqrt{77}}{363}$。再者扭率亦可利用公式計算如下：
$\displaystyle\begin{aligned}\tau\left(t\right)=&\frac{\left[\left(x'\left(t\right),y'\left(t\right),z'\left(t\right)\right)\times\left(x''\left(t\right),y''\left(t\right),z''\left(t\right)\right)\right]\cdot\left(x'''\left(t\right),y'''\left(t\right),z'''\left(t\right)\right)}{\lvert\left(x'\left(t\right),y'\left(t\right),z'\left(t\right)\right)\times\left(x''\left(t\right),y''\left(t\right),z''\left(t\right)\right)\rvert^2}\\=&\frac{\left[\left(1,2t,t^2\right)\times\left(0,2,2t\right)\right]\cdot\left(0,0,2\right)}{\lvert\left(1,2t,t^2\right)\times\left(0,2,2t\right)\rvert^2}\\=&\frac{\left(2t^2,-2t,2\right)\cdot\left(0,0,2\right)}{\lvert\left(2t^2,-2t,2\right)\rvert^2}\\=&\frac{2}{\sqrt{t^4+t^2+1}}\end{aligned}$
故 $\displaystyle\tau\left(2\right)=\frac{2}{\sqrt{21}}$。訣竅
利用不等式重新估計大小排列順序即可。解法
由原先的積分範圍容易得到 $0\leq x\leq y\leq\sqrt{1-z^2}\leq1$。如此我們有$0\leq x\leq y$
依此可以推得 $0\leq y\leq\sqrt{1-z^2}$，最終可得 $0\leq z\leq1$。因此我們可以將原三重積分改寫如下$\displaystyle\int_0^1\int_0^{\sqrt{1-z^2}}\int_0^yf\left(x,y,z\right)dxdydz$.
訣竅
我們應對本題的老鼠的出發位置與行為做些假定，接著列出對應的微分方程後求解即可。解法
不妨設老鼠的起始位置為 $\left(a,0\right)$，其中 $a\geq0$，且移動方向為 $x$ 軸正向。另一方面我們考慮貓咪移動軌跡的函數為 $y=f\left(x\right)$，根據初始條件有 $f\left(0\right)=10$、$\displaystyle f'\left(0\right)=-\frac{10}a$。假設貓咪的座標為 $\left(x_0,f\left(x_0\right)\right)$，由此可得切線方程式為 $y-f\left(x_0\right)=f'\left(x_0\right)\left(x-x_0\right)$，因此與 $x$ 軸的交點座標為 $\displaystyle x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$。
由上一段所得的資訊以及貓咪的速度為老鼠的兩倍快，從而我們有等式：
$\displaystyle2\left(x-\frac{f\left(x\right)}{f'\left(x\right)}\right)=\int_0^x\sqrt{1+f'^2\left(t\right)}dt$
同時將左右兩式對 $x$ 微分可得$\displaystyle\frac{2f\left(x\right)f''\left(x\right)}{\left(f'\left(x\right)\right)^2}=\sqrt{1+f'^2\left(x\right)}$
為了方便起見，我們將 $f\left(x\right)$ 簡記為 $y$，如此即解下列的初始值微分方程問題：$\displaystyle\left\{\begin{aligned} &2yy''=y'^2\sqrt{1+y'^2}\\&y\left(0\right)=10,\quad y'\left(0\right)=-\frac{10}{a}\end{aligned}\right.$
令 $\displaystyle p\left(y\right)=\frac{dy}{dx}$，則有 $\displaystyle\frac{d^2y}{dx^2}=p\frac{dp}{dy}$，因此原微分方程式改寫為$\displaystyle2yp\frac{dp}{dy}=p^2\sqrt{1+p'^2}$ 或寫為 $\displaystyle\frac{2dp}{p\sqrt{1+p^2}}=\frac{dy}{y}$
