- (10%) Find the horizontal and vertical asymptotes of $y=f\left(x\right)$, where
$\displaystyle f\left(x\right)=\frac{\sqrt{x^4+x^3+1}-\sqrt{x^4-x^3+1}}{\left|x\right|}+\frac{\ln\left|x+1\right|}x$
- 先計算水平漸近線如下
$\displaystyle\lim_{x\to\pm\infty}f\left(x\right)=\lim_{x\to\pm\infty}g\left(x\right)+\lim_{x\to\pm\infty}h\left(x\right)=\pm\frac12$
故水平線 $\displaystyle y=\frac12$ 與 $\displaystyle y=-\frac12$ 皆為水平漸近線。 - 另一方面可以注意到
$\displaystyle\lim_{x\to-1}f\left(x\right)=\lim_{x\to-1}g\left(x\right)+\lim_{x\to-1}h\left(x\right)=\infty$
故 $x=-1$ 為鉛直漸近線。此外，由於$\displaystyle\lim_{x\to0}f\left(x\right)=\lim_{x\to0}g\left(x\right)+\lim_{x\to0}h\left(x\right)=\frac0{2}+\lim_{x\to0}\frac{1/\left(x+1\right)}1=1$
故 $x=0$ 並非鉛直漸近線。 - (4%) Show that $\displaystyle f\left(x\right)=1+x+\int_{-x}^xe^{-t^2}dt$ is one-to-one.
- (6%) Let $g\left(x\right)=f^{-1}\left(x\right)$. Find $g\left(1\right)$, $g'\left(1\right)$, and $g''\left(1\right)$.
- 利用微積分基本定理與連鎖律可知
$\displaystyle f'\left(x\right)=1+e^{-x^2}\cdot1-e^{-\left(-x\right)^2}\cdot\left(-1\right)=1+2e^{-x^2}>1$
因此若由反證法，設函數 $f$ 並非單射，即存在 $x_1<x_2$ 滿足 $f\left(x_1\right)=f\left(x_2\right)$，那麼由均值定理可知存在 $c\in\left(x_1,x_2\right)$ 滿足$\displaystyle f'\left(c\right)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=0$
這便與 $f'>1$ 矛盾，故 $f$ 單射。 - 按照反函數的定義可知 $g\left(f\left(x\right)\right)=x$，那麼由 $f\left(0\right)=1$ 可取 $x=0$ 得 $g\left(1\right)=0$。再者求導兩次有
$g'\left(f\left(x\right)\right)f'\left(x\right)=1$, $g''\left(f\left(x\right)\right)\left[f'\left(x\right)\right]^2+g'\left(f\left(x\right)\right)f''\left(x\right)=0$
皆取 $x=0$ 可知$g'\left(1\right)f'\left(0\right)=1$, $g''\left(1\right)\left[f'\left(0\right)\right]^2+g'\left(1\right)f''\left(0\right)=0$
又留意 $f''\left(x\right)=-4xe^{-x^2}$，因此所求為$\displaystyle g'\left(1\right)=\frac1{f'\left(0\right)}=\frac13$, $g''\left(1\right)=0$。
- (10%) Consider $\displaystyle\sum_{i=1}^{1000}\sqrt[3]{i}$ and $\displaystyle\sum_{i=1}^{999}\sqrt[3]{i}$. Use an integral to give $\displaystyle\sum_{i=1}^{1000}\sqrt[3]{i}$ an upper bound and a lower bound.
- Evaluate the integral or show that it diverges.
- (6%) $\displaystyle\int_1^{\infty}\frac{\text{arctan}\left(x\right)}{x^2}dx$.
- (6%) $\displaystyle\int_0^1\frac1{x\sqrt{1+\left(\ln x\right)^2}}dx$.
