- Find $dy/dx$ and $d^2y/dx^2$ at the given point $\left(1/2,\pi/4\right)$ for the equation $x=\sin^2y$. ($10\%$)
- Find the critical numbers, points of inflection, and vertical and horizontal asymptotes of $\displaystyle f\left(x\right)=\frac{x^{1/3}}{x^{2/3}-4}$ (if any), describe the concavity of $f$, and sketch the graph of $f$. ($20\%$)
- Sketch the polar curves $r=2\sin\theta$ and $\displaystyle r=\frac32-\sin\theta$, and calculate the area enclosed by them. ($10\%$)
- Calculate $\displaystyle\int\frac{x^3+4x^2-4x-1}{\left(x^2+1\right)^2}dx$. ($10\%$)
- Determine whether the following series converges or diverges: (a) $\displaystyle\sum\ln\left(\frac{k+1}k\right)$ (b) $\displaystyle\sum\left(\sqrt{k}-\sqrt{k-1}\right)^k$ ($10\%$)
- 計算部分和並取極限如下
$\sum\ln\left(\frac{k+1}k\right)=\lim_{n\to\infty}\sum_{k=k_0}^n\left[\ln\left(k+1\right)-\ln k\right]=\lim_{n\to\infty}\left[\ln\left(n+1\right)-\ln k_0\right]=\infty$
故給定的級數發散。 - 設 $a_k=\left(\sqrt{k}-\sqrt{k-1}\right)^k$，那麼計算極限
$\displaystyle\lim_{k\to\infty}\sqrt{\left|a_k\right|}=\lim_{k\to\infty}\frac1{\sqrt{k}+\sqrt{k-1}}=0<1$
因此由根式審歛法知給定的級數收斂。 - Expand $\sin\left(x\right)$ in power of $x-\pi$ and specify the values of $x$ for which the expansion is valid. ($10\%$)
- Find the length of the given curve $\vec{r}\left(t\right)=\left(\ln t\right)\vec{i}+2t\vec{j}+t^2\vec{k}$ from $t=1$ to $t=e$. ($10\%$)
- Find the area of the surface $z^2=x^2+y^2$ from $z=0$ to $z=1$. ($10\%$)
- Let $T$ be the solid ellipsoid $\displaystyle\left(\frac{x}a\right)^2+\left(\frac{y}b\right)^2+\left(\frac{z}c\right)^2\leq1$.
Calculate the volume of $T$ and $\displaystyle\iiint_T\left[\left(\frac{x}a\right)^2+\left(\frac{y}b\right)^2+\left(\frac{z}c\right)^2\right]dxdydz$. ($10\%$)
訣竅
將 $y$ 在給定點處的表達式明確寫出後求導即可；亦可運用隱函數微分求解。解法一
整理可知 $y=\sin^{-1}\sqrt{x}$，如此求一階與二階導函數有$\displaystyle\frac{dy}{dx}=\frac1{\sqrt{1-\sqrt{x}^2}}\cdot\frac1{2\sqrt{x}}=\frac12\left(x-x^2\right)^{-1/2}$, $\displaystyle y''=-\frac14\left(x-x^2\right)^{-3/2}\left(1-2x\right)$
那麼代入 $\displaystyle x=\frac12$ 可得 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(\frac12,\frac\pi4\right)}=1$ 與 $\displaystyle\left.\frac{d^2y}{dx^2}\right|_{\left(x,y\right)=\left(\frac12,\frac\pi4\right)}=0$。解法二
使用隱函數微分求導有$\displaystyle1=2\sin y\cos y\frac{dy}{dx}$
那麼取 $\displaystyle\left(x,y\right)=\left(\frac12,\frac\pi4\right)$ 有 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(\frac12,\frac\pi4\right)}=1$。繼續求導有$\displaystyle0=2\cos^2y\left(\frac{dy}{dx}\right)^2-2\sin^2y\left(\frac{dy}{dx}\right)^2+2\sin y\cos y\frac{d^2y}{dx^2}$
