※ Please write down the detailed calculation process.
- ($10\%$) A one-story storage building is to have a volume of $250,000$ cubic feet. The roof costs $\$32$ per square foot, the walls cost $\$10$ per square foot, and the floor costs $\$8$ per square foot. Find the dimensions that minimize the cost of the building.
- ($10\%$) Consider the double integral
$\displaystyle\int_0^1\int_{2x-1}^{\sqrt{x}}2xydydx$.
- ($5\%$) Rewrite the above double integral by exchanging the order of integration.
- ($5\%$) Evaluate the double integral derived in (a).
- 原先的積分區域 $\left\{\begin{aligned}&0\leq x\leq1\\&2x-1\leq y\leq\sqrt{x}\end{aligned}\right.$ 可改寫為 $\left\{\begin{aligned}&y^2\leq x\leq\frac{y+1}2\\&0\leq y\leq1\end{aligned}\right.$ 聯集 $\left\{\begin{aligned}&0\leq x\leq\frac{y+1}2\\&-1\leq y\leq0\end{aligned}\right.$，如此給定的重積分可改寫如下
$\displaystyle\int_0^1\int_{2x-1}^{\sqrt{x}}2xydydx=\int_0^1\int_{y^2}^{\frac{y+1}2}2xydxdy+\int_{-1}^0\int_0^{\frac{y+1}2}2xydxdy$
- 直接演算可知
$\displaystyle\begin{aligned}\int_0^1\int_{y^2}^{\frac{y+1}2}2xydxdy+\int_{-1}^0\int_0^{\frac{y+1}2}2xydxdy&=\int_0^1x^2y\Big|_{y^2}^{\frac{y+1}2}dy+\int_{-1}^0x^2y\Big|_0^{\frac{y+1}2}dy\\&=\int_0^1\left(\frac{y^3+2y^2+y}4-y^5\right)dy+\int_{-1}^0\frac{y^3+2y^2+y}4dy\\&=\left.\left(\frac{y^4}{16}+\frac{y^3}6+\frac{y^2}8-\frac{y^6}6\right)\right|_0^1+\left.\left(\frac{y^4}{16}+\frac{y^3}6+\frac{y^2}8\right)\right|_{-1}^0=\frac16\end{aligned}$
- ($10\%$) Find the least squares approximation $h\left(x\right)=ax+ax^2$ for $f\left(x\right)=e^{-x}$, $0\leq x\leq1$. [Hint: Solve $a$ by minimizing the integral of the squared difference between $f\left(x\right)$ and $h\left(x\right)$.]
- ($14\%$) Suppose that you currently have $\$40,000$ invested in a savings account earning $3\%$ per year compounded continuously. Market conditions are improving and you decide to continuously transfer $15\%$ of the balance of your savings account into a stock fund earning $12\%$ per year. You also withdraw $\$5,000$ from the savings account each year for expenses.
- ($10\%$) Set up and solve a linked pair of differential equations to determine respectively the amount of money in the savings account and stock fund $t$ years from now.
- ($4\%$) How long does it take for the savings account to be exhausted? How much money is in the stock fund at this time point?
- 設 $y\left(t\right)$ 與 $z\left(t\right)$ 分別表自現在過 $t$ 年後儲蓄帳戶金額與股票金額，且按題意有 $y\left(0\right)=40000$ 與 $z\left(0\right)=0$。再者，由條件有微分方程式
$\displaystyle y'\left(t\right)=-0.12y\left(t\right)-5000$, $z'\left(t\right)=0.15y\left(t\right)+0.12z\left(t\right)$.
