這張考卷的題目不少，需要很熟練的技巧，像散度定理遇到有缺口的表面要如何處理…等。如果你的目標科系在台大理工學院，務必要精熟微積分全部內容與微分方程，練習時可以留意題目的作答順序與時間分配。
本解答由張旭老師及南享老師共同完成。
建議使用電腦瀏覽，手機或平板的算式可能會超出畫面。
Part I 選擇題 Multiple Choice
第一題 (5 分)
Let $f$ be a smooth function on $\mathbb{R}$
(a) If $f^{\prime\prime}>0$ on $\mathbb{R}$, then $f’$ must be increasing on $\mathbb{R}$.
(b) If $f^{\prime\prime}>0$ on $\mathbb{R}$, then $f$ must be concave upward on $\mathbb{R}$.
(c) If $f^{\prime\prime}(a)=0$, then the point $(a,f(a))$ must be an inflection point.
Among the above three statements, how many of them are true ?
A) Only one B) Only two C) All of them D) None of them
提示
這題考一次微分檢驗法、二次微分檢驗法以及反曲點的觀念，特別要注意的是反曲點的觀念，$f^{\prime\prime}(a)=0$ 並無法保證 $(a,f(a))$ 為反曲點。
第二題 (5 分)
A function $f$ on $\mathbb{R}$ is odd if $f(-x)=-f(x)$ for all $x\in\mathbb{R}$ while it is even if $f(-x)=f(x)$ for all $x\in\mathbb{R}$. Suppose that $g$ and $h$ are two functions on $\mathbb{R}.$
(a) If $g$ is an even function $h\circ g$ must be even.
(b) If $g$ is an odd function and $\mathop {\lim }\limits_{x \to 0^ + } g(x) = 3$, then the limit $\mathop {\lim }\limits_{x \to 0 } g(x) $ must be $3$.
(c) If $g$ is an even function and $\mathop {\lim }\limits_{x \to 0^ + } g(x) = 3$, then the limit $\mathop {\lim }\limits_{x \to 0 } g(x)$ must be $3$.
Among the above three statements, how many of them are true ?
A) Only one B) Only two C) All of them D) None of them
提示
奇函數的特性會將負號放出來，而偶函數會把負號吸收，由這兩種性質即可快速做出判斷。
解答
(a) $\because (h\circ g)(-x)=h(g(-x))=h(h(x))=(h\circ g)(x)$
$\therefore h\circ g$ is even.
(b) Let $x=-t,\ \mathop {\lim }\limits_{x \to 0^ – } g(x) = \mathop {\lim }\limits_{t \to 0^ + } g( – t) = \mathop {\lim }\limits_{t \to 0^ + } ( – g(t)) = – 3$
(c) Let $x=-t,\ \mathop {\lim }\limits_{x \to 0^ – } g(x) = \mathop {\lim }\limits_{t \to 0^ + } g( – t) = \mathop {\lim }\limits_{t \to 0^ + } g(t) = 3$
有三個敘述正確，因此選 C) ∎
此題考點：極限的直觀定義
第三題 (5 分)
Consider the integral $I_{p,r} : =\iint_D \frac{1}{{(x^2 + y^2 )^{p/2} }}dA$, where $D$ is the region bounded by two concentric circles centred at the origin with radii $r$ and $1$ respectively, $0<r<1$. Let $J_p : = \mathop {\lim }\limits_{r \to 0^ + } I_{p,r}$
(a) $J_p$ is convergent if $p<1$
(b) $J_p$ is convergent if $1<p<2$.
(c) $J_p$ is convergent if $2<p$.
Among the above three statements, how many of them are true ?
A) Only one B) Only two C) All of them D) None of them
提示
本題看起來是二重積分，稍微推導後就變成分母為指數型的積分，這時發現形式和P級數有關。
解答
$\begin{array}{cl} 1^{\circ}\ \iint_D\frac{1}{{(x^2 + y^2 )^{\frac{p}{2}} }}dA & =\iint_D\frac{1}{{\rho ^p }}\cdot\rho d\rho d\theta\\ &=\int_0^{2\pi } {\int_r^1 {\frac{1}{{\rho ^{p – 1} }}} } d\rho d\theta \\ & =2\pi \int_r^1 {\frac{1}{{\rho ^{p – 1} }}} d\rho \end{array}$
$\Rightarrow J_p = \mathop {\lim }\limits_{r \to 0^ + } I_{p,r} = \mathop {\lim }\limits_{r \to 0^ + } 2\pi \int_r^1 {\frac{1}{{\rho ^{p – 1} }}} d\rho ( = 2\pi \int_0^1 {\frac{1}{{\rho ^{p – 1} }}} d\rho )$.