如此代入 $\displaystyle\left(x,y\right)=\left(\frac12,\frac\pi4\right)$ 與 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(\frac12,\frac\pi4\right)}=1$ 可得 $\displaystyle\left.\frac{d^2y}{dx^2}\right|_{\left(x,y\right)=\left(\frac12,\frac\pi4\right)}=0$。訣竅
透過計算極限、一階與二階導函數等尋求極值、反曲點、遞增遞減區間、凹凸區間以及漸近線等繪圖資訊。解法
給定的函數在 $\mathbb{R}-\left\{-8,8\right\}$ 有定義，且容易發現 $\displaystyle\lim_{x\to\pm\infty}f\left(x\right)=0$ 且 $\displaystyle\lim_{x\to\pm8}\left|f\left(x\right)\right|=\infty$，故有水平漸近線 $y=0$ 與鉛直漸近線 $x=8$ 與 $x=-8$。為了求出極值，我們求一階導函數為
$\displaystyle f'\left(x\right)=\frac13x^{-2/3}\left(x^{2/3}-4\right)^{-1}-\frac23\left(x^{2/3}-4\right)^{-2}=-\frac{\left(x^{2/3}-4\right)^{-2}\left(1+4x^{-2/3}\right)}3$
其中 $x\neq\pm8$ 且 $x\neq0$。又 $1+4x^{-2/3}$、$\left(x^{2/3}-4\right)^{-2}$ 皆恆正，故 $f'\left(x\right)<0$，即函數 $f$ 恆遞減，因此無局部極值。現計算二階導函數為
$\displaystyle\begin{aligned}f''\left(x\right)&=\frac{2\left(x^{2/3}-4\right)^{-3}\cdot\frac23x^{-1/3}\left(1+4x^{-2/3}\right)+\left(x^{2/3}-4\right)^{-2}\cdot\frac83x^{-5/3}}3\\&=\frac49\left(x^{2/3}-4\right)^{-3}x^{-5/3}\left(x^{4/3}+6x^{2/3}-8\right)\end{aligned}$
其中 $x\neq\pm8$ 且 $x\neq0$。而解方程式 $f''\left(x\right)=0$，即解 $x^{4/3}+6x^{2/3}-8=0$，可得 $x^{2/3}=-3+\sqrt{17}$，因此有 $x=\pm\sqrt{\left(-3+\sqrt{17}\right)^3}:=x_\pm$。那麼容易看出當 $x\in\left(-8,x_-\right)\cup\left(0,x_+\right)\cup\left(8,\infty\right)$ 時有 $f''\left(x\right)>0$，即 $f$ 在此區間上凹口向上；而當 $x\in\left(-\infty,8\right)\cup\left(x_-,0\right)\cup\left(x_+,8\right)$ 時有 $f''\left(x\right)<0$，即 $f$ 在此區間上凹口向下。據此將 $y=f\left(x\right)$ 的函數圖形繪製如下
訣竅
描點繪製草圖並計算交點位置後，運用極座標下的面積公式計算即可。解法
藉由描點繪製草圖如下
$\displaystyle\begin{aligned}A&=2\left[\frac12\int_0^{\frac\pi6}\left(2\sin\theta\right)^2d\theta+\frac12\int_{\frac\pi6}^{\frac\pi2}\left(\frac32-\sin\theta\right)^2d\theta\right]\\&=4\int_0^{\frac\pi6}\sin^2\theta d\theta+\int_{\frac\pi6}^{\frac\pi2}\left(\frac94-3\sin\theta+\sin^2\theta\right)d\theta\\&=2\int_0^{\frac\pi6}\left(1-\cos2\theta\right)d\theta+\frac14\int_{\frac\pi6}^{\frac\pi2}\left(11-12\sin\theta-2\cos2\theta\right)d\theta\\&=\left(2\theta-\sin2\theta\right)\Big|_0^{\frac\pi6}+\left.\frac{11\theta+12\cos\theta-\sin2\theta}4\right|_{\frac\pi6}^{\frac\pi2}\\&=\frac\pi3-\frac{\sqrt3}2+\frac14\left[\left(\frac{11\pi}2+0-0\right)-\left(\frac{11\pi}6+6\sqrt3-\frac{\sqrt3}2\right)\right]=\frac{10\pi-15\sqrt3}8\end{aligned}$
