※注意：請於試卷上「非選擇題作答區」內依序作答，並應註明作答之大題及其題號。
Any device with computer algebra system is prohibited during the exam. Solve the following problems. You need to write down your reasoning.
- Suppose that $f\left(x\right)$ satisfies the equation $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+xy-x^2y-xy^2$ for all $x,y\in R$ and $\displaystyle\lim_{x\to0}\frac{f\left(x\right)}x=2$.
- (6 pts) Find $f\left(0\right)$ and $f'\left(x\right)$.
- (6 pts) Sketch the graph of $f\left(x\right)$, indicating intervals of increasing/decreasing, and concavity.
- 取 $x=y=0$ 代入便有 $f\left(0\right)=2f\left(0\right)$，因此 $f\left(0\right)$。如此取 $y=h$ 並移項整理取極限（$h\to0$）有
$\displaystyle f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}h=\lim_{h\to0}\frac{f\left(h\right)+xh-x^2h-xh^2}h=\lim_{h\to0}\left[\frac{f\left(h\right)}h+x-x^2-xh\right]=2+x-x^2$.
- 解不等式 $f'\left(x\right)>0$，即 $x^2-x-2=\left(x-2\right)\left(x+1\right)<0$，可得 $-1<x<2$，故 $f$ 在 $\left(-1,2\right)$ 上遞增而在 $\left(-\infty,-1\right)\cup\left(2,\infty\right)$ 上遞減。又計算二階導函數有 $f''\left(x\right)=1-2x$，因此在 $\left(-\infty,1/2\right)$ 上凹口向上而在 $\left(1/2,\infty\right)$ 上凹口向下。結合以上資訊可繪製圖形如下
- A man runs twice as fast as he swims. He is at point $A$ on the edge of a circular pool with radius $20$ meters and he wants to get to the diametrically opposite point $B$ as quickly as possible. He can run around the edge to a point $C$ and then swim directly from $C$ to $B$.
- (10 pts) How should he choose the point $C$ to minimize the total time?
- (6 pts) If he runs $m$ times as fast as he swims, how will his best strategy be modified as $m$ varies ($m\geq1$)?
- 設圓池圓心為 $O$，設跑至點 $C$ 時 $\angle AOC=\theta$ 以及游泳速度為 $v$，那麼所需時間為
$\displaystyle f\left(\theta\right)=\frac{20\theta}{2v}+\frac{40\cos\left(\theta/2\right)}v$
其中 $\theta\in\left[0,\pi\right]$。據此求導有$\displaystyle f'\left(\theta\right)=\frac{10}v-\frac{20\sin\left(\theta/2\right)}v$
那麼解方程式 $f'\left(\theta\right)=0$ 有 $\displaystyle\theta=\frac\pi3$。又計算二階導函數有 $\displaystyle f''\left(\theta\right)=-\frac{10\cos\left(\theta/2\right)}v\leq0$，因此 $f$ 在 $\pi/3$ 處為局部極大值；且知 $f'$ 遞減，故 $f$ 在 $\pi/3$ 處為絕對極大值。又計算 $f\left(0\right)=40/v$ 且 $f\left(\pi\right)=10\pi/v$。因此直接沿著圓周跑到 $B$ 點是最節省時間的辦法，即取 $C=B$。 - 依據相同的道理，考慮函數 $f_m\left(\theta\right)$ 如下
$\displaystyle f_m\left(\theta\right)=\frac{20\theta}{mv}+\frac{40\cos\left(\theta/2\right)}v$
其中 $\theta\in\left[0,\pi\right]$。求導可得$\displaystyle f_m'\left(\theta\right)=\frac{20}{mv}-\frac{20\sin\left(\theta/2\right)}v$, $\displaystyle f_m''\left(\theta\right)=-\frac{10\cos\left(\theta/2\right)}v\leq0$.
這表示最小值必發生在邊界。容易知道 $\displaystyle f_m\left(0\right)=\frac{40}v$ 而 $\displaystyle f_m\left(\pi\right)=\frac{20\pi}{mv}$。那麼當 $\displaystyle m<\frac\pi2$ 時直接游泳直線過去會比較快，而當 $\displaystyle m>\frac\pi2$ 時直接沿著圓周跑過去比較快。而當 $\displaystyle m=\frac\pi2$ 時可沿圓周跑過去亦可游泳直線過去。 - Investigate the integral $\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx$.
- (7 pts) Show that the improper integral $\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx$ converges. Show that
$\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx=-\int_1^{\infty}\frac{\ln x}{1+x^2}dx$.
- (4 pts) Write $\displaystyle\frac1{1+x^2}$ as the sum of a power series $\displaystyle\sum a_nx^n$.