$2^{\circ}\ \int_0^1 {\frac{1}{{\rho ^{p – 1} }}} d\rho \left\{ {\begin{array}{cc} \text{convergent,} & {p – 1 < 1} \\ {\infty ,} & {p – 1 = 1} \\ {\infty ,} & {p – 1 > 1} \end{array}} \right.$
$\Rightarrow \int_0^1 {\frac{1}{{\rho ^{p – 1} }}} d\rho$ convergent $\Leftrightarrow p < 2$
$\Rightarrow J_p = \mathop {\lim }\limits_{r \to 0^ + } I_{p,r} = 2\pi \int_0^1 {\frac{1}{{\rho ^{p – 1} }}} d\rho$ convergent $\Leftrightarrow p < 2$
$3^{\circ}$ 由$1^{\circ},\ 2^{\circ}$ 得知 (a), (b) 正確，因此選 B) ∎
第四題 (5 分)
Let $f(x,y) = \left\{ {\begin{array}{cc} {\frac{{x\sin y^2 }}{{1 – e^{x^2 + y^4 } }},} & {{\text{if }}(x,y) \ne (0,0)} \\ {0,} & {{\text{if }}(x,y) = (0,0)} \end{array}} \right.$
(a) $\mathop {\lim }\limits_{x \to 0} \mathop {\lim }\limits_{y \to 0} f(x,y) = 0$
(b) $\mathop {\lim }\limits_{x \to 0} \mathop {\lim }\limits_{y \to 0} f(x,y) = \mathop {\lim }\limits_{y \to 0} \mathop {\lim }\limits_{x \to 0} f(x,y)$.
(c) $\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = 0$.
Among the above three statements, how many of them are true ?
A) Only one B) Only two C) All of them D) None of them
提示
本題考的是多變量函數求極限的過程。求二變數的極限和逼近路徑有關，要注意並非沿 $x$ 軸、$y$ 軸得到的極限存在且相等，就代表求出極限了，可以由函數 $x,\ y$ 的次方關係，討論路徑 $y^m=x^n$。
解答
(a) $\mathop {\lim }\limits_{x \to 0} \mathop {\lim }\limits_{y \to 0} \frac{{x\sin y^2 }}{{1 – e^{x^2 + y^4 } }} = \mathop {\lim }\limits_{x \to 0} \frac{0}{{1 – e^{x^2 } }} = 0$
(b) $\mathop {\lim }\limits_{y \to 0} \mathop {\lim }\limits_{x \to 0} \frac{{x\sin y^2 }}{{1 – e^{x^2 + y^4 } }} = \mathop {\lim }\limits_{y \to 0} \frac{0}{{1 – e^{y^4 } }} = 0$.
(c) Along $y^2=x$
$\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = \mathop {\lim }\limits_{x \to 0} \frac{{x\sin x}}{{1 – e^{2x^2 } }}\mathop = \limits^{\text{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\sin x + x\cos x}}{{ – e^{2x^2 } \cdot 4x}} = \frac{1}{{ – 1}}(\frac{1}{4} + \frac{1}{4}) = – \frac{1}{2}$
$\Rightarrow \mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y)$ D.N.E.
有兩個敘述正確，因此選 B) ∎
附註：$\mathop = \limits^{\rm{L}}$ 表示此等號使用羅必達法則 (L’Hôpital’s rule)
此題考點：二變數函數的極限
第五題 (5 分)
(a) The series $\sum\limits_{n = 1}^\infty {n\cos (n\pi )\sin \frac{1}{n}}$ is absolutely convergent.
(b) The series $\sum\limits_{n = 1}^\infty {\sin (n)\frac{{n^2 + 1}}{{5^n }}}$ is absolutely convergent.
(c) The series $\sum\limits_{n = 2}^\infty {\frac{{( – 1)^n }}{{n(\ln n)^{\sqrt 2 – 1} }}}$ is absolutely convergent.
Among the above three statements, how many of them are true ?
