國立臺灣大學九十一學年度研究所碩士班入學考試試題：應用微積分

※注意：請於答案卷上依序作答，並標明題號，不必列出計算式。

1. 1. Find the length of the astroid defined as:

      $x=\cos^3t$, $y=\sin^3t$, $0\leq t\leq2\pi$ shown below, ($10\%$)

   2. Find the coordinates of the centroid of the first quadrant arc of the astroid defined in part (a). ($8\%$)
訣竅
第一小題可使用曲線弧長公式計算；第二小題可由對稱性觀察知道重心落在直線 $y=x$ 上，故僅需計算重心的其中一個分量。
解法
1. 使用曲線弧長公式以及對稱性可計算如下

   $\displaystyle\begin{aligned}s&=4\int_0^{\frac\pi2}\sqrt{x'^2\left(t\right)+y'^2\left(t\right)}dt=4\int_0^{\frac\pi2}\sqrt{\left(-3\cos^2t\sin t\right)^2+\left(3\sin^2t\cos t\right)^2}dt\\&=12\int_0^{\frac\pi2}\cos t\sin tdt=6\int_0^{\frac\pi2}\sin2tdt=-3\cos2t\Big|_0^{\frac\pi2}=6\end{aligned}$

2. 由對稱性可知星形線在第一象限的重心必落在直線 $y=x$ 上，故我們僅需計算重心的 $x$ 分量。按照定義可列式並計算如下

   $\displaystyle \bar{x}=\frac{\displaystyle\int_0^{\frac\pi2}x\left(t\right)\sqrt{x'^2\left(t\right)+y'^2\left(t\right)}dt}{\displaystyle\int_0^{\frac\pi2}\sqrt{x'^2\left(t\right)+y'^2\left(t\right)}dt}=\frac{\displaystyle\int_0^{\frac\pi2}\cos^3t\cdot3\cos t\sin tdt}{\displaystyle\frac64}=2\int_0^{\frac\pi2}\cos^4t\sin tdt=\left.-\frac25\cos^5t\right|_0^{\frac\pi2}=\frac25$

   因此所求的重心為 $\displaystyle\left(\frac25,\frac25\right)$。


2. Find the particular solution of the differential equation

   $\displaystyle\frac{d^2y}{dx^2}-2\frac{dy}{dx}-2y=0$

   that satisfies the conditions $y=2$ and $\displaystyle\frac{dy}{dx}=5$ when $x=0$. ($10\%$)
訣竅
此為二階常係數微分方程，可運用特徵方程來求解。
解法

考慮此微分方程所對應的特徵方程 $\lambda^2-2\lambda-2=0$，可解得 $\lambda=1\pm\sqrt3$，如此該微分方程的通解為

$y\left(x\right)=C_1e^{\left(1+\sqrt3\right)x}+C_2e^{\left(1-\sqrt3\right)x}$

現使用初始條件可知

$\left\{\begin{aligned}&y\left(0\right)=C_1+C_2=2\\&y'\left(0\right)=C_1\left(1+\sqrt3\right)+C_2\left(1-\sqrt3\right)=5\end{aligned}\right.$

將第二式減去第一式可得 $\sqrt3\left(C_1-C_2\right)=3$，即 $C_1-C_2=\sqrt3$，從而有

$\displaystyle C_1=\frac{2+\sqrt3}2,\quad C_2=\frac{2-\sqrt3}2$

因此所求的特解為

$\displaystyle y\left(x\right)=\frac{\left(2+\sqrt3\right)e^{\left(1+\sqrt3\right)x}+\left(2-\sqrt3\right)e^{\left(1-\sqrt3\right)x}}2$



3. Please find the sum of the infinite series $\displaystyle\sum_{n=2}^{\infty}\left(\left(\frac67\right)^n-\frac4{2^n}\right)$. ($10\%$)
訣竅
運用無窮等比級數和公式計算求解即可。
解法