- (6 pts) Write $\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx=\int_0^1\big(\sum a_nx^n\big)\ln x\,dx$ as the sum of a series. Thus, we can use its partial sums to estimate the integral.
- (7 pts) Show that the improper integral $\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx$ converges. Show that
- 由於 $\displaystyle\frac1{1+x^2}\leq1$ 且 $\displaystyle\ln x\leq0$，因此有 $\displaystyle\frac{\ln x}{1+x^2}\geq\ln x$。容易發現
$\displaystyle\int_0^1\ln x\,dx=\lim_{t\to0^+}\int_t^1\ln x\,dx=\lim_{t\to0^+}\left[x\ln x-x\right]_t^1=\lim_{t\to0^+}\left(-1-t\ln t+t\right)=-1$.
因此藉由比較可知瑕積分 $\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx$ 收斂。令 $u=1/x$，那麼
- 當 $x\to0^+$ 時有 $u\to\infty$；
- 當 $x=1$ 時有 $u=1$；
- 函數 $\ln x=\ln\left(1/u\right)=-\ln u$，且求導有 $dx=-u^{-2}du$。
$\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx=\int_\infty^0\frac{-\ln u}{1+\left(1/u\right)^2}\cdot-\frac1{u^2}du=-\int_0^\infty\frac{\ln u}{1+u^2}du$.
最後透過啞變數變換（令 $u=x$）便完成證明。 - 視 $-x^2$ 為公比，那麼有
$\displaystyle\frac1{1+x^2}=\frac1{1-\left(-x^2\right)}=\sum_{n=0}^\infty\left(-x^2\right)^n=\sum_{n=0}^\infty\left(-1\right)^nx^{2n}$.
- 利用 (b) 的結果與分部積分法有
$\displaystyle\begin{aligned}\int_0^1\frac{\ln x}{1+x^2}dx&=\int_0^1\sum_{n=0}^\infty\left(-1\right)^nx^{2n}\ln x\,dx=\sum_{n=0}^\infty\left(-1\right)^n\int_0^1x^{2n}\ln x\,dx\\&=\sum_{n=0}^\infty\frac{\left(-1\right)^n}{2n+1}\left[x^{2n+1}\ln x\Big|_0^1-\int_0^1x^{2n}\,dx\right]\\&=\sum_{n=0}^\infty\frac{\left(-1\right)^{n+1}}{\left(2n+1\right)^2}.\end{aligned}$
如此對於給定的誤差 $\varepsilon>0$，我們可取正整數 $\displaystyle N>\frac1{2\sqrt{\varepsilon}}$ 則知級數 $\displaystyle\sum_{n=0}^N\frac{\left(-1\right)^{n+1}}{\left(2n+1\right)^2}$ 與瑕積分 $\displaystyle\int_0^1\frac{\ln x}{1+x^2}dx$ 之差不超過 $\varepsilon$。 - (10 pts) Find the twice differentiable function $f\left(x\right)$ such that
$\displaystyle f'\left(x\right)=\int_0^x\sqrt{1+\left(f'\left(u\right)\right)^2}\,du$, $f\left(0\right)=2$.
- (5 pts) Let $\displaystyle f\left(x,y,z\right)=\int_{x^z}^{\sqrt y}\sin\left(t^2\right)\,dt$. Find $\nabla f$, the gradient of $f$.
- (5 pts) $\displaystyle f\left(x,y\right)=\frac{\sin\left(xy^2\right)}{x^2+y^2}$ for $\left(x,y\right)\neq\left(0,0\right)$ and $f\left(0,0\right)=0$. Compute the directional derivative of $f$ along ${\bf u}=\left(\cos\theta,\sin\theta\right)$ at $\left(0,0\right)$.
- 按照梯度的定義並熟用微積分基本定理及連鎖律可得
$\displaystyle\nabla f\left(x,y,z\right)=\left(f_x\left(x,y,z\right),f_y\left(x,y,z\right),f_z\left(x,y,z\right)\right)=\left(-\sin\left(x^{2z}\right)\cdot zx^{z-1},\sin\left(y\right)\cdot\frac1{2\sqrt y},-\sin\left(x^{2z}\right)\cdot x^z\ln x\right)$.
- 由方向導數的定義可知
$\displaystyle D_{\bf u}f\left(0,0\right)=\lim_{h\to0}\frac{f\left(h\cos\theta,h\sin\theta\right)-f\left(0,0\right)}h=\lim_{h\to0}\frac{\sin\left(h^3\cos\theta\sin^2\theta\right)}{h^3}=\cos\theta\sin^2\theta$.
