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需列計算過程，否則不予計分
填充計算題（總計 $10$ 題，每題 $10$ 分）
- Is there a number $m$ such that $\displaystyle\lim_{x\to-2}\frac{3x^2+mx+m+3}{x^2+x-2}$ exists? If so, find the value of $m$ and the value of the limit.
- How many tangent lines to the curve $y=x/\left(x+1\right)$ pass through the point $\left(1,2\right)$? At which points to these tangent lines tough the curve?
- Differentiate the function: $\displaystyle g\left(x\right)=\frac{\ln x}{1+\ln\left(2x\right)}$.
- Find the local maximum and minimum values and saddle point(s), if any:
$f\left(x,y\right)=\left(x^2+y^2\right)e^{y^2-x^2}$. - 因為 $D\left(\pm1,0\right)=-16e^{-2}<0$，故 $\left(\pm1,0\right)$ 為鞍點。
- 因為 $D\left(0,0\right)=4>0$ 且 $f_{xx}\left(0,0\right)$ 與 $f_{yy}\left(0,0\right)$ 皆為正數，故 $\left(0,0\right)$ 為極小點。【事實上 $f\left(0,0\right)=0$ 且函數 $f$ 非負，故 $f$ 在 $\left(0,0\right)$ 達到絕對極小值。】
- An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $\$400,000/\mbox{km}$ over land to a point $P$ on the north bank and $\$800,000/\mbox{km}$ under the river to the tanks. To minimize the cost of the pipeline, where should $P$ be located?
- Evaluate the following equation: $\displaystyle\int_0^1\log\frac1{1-x}dx=$ .
- 當 $x=0$ 時有 $u=1$；
- 當 $x\to1^-$ 時有 $u\to0^+$；
- 求導有 $dx=-du$。
- Evaluate the following equation: $\displaystyle\int_0^{\infty}\frac{\sin x\cos x}xdx=$ .
- Find the mass of the portion of the plane $x+y+z=1$ in the first octane if the area density at any point $\left(x,y,z\right)$ on the surface is $kx^2$ kilograms per square meter, where $k$ is a constant.
- Evaluate the following equation: $\displaystyle\int_0^{\infty}\frac{\left(\mbox{arctan}\,x^2\right)}{x^2}dx=$ .
- Find the volume of the solid bounded by the cylinder $x^2+y^2=25$, the plane $x+y+z=8$, and the $xy$ plane.
訣竅
運用極限的四則運算定理求解。解法
假若該極限存在，記其值為 $L$，那麼$\displaystyle15-m=\lim_{x\to-2}\left(3x^2+mx+m+3\right)=\lim_{x\to-2}\left[\frac{3x^2+mx+m+3}{x^2+x-2}\cdot\left(x^2+x-2\right)\right]=L\cdot0=0$
故 $m=15$。此時我們可計算該極限值如下$\displaystyle\lim_{x\to-2}\frac{3x^2+15x+18}{x^2+x-2}=\lim_{x\to-2}\frac{3\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x-1\right)}=\lim_{x\to-2}\frac{3\left(x+3\right)}{x-1}=\frac{3\cdot\left(-2+3\right)}{-2-1}=-1$
故確實存在一數 $m$ 使極限存在。該 $m$ 值為 $15$，而其極限值為 $-1$。訣竅
給定曲線上一點後寫出切線方程式並考慮通過給定點，隨後求出切線座標。解法
設切點 $P$ 為 $\left(a,f\left(a\right)\right)$，如此切線方程式為$y-f\left(a\right)=f'\left(a\right)\left(x-a\right)$
又此切線通過 $\left(1,2\right)$，故有$\displaystyle2-\frac{a}{a+1}=\frac1{\left(a+1\right)^2}\left(1-a\right)$
兩邊同乘以 $\left(a+1\right)^2$ 可得 $a^2+4a+1=0$，那麼解得 $\displaystyle a=-2\pm\sqrt3$。故有兩條切線方程式會通過 $\left(1,2\right)$。訣竅
運用微分公式求解即可。解法
使用微分公式求解可知$\displaystyle g'\left(x\right)=\frac{\displaystyle\frac1x\cdot\left(1+\ln\left(2x\right)\right)+\ln x\cdot\frac1{2x}\cdot2}{\left(1+\ln\left(2x\right)\right)^2}=\frac{1+\ln2+2\ln x}{x\left(1+\ln\left(2x\right)\right)^2}$