使用無窮等比級數和公式可知

$\displaystyle\sum_{n=2}^{\infty}\left(\left(\frac67\right)^n-\frac4{2^n}\right)=\sum_{n=2}^{\infty}\left(\frac67\right)^n-4\sum_{n=2}^\frac1{2^n}=\frac{\displaystyle\frac{36}{49}}{\displaystyle1-\frac67}-4\cdot\frac{\displaystyle\frac14}{\displaystyle1-\frac12}=\frac{36}7-2=\frac{22}7$



4. Please find out the local extrema and reflection point(s) of $f\left(x\right)=x^4-4x^3+10$. ($12\%$)
訣竅
利用一階與二階導函數來找出極值與反曲點的位置。
解法

求導可得

$f'\left(x\right)=4x^3-12x^2,\quad f''\left(x\right)=12x^2-24x$

為了求出極值候選點，我們解方程式 $f'\left(x\right)=0$，即 $4x^2\left(x-3\right)=4x^3-12x^2=0$，可得 $x=0$ 或 $x=3$。

由於在 $x=0$ 附近處， $f'\left(x\right)$ 恆負，故 $f$ 在 $x=0$ 處不為極值。而在 $x=3$ 處可由 $f''\left(3\right)=36>0$ 可知為局部極小，其值為 $f\left(3\right)=-17$。

為了找出反曲點，應解方程式 $f''\left(x\right)=0$，即 $12x\left(x-2\right)=12^2-24x=0$，可得 $x=0$ 或 $x=2$。可以容易檢驗確認出兩者皆為反曲產生的位置，故有兩個反曲點為 $\left(0,10\right)$ 與 $\left(2,6\right)$。



5. Three hundred students attended the dedication ceremony of a new building on a college campus. The president of the traditional female college announced a new expansion program, which included plans to make the college coeducational. The number of students who learned of the new program $t$ hour later is given by the function

   $\displaystyle f\left(t\right)=\frac{3000}{1+Be^{-kt}}$.

   If $600$ students on campus had heard about the new program $2$ hours after the ceremony, how fast was the rumor spreading $4$ hours after the ceremony? ($8\%$)
訣竅
本題出自Soo T. Tan 所著的 Applied Calculus for the Managerial, Life, and Social Sciences (第九版)，第五章第六節第二十五題。按照題意可求出給定函數中的參數，而為了計算傳播速率，我們對函數求導。
解法

由題幹可知一開始有 $300$ 名學生聽聞公告事項，此即 $\displaystyle f\left(0\right)=\frac{3000}{1+B}=300$，因此 $B=9$。另一方面，兩個小時後有 $600$ 名學生聽聞此事項，這表明

$\displaystyle f\left(2\right)=\frac{3000}{1+9e^{-2k}}=600$

即有 $1+9e^{-2k}=5$，或 $9e^{-2k}=4$，可解得 $\displaystyle k=\ln\frac32$。

先求 $f$ 的導函數如下

$\displaystyle f'\left(t\right)=-\frac{3000}{\left[1+9\left(2/3\right)^t\right]^2}\cdot9\left(\frac23\right)^t\ln\frac23$

故所求為 $\displaystyle f'\left(4\right)=\frac{3456\ln\left(3/2\right)}5$。



6. According to a joint study conducted by Oxnard's Environmental Management Department and a state government agency, the concentration of carbon monoxide (CO) in the air due to automobile exhaust is increasing at the rate given by

   $\displaystyle f\left(t\right)=\frac{8\left(0.1t+1\right)}{300\left(0.2t^2+4t+64\right)^{1/3}}$

   parts per million (ppm) per years $t$. Currently, the CO concentration due to automobile exhaust is 0.16 ppm. Find the expression giving the CO concentration $t$ year from now. ($8\%$)
訣竅
本題出自Soo T. Tan 所著的 Applied Calculus for the Managerial, Life, and Social Sciences (第九版)，第六章第二節自我練習第四題。利用積分將變化率還原回所求的量。
解法

