台聯大轉學考微積分 107 A3,A4,A7 卷解答

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。

甲. 簡答題

第一題 (8 分)
Find the critical number of $y = 1 – \frac{4}{{\pi ^2 }}(\tan ^{ – 1} x)^2$.

提示

這題要求臨界點,因此只要找其導函數等於 0 的 $x$ 值。過程需注意連鎖律與 $\tan^{-1}x$ 的微分。

解答

$y’ = – \frac{4}{{\pi ^2 }} \cdot 2 \cdot (\tan ^{ – 1} x) \cdot \frac{1}{{1 + x^2 }}\mathop = \limits^{{\text{Let}}} 0$
$\Rightarrow \tan ^{ – 1} x = 0 \Rightarrow x = 0 ∎$

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此題考點微分求極值反三角函數的導函數

第二題 (8 分)
Determine the limits of integration where $a\leq b$ such that $\int_a^b {(x^2 – 16)} dx$ has minimal value.

提示

此題要求積分式的最小值,可以從積分的基本觀念著手。

解答

被積函數為開口向下之拋物線,由面積觀念得積分最大值為 $x^2-16 \ge 0$ 處
$\begin{array}{cl} \int_a^b(x^2-16)dx & =\int_{{\rm{ – 4}}}^{\rm{4}} {x^2 – {\rm{16}}} dx \\ & =2\int_0^4 {x^2 – 16} dx \\ & =2\left[ {\frac{{x^3 }}{3} – 16x} \right]_{x = 0}^{x = 4}\\&=2(\frac{{64}}{3} – 64) = – \frac{{256}}{3} ∎ \end{array}$

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此題考點定積分直觀觀念

第三題 (8 分)
Evaluate $\int_{ – \infty }^\infty {\frac{{e^x }}{{1 + e^{2x} }}} dx$.

提示

此題為定積分基本題,一定要熟悉四大積分法。

解答

Let $u = e^x \Rightarrow e^x dx = du$
$\begin{array}{cl} \int_{ – \infty }^\infty {\frac{{e^x }}{{1 + e^{2x} }}} dx=\int_0^\infty {\frac{1}{{1 + u^2 }}} du &= \left. {\tan ^{ – 1} u} \right|_{u = 0}^{u = \infty } \\ & =\frac{\pi }{2} – 0 = \frac{\pi }{2} ∎ \end{array}$

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此題考點變數變換法

第四題 (8 分)
Find the slope of the surface $f{\rm{(}}x,y{\rm{) = (}}x^3 + y^3 )^{\frac{1}{3}}$ at the point $(0,0)$ in the $y$-direction.

提示

此題要求空間中曲面沿 $y$ 軸方向的切線斜率,使用方向導數的定義即可。

解答

所求$= \mathop {\lim }\limits_{t \to 0} \frac{{f((0,0) + t(0,1)) – f(0,0)}}{t}$
所求$=\mathop {\lim }\limits_{t \to 0} \frac{{f(0,t) – 0}}{t} = \mathop {\lim }\limits_{t \to 0} \frac{{(t^3 )^{\frac{1}{3}} }}{t} = \mathop {\lim }\limits_{t \to 0} 1 = 1 ∎$

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此題考點方向導數

第五題 (8 分)
Find the surface area of the portion of the plane $z=4-2x-2y$ that lies above the circle $x^2 + y^2 \le 1$ in the first quadrant.

提示

此題為求空間中曲面面積的基本題,要注意到積分範圍只有取第一象限。

解答

◎ 曲面面積公式:
(一般型) $\text{A}=\iint_{\Omega}\sqrt {1 + f_x ^2 + f_y ^2 }dA$
(參數型) $\text{A}=\iint_{\Omega}\parallel \overset{\rightharpoonup}{u_x}\times \overset{\rightharpoonup}{u_y} \parallel dA$

所求$=\iint_{\Omega}\sqrt {1 + (-2) ^2 +(-2) ^2 }dA=3\iint_{\Omega}dA=3\cdot \frac{\pi}{4}=\frac{3\pi}{4} ∎$

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此題考點曲面分析與面積分

第六題 (8 分)
Find an equation of the tangent plane to the paraboloid $\textbf{r}(u,v)=u\textbf{i}+v\textbf{j}+(u^2+v^2)\textbf{k}$ at the point $(1,2,5)$.

