- ($10\%$) Evaluate the integrals.
(a) $\displaystyle\int_0^{\infty}\frac{\ln x}{1+x^2}dx$ (b) $\displaystyle\int\frac{x}{1-x^2+\sqrt{1-x^2}}dx$
- 首先確認瑕積分的收斂性，我們將該瑕積分分為兩段：
$\displaystyle\int_0^{\infty}\frac{\ln x}{1+x^2}dx=\int_0^1\frac{\ln x}{1+x^2}dx+\int_1^{\infty}\frac{\ln x}{1+x^2}dx$
對於第一個瑕積分，觀察到當 $x\in\left(0,1\right]$ 時有 $\displaystyle\frac{\ln x}{1+x^2}\geq\ln x$，因此$\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx\geq\int_0^1\ln xdx=\left(x\ln x-x\right)\Big|_0^1=-1$
故第一個瑕積分收斂。另一方面，對於第二個瑕積分，當 $x\in\left[1,\infty\right)$ 時有 $\displaystyle\frac{\ln x}{1+x^2}\leq\frac{\ln x}{x^2}$，因此$\displaystyle\int_1^{\infty}\frac{\ln x}{1+x^2}dx\leq\int_1^{\infty}\frac{\ln x}{x^2}dx=\left.-\frac{\ln x}x\right|_1^{\infty}+\int_1^{\infty}\frac{dx}{x^2}=\left.-\frac1x\right|_1^{\infty}=1$
結合兩者可知給定的瑕積分收斂。
現在，對於第一個瑕積分，我們考慮變數變換，令 $\displaystyle u=\frac1x$，那麼- 當 $x\to0^+$ 時有 $u\to\infty$；
- 當 $x=1$ 時有 $u=1$；
- 當 $\displaystyle du=-\frac1{x^2}dx$。
$\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx=\int_{\infty}^1\frac{\ln\left(1/u\right)}{1+\left(1/u\right)^2}\cdot-\frac1{u^2}du=-\int_1^{\infty}\frac{\ln u}{1+u^2}du=-\int_1^{\infty}\frac{\ln x}{1+x^2}dx$
恰為第二個瑕積分的負值，從而所求為零。 - 令 $u=\sqrt{1-x^2}$，那麼有 $1-x^2=u^2$ 且 $-xdx=udu$，因此所求的不定積分可改寫並計算如下
$\displaystyle\int\frac{x}{1-x^2+\sqrt{1-x^2}}dx=-\int\frac{udu}{u^2+u}=-\int\frac{du}{1+u}=-\ln\left|1+u\right|+C=-\ln\left(1+\sqrt{1-x^2}\right)+C$
- ($10\%$) A heavy rope, $50$ ft long, weighs $0.5$ lb/ft and hangs over the edge of a building $120$ ft high. How much work is done in pulling the rope to the top of the building.
- ($10\%$) Solve the differential equation $x^2y'+xy=1$, $x>0$.
- ($10\%$) Let $n$ be a positive integer.
- Maximize $\displaystyle\sum_{i=1}^nx_iy_i$ subject to the constraints $\displaystyle\sum_{i=1}^nx_i^2=1$ and $\displaystyle\sum_{i=1}^ny_i^2=1$.
- Use the result of part (a) to show that
$\displaystyle\left(\sum_{i=1}^na_ib_i\right)^2\leq\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2$
for any numbers $a_1,\cdots,a_n,b_1,\cdots,b_n$. This inequality is known as the Cauchy-Schwarz inequality.