對第一式運用積分因子法有$\left(e^{0.12t}y\left(t\right)\right)'=e^{0.12t}y'\left(t\right)+0.12e^{0.12t}y\left(t\right)=-5000e^{0.12t}$
取積分有$\displaystyle e^{0.12t}y\left(t\right)-40000=\frac{125000}3\left(1-e^{0.12t}\right)$
因此有 $\displaystyle y\left(t\right)=\frac{5000}3\left(49e^{-0.12t}-25\right)$。而對於 $z$ 則同樣使用積分因子法有$\left(e^{-0.12t}z\left(t\right)\right)'=e^{-0.12t}z'\left(t\right)-0.12e^{-0.12t}z\left(t\right)=0.15e^{-0.12t}y\left(t\right)=6250\left(e^{-0.12t}-1\right)$
取積分有$\displaystyle e^{-0.12t}z\left(t\right)=6250\left(\frac{1-e^{-0.12t}}{0.12}-t\right)$
故得 $\displaystyle z\left(t\right)=\frac{156250}3\left(e^{0.12t}-\frac3{25}te^{0.12t}-1\right)$。 - 那麼當 $y\left(t\right)=0$ 時有 $\displaystyle e^{0.12t}=\frac{49}{25}$，這表明 $\displaystyle t=\frac{50}3\ln\frac75$。從而
$\displaystyle z\left(\frac{50}3\ln\frac75\right)=\frac{156250}3\left(\frac{49}{25}-\frac3{25}\cdot\frac{49}{25}\cdot\frac{50}3\ln\frac75-1\right)=\frac{6250}3\left(24-98\ln\frac75\right)$
- ($6\%$) If $\displaystyle\lim_{x\to0}\left(x^{-3}\sin\left(3x\right)+rx^{-2}+s\right)=0$, what are the values of $r$ and $s$?
- ($10\%$) Find $\displaystyle\frac{d}{dx}\ln\sec5x$.
- ($10\%$) Evaluate the definite integral $\displaystyle\int_0^1\sqrt{4-x^2}xdx$.
- ($10\%$) Find the area of an ellipse by using the parametric equations $x=5\cos\theta$ and $y=6\sin\theta$.
- ($10\%$) Find the interval of convergence for the power series $\displaystyle x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots$.
- ($10\%$) Solve the differential equation $x^2dy+2xydx+2xdx+dy=0$.
訣竅
依據題意列出成本函數，隨後透過初等不等式求解即可。解法
設長寬與高分別為 $x,y,z$ 英尺，那麼限制條件為 $xyz=250000$，而成本函數為 $C\left(x,y,z\right)=40xy+20yz+20zx$。那麼運用算術幾何不等式可知$C\left(x,y,z\right)\geq3\sqrt[3]{40\cdot20\cdot20x^2y^2z^2}=60\sqrt[3]{2\cdot250000^2}=300000$
此時等號成立條件為 $40xy=20yz=20zx$，即 $2x=2y=z$，又因 $xyz=250000$，故得 $x=y=50\sqrt[3]4$、$z=100\sqrt[3]4$。訣竅
首先釐清積分區域並改寫為適當的形式，隨後演算該積分。解法
$\displaystyle\begin{aligned}\int_0^1\int_{2x-1}^{\sqrt{x}}2xydydx&=\int_0^1xy^2\Big|_{2x-1}^{\sqrt{x}}dx=\int_0^1\left[x^2-x\left(4x^2-4x+1\right)\right]dx\\&=\left.\left(\frac{x^3}3-x^4+\frac{4x^3}3-\frac{x^2}2\right)\right|_0^1=\frac16\end{aligned}$
訣竅
按照題意演算後獲得涉及參數的函數，運用求導獲得極小值。解法
設差平方的積分函數為$\displaystyle\begin{aligned}g\left(a\right)&=\int_0^1\left[f\left(x\right)-h\left(x\right)\right]^2dx=\int_0^1\left(e^{-x}-ax-ax^2\right)dx\\&=\int_0^1\left(e^{-2x}-2a\left(x+x^2\right)e^{-x}+a^2\left(x+x^2\right)^2\right)dx\end{aligned}$