A) Only one B) Only two C) All of them D) None of them
提示
這一題很明顯的是考級數的斂散性問題，要立刻想到八大審斂法。
解答
(a) $\sum\limits_{n = 1}^\infty {\left| {n\underbrace {\cos (n\pi )}_{( – 1)^n }\sin \frac{1}{n}}\right|} = \sum\limits_{n = 1}^\infty {\left| {n\sin \frac{1}{n}} \right|}$
$\because \mathop {\lim }\limits_{n \to \infty } n\sin \frac{1}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{t \to 0^ + } \frac{{\sin t}}{t} = 1$ (Let $t=\frac{1}{n}$)
$\therefore \mathop {\lim }\limits_{n \to \infty } \left| {n\sin \frac{1}{n}} \right| = 1 \ne 0$
$\Rightarrow \sum\limits_{n = 1}^\infty {\left| {n\cos (n\pi )\sin \frac{1}{n}} \right|}$ divergent
(b) ($\mathop {\lim }\limits_{n \to \infty } \sin n \cdot \frac{{n^2 + 1}}{{5^n }} = 0 \Rightarrow$ 無法推得斂散性)
$\sum\limits_{n = 1}^\infty {\left| {\sin (n)\frac{{n^2 + 1}}{{5^n }}} \right|} \le \sum\limits_{n = 1}^\infty {\left| {\sin (n)} \right|} \frac{{n^2 + 1}}{{5^n }} \le \sum\limits_{n = 1}^\infty {\frac{{n^2 + 1}}{{5^n }}}$
$\because \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{(n + 1)^2 + 1}}{{5^{n + 1} }}}}{{\frac{{n^2 + 1}}{{5^n }}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n^2 + 2n + 2}}{{5(n^2 + 1)}} = \frac{1}{5} < 1$
$\therefore \sum\limits_{n = 1}^\infty {\frac{{n^2 + 1}}{{5^n }}}$ convergent. $\Rightarrow \sum\limits_{n = 1}^\infty {\left| {\sin (n)\frac{{n^2 + 1}}{{5^n }}} \right|}$ convergent.
(c) $\sum\limits_{n = 2}^\infty {\left| {\frac{{( – 1)^n }}{{n(\ln n)^{\sqrt 2 – 1} }}} \right|} = \sum\limits_{n = 2}^\infty {\frac{1}{{n(\ln n)^{\sqrt 2 – 1} }}}$
$\because \int_2^\infty {\frac{1}{{x(\ln x)^{\sqrt 2 – 1} }}} dx = \int_{\ln 2}^\infty {\frac{1}{{t^{\sqrt 2 – 1} }}} dt$ divergent (p-series)
$\therefore \sum\limits_{n = 2}^\infty {\frac{{( – 1)^n }}{{n(\ln n)^{\sqrt 2 – 1} }}}$ divergent.
有一個敘述正確，因此選 A) ∎
Part II 填充題 Fill in the blanks
第六題 (5 分)
$\mathop {\lim }\limits_{n \to \infty } \frac{{(1^2 + 2^2 + \cdots + n^2 )}}{{(\sqrt 1 + \sqrt 2 + \cdots + \sqrt n )^2 }} = ?$
提示
這題極限題，只是分子分母都是級數因此想到黎曼和，首先把這兩個級數化成黎曼和的型式，也就是整理出 $\frac{1}{n}$ 與 $\frac{k}{n}$ ，再搭配一些極限運算性質就是基本的積分。
解答
$1^{\circ}$ 原式$= \mathop {\lim }\limits_{n \to \infty } \frac{{(\frac{1}{n})^2 + (\frac{2}{n})^2 + \cdots + (\frac{n}{n})^2 }}{{\frac{1}{n}(\sqrt {\frac{1}{n}} + \sqrt {\frac{2}{n}} + \cdots + \sqrt {\frac{n}{n}} )^2 }}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{n}\sum\limits_{k = 1}^n {(\frac{k}{n})^2 } }}{{\frac{1}{{n^2 }}(\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} } )^2 }}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{n}\sum\limits_{k = 1}^n {(\frac{k}{n})^2 } }}{{(\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} } )^2 }}$
$= \frac{{\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {(\frac{k}{n})^2 } }}{{\mathop {\lim }\limits_{n \to \infty } (\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} } )^2 }} = \frac{{\int_0^1 {x^2 } dx}}{{(\int_0^1 {\sqrt x } dx)^2 }} = \frac{{\left. {\frac{1}{3}x} \right|_{x = 0}^{x = 1} }}{{\left[ {\left.{\frac{{x^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{x = 0}^{x = 1} } \right]^2 }} = \frac{{\frac{1}{3}}}{{(\frac{2}{3})^2 }} = \frac{3}{4}$ ∎
此題考點：定積分的直觀觀念
第七題 (5 分)
Let $f(x) = \int_0^{\frac{\pi }{2} + 2\sin x} {\sin (y + x\cos y)} dy$. It follow that $f'(0)=$?