- 使用分部積分法可知
$\displaystyle\begin{aligned}\int_1^{\infty}\frac{\text{arctan}\left(x\right)}{x^2}dx&=\left.-\frac{\tan^{-1}x}x\right|_1^{\infty}+\int_1^{\infty}\frac{dx}{x\left(x^2+1\right)}\\&=\frac\pi4+\int_1^{\infty}\left(\frac1x-\frac{x}{x^2+1}\right)dx=\frac\pi4+\left.\left(\ln\left|x\right|-\frac12\ln\left(x^2+1\right)\right)\right|_1^{\infty}\\&=\frac\pi4+\left.\ln\frac{x}{\sqrt{x^2+1}}\right|_1^{\infty}=\frac\pi4+\frac{\ln2}2\end{aligned}$
- 令 $\theta=\tan^{-1}\left(\ln x\right)$，那麼
- 當 $x\to0^+$ 時有 $\displaystyle\theta\to-\frac\pi2$；
- 當 $x=1$ 時有 $\theta=0$；
- 容易注意到 $x=e^{\tan\theta}$，故 $1+\left(\ln x\right)^2=\sec^2\theta$，且知 $dx=e^{\tan\theta}\sec^2\theta d\theta$。
$\displaystyle\begin{aligned}\int_0^1\frac1{x\sqrt{1+\left(\ln x\right)^2}}dx&=\int_{-\frac\pi2}^0\frac1{e^{\tan\theta}\cdot\sec\theta}\cdot e^{\tan\theta}\sec^2\theta d\theta=\int_{-\frac\pi2}^0\sec\theta d\theta\\&=\ln\left|\sec\theta+\tan\theta\right|\Big|_{-\frac\pi2}^0=-\lim_{\theta\to\left(-\frac\pi2\right)^+}\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|=-\lim_{\theta\to\left(-\frac\pi2\right)^+}\ln\left|\frac{\cos\theta}{1-\sin\theta}\right|=\infty\end{aligned}$
故此瑕積分發散。 - Consider the flow of blood through a blood vessel with cross-section $D=\left\{\left(x,y\right)\mid\,x^2+y^2\leq R^2\right\}$, a small disc with radius $R>0$. Suppose that the velocity of the blood, $v$, through each point of the cross-section is $v\left(x,y\right)=cP\left(R^2-x^2-y^2\right)$, where $c$ is a constant and $P$ is the pressure difference between the ends of the vessel. We define the flux of the blood, $F$, as $\displaystyle F=\iint_Dv\left(x,y\right)dA$.
- (4%) Derive the formula of $F$ in terms of $R$ and $P$.
- (8%) Assume that the flux of the blood is constant, and the pressure $P=4000$ dynes/cm$^2$ when $R=0.008$ cm. Now the radius of the cross-section decreases at a constant rate $dR/dt=-0.0002$ cm/year. Find the rate of increase of $P$ when $R=0.006$ cm.
- 由極座標變換 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，那麼 $\left\{\begin{aligned}&0\leq r\leq R\\&0\leq\theta\leq2\pi\end{aligned}\right.$，如此所求之通量的重積分可改寫並計算如下
$\displaystyle R=\int_0^{2\pi}\int_0^RcP\left(R^2-r^2\right)rdrd\theta=\left.2cP\pi\int_0^R\left(R^2r-r^3\right)dr=2cP\pi\left(\frac{R^2r^2}2-\frac{r^4}4\right)\right|_0^R=\frac{cPR^4\pi}2$
- 由於 $F$ 為常數，故 $PR^4$ 為常數 $C$，那麼按照題意我們先求當 $R=0.006$ 公分時的壓力，即
$4000\cdot0.008^4=P\cdot0.006^4$
可知此時的 $\displaystyle P=4000\cdot\left(\frac43\right)^4=\frac{1024000}{81}$。又將方程式 $PR^4=C$ 同時對 $t$ 求導可得$\displaystyle R^4\frac{dP}{dt}+4PR^3\frac{dR}{dt}=0$
那麼有$\displaystyle\frac{dP}{dt}=-\frac{4P}{R}\frac{dR}{dt}=-\frac{16000\left(4/3\right)^4}{0.006}\cdot-0.0002=\frac{409600}{243}$
- Consider the ellipse $x^2+xy+y^2=3$.
- (6%) Find the points on which the ellipse obtains the maximum and the minimum $y$ coordinate.
- (6%) Find the equations of the two tangent lines to the ellipse that pass through the point $\left(4,-2\right)$.