訣竅
先求一階偏導的點，隨後由二階行列式判定極值性質。解法
為了找出極值候選點，我們先解方程組
$\left\{\begin{aligned}&f_x\left(x,y\right)=2xe^{y^2-x^2}-2x\left(x^2+y^2\right)e^{y^2-x^2}=0\\&f_y\left(x,y\right)=2ye^{y^2-x^2}+2y\left(x^2+y^2\right)e^{y^2-x^2}=0\end{aligned}\right.$
兩式皆除以 $e^{y^2-x^2}$ 可得$\left\{\begin{aligned}&2x-2x\left(x^2+y^2\right)=0\\&2y+2y\left(x^2+y^2\right)=0\end{aligned}\right.$
對於第二式除以 $x^2+y^2+1$ 得 $y=0$，那麼由第一式有 $2x-2x^3=0$，因此 $x=-1,0,1$。現計算二階行列式如下
$D\left(x,y\right)=\left|\begin{matrix}f_{xx}\left(x,y\right)&f_{yx}\left(x,y\right)\\f_{xy}\left(x,y\right)&f_{yy}\left(x,y\right)\end{matrix}\right|=\left|\begin{matrix}\left(4x^2y^2+4x^4-10x^2-2y^2+2\right)e^{y^2-x^2}&-4xy\left(x^2+y^2\right)e^{y^2-x^2}\\-4xy\left(x^2+y^2\right)e^{y^2-x^2}&\left(4y^4+4x^2y^2+10y^2+2x^2+2\right)e^{y^2-x^2}\end{matrix}\right|$
據此可知訣竅
按照題意進行假設，並列出成本函數，隨後運用求導找出最小成本。解法
設 $P$ 距離煉油廠 $x$ 公尺，那麼成本函數可設為$C\left(x\right)=400000x+800000\sqrt{2^2+\left(6-x\right)^2}$
其中 $x\in\left[0,6\right]$。求導有$\displaystyle C'\left(x\right)=400000-800000\cdot\frac{6-x}{\sqrt{4+\left(6-x\right)^2}}$
方程 $C'\left(x\right)=0$ 可得 $\sqrt{4+\left(6-x\right)^2}=12-2x$，平方有 $3x^2-36x+104=0$，故所求為 $\displaystyle x=6\pm\frac2{\sqrt3}$。因為 $x\in\left[0,6\right]$，故得 $\displaystyle x=6-\frac2{\sqrt3}$。容易檢驗知$\displaystyle C\left(0\right)=1600000\sqrt{10},\quad,C\left(6-\frac2{\sqrt3}\right)=800000\left(3+\sqrt3\right),\quad C\left(6\right)=4000000$
因此在距離煉油廠 $\displaystyle x=6-\frac2{\sqrt3}$ 公里處有最小成本。訣竅
運用變數代換後使用分部積分法求解。本題亦可於 Mathematical Analysis (Apostol 著) 第十章習題第六題 a 小題見到。解法
令 $u=1-x$，那麼有$\displaystyle\int_0^1\log\frac1{1-x}dx=-\int_1^0\log u\cdot-du=-\int_0^1\log udu=-\left(u\log u-u\right)\Big|_0^1=1$
訣竅
運用倍角公式與變數變換以及經典的瑕積分結果求解。本題亦可於 Mathematical Analysis (Apostol 著) 第十章習題第十題 a小題見到。解法
首先由倍角公式可知$\displaystyle\int_0^{\infty}\frac{\sin x\cos x}xdx=\int_0^{\infty}\frac{\sin2x}{2x}dx$
令 $u=2x$，那麼 $du=2dx$，據此所求可改寫並計算如下$\displaystyle\int_0^{\infty}\frac{\sin x\cos x}xdx=\frac12\int_0^{\infty}\frac{\sin u}udu=\frac12\cdot\frac\pi2=\frac\pi4$
訣竅
將曲面參數化後積分求解。解法
將在第一象限的曲面 $x+y+z=1$ 參數化為${\bf r}\left(u,v\right)=\left(u,v,1-u-v\right)$
其中變數範圍為 $\left\{\begin{aligned}&0\leq u\leq1\\&0\leq v\leq1-u\end{aligned}\right.$。如此所求為$\displaystyle m=\int_0^1\int_0^{1-u}ku^2\left|{\bf r}_u\times{\bf r}_v\right|dvdu=k\int_0^1\int_0^{1-u}\sqrt3dvdu=k\sqrt3\int_0^1\left(1-u\right)du=\frac{k\sqrt3}2$
訣竅
運用分部積分法與有理函數的積分技巧處理。解法
使用分部積分法直接計算可知$\displaystyle\begin{aligned} &\int_0^{\infty}\frac{\left(\mbox{arctan}\,x^2\right)}{x^2}dx=-\int_0^{\infty}\tan^{-1}\left(x^2\right)d\frac1x=\left.