提示
這一題要算積分型式函數對 $x$ 微分，而且上下界有未知數，但原本的被積函數也包含 $x$ ，不太像微積分基本定理，而是萊布尼茲微分公式。
解答
$1^{\circ}$ 萊布尼茲微分公式：$\frac{d}{{dx}}\int_a^{g(x)} {f(x,t)} dt$
$= f(x,g(x))g'(x) + \int_a^{g(x)} {\frac{\partial }{{\partial x}}f(x,t)} dt$
$\begin{array}{cl} f'(x)= & \sin (\frac{\pi }{2} + 2\sin x + x\cos (\frac{\pi }{2} + 2\sin x)) \cdot 2\cos x \\ & +\int_0^{\frac{\pi }{2} + 2\sin x} {\cos (y + x\cos y)\cos y} dy \end{array}$
$\begin{array}{cl} \Rightarrow f'(0) & =\sin (\frac{\pi }{2}) \cdot 2\cos (0) + \int_0^{\frac{\pi }{2}} {\cos ^2 y} dy \\ & =2 + \int_0^{\frac{\pi }{2}} {\frac{{1 + \cos 2y}}{2}} dy \\ & =2 + \frac{1}{2}\left[ {y + \frac{{\sin y}}{2}} \right]_{y = 0}^{y = \frac{\pi }{2}} = 2 + \frac{\pi }{4} ∎\end{array}$
此題考點：萊布尼茲微分公式、微積分基本定理 II
第八題 (5 分)
Let $f(x) = \arctan (x^2 ) + \frac{{x^2 – 1}}{{x^2 + 1}}$. The graph of $f$ has a horizontal asymptote respresented by the equation and the global minimum value of $f$ is .
提示
這題要求 $f(x)$ 的水平漸近線與全域最小值。水平漸近線分兩種，一種是當 $x\to\infty$ ，另一種是 $x\to-\infty$ 我們要看的就是在這兩種狀態時，函數圖形是不是會越來越靠近某個水平線。而全域最小值並不是只算出微分等於0的點就可以了，只能知道它們是局部極值的候選人，因此算完還要再判斷。
解答
$1^{\circ}$ 萊布尼茲微分公式：$\frac{d}{{dx}}\int_a^{g(x)} {f(x,t)} dt$
$= f(x,g(x))g'(x) + \int_a^{g(x)} {\frac{\partial }{{\partial x}}f(x,t)} dt$
$\begin{array}{cl} f'(x)= & \sin (\frac{\pi }{2} + 2\sin x + x\cos (\frac{\pi }{2} + 2\sin x)) \cdot 2\cos x \\ & +\int_0^{\frac{\pi }{2} + 2\sin x} {\cos (y + x\cos y)\cos y} dy \end{array}$
$\begin{array}{cl} \Rightarrow f'(0) & =\sin (\frac{\pi }{2}) \cdot 2\cos (0) + \int_0^{\frac{\pi }{2}} {\cos ^2 y} dy \\ & =2 + \int_0^{\frac{\pi }{2}} {\frac{{1 + \cos 2y}}{2}} dy \\ & =2 + \frac{1}{2}\left[ {y + \frac{{\sin y}}{2}} \right]_{y = 0}^{y = \frac{\pi }{2}} = 2 + \frac{\pi }{4} ∎\end{array}$
此題考點：萊布尼茲微分公式、微積分基本定理 II
第九題 (5 分)
$\int_0^1 {\frac{1}{{(x^2 + 1)^2 }}} dx =$?