訣竅
運用有理數積分法，先寫為部分分式後逐項積分即可。解法
將被積分函數適當改寫有$\displaystyle\begin{aligned}\int\frac{x^3+4x^2-4x-1}{\left(x^2+1\right)^2}dx&=\int\frac{\left(x^2+1\right)\left(x+4\right)-5x-5}{\left(x^2+1\right)^2}dx\\&=\int\left(\frac{x+4}{x^2+1}-\frac{5x}{\left(x^2+1\right)^2}-\frac5{\left(x^2+1\right)^2}\right)dx\\&=\frac12\ln\left(x^2+1\right)+4\tan^{-1}x+\frac5{2\left(x^2+1\right)}-5\int\frac{dx}{\left(x^2+1\right)^2}\end{aligned}$
對於最後一項，我們令 $x=\tan\theta$，那麼有 $dx=\sec^2\theta d\theta$，如此可得$\displaystyle\begin{aligned}5\int\frac{dx}{\left(x^2+1\right)^2}&=5\int\frac{\sec^2\theta d\theta}{\sec^4\theta}=5\int\cos^2\theta d\theta=\frac52\int\left(1+\cos2\theta\right)d\theta\\&=\frac52\left(\theta+\frac{\sin2\theta}2\right)+C=\frac52\left(\theta+\sin\theta\cos\theta\right)+C=\frac52\left(\tan^{-1}x+\frac{x}{x^2+1}\right)+C\end{aligned}$
如此所求之不定積分為$\displaystyle\int\frac{x^3+4x^2-4x-1}{\left(x^2+1\right)^2}dx=\frac12\ln\left(x^2+1\right)+\frac32\tan^{-1}x+\frac5{2\left(x^2+1\right)}-\frac{5x}{x^2+1}-C$
訣竅
利用直接求部分和取極限以及根式審歛法判別歛散性即可。解法
訣竅
利用三角恆等式以及正弦函數的展開式表達之，隨後計算收斂區間；也可按照 Taylor 級數的定義逐項係數計算求解。解法一
直接計算可知$\displaystyle\sin\left(x\right)=\sin\left(\pi-x\right)=-\sin\left(x-\pi\right)=\sum_{n=0}^{\infty}\left(-1\right)^{n+1}\frac{\left(x-\pi\right)^{2n+1}}{\left(2n+1\right)!}$
容易計算其收斂半徑如下$\displaystyle\lim_{n\to\infty}\left|\frac{\left(x-\pi\right)^{2n+3}}{\left(2n+3\right)!}\div\frac{\left(x-\pi\right)^{2n+1}}{\left(2n+1\right)!}\right|<1$
展開有 $0<1$，因此整個實數線上皆收斂，即該表達式在實數線上皆有效。解法二
設 $f\left(x\right)=\sin\left(x\right)$，那麼求導可知$f^{\left(n\right)}\left(x\right)=\begin{cases}\cos x,&\text{if}~n=4k+1,\\-\sin x,&\text{if}~n=4k+2,\\-\cos x,&\text{if}~n=4k+3,\\\sin x,&\text{if}~n=4k,\end{cases}$
其中 $k\in\mathbb{N}\cup\left\{0\right\}$。如此有$f^{\left(n\right)}\left(\pi\right)=\begin{cases}-1,&\text{if}~n=4k+1,\\0,&\text{if}~n=4k+2,\\1,&\text{if}~n=4k+3,\\0,&\text{if}~n=4k,\end{cases}$
因此所求的級數為$\displaystyle\sum_{n=0}^{\infty}\frac{f^{\left(n\right)}\left(\pi\right)}{n!}\left(x-\pi\right)^n=\sum_{k=0}^{\infty}\frac{-1}{\left(4k+1\right)!}\left(x-\pi\right)^{4k+1}+\sum_{k=0}^{\infty}\frac1{\left(4k+3\right)!}\left(x-\pi\right)^{4k+3}=\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k+1}}{\left(2k+1\right)!}\left(x-\pi\right)^{2k+1}$
同樣由比值審歛法的概念可知其收斂半徑為無窮，故該級數式在整個實數線上皆有效。訣竅
運用參數曲線的弧長公式計算即可。解法
運用曲線弧長的公式即有$\displaystyle s=\int_1^e\left\|\vec{r}'\left(t\right)\right\|dt=\int_1^e\sqrt{\left(\frac1t\right)^2+2^2+\left(2t\right)^2}dt=\int_1^e\left(\frac1t+2t\right)dt=\left(\ln t+t^2\right)\Big|_1^e=e^2$