- 考慮拉格朗日乘子函數如下
$\displaystyle F\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=\sum_{i=1}^nx_iy_i+\lambda_1\left(\sum_{i=1}^nx_i^2-1\right)+\lambda_2\left(\sum_{i=1}^ny_i^2-1\right)$
據此解聯立方程組如下$\left\{\begin{aligned}&F_{x_1}\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=y_1+2\lambda_1x_1=0\\&\vdots\\&F_{x_n}\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=y_n+2\lambda_1x_n=0\\&F_{y_1}\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=x_1+2\lambda_2y_1=0\\&\vdots\\&F_{y_n}\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=x_n+2\lambda_2y_n=0\\&F_{\lambda_1}\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=\sum_{i=1}^nx_i^2-1=0\\&F_{\lambda_2}\left(x_1,\cdots,x_n,y_1,\cdots,y_n,\lambda_1,\lambda_2\right)=\sum_{i=1}^ny_i^2-1=0\end{aligned}\right.$
容易看出對於 $k\in\left\{1,\cdots,n\right\}$ 有 $y_k=-2\lambda_1x_k=-2\lambda_1\cdot\left(-2\lambda_2y_k\right)$，即 $y_k=4\lambda_1\lambda_2y_k$。同取平方後並對 $k$ 求和則有 $1=16\lambda_1^2\lambda_2^2$，故 $4\lambda_1\lambda_2=\pm1$。
假若 $4\lambda_1\lambda_2=-1$，那麼對於所有 $k$ 必有 $y_k=0$，這與最後一式矛盾。故我們有 $4\lambda_1\lambda_2=1$。現在對於前 $n$ 式分別同乘以 $y_1,\cdots,y_n$ 並求和，那麼便有$\displaystyle\sum_{i=1}^nx_iy_i=-\frac1{2\lambda_1}\sum_{i=1}^ny_i^2=-\frac1{2\lambda_1}$
另一方面，類似的也有$\displaystyle\sum_{i=1}^nx_iy_i=-\frac1{2\lambda_2}\sum_{i=1}^nx_i^2=-\frac1{2\lambda_2}$
至此可知 $\lambda_1=\lambda_2$，故由 $4\lambda_1\lambda_2=1$ 而得 $\displaystyle\lambda_1=\lambda_2=\pm\frac12$。檢驗可知 $\displaystyle\sum_{i=1}^nx_iy_i$ 的最大值為 $1$ 而最小值為 $-1$。 - 承前一小題，我們知道
$\displaystyle\left(\sum_{i=1}^nx_iy_i\right)^2\leq1$
現在取 $\displaystyle x_i=\frac{a_i}{\sqrt{\sum_{k=1}^na_k^2}}$ 與 $\displaystyle y_i=\frac{b_i}{\sqrt{\sum_{k=1}^nb_k^2}}$，那麼容易檢驗知道 $\displaystyle\sum_{i=1}^nx_i^2=\sum_{i=1}^ny_i^2=1$，從而有$\displaystyle\frac1{\sum_{k=1}^na_k^2b_k^2}\left(\sum_{i=1}^na_ib_i\right)^2=\left(\sum_{i=1}^n\frac{a_ib_i}{\sqrt{\sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2}}\right)^2\leq1$
移項即獲證。 - ($10\%$) Evaluate the integral
$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^{2-x^2-y^2}\left(x^2+y^2\right)^{3/2}dzdydx$
- ($10\%$) Evaluate the line integral
$\displaystyle\oint_C\left(3y^2+\tan^{-1}x\right)dx+\left(3x+\sin y\right)dy$
where $C$ is the boundary of the region enclosed by the parabola $y=x^2$ and the line $y=4$. - ($8\%$) A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is $0.6$ mile north of the intersection and the car is $0.8$ mile to the east, the police determine with radar that the distance between them and the car is increasing at $20$ mile per hour (mph). If the cruiser is moving at $60$ mph at the instant of measurement, what is the speed of the car?
Answer: Speed $=$ mph - ($8\%$) A rectangle is to be inscribed in a semicircle of radius $2$. What is the largest area the rectangle can have, and what are its dimensions?
Answer: Area $=$ , Height $=$ , Length $=$ . - ($8\%$) $\displaystyle\lim_{x\to\infty}x^{1/x}=$ .
- ($8\%$) Find the length of the cardioid $r=1-\cos\theta$
Answer: Length $=$ . - ($8\%$) Find $\partial w/\partial x$ at the point $\left(x,y,z\right)=\left(2,-1,1\right)$, if $w=x^2+y^2+z^2$, $z^3-xy+yz+y^3=1$, and $x$ and $y$ are the independent variables.
Answer: $\partial w/\partial x=$ .