- 【方法一】 使用隱函數微分可得
$\displaystyle2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$
若在 $\left(a,b\right)$ 處產生極值，那麼 $\displaystyle\frac{dy}{dx}=0$，故 $\left(a,b\right)$ 滿足 $a^2+ab+b^2=3$ 且 $2a+b=0$。利用代入消去法得 $a^2+a\cdot\left(-2a\right)+\left(-2a\right)^2=3$，因此 $a^2=1$，即 $a=\pm1$，而有 $b=\mp2$。故產生最高點與最低點的位置分別為 $\left(-1,2\right)$ 與 $\left(1,-2\right)$。【方法二】 設拉格朗日乘子函數為
$F\left(x,y,\lambda\right)=y+\lambda\left(x^2+xy+y^2-3\right)$
據此解聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=\lambda\left(2x+y\right)=0\\&F_y\left(x,y,\lambda\right)=1+\lambda\left(x+2y\right)=0\\&F_{\lambda}\left(x,y,\lambda\right)=x^2+xy+y^2-3=0\end{aligned}\right.$
由於明顯 $\lambda\neq0$，故 $2x+y=0$，代入第三式可循方法一同樣解得 $\displaystyle x=\pm1$ 而 $y=\mp2$，故最高點為 $\left(-1,2\right)$，而最低點為 $\left(1,-2\right)$。【方法三】 直接解得 $\displaystyle y=\frac{-x\pm\sqrt{12-3x^2}}2$，其中 $x\in\left[-2,2\right]$，求導有
$\displaystyle\frac{dy}{dx}=\displaystyle-\frac12\mp\frac{3x}{2\sqrt{12-3x^2}}$
若解方程式 $\displaystyle\frac{dy}{dx}=0$ 可得 $x=\pm1$，從而得 $y=\mp2$。再者當 $x=\pm2$ 時，$y=\mp1$。因此最高點為 $\left(-1,2\right)$、最低點為 $\left(1,-2\right)$。 - 設橢圓上之切點為 $\left(x_0,y_0\right)$，那麼由前一小題的解法一可知在該處的切線斜率為 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(x_0,y_0\right)}=-\frac{2x_0+y_0}{x_0+2y_0}$，故切線方程式能寫為
$\displaystyle y-y_0=-\frac{2x_0+y_0}{x_0+2y_0}\left(x-x_0\right)$
由於通過 $\left(4,-2\right)$，因此有$\left(2+y_0\right)\left(x_0+2y_0\right)=\left(2x_0+y_0\right)\left(4-x_0\right)$
即 $2x_0^2+2x_0y_0+2y_0^2-6x_0=0$，故得 $x_0=1$。代入該方程式中有 $y_0^2+y_0-2=0$，因而 $y_0=1$ 或 $y_0=-2$，如此有兩條切線方程式 $x+y=2$ 與 $y=-2$ 滿足條件。 - (10%) Use Taylor series to approximate $\displaystyle\int_0^1\cos\left(x^5\right)dx$ with error smaller than $10^{-4}$.
- Let
$f\left(x,y\right)=\begin{cases}\displaystyle\frac{x^2y}{x^2+y^2}&\mbox{if}~\left(x,y\right)\neq\left(0,0\right)\\0&\mbox{if}~\left(x,y\right)=\left(0,0\right)\end{cases}$
- (4%) Compute the directional derivative $D_{\vec{u}}f\left(0,0\right)$, where $\vec{u}=\left(\cos\theta,\sin\theta\right)$. Is $\left(0,0\right)$ a critical point of $f$?
- (8%) Find the maximum and minimum values of $f$ on the unit disc $D=\left\{\left(x,y\right)\mid\,x^2+y^2\leq1\right\}$.