- (10 pts) Find the critical points of $f\left(x,y\right)$, where $z=f\left(x,y\right)$ satisfies the equation $yz+x\ln y=z^2$. Are these critical points local maximum, local minimum, or saddle points?
- 對於臨界點 $\left(x,y,z\right)=\left(0,1,0\right)$ 有 $f_{xx}\left(0,1\right)=0$、$f_{xy}\left(0,1\right)=-1$、$f_{yy}\left(0,1\right)=0$，如此其判別式
$D=f_{xx}\left(0,1\right)f_{yy}\left(0,1\right)-f_{xy}^2\left(0,1\right)=-1<0$
因此臨界點 $\left(0,1,0\right)$ 為鞍點。 - 對於臨界點 $\left(x,y,z\right)=\left(-1,1,1\right)$ 有 $f_{xx}\left(-1,1\right)=0$、$f_{xy}\left(-1,1\right)=1$、$f_{yy}\left(-1,1\right)=1$，如此其判別式為
$D=f_{xx}\left(-1,1\right)f_{yy}\left(-1,1\right)-f_{xy}^2\left(-1,1\right)=-1<0$
因此臨界點 $\left(-1,1,1\right)$ 為鞍點。 - Evaluate the following multiple integrals.
- (7 pts) $\displaystyle\int_0^{\frac{\sqrt3}2}\int_{\sqrt{1-y^2}}^{\sqrt{4-y^2}}e^{x^2+y^2}\,dxdy+\int_{\frac{\sqrt3}2}^{\sqrt3}\int_{\frac{y}{\sqrt3}}^{\sqrt{4-y^2}}e^{x^2+y^2}\,dxdy$.
- (8 pts) $\displaystyle\iiint_E\frac1{1+z}\,dV$, where $E=\left\{\left(x,y,z\right)\mid\,x^2+y^2+z^2\leq1,\,x\geq0,\,y\geq0,\,z\geq0\right\}$.
- 將兩項重積分區域繪圖如下令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，那麼由圖形所指示可知變數範圍為 $\left\{\begin{aligned}&1\leq r\leq2\\&0\leq\theta\leq\frac\pi3\end{aligned}\right.$，如此所求能表達並計算如下
$\displaystyle\int_0^{\frac\pi3}\int_1^2e^{r^2}r\,dr\,d\theta=\frac\pi3\int_1^2re^{r^2}dr=\left.\frac\pi6e^{r^2}\right|_1^2=\frac{\pi\left(e^4-e\right)}6.$
- 令 $\left\{\begin{aligned}&x=\rho\cos\theta\sin\phi\\&y=\rho\sin\theta\sin\phi\\&z=\rho\cos\phi\end{aligned}\right.$，其中變數範圍為 $\left\{\begin{aligned}&0\leq\rho\leq1\\&0\leq\theta\leq\frac\pi2\\&0\leq\phi\leq\frac\pi2\end{aligned}\right.$，如此所求的三重積分可改寫並計算如下
$\displaystyle\begin{aligned}\iiint_E\frac1{1+z}\,dV&=\int_0^{\frac\pi2}\int_0^{\frac\pi2}\int_0^1\frac1{1+\rho\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\&=\frac\pi2\int_0^1\int_0^{\frac\pi2}\frac{\rho^2\sin\phi}{1+\rho\cos\phi}d\phi\,d\rho\\&=-\frac\pi2\int_0^1\rho\ln\left(1+\rho\cos\phi\right)\Big|_{\phi=0}^{\phi=\frac\pi2}d\rho\\&=\frac\pi2\int_0^1\rho\ln\left(1+\rho\right)d\rho\\&=\frac\pi4\left(\rho^2\ln\left(1+\rho\right)\Big|_0^1-\int_0^1\frac{\rho^2}{1+\rho}d\rho\right)\\&=\frac\pi4\left[\ln2-\int_0^1\left(\rho-1+\frac1{1+\rho}\right)d\rho\right]\\&=\frac\pi4\left[\ln2-\left(\frac{\rho^2}2-\rho+\ln\left(1+\rho\right)\right)\Big|_0^1\right]=\frac\pi8.\end{aligned}$
- (10 pts) Let $S$ be the part of the cylinder $x^2+y^2=2y$ that lies in the sphere $x^2+y^2+z^2=4$ and inside the first quadrant. Compute $\displaystyle\iint_Sz\,dS$.