訣竅
針對第一小題的瑕積分，我們先驗證其收斂性後分段使用變數變換計算；第二小題則運用變數代換法改寫被積分函數後求解即可。解法
訣竅
考慮變力做功的計算。解法
設拉上 $x$ 英呎時所需要的靜拉力為 $F\left(x\right)$，可以知道 $F\left(x\right)=0.5\left(50-x\right)=25-0.5x$ 磅，故需要$\displaystyle W=\int_0^{50}\left(25-0.5x\right)dx=\left.\left(25x-\frac{x^2}4\right)\right|_0^{50}=625~\mbox{英呎．磅}$
的功把繩子拉至樓頂。訣竅
運用積分因子改寫後取積分即可求解。解法
將兩邊同除以 $x^2$ 後有$\displaystyle y'+\frac1xy=\frac1{x^2}$
那麼其所對應的積分因子為 $e^{\int\frac{dx}x}=e^{\ln x}=x$，故再同乘以 $x$ 可得$\displaystyle\left(xy\right)'=xy'+y=\frac1x$
同取積分可得$xy=\ln x+C$
因此所求為 $\displaystyle y\left(x\right)=\frac{\ln x+C}x$，此處 $C$ 為任意常數。訣竅
第一小題可運用拉格朗日乘子法求解；第二小題可對於第一小題取適當的 $x_i$ 和 $y_i$ 以獲得所需的結論。解法
訣竅
先計算一層迭代積分後發現積分範圍與被積分函數適合使用極座標變換處理。解法
先作一次積分可得$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^{2-x^2-y^2}\left(x^2+y^2\right)^{3/2}dzdydx=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left(x^2+y^2\right)^{3/2}\left(2-2x^2-2y^2\right)dydx$
使用極座標變換，令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，如此變數範圍為 $\left\{\begin{aligned}&0\leq r\leq1\\&0\leq\theta\leq2\pi\end{aligned}\right.$，因此所求可改寫並計算如下$\displaystyle\begin{aligned}\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^{2-x^2-y^2}\left(x^2+y^2\right)^{3/2}dzdydx&=\int_0^{2\pi}\int_0^1\left(r^2\right)^{3/2}\left(2-2r^2\right)rdrd\theta\\&=4\pi\int_0^1\left(r^4-r^6\right)dr=\left.4\pi\left(\frac{r^5}5-\frac{r^7}7\right)\right|_0^1=\frac{8\pi}{35}\end{aligned}$
訣竅
運用 Green 定理計算此線積分，隨後可直接計算迭代積分。解法
設 $D$ 為 $C$ 所包圍的區域，即$D=\left\{\left(x,y\right)\in\mathbb{R}^2:~-2\leq x\leq2,~x^2\leq y\leq4\right\}$
那麼使用 Green 定理，所求可以改寫並計算如下$\displaystyle\begin{aligned}\oint_C\left(3y^2+\tan^{-1}x\right)dx+\left(3x+\sin y\right)dy&=\iint_D\left[\frac{\partial}{\partial x}\left(3x+\sin y\right)-\frac{\partial}{\partial y}\left(3y^2+\tan^{-1}x\right)\right]dA\\&=\iint_D\left(3-6y\right)dA=\int_{-2}^2\int_{x^2}^4\left(3-6y\right)dydx=\int_{-2}^2\left(3y-3y^2\right)\Big|_{x^2}^4dx\\&=\int_{-2}^2\left(-36-3x^2+3x^4\right)dx=6\int_0^2\left(x^4-x^2-12\right)dx\\&=\left.6\left(\frac{x^5}5-\frac{x^3}3-12x\right)\right|_0^2=-\frac{608}5\end{aligned}$
訣竅
設定警車與逃逸車輛與十字路口對應的距離函數，由畢氏定理產生出兩車間的關係，接著對時間求導後列出題意中所有的條件後可看出所求。解法
設警車與逃逸車輛和十字路口的距離分別為時間的函數 $x\left(t\right)$ 與 $y\left(t\right)$，那麼兩車的距離為 $d\left(t\right)=\sqrt{x^2\left(t\right)+y^2\left(t\right)}$。按照題意，假定測量時刻為 $t_0$，那麼有 $x\left(t_0\right)=0.6$、$y\left(t_0\right)=0.8$、$\displaystyle\frac{dz}{dt}\left(t_0\right)=20$ 以及 $\displaystyle\frac{dx}{dt}=-60$。那麼由連鎖律可知$\displaystyle\frac{dz}{dt}=\frac{x\left(t\right)x'\left(t\right)+y\left(t\right)y'\left(t\right)}{\sqrt{x^2\left(t\right)+y^2\left(t\right)}}$