-\frac{\tan^{-1}x^2}x\right|_0^{\infty}+\int_0^{\infty}\frac1x\cdot\frac1{1+x^4}\cdot2xdx\\&=\int_0^{\infty}\frac2{\left(1+x^2\right)^2-\left(\sqrt2x\right)^2}dx=\int_0^{\infty}\left[\frac{\displaystyle\frac{\sqrt2}2x+1}{\displaystyle\left(x+\frac{\sqrt2}2\right)^2+\left(\frac{\sqrt2}2\right)^2}+\frac{\displaystyle-\frac{\sqrt2}2x+1}{\displaystyle\left(x-\frac{\sqrt2}2\right)^2+\left(\frac{\sqrt2}2\right)^2}\right]dx\\&=\int_0^{\infty}\left[\frac{\sqrt2}2\frac{\displaystyle x+\frac{\sqrt2}2}{x^2+\sqrt2x+1}+\frac12\frac1{\displaystyle\left(x+\frac{\sqrt2}2\right)^2+\left(\frac{\sqrt2}2\right)^2}-\frac{\sqrt2}2\frac{\displaystyle x-\frac{\sqrt2}2}{x^2-\sqrt2x+1}+\frac12\frac1{\displaystyle\left(x-\frac{\sqrt2}2\right)^2+\left(\frac{\sqrt2}2\right)^2}\right]dx\\&=\left.\left[\frac{\sqrt2}4\ln\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}+\frac{\sqrt2}2\tan^{-1}\left(\sqrt2x+1\right)+\frac{\sqrt2}2\tan^{-1}\left(\sqrt2x-1\right)\right]\right|_0^{\infty}\\&=\left(\frac{\sqrt2}2\cdot\frac\pi2+\frac{\sqrt2}2\cdot\frac\pi2\right)-\left(\frac{\sqrt2}2\cdot\frac\pi4-\frac{\sqrt2}2\cdot\frac\pi4\right)=\frac{\pi\sqrt2}2\end{aligned}$
【註記】 筆者懷疑本題應當是出自 Mathematical Analysis (Apostol 著)，第十章習題第二十六題，題目如下：
$\displaystyle\int_0^{\infty}\frac{\left(\mbox{arctan}\,x\right)^2}{x^2}dx$
這也能夠說明為何分子之處有括號。解法 首先考慮雙變數函數 $f$ 如下：
$\displaystyle f\left(x,y\right)=\int_0^{\infty}\frac{dt}{\left(1+x^2t^2\right)\left(1+y^2t^2\right)}$
直接計算可以注意到$\displaystyle f\left(x,y\right)=\frac1{x^2-y^2}\int_0^{\infty}\left(\frac{x^2}{1+x^2t^2}-\frac{y^2}{1+y^2t^2}\right)dt=\left.\frac{x\tan^{-1}\left(xt\right)-y\tan^{-1}\left(yt\right)}{x^2-y^2}\right|_0^{\infty}=\frac{\pi}{2\left(x+y\right)}$
那麼計算重積分可以得到$\displaystyle\begin{aligned}\int_0^1\int_0^1f\left(x,y\right)dxdy&=\frac\pi2\int_0^1\int_0^1\frac1{x+y}dxdy=\frac\pi2\int_0^1\left[\ln\left(1+y\right)-\ln y\right]dy\\&=\left.\frac\pi2\left[y\ln\left(1+y\right)+\ln\left(1+y\right)-y\ln y\right]\right|_0^1=\pi\ln2\end{aligned}$
另一方面藉由交換積分次序則有$\displaystyle\begin{aligned}\int_0^1\int_0^1f\left(x,y\right)dxdy&=\int_0^{\infty}\int_0^1\int_0^1\frac1{\left(1+x^2t^2\right)\left(1+y^2t^2\right)}dxdydt\\&=\int_0^{\infty}\left(\int_0^1\frac{dx}{1+x^2t^2}\right)\left(\int_0^1\frac{dy}{1+y^2t^2}\right)dt\\&=\int_0^{\infty}\frac{\tan^{-1}t}t\cdot\frac{\tan^{-1}t}tdt=\int_0^{\infty}\frac{\left(\mbox{arctan}\,t\right)^2}{t^2}dt\end{aligned}$
如此所求計算完畢。訣竅
先利用重積分列式，隨後使用極座標變換計算求解。解法
設 $D=\left\{\left(x,y\right)\in\mathbb{R}^2:~x^2+y^2\leq25\right\}$，那麼所求的體積為$\displaystyle V=\iint_D\left[\left(8-x-y\right)-0\right]dA$
使用極座標變換，令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，那麼 $\left\{\begin{aligned}&0\leq r\leq5\\&0\leq\theta\leq2\pi\end{aligned}\right.$，如此所求的體積可改寫並計算如下$\displaystyle\begin{aligned}V&=\int_0^{2\pi}\int_0^5\left(8-r\cos\theta-r\sin\theta\right)rdrd\theta\\&=\int_0^{2\pi}\left.\left(4r^2-\frac{r^3\left(\cos\theta+\sin\theta\right)}3\right)\right|_0^5d\theta\\&=\int_0^{2\pi}\left(100-\frac{125}3\left(\cos\theta+\sin\theta\right)\right)d\theta\\&=\left.\left[100\theta-\frac{125}3\left(\sin\theta-\cos\theta\right)\right]\right|_0^{2\pi}=200\pi\end{aligned}$
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