- 按照定義求方向導數有
$\displaystyle D_{\vec{u}}f\left(0,0\right)=\lim_{h\to0}\frac{f\left(h\cos\theta,h\sin\theta\right)-f\left(0,0\right)}h=\lim_{h\to0}\frac{h^3\cos^2\theta\sin\theta}{h^3}=\cos^2\theta\sin\theta$
當 $\theta=0$ 或 $\pi$ 或 $\displaystyle\pm\frac\pi2$ 其方向導數皆為零，故 $f$ 在 $\left(0,0\right)$ 的兩個偏導函數值皆為零。因此 $\left(0,0\right)$ 為臨界點。 【方法一】容易由算術幾何不等式注意到
$\displaystyle\frac{x^2+y^2}3=\frac{x^2+y^2/2+y^2/2}3\geq\sqrt[3]{\frac{x^4y^2}4}=\frac{\left(x^2y\right)^{2/3}}{\sqrt[3]{4}}$
立方後開根號可得$\displaystyle\frac{\left(x^2+y^2\right)\sqrt{x^2+y^2}}{3\sqrt3}\geq\frac{x^2y}2\geq-\frac{\left(x^2+y^2\right)\sqrt{x^2+y^2}}{3\sqrt3}$
即$\displaystyle\frac{2\sqrt3}9\geq\frac{2\sqrt{x^2+y^2}}{3\sqrt3}\geq\frac{x^2y}{x^2+y^2}\geq-\frac{2\sqrt{x^2+y^2}}{3\sqrt3}\geq-\frac{2\sqrt3}9$
故最大值與最小值分別為 $\displaystyle\pm\frac{2\sqrt3}9$。【方法二】 若在內點但不為原點處，可容易求偏導得
$\displaystyle f_x\left(x,y\right)=\frac{2xy^3}{\left(x^2+y^2\right)^2}$, $\displaystyle f_y\left(x,y\right)=\frac{x^4-x^2y^2}{\left(x^2+y^2\right)^2}$
明顯可知無處為零，又由前一小題的結果能知在圓盤內點處僅當在原點處有偏導為零，但透過方向導數的型態可知在該處不產生極值。為此，我們考慮限制條件為 $x^2+y^2=1$，此時欲求極值之函數可寫為 $x^2y$，據此設拉格朗日乘子函數如下$F\left(x,y,\lambda\right)=x^2y+\lambda\left(x^2+y^2-1\right)$
如此我們解下列的聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=2xy+2\lambda x=0\\&F_y\left(x,y,\lambda\right)=x^2+2\lambda y=0\\&F_{\lambda}\left(x,y,\lambda\right)=x^2+y^2-1=0\end{aligned}\right.$
由第一式可知 $x=0$ 或 $\lambda=-y$。- 若 $x=0$，那麼可由第三式知 $y=\pm1$；
- 若 $\lambda=-y$，那麼由第二式可知 $x^2=2y^2$，由第三式便得 $\displaystyle y=\pm\frac1{\sqrt3}$，而 $\displaystyle x=\pm\frac{\sqrt2}{\sqrt3}$（共計四個點）。
- (12%) Evaluate the integral $\displaystyle\iiint_Eze^{x^2+y^2}dV$, where $E$ is the portion of the unit ball $x^2+y^2+z^2\leq1$ that lies in the first octant and above the cone $z=\sqrt{x^2+y^2}$.
訣竅
按照水平漸近線與鉛直漸近線的定義計算即可。解法
為了方便分析，設$\displaystyle\begin{aligned}&g\left(x\right)=\frac{\sqrt{x^4+x^3+1}-\sqrt{x^4-x^3+1}}{\left|x\right|}=\frac{x\left|x\right|}{\sqrt{x^4+x^3+1}+\sqrt{x^4-x^3+1}}\\&h\left(x\right)=\frac{\ln\left|x+1\right|}{x}\end{aligned}$
那麼 $f=g+h$。訣竅
利用使用微積分基本定理求導後確認函數的單調性，從而說明該函數為單射；隨後利用反函數微分法求解。解法
訣竅
利用積分所得的面積估計其上下界。解法
藉由觀察函數 $f\left(x\right)=\sqrt[3]{x}$ 在 $\left[1,1000\right]$ 與 $\left[1,999\right]$ 的面積可知$\displaystyle\sum_{i=1}^{999}\sqrt[3]{i}<\int_1^{1000}f\left(x\right)dx<\sum_{i=2}^{1000}\sqrt[3]{i}$
如此便得$\displaystyle\frac{4+3\left(10^4-1\right)}4=1+\int_1^{1000}\sqrt[3]{x}dx<\sum_{i=1}^{1000}\sqrt[3]{i}<10+\int_1^{1000}\sqrt[3]{x}dx=\frac{40+3\left(10^4-1\right)}4$
訣竅
分別運用分部積分法與變數代換法處理即可。解法
訣竅
依照題意使用極座標計算重積分即可獲得 $F$ 的表達式；隨後使用相關變率的概念求導即可。解法
訣竅
利用隱函數微分找出滿足極值的條件，又或使用拉格朗日乘子法找出極值，又或解出後求導得極值；先表達出切線方程式後代入座標以求得切點。解法