設一氧化碳距現在起 $t$ 年後的濃度為 $F\left(t\right)$，那麼由微積分基本定理可知

$\displaystyle\begin{aligned}F\left(t\right)&=F\left(0\right)+\int_0^tf\left(s\right)ds=0.16+\int_0^t\frac{8\left(0.1s+1\right)}{300\left(0.2s^2+4s+64\right)^{1/3}}ds\\&=0.16+\frac1{150}\int_0^t\frac{0.4s+4}{\left(0.2s^2+4s+64\right)^{1/3}}ds\\&=0.16+\frac1{150}\int_0^t\frac{d\left(0.2s^2+4s+64\right)}{\left(0.2s^2+4s+64\right)^{1/3}}\\&=0.16+\frac1{150}\left[\left(0.2t^2+4t+64\right)^{2/3}-16\right]\end{aligned}$



7. In an endeavor to curb population growth in a Southeast Asia island state, the government has decided to launch an extensive propaganda campaign. Without curbs, the government expects the rate of population growth to have been $60e^{0.02t}$ thousand people/year $t$ yr from now, over next $5$ year. However, successful implementation of the proposed campaign is expected to result in a population growth rate of $-t^2+60$ thousand people/year $t$ yr, over next $5$ year. Assuming that the campaign is mounted, how many fewer people will be there in that country $5$ year from now than there would have been if no curbs had been imposed? ($8\%$)
訣竅
本題出自Soo T. Tan 所著的 Applied Calculus for the Managerial, Life, and Social Sciences (第九版)，第十一章第六節習題第五十三題。按照題意，將兩個變化率之差進行積分計算即可看出總量產生的差異。
解法

按照題意可知所求為

$\displaystyle\int_0^5\left[60e^{0.02t}-\left(-t^2+60\right)\right]dt=\left.\left(3000e^{0.02t}+\frac{t^3}3-60t\right)\right|_0^5=3000e^{0.1}-\frac{9775}3$



8. The Robertson Controls Company manufactures two basic models of setback thermostats: a standard mechanical thermostat and a deluxe electronic thermostat. Robertson's monthly revenue (in hundreds of dollars) is

   $\displaystyle R\left(x,y\right)=-\frac18x^2-\frac12y^2-\frac14xy+20x+60y$

   where $x$ (in units of hundred) denotes the number of mechanical thermostats manufactured and $y$ (in units of hundred) denotes the number of electronic thermostats manufactured per month. The total monthly cost incurred in producing these thermostats is $c\left(x,y\right)=7x+20y+280$ hundred dollars. Find how many thermostats of each model Robertson should manufacture per month in order to maximize its profit? ($8\%$)
訣竅
本題出自Soo T. Tan 所著的 Applied Calculus for the Managerial, Life, and Social Sciences (第九版)，第八章第三節自我練習第二題。為了找出極值可直接解一階偏導為零的位置即可。
解法
設利潤函數 $P$ 為收入減去成本，即
$\displaystyle P\left(x,y\right)=R\left(x,y\right)-c\left(x,y\right)=-\frac18x^2-\frac12y^2-\frac14xy+13x+40y-280$
為了求出利潤函數的最大值，我們解下列的聯立方程組
$\displaystyle\left\{\begin{aligned}&P_x\left(x,y\right)=-\frac{x}4-\frac{y}4+13=0\\&P_y\left(x,y\right)=-y-\frac{x}4+40=0\end{aligned}\right.$
即
$\left\{\begin{aligned}&x+y=52\\&x+4y=160\end{aligned}\right.$
可解得 $x=16$、$y=36$。
現在進一步計算二階判別式如下
$\displaystyle D\left(x,y\right)=\left|\begin{matrix}P_{xx}\left(x,y\right)&P_{xy}\left(x,y\right)\\P_{yx}\left(x,y\right)&P_{yy}\left(x,y\right)\end{matrix}\right|=\left|\begin{matrix}\displaystyle-\frac14&\displaystyle-\frac14\\\displaystyle-\frac14&-1\end{matrix}\right|=\frac3{16}>0$
且 $P_{xx}$ 與 $P_{yy}$ 恆負，故 $P$ 於 $\left(16,36\right)$ 處達到最大值。
故一般標準型恆溫器應生產 1600 台，而電子恆溫器則應生產 3600 台。