提示
單變數的積分馬上想到四大積分法，看到被積函數的分子是1，分母可以看成兩個平方和，因此可以快速篩選出三角置換法。
解答
$1^{\circ}$ 令 $x = \tan u \Rightarrow dx = \sec ^2 udu$
$2^{\circ}$ 原式$= \int_0^{\frac{\pi }{4}} {\frac{1}{{(\sec u)^4 }} \cdot \sec ^2 u} du$
$= \int_0^{\frac{\pi }{4}} {\cos ^2 u} du$
$= \int_0^{\frac{\pi }{4}} {\frac{{1 + \cos u}}{2}} du$
$= \frac{1}{2}\left[ {u + \frac{{\sin 2u}}{2}} \right]_{u = 0}^{u = \frac{\pi }{4}} = \frac{1}{2}(\frac{\pi }{4} + \frac{1}{2}) = \frac{\pi }{8} + \frac{1}{4} ∎$
此題考點：三角置換法
第十題 (10 分，每格 5 分)
The third term of the Maclaurin series (i.e. the Taylor series centred at $x=0$) of $\arctan(3x)$ is (note that the answer should be monomial in $x$ ; the term of $x^0$ is counted as the 0-th term). The radius of convergence of the series is ?
提示
這題要求 $\arcsin 3x$ 的馬克勞林級數及其收斂區間，也就是要求在 $x=0$ 的泰勒展開式。$\arcsin 3x$ 的處理方式要有一點技巧，我們先將它微分後可以得到 $3(1-9x^2)^{-\frac{1}{2}}$，這種類型就可以用二項式定理的推廣版處理，之後小心計算寫到 $x^3$ 項即可完成前半部分，後半部分要求收斂半徑就使用公式，依題型可以使用根值或比值審斂法。
解答
$1^{\circ}\ \left[ {\arcsin (3x)} \right]^\prime=\frac{3}{{\sqrt {1 – (3x)^2 } }} = 3(1 – 9x^2 )^{ – \frac{1}{2}}$
$=3\left[ {1^{ – \frac{1}{2}} ( – 9x^2 )^0 + ( – \frac{1}{2}) \cdot 1^{ – \frac{3}{2}} ( – 9x^2 )^1 + \frac{{( – \frac{1}{2})( – \frac{3}{2})}}{{2!}}1^{ – \frac{5}{2}} ( – 9x^2 )^2 + \cdots } \right] $
$=3\left[ {\sum\limits_{k = 1}^\infty {\frac{{( – \frac{1}{2})( – \frac{3}{2})( – \frac{5}{2}) \cdots ( – \frac{{2k – 1}}{2})}}{{k!}}\underbrace {( – 9x^2 )^k }_{( – 1)^k 3^{2k} x^{2k} }} + 1} \right]$
$\begin{array}{cl} \Rightarrow (\arcsin (3x))’ & =3\left[ {\sum\limits_{k = 1}^\infty {\frac{{( – 1)^k (2k)!}}{{(k!2^k )^2 }}( – 1)^k 3^{2k} x^{2k} } + 1} \right] \\ & =3 + \sum\limits_{k = 1}^\infty {\frac{{(2k)!3^{2k + 1} }}{{(k!2^k )^2 }}} x^{2k} \end{array}$
$\begin{array}{cl} \Rightarrow \arcsin(3x) & =3x + \sum\limits_{k = 1}^\infty {\frac{{(2k)!3^{2k + 1} }}{{(k!)}}} \cdot \frac{{x^{2k + 1} }}{{2k + 1}} + \underbrace {\arcsin (3 \cdot 0)}_0 \\ & =3x + \frac{{2! \times 3^3 }}{{(1! \times 2)^2 }} \cdot \frac{{x^3 }}{3} + \cdots\\&=3x + \frac{9}{2}x^3 + \cdots ∎\end{array}$
$\begin{array}{cl} 2^{\circ}\ a_k & =\frac{{(2k)! \times 3^{2k + 1} }}{{(k! \times 2^k )^2 }} \cdot \frac{1}{{2k + 1}} \\ & =\Rightarrow \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{\frac{{(2k + 2)! \times 3^{2k + 3} }}{{((k + 1)! \times 2^{k + 1} )^2 }}}}{{\frac{{(2k)! \times 3^{2k + 1} }}{{(k! \times 2^k )^2 }} \cdot \frac{{x^{2k + 1} }}{{2k + 1}}}}} \right| = x^2 \mathop {\lim }\limits_{k \to \infty } \frac{{(2k + 1)(2k + 2)}}{{(k + 1)^2 \cdot 2^2 }} \cdot \frac{{2k + 1}}{{2k + 3}} \\ & =x^2 \mathop {\lim }\limits_{k \to \infty } \frac{{(2k + 1)(2k + 2) \cdot 3^2 }}{{(k + 1)^2 \cdot 2^2 }} \cdot \frac{{2k + 1}}{{2k + 3}} = 9x^2 < 1\\&=\Rightarrow x^2 < \frac{1}{9} \Rightarrow – \frac{1}{3} < x < \frac{1}{3} \end{array}$
收斂半徑 $r=\frac{1}{3} ∎$
第十一題 (5 分)
If $y=y(x)$ satisfies the differential equation $x^2y’+xy=2\ln x$ for $x>0$ with $y(1)=2$, then $y(x)=$?