求一階與二階導函數有$\displaystyle g'\left(a\right)=-2\int_0^1\left(x+x^2\right)e^{-x}dx+2a\int_0^1\left(x+x^2\right)^2dx$, $\displaystyle g''\left(a\right)=2\int_0^1\left(x+x^2\right)^2dx>0$
故函數 $g$ 在$\displaystyle\begin{aligned}a&=\frac{\displaystyle\int_0^1\left(x+x^2\right)e^{-x}dx}{\int_0^1\left(x+x^2\right)^2dx}=\frac{\displaystyle-e^{-x}\left(x+x^2\right)\Big|_0^1+\int_0^1\left(2x+1\right)e^{-x}dx}{\displaystyle\int_0^1\left(x^2+2x^3+x^4\right)dx}\\&=\frac{\displaystyle-2e^{-1}-e^{-x}\left(2x+1\right)\Big|_0^1+2\int_0^1e^{-x}dx}{\displaystyle\left.\frac{x^3}3+\frac{x^4}2+\frac{x^5}5\right|_0^1}=\frac{90-210e^{-1}}{31}\end{aligned}$
處有絕對極小值。訣竅
依據題意的描述設定對應的微分方程，透過代入消去化為其中一個未知函數來求解；根據獲得的函數先瞭解何時儲蓄帳戶耗盡金額，並將該時刻代入股票金額函數即可。解法
訣竅
利用經典的極限結果、極限的四則運算定理與羅必達法則即可求解。解法
運用四則運算定理可知$\displaystyle0=\lim_{x\to0}\left[x^2\cdot\left(x^{-3}\left(3x\right)+rx^{-2}+s\right)\right]=\lim_{x\to0}\left(\frac{\sin3x}x+r+sx^2\right)=r+3$
故得 $r=-3$。又同樣使用四則運算定理與羅必達法則便有$\displaystyle s=-\lim_{x\to0}\left(\frac{\sin3x}{x^3}-\frac3{x^2}\right)=\lim_{x\to0}\frac{3x-\sin3x}{x^3}=\lim_{x\to0}\frac{3-3\cos3x}{3x^2}=\lim_{x\to0}\frac{9\sin3x}{6x}=\frac92$
訣竅
直接使用連鎖律與基本導函數求解。解法
直接求導有$\displaystyle\frac{d}{dx}\ln\sec5x=\frac1{\sec5x}\cdot\sec5x\tan5x\cdot5=5\tan5x$
訣竅
運用變數代換法的概念直接計算求解。解法
直接計算有$\displaystyle\begin{aligned}\int_0^1\sqrt{4-x^2}xdx&=\frac12\int_0^1\left(4-x^2\right)^{1/2}dx^2=-\frac12\int_0^1\left(4-x^2\right)^{1/2}d\left(4-x^2\right)\\&=\left.-\frac13\left(4-x^2\right)^{3/2}\right|_0^1=\frac{8-3\sqrt3}3\end{aligned}$
訣竅
運用參數下的面積公式求解。解法
使用參數化的面積公式列式並透過二倍角公式改寫可知$\displaystyle A=2\int_{x=-5}^{x=5}y\left(\theta\right)dx\left(\theta\right)=2\int_0^{\pi}5\cdot6\sin^2\theta d\theta=30\int_0^{\pi}\left(1-\cos2\theta\right)d\theta=\left.30\left(\theta-\frac{\sin2\theta}2\right)\right|_0^{\pi}=30\pi$
訣竅
運用比值審歛法的概念求收斂半徑，隨後檢驗端點的歛散性。解法
給定的冪級數為 $\displaystyle\sum_{n=0}^{\infty}\left(-1\right)^n\frac{x^{2n+1}}{2n+1}$。那麼運用比值審歛法可知收斂的條件為$\displaystyle\lim_{n\to\infty}\left|\left(-1\right)^{n+1}\frac{x^{2n+3}}{2n+3}\div\left(-1\right)^n\frac{x^{2n+1}}{2n+1}\right|<1$
此即 $\left|x\right|^2<1$，因此收斂半徑為 $1$。現檢查端點可知當 $x=\pm1$ 時冪級數能寫為 $\displaystyle\pm\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{2n+1}$，那麼由交錯級數審歛法可知其收斂，因此收歛區間為 $\left[-1,1\right]$。訣竅
運用分離變量法求解即可。解法
移項整理有$\displaystyle\frac{dy}{y+1}=-\frac{2xdx}{x^2+1}$
兩邊同取不定積分便有 $\ln\left|y+1\right|=-\ln\left|x^2+1\right|+\ln C$，或寫為$\displaystyle y=-1+\frac{C}{x^2+1}$
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