9. The operator of the Viking Princess, a luxury cruise liner, are contemplating the addition of another swimming pool to the ship. The chief engineer has suggested that an area in form of an ellipse located in the rear of the promenade deck would be suitable for this purpose. This location would provide a poolside area with sufficient space for passenger movement and placement of deck chairs. It has been determined that the shape of the ellipse may be described by the equation $x^2+4y^2=3600$, where $x$ and $y$ are measured in feet. Viking's operators would like to know the dimensions of the rectangular pool with the largest possible area that would meet these requirements. What are the dimensions of the pool with maximum area? ($8\%$)
訣竅
本題出自Soo T. Tan 所著的 Applied Calculus for the Managerial, Life, and Social Sciences (第九版)，第十二章第一節應用例題五。根據題意設定長方形的頂點從而獲得其對應的長與寬，應用初等不等式即可求其面積最大值。
解法
設長方形在第一象限中的頂點為 $\left(x,y\right)$，那麼可以看出其對應的長與寬分別為 $2x$ 與 $2y$，從而面積表達為 $4xy$。容易由算術幾何不等式發現
$3600=x^2+4y^2\geq2\sqrt{x^2\cdot4y^2}=4xy$
故面積的最大值為 $3600$ 平方英尺。此時等號成立的條件為 $x^2=4y^2$，即 $x=2y$，故 $x=30\sqrt2$，而 $y=15\sqrt2$。


10. Consider the homogeneous linear system

    $\displaystyle\left\{\begin{aligned}&\frac{dx_1}{dt}=3x_1+x_2-x_3,\\&\frac{dx_2}{dt}=x_1+3x_2-x_3,\\&\frac{dx_3}{dt}=3x_1+3x_2-x_3.\end{aligned}\right.$

    What is the general solution? ($10\%$)
訣竅
試湊簡化算式，將多個函數視為一個函數來求微分方程。
解法

第一式減去第二式有

$\displaystyle\frac{d\left(x_1-x_2\right)}{dt}=2\left(x_1-x_2\right)$

故有

$\displaystyle\frac{d\left(x_1-x_2\right)}{x_1-x_2}=2dt$

取不定積分可得 $\ln\left|x_1-x_2\right|=2t+k$，即 $x_1=x_2+C_1e^{2t}$，其中 $C_1=e^k$。此時注意後兩式可寫為

$\left\{\begin{aligned}&\frac{dx_2}{dt}=4x_2-x_3+C_1e^{2t}\\&\frac{dx_3}{dt}=6x_2-x_3+3C_1e^{2t}\end{aligned}\right.$

將前者乘以 $3$ 減去後者可得

$\displaystyle\frac{d\left(3x_2-x_3\right)}{dt}=2\left(3x_2-x_3\right)$

使用同樣的方式可得 $3x_2=x_3+C_2e^{2t}$。至此原題的第二道微分方程可寫為

$\displaystyle\frac{dx_2}{dt}=x_2+\left(C_1+C_2\right)e^{2t}$

移項後兩邊同乘以 $e^{-t}$ 有

$\displaystyle\frac{d}{dt}\left(e^{-t}x_2\left(t\right)\right)=e^{-t}\frac{dx_2}{dt}-e^{-t}x_2=\left(C_1+C_2\right)e^t$

最後取一次不定積分有

$x_2\left(t\right)=C_1e^{2t}+C_2e^{2t}+C_3e^t$

代回先前的方程中可解得

$\left\{\begin{aligned}&x_1\left(t\right)=2C_1e^{2t}+C_2e^{2t}+C_3e^t\\&x_2\left(t\right)=C_1e^{2t}+C_2e^{2t}+C_3e^t\\&x_3\left(t\right)=3C_1e^{2t}+2C_2e^{2t}+3C_3e^t\end{aligned}\right.$