取 $t=t_0$ 並代入上述資訊可得$\displaystyle20=\frac{0.6\cdot\left(-60\right)+0.8\cdot y'\left(t_0\right)}{\sqrt{0.6^2+0.8^2}}$
可求得 $y'\left(t_0\right)=70$。訣竅
依據題意設定長方形的頂點並表達其面積，隨後運用微分求其極值。解法
設上半圓方程式為 $y=\sqrt{4-x^2}$，而內接於半圓的長方形四個頂點分別為 $\left(\pm a,0\right),\left(\pm a,\sqrt{4-a^2}\right)$，那麼長方形面積可表達為 $a$ 的函數如下$f\left(a\right)=2a\sqrt{4-a^2},\quad0<a<2$
為了求出最大值，解方程式 $f'\left(a\right)=0$，即$\displaystyle2\sqrt{4-a^2}+2a\cdot\frac{-a}{\sqrt{4-a^2}}=0$
同乘以 $\sqrt{4-a^2}$ 可得 $2\left(4-a^2\right)-2a^2=0$，因此 $a=\sqrt2$。檢驗可知 $f\left(0\right)=f\left(2\right)=0$，故 $f$ 在 $a=\sqrt2$ 處達到最大值，其值為 $f\left(\sqrt2\right)=4$。此時長與高分別為 $2\sqrt2$ 與 $\sqrt2$。訣竅
換底後使用羅必達法則求極限即可。解法
換底後使用羅必達法則計算如下$\displaystyle\lim_{x\to\infty}x^{1/x}=\lim_{x\to\infty}\exp\left(\frac{\ln x}x\right)=\exp\left(\lim_{x\to\infty}\frac{\ln x}x\right)=\exp\left(\lim_{x\to\infty}\frac{1/x}1\right)=\exp\left(0\right)=1$
訣竅
使用極座標下的曲線弧長公式計算即可。解法
使用曲線下的弧長公式計算如下$\displaystyle\begin{aligned}s&=\int_0^{2\pi}\sqrt{r^2\left(\theta\right)+\left(\frac{dr}{d\theta}\left(\theta\right)\right)^2}d\theta=\int_0^{2\pi}\sqrt{\left(1-\cos\theta\right)^2+\left(\sin\theta\right)^2}d\theta\\&=\int_0^{2\pi}\sqrt{2-2\cos\theta}d\theta=2\int_0^{2\pi}\left|\sin\frac{\theta}2\right|d\theta=\left.\left(-4\cos\frac{\theta}2\right)\right|_0^{2\pi}=8\end{aligned}$
訣竅
使用多變函數的連鎖律計算之。解法
由於 $x,y$ 為兩獨立變量，故 $z=z\left(x,y\right)$，此表明$w\left(x,y\right)=x^2+y^2+\left[z\left(x,y\right)\right]^2$
因此求 $w$ 對 $x$ 的偏導可知$\displaystyle\frac{\partial w}{\partial x}=2x+2z\left(x,y\right)\frac{\partial z}{\partial x}\left(x,y\right)$
取 $\left(x,y\right)=\left(2,-1\right)$ 以及 $z\left(2,-1\right)=1$ 可得$\displaystyle\left.\frac{\partial w}{\partial x}\right|_{\left(x,y,z\right)=\left(2,-1,1\right)}=4+2\frac{\partial z}{\partial x}\left(2,-1\right)$
另一方面，對於給定的第二式求對 $x$ 的偏導則有$\displaystyle3z^2\left(x,y\right)\frac{\partial z}{\partial x}\left(x,y\right)-y+y\frac{\partial z}{\partial x}\left(x,y\right)=0$
此時取 $\left(x,y\right)=\left(2,-1\right)$ 與 $z\left(2,-1\right)=1$ 可得 $\displaystyle\frac{\partial z}{\partial x}\left(2,-1\right)=-\frac12$，從而所求為 $\displaystyle\left.\frac{\partial w}{\partial x}\right|_{\left(x,y,z\right)=\left(2,-1,1\right)}=3$。
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