提示
本題是常微分方程，如果看到題目可以整理成 $y’+P(x)y+Q(x)$ ，則可以想到公式解，也就是利用 $P(x)$ 求出積分因子 $I(x)$ ，之後可以將題目化成 $\left[ {I(x)y} \right]^\prime = Q(x)$ 。
解答
$1^{\circ}$ 原式$\Rightarrow y’ + \frac{3}{x}y = \frac{{2\ln x}}{{x^2 }} = Q(x)$
令 $u(x) = e^{\int {\frac{3}{x}} dx} = e^{3\ln \left| x \right| + C}$，取 $u(x) = e^{3\ln x} = x^3$
$2^{\circ}\ \underbrace {x^3 y’ + 3x^2 y}_{(x^3 y)’} = 2x\ln x$
$\begin{array}{cl} \Rightarrow x^3 y = \int {2x\ln x} dx & =x^2 \ln x – \int x dx \\ & =x^2 \ln x – \frac{x}{2} + C\end{array}$
$\Rightarrow y = \frac{{\ln x}}{x} – \frac{1}{{2x}} + \frac{C}{{x^3 }}$
$3^{\prime}\ \because y(1)=2$
$\therefore \frac{1}{2} + C = 2 \Rightarrow C = \frac{5}{2}$
$\Rightarrow y = \frac{{\ln x}}{x} – \frac{1}{{2x}} + \frac{5}{{2x^3 }} ∎$
此題考點：微分方程：公式解
第十二題 (15 分，每格5分)
Let $C$ be a variable path in the $xy$-plane of arc-length 1 starting at the point $(\sqrt{3},1)$ and ending at the point $(a,b)$. Suppose that $G(a,b): = \int_C {\nabla f} \cdot dr$, where $f(x,y): = \arctan \frac{y}{x}$, is a function $a$ and $b$. Then, $G$ attains its maximum at $a=$ and $b=$ , and the maximum value of $G$ is .
提示
要找出起始點是 $(\sqrt{3},1)$ 且路徑長度是 $1$ 時線積分的最大值與發生的點， $G(a,b)$ 定義成 $\nabla f$ 在這段路徑上的線積分，如果你熟悉線積分可以發現 $f(x,y)$ 其實就是這個線積分的potential function，如此一來這個線積分的結果就是 $f(x,y)$ 代入起始點減掉終點，這道題瞬間就變簡單了，長度是 1 的路徑必定在以 $(\sqrt{3},1)$ 為圓心，半徑為 1 的一個圓內，因此這個極值問題就可以畫圖來解決。
解答
$1^{\circ}$ 依提示作圖，則 $\arctan^{-1}\frac{b}{a}$ 的最大值
為 $\tan ^{ – 1} \frac{{\frac{3}{2}}}{{\frac{{\sqrt 3 }}{2}}} – \tan ^{ – 1} \frac{1}{{\sqrt 3 }} = \frac{\pi }{3} – \frac{\pi }{6} = \frac{\pi }{6} ∎$
此題考點：線積分的微積分基本定理
第十三題 (5 分)
If $a$ and $b$ are positive constants and if $\max\{p,q\}$ denotes the maximum between the numbers $p$ and $q$, the iterated integral $\int_0^a {\int_0^b {e^{\max \left\{ {b^2 x^2 ,a^2 y^2 } \right\}} } dy} dx =$ .