訣竅
將曲面參數化後計算曲面積分獲得表面積；亦可視為旋轉體表面積。解法一
將曲面參數化為 ${\bf r}\left(x,y\right)=\left(x,y,\sqrt{x^2+y^2}\right)$，因 $0\leq z\leq1$，故考慮 $D=\left\{\left(x,y\right)\in\mathbb{R}^2:\,x^2+y^2\leq1\right\}$。因此所求的表面積為$\displaystyle\begin{aligned}A&=\iint_D\left|{\bf r}_x\times{\bf r}_y\right|dA=\iint_D\left|\left(1,0,\frac{x}{\sqrt{x^2+y^2}}\right)\times\left(0,1,\frac{y}{\sqrt{x^2+y^2}}\right)\right|dA\\&=\iint_D\sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1}dA=\iint_D\sqrt2dA=\pi\sqrt2\end{aligned}$
解法二
可視此圖形為 $z=x$ 繞 $z$ 軸旋轉所得的曲面，因此由旋轉體表面積公式可知$\displaystyle A=\int_0^12\pi z\sqrt{1+\left(\frac{dz}{dx}\right)^2}dx=2\pi\sqrt2\int_0^1xdx=\pi\sqrt2$
訣竅
運用橢球座標變換處理之。解法
令 $\left\{\begin{aligned}&x=a\rho\cos\theta\sin\phi\\&y=b\rho\sin\theta\sin\phi\\&z=c\rho\cos\phi\end{aligned}\right.$，那麼變數範圍為 $\left\{\begin{aligned}&0\leq\rho\leq1\\&0\leq\theta\leq2\pi\\&0\leq\phi\leq\pi\end{aligned}\right.$，而所對應的 Jacobian 行列式為$\displaystyle\left|J\right|=\left|\frac{\partial\left(x,y,z\right)}{\partial\left(\rho,\theta,\phi\right)}\right|=\Big|\begin{vmatrix}\displaystyle\frac{\partial x}{\partial\rho}&\displaystyle\frac{\partial x}{\partial\theta}&\displaystyle\frac{\partial x}{\partial\phi}\\\displaystyle\frac{\partial y}{\partial\rho}&\displaystyle\frac{\partial y}{\partial\theta}&\displaystyle\frac{\partial y}{\partial\phi}\\\displaystyle\frac{\partial z}{\partial\rho}&\displaystyle\frac{\partial z}{\partial\theta}&\displaystyle\frac{\partial z}{\partial\phi}\end{vmatrix}\Big|=\Big|\begin{vmatrix}a\cos\theta&-a\rho\sin\theta&a\rho\cos\theta\cos\phi\\b\sin\theta\sin\phi&b\rho\cos\theta\sin\phi&b\rho\sin\theta\cos\phi\\c\cos\phi&0&-c\rho\sin\phi\end{vmatrix}\Big|=abc\rho^2\sin\phi$
故所求的體積為$\displaystyle V=\iiint_T1dV=\int_0^{\pi}\int_0^{2\pi}\int_0^1abc\rho^2\sin\phi d\rho d\theta d\phi=abc\left(\int_0^\pi\sin\phi d\phi\right)\left(\int_0^{2\pi}d\theta\right)\left(\int_0^1\rho^2d\rho\right)=\frac{4abc\pi}3$
且有$\displaystyle\begin{aligned}\iiint_T\left[\left(\frac{x}a\right)^2+\left(\frac{y}b\right)^2+\left(\frac{z}c\right)^2\right]dxdydz&=\int_0^{\pi}\int_0^{2\pi}\int_0^1\rho^2\cdot abc\rho^2\sin\phi d\rho d\theta d\phi\\&=abc\left(\int_0^{\pi}\sin\phi d\phi\right)\left(\int_0^{2\pi}d\theta\right)\left(\int_0^1\rho^4d\rho\right)=\frac{4abc\pi}5\end{aligned}$
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