再令 $s\left(y\right)=\sqrt{1+p^2\left(y\right)}$，於是有 $p\left(y\right)=\sqrt{s^2\left(y\right)-1}$、$\displaystyle dp\left(y\right)=\frac{sds}{\sqrt{s^2-1}}$，因此微分方程式化為如下$\displaystyle\left(\frac{1}{s-1}-\frac{1}{s+1}\right)ds=\frac{2ds}{s^2-1}=\frac{dy}{y}$
此時對左右兩式同取積分，如此可得：$\displaystyle\ln\frac{\sqrt{1+y'^2}-1}{\sqrt{1+y'^2}+1}=\ln\frac{s-1}{s+1}=\ln y+\ln C_1$
取 $x=0$ 代入後可得 $\displaystyle C_1=\frac{\sqrt{a^2+100}-a}{10\left(\sqrt{a^2+100}+a\right)}$。透過整理，我們可以解得$\displaystyle p\left(y\right)=\frac{dy}{dx}=\frac{2\sqrt{C_1}}{1-C_1y}\sqrt{y}$ 或寫為 $\displaystyle\left(y^{-\frac{1}{2}}-C_1y^{\frac{1}{2}}\right)dy=2\sqrt{C_1}dx$
同取積分可得$\displaystyle2\sqrt{y}-\frac{2C_1}{3}y^{\frac{3}{2}}=2\sqrt{C_1}x+C_2$
取 $x=0$ 代入後可得 $\displaystyle C_2=2\sqrt{10}-\frac{20\sqrt{10}C_1}{3}$。而當 $y=0$ 時即為貓咪抓到老鼠的位置，即 $\displaystyle x=-\frac{C_2}{2\sqrt{C_1}}$。訣竅
注意到擺線的等時性以及物理與微積分的基本應用是本題的解題關鍵。解法
首先為了簡便起見，我們將參數 $t$ 變為參數 $\theta$ 以便與時間做出區隔，並且記 $T_k$ 表示此球乃在擺線上 $y=-k$ 公分處開始來回滑動的週期。容易注意到週期為小球滑至擺線底部時間之四倍，因此我們僅需比較滑至底部的所需時間即可。
首先我們記 $\theta_{0.2}$ 為 $y=-0.2$ 時所對應位置的參數，其中 $0<\theta_{0.2}<\pi$。而當小球下滑到參數 $\theta$ 時根據力學能守恆可知質點在該處的瞬時速率為 $v\left(\theta\right)=\sqrt{2g\left(\cos\theta_{0.2}-\cos\theta\right)}$。再者弧長所對應的微分如下
$\displaystyle ds=\sqrt{\left(dx\right)^2+\left(dy\right)^2}=2\sin\frac{\theta}{2}d\theta$
因此滑到最低點所需的時間為$\displaystyle\int_{\theta_{0.2}}^{\pi}\frac{2\sin\frac{\theta}{2}d\theta}{\sqrt{2g\left(\cos\theta_{0.2}-\cos\theta\right)}}$
利用半角公式 $\displaystyle\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$，因此所需時間可改寫並計算如下：$\displaystyle\frac{1}{\sqrt{g}}\int_{\theta_{0.2}}^{\pi}\sqrt{\frac{1-\cos\theta}{\cos\theta_{0.2}-\cos\theta}}d\theta$
令 $u=\cos\theta$，則$\displaystyle\frac{1}{\sqrt{g}}\int_{-1}^{\cos\theta_{0.2}}\frac{du}{\sqrt{\left(\cos\theta_{0.2}-u\right)(1+u)}}=\frac{1}{\sqrt{g}}\int_{-1}^{\cos\theta_{0.2}}\frac{du}{\sqrt{\frac{\left(\cos\theta_{0.2}+1\right)^2}{4}-\left(u-\frac{\cos\theta_{0.2}-1}{2}\right)^2}}$
因此我們令 $\displaystyle s=\frac{2}{\cos\theta_{0.2}+1}\left(u-\frac{\cos\theta_{0.2}-1}{2}\right)$，則$\displaystyle\frac{1}{\sqrt{g}}\int_{-1}^{1}\frac{ds}{\sqrt{1-s^2}}=\frac{\pi}{\sqrt{g}}$
運用相同的方法可以知道自 $y=-0.3$ 公分處滑至底部的時間也相同，從而題目所問的兩項週期相等。
"因此原微分方程式改寫為"的下一行之p' 要改成p
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