提示
這一題二重積分很明顯跟以往不太一樣，被積函數並不是一個固定的形式，遇到這種題目先把題意釐清。積分區域是長寬為 $a$ 跟 $b$ 的矩形，指數部分選擇 $b^2x^2$ 與 $a^2y^2$ 之間較大的，就和往常的習慣一樣，我們畫圖分析，可以將積分區域沿 $ay=bx$ 分成兩個部分，恰好兩個區域都能快速找到各自較大的選項，這樣一來題目就破解了，利用對稱性分別將兩個區域積分完加總就是所求了。
解答
$1^{\circ}$ 原式$=\iint_{\text{I}}e^{b^2x^2}dydx+\iint_{\text{II}}e^{a^2y^2}dydx$
$\begin{array}{cl} \iint_{\text{I}}e^{b^2x^2}dydx & =\int_0^a {\int_0^{\frac{b}{a}x} {e^{b^2 x^2 } } } dydx \\ & =\int_0^a {e^{b^2 x^2 } } \frac{b}{a}xdx\ (\text{Let }u = b^2 x^2 \Rightarrow du = 2b^2 xdx) \\ & =\frac{b}{a}\int_0^{a^2 b^2 } {e^u } \frac{{du}}{{2b^2 }} = \frac{1}{{2ab}}\int_0^{a^2 b^2 } {e^u } du\\&=\left. {\frac{1}{{2ab}}e^u } \right|_0^{a^2 b^2 } = \frac{1}{{2ab}}(e^{a^2 b^2 } – 1) \end{array}$
$\begin{array}{cl} \iint_{\text{II}}e^{a^2y^2}dxdy & =\int_0^b {e^{a^2 y^2 } } ydy \\ & =\int_0^a {e^{a^2 y^2 } } \frac{b}{a}xdx\ (\text{Let }u = a^2 y^2 \Rightarrow du = 2a^2 ydy) \\ & =\frac{a}{b}\int_0^{a^2 b^2 } {e^u } \cdot \frac{{du}}{{2a^2 }} = \frac{1}{{2ab}}(e^{a^2 b^2 } – 1) \end{array}$
$2^{\circ}$ 所求$= \frac{1}{{ab}}(e^{a^2 b^2 } – 1) ∎$
此題考點：二變數函數的積分
第十四題 (15 分，每小題 5 分)
Let $E$ be a tetrahedron (四面體) in $\mathbb{R}^3$ bounded by the planes $x+y+z=3,\ x=2z,\ y=0$ and $z=0$. Let also $\textbf{F}:=(x-y)\textbf{i}+(y^2+z^2)\textbf{j}+e^{x^2}\textbf{k}$.
(a) curl$(\textbf{F})=$?
(b) If $S_1$ is the boundary surface of $E$ (including all faces) endowed with the outward orientation, one has $\iint_{S_1}\text{curl}\textbf{F}\cdot d\textbf{S}=$?
(c) If $S_2$ is the surface obtained from $S_1$ by removing the face in the $xy$-plane while keeping the orientation from $S_1$ on all other faces, one then has $\iint_{S_2}\text{curl}\textbf{F}\cdot d\textbf{S}=$?
提示
本卷的最後一題沒有意外的是考通量。第一小題基本上送分，只要知道旋度的公式就可以直接算，第二小題將這個旋度當作向量場求這個區域內的散度，而這題的架構剛好符合高斯散度定理的使用時機，對於熟悉的人來說也是送分，第三小題則是將這個封閉的區域挖掉一個面再求通量，這裡就需要用到散度定理的常用技巧，將不滿足的地方補上並且求出通量後，再扣掉不需要的通量就完成了。
解答
(a) $\text{curl}\textbf{F} = \left| {\begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k} \\ {\partial _x } & {\partial _y } & {\partial _z } \\ {x – y} & {y^2 + z^2 } & {e^{x^3 } } \end{array}} \right| = ( – 2z)\textbf{i} – (3x^2 e^{x^3 } )\textbf{j} +\textbf{ k}$
$\text{curl}\textbf{F} =(-2z,-3x^2e^{x^3},1) ∎$
$\begin{array}{cl} (b) \iint_{S_1}\text{curl}\textbf{F}dS & =\iiint_E\text{div(curl}\textbf{F}) \\ & =\iiint_E 0+0+0dV=0 ∎\end{array}$
$\begin{array}{cl} (c) \iint_{S_2}\text{curl}\textbf{F}ds & =\iint_{S_1}\text{curl}\textbf{F}dS-\iint_{E_1}\text{curl}\textbf{F}dS \\ & =-\iint_{\begin{array}{l} z = 0 \\ x + y \le 3 \\ x \ge 0,y \ge 0 \end{array}} ( – 2z, – 3x^2 e^{x^3 } ,1) \cdot (0,0, – 1)dxdy\\&=\iint_{E_1}dxdy=\frac{9}{2} ∎ \end{array}$
(令 $E_1$ 為四面體的底面)
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