訣竅
由基本的泰勒展開式計算並透過交錯級數的誤差估計獲得近似值。解法
由餘弦函數的泰勒展開式可知$\displaystyle\cos\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!}x^{2n}=1-\frac{x^2}2+\frac{x^4}{24}-\cdots$
用 $x^5$ 取代 $x$ 可知$\displaystyle\cos\left(x^5\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!}x^{10n}=1-\frac{x^{10}}2+\frac{x^{20}}{24}-\cdots$
故同取定積分能得$\displaystyle\int_0^1\cos\left(x^5\right)dx=\int_0^1\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!}x^{10n}dx=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!}\int_0^1x^{10n}dx=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{\left(2n\right)!\left(10n+1\right)}$
因為當 $n=3$ 時有 $\displaystyle\left|\frac{\left(-1\right)^3}{6!\cdot31}\right|=\frac1{22320}<10^{-4}$，故得該定積分在不超過 $10^{-4}$ 為誤差的近似值為$\displaystyle\int_0^1\cos\left(x^5\right)dx\approx\sum_{n=0}^2\frac{\left(-1\right)^n}{\left(2n\right)!\left(10n+1\right)}=1-\frac1{22}+\frac1{504}$
訣竅
按照定義計算方向導數；隨後使用初等不等式或拉格朗日乘子法求其極值。解法
訣竅
直接計算後使用極座標；亦可直接使用球面座標計算。解法一
設 $D=\left\{\left(x,y\right)\in\mathbb{R}^2:\,x^2+y^2\leq1/2,\,x,y\geq0\right\}$，如此所求可先初步計算如下$\displaystyle\iiint_Eze^{x^2+y^2}dV=\iint_D\int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}ze^{x^2+y^2}dzdA=\frac12\iint_De^{x^2+y^2}\left(1-2x^2-2y^2\right)dA$
於是令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，如此 $\left\{\begin{aligned}&0\leq r\leq1/\sqrt2\\&0\leq\theta\leq\frac\pi2\end{aligned}\right.$，所求之三重積分可改寫並計算如下$\displaystyle\iiint_Eze^{x^2+y^2}dV=\frac12\int_0^{\frac\pi2}\int_0^{1/\sqrt2}e^{r^2}\left(1-2r^2\right)rdrd\theta=\frac\pi4\int_0^{1/\sqrt2}e^{r^2}\left(1-2r^2\right)rdr$
令 $u=r^2$，如此可使用變數代換法後再由分部積分法計算如下$\displaystyle\iiint_Eze^{x^2+y^2}dV=\frac\pi8\int_0^{1/2}e^u\left(1-2u\right)du=\left.\frac\pi8\left(3e^u-2ue^u\right)\right|_0^{1/2}=\frac\pi8\left(2e^{1/2}-3\right)$
解法二
令 $\left\{\begin{aligned}&x=\rho\cos\theta\sin\phi\\&y=\rho\sin\theta\sin\phi\\&z=\rho\cos\phi\end{aligned}\right.$，那麼按照題意所描述的區域可知 $\left\{\begin{aligned}&0\leq\rho\leq1\\&0\leq\theta\leq\frac\pi2\\&0\leq\phi\leq\frac\pi4\end{aligned}\right.$，如此所求的三重積分可改寫並計算如下$\displaystyle\begin{aligned}\iiint_Eze^{x^2+y^2}dV&=\int_0^{\frac\pi2}\int_0^{\frac\pi4}\int_0^1\rho\cos\phi e^{\rho^2\sin^2\phi}\rho^2\sin\phi d\rho d\phi d\theta\\&=\frac\pi2\int_0^1\int_0^{\frac\pi4}e^{\rho^2\sin^2\phi}\rho^3\cos\phi\sin\phi d\phi d\rho=\frac\pi4\int_0^1\rho e^{\rho^2\sin^2\phi}\Big|_0^{\pi/4}d\rho\\&=\frac\pi4\int_0^1\left(\rho e^{\rho^2/2}-\rho\right)d\rho=\left.\frac\pi4\left(e^{\rho^2/2}-\frac{\rho^2}2\right)\right|_0^1=\frac{\pi}4\left(e^{1/2}-\frac32\right)=\frac\pi8\left(2e^{1/2}-3\right)\end{aligned}$
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