訣竅
對於第一小題直接按定義計算即可；第二小題則運用導函數的資訊求解。解法
訣竅
利用圓心角來描述點的位置，從而建立關於時間的函數並利用導函數找出最小值。解法
訣竅
首先藉由比較審歛法證明收斂性，隨後運用變數代換法證明該等式；之後由經典的無窮等比級數表達之，最後利用常見的分部積分法表達其總和並給出誤差估計。解法
訣竅
利用微積分基本定理求導後獲得二階微分方程，並利用分離變量法求解即可。解法
取 $x=0$ 代入有 $f'\left(0\right)=0$。隨後求導可得$\displaystyle f''\left(x\right)=\sqrt{1+\left(f'\left(x\right)\right)^2}$, $f\left(0\right)=2$, $f'\left(0\right)=0$.
先記 $g\left(x\right)=f'\left(x\right)$，那麼移項便有$\displaystyle\frac{dg}{\sqrt{1+g^2}}=dx$
同取積分並利用 $g\left(0\right)=f'\left(0\right)=0$ 便得$\ln\left|g+\sqrt{1+g^2}\right|=x$
或寫為 $g+\sqrt{1+g^2}=e^x$。藉由同乘以 $g-\sqrt{1+g^2}$ 可導出 $g-\sqrt{1+g^2}=-e^{-x}$，故兩式相加除以二可得$\displaystyle f'\left(x\right)=g\left(x\right)=\frac{e^x-e^{-x}}2$
最後再取一次積分並使用 $f\left(0\right)=2$ 得$\displaystyle f\left(x\right)=\frac{e^x+e^{-x}}2+1=\frac{\left(e^{x/2}+e^{-x/2}\right)^2}2=\frac12\cosh^2\left(\frac{x}2\right)$.
訣竅
按梯度與方向導數的定義計算即可。解法
訣竅
運用隱函數微分計算一階與二階偏導函數，並利用二階判別式檢驗臨界點的性質。解法
首先將給定的方程式對 $x$ 與 $y$ 求偏導可得$yz_x+\ln y=2zz_x$, $\displaystyle z+yz_y+\frac{x}y=2zz_y$.
若 $\left(x_0,y_0\right)$ 為臨界點且記 $z_0:=f\left(x_0,y_0\right)$，那麼有 $z_x=f_x\left(x_0,y_0\right)=0$ 且 $z_y=f\left(x_0,y_0\right)=0$，則有聯立方程為$\left\{\begin{aligned}&\ln y_0=0,\\&z_0+\frac{x_0}{y_0}=0,\\&y_0z_0+x_0\ln y_0=z_0^2.\end{aligned}\right.$
由第一式可知 $y_0=1$，那麼代入第三式便有 $z_0=z_0^2$，即得 $z_0=0$ 或 $z_0=1$。這樣對於第二式則可獲得兩個臨界點為 $\left(x_0,y_0,z_0\right)=\left(0,1,0\right)$ 或 $\left(x_0,y_0,z_0\right)=\left(-1,1,1\right)$。進一步，計算二階偏導函數有
$yz_{xx}=2z_x^2+2zz_{xx}$, $\displaystyle z_x+yz_{xy}+\frac1y=2z_xz_y+2zz_{xy}$, $\displaystyle 2z_y+yz_{yy}-\frac{x}{y^2}=2z_y^2+2zz_{yy}$.
那麼訣竅
分別運用極座標變換與球面座標變換求解即可。解法
訣竅
將曲面參數化後代入計算即可。解法
將 $S$ 參數化如下${\bf u}\left(r,\theta\right)=\left(r\cos\theta,r\sin\theta,\sqrt{4-r^2}\right)$
其中參數 $r$ 與 $\theta$ 的範圍為 $\left\{\begin{aligned}&0\leq r\leq2\sin\theta\\&0\leq\theta\leq\frac\pi2\end{aligned}\right.$。再者有$\displaystyle\begin{aligned}dS&=\left|{\bf u}_r\times{\bf u}_\theta\right|dr\,d\theta=\left|\left(\cos\theta,\sin\theta,-\frac{r}{\sqrt{4-r^2}}\right)\times\left(-r\sin\theta,r\cos\theta,0\right)\right|dr\,d\theta\\&=\left|\left(\frac{r^2\cos\theta}{\sqrt{4-r^2}},\frac{r^2\sin\theta}{\sqrt{4-r^2}},r\right)\right|dr\,d\theta=\frac2{\sqrt{4-r^2}}dr\,d\theta.\end{aligned}$
據此所求的曲面積分可表達並計算如下$\displaystyle\iint_Sz\,dS=\int_0^{\frac\pi2}\int_0^{2\sin\theta}\sqrt{4-r^2}\cdot\frac2{\sqrt{4-r^2}}dr\,d\theta=\int_0^{\frac\pi2}4\sin\theta\,d\theta=-4\cos\theta\Big|_0^{\frac\pi2}=4.$
沒有留言:
張貼留言