提示

此題於拋物面上求切平面,題目給了參數式來定義曲面,可用兩個不同方向的偏導數求得法向量,即可找到切平面方程式;也可利用等值面的想法求得。

解答


法一 由曲面基本觀念
$\left\{ {\begin{array}{l} {\overset{\rightharpoonup}{\textbf{r}_u} = (1,0,2u)} \\ {\overset{\rightharpoonup}{\textbf{r}_v} = (0,1,2v)} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{l} {\overset{\rightharpoonup}{\textbf{r}_u} = (1,0,4)} \\ {\overset{\rightharpoonup}{\textbf{r}_v} (1,2) = (0,1,4)}\end{array}} \right. \Rightarrow$ 取 $\overset{\rightharpoonup}{\textbf{r}_u}\times \overset{\rightharpoonup}{\textbf{r}_v}=(2,4,1)$
$\Rightarrow$ 切平面方程式為:$2x+4y-z=5 ∎$

法二 由等值面觀念
曲面方程式可寫成 $z = x^2 + y^2 \Rightarrow f(x,y,z) = x^2 + y^2 – z = 0$
等位面 $f(x,y,z)=k$ 上一點 $(a,b,c)$ 的法向量為 $\nabla f(a,b,c)$
$\begin{array}{cl} \Rightarrow \overset{\rightharpoonup}{n} & =\nabla f(1,2,5) \\ & =\left. {(2x,2y, – 1)} \right|_{(1,2,5)} = (2,4, – 1) \end{array}$
$\Rightarrow$ 切平面方程式為:$2x+4y-z=5 ∎$

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此題考點等值面與切平面曲面分析與面積分

第七題 (8 分)
Evaluate the integral $\int_0^\infty {\int_0^\infty {\frac{1}{{(1 + x^2 + y^2 )^2 }}} } dxdy$.

提示

不易直接計算的二重積分可以嘗試交換積分次序或極座標變換,由於此題被積函數有 $x^2+y^2$ 因此使用極座標變換。

解答

$\begin{array}{cl} \int_0^\infty {\int_0^\infty {\frac{1}{{(1 + x^2 + y^2 )^2 }}} } dxdy & =\int_0^{\frac{\pi }{2}} {\int_0^\infty {\frac{1}{{(1 + r^2 )^2 }}} } \cdot rdrd\theta \\ & =\int_0^{\frac{\pi }{2}} {\int_0^\infty {\frac{1}{{u^2 }} \cdot \frac{{du}}{2}} } d\theta \\ & =\frac{1}{2}\int_0^{\frac{\pi }{2}} {\left. { – \frac{1}{u}} \right|} _{u = 1}^{u = \infty } d\theta\\&=\frac{1}{2}\int_0^{\frac{\pi }{2}} {0 – ( – 1)} d\theta = \frac{1}{2}\underbrace {\int_0^{\frac{\pi }{2}} {d\theta } }_{\frac{\pi }{2}} = \frac{\pi }{4}  ∎ \end{array}$

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此題考點二重積分座標變換

第八題 (8 分)
Use a change of variable to find the volume of the solid region lying below the surface $z = \sqrt {(x + 4y)(x – y)}$ and above the plane region $R$: region bounded by the parallelogram with vertices $(0,0),\ (1,1),\ (5,0)$ and $(4,-1)$.

提示

本題算體積分,由定義列出二重積分後,由於積分區域為一平行四邊形,使用座標變換後計算會容易許多。

解答

Let $\left\{ {\begin{array}{c} {u = x + 4y} \\ {v = x – y} \end{array}} \right. \Rightarrow \frac{{dudv}}{{dxdy}} = \left| {\begin{array}{l} 1 & 4 \\ 1 & { – 1}\end{array}} \right| = – 5$
$\Rightarrow \left| {\frac{{dudv}}{{dxdy}}} \right| = 5 \Rightarrow \left| {\frac{{dxdy}}{{dudy}}} \right| = \frac{1}{5}$

$\begin{array}{cl} V & =\iint_{R}\sqrt{(x+4y)(x-y)}dxdy \\ & =\iint_{R^{\prime}}\sqrt {uv} \cdot \left| {\frac{{dxdy}}{{dudv}}} \right| dudv \\ & =\frac{1}{5}\int_0^5 {\int_0^5 {\sqrt {uv} } } dudv\\&=\frac{1}{5}\int_0^5 {\sqrt v } (\int_0^5 {\sqrt u } du)dv = \frac{1}{5}(\int_0^5 {\sqrt u } du)(\int_0^5 {\sqrt v } dv)\\&=\frac{1}{5}(\int_0^5 {\sqrt u } du)^2 = \frac{1}{5}(\left. {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{u = 0}^{u = 5} )^2\\&=\frac{1}{5} \cdot \frac{4}{9} \cdot (5\sqrt 5 )^2 = \frac{{100}}{9} ∎  \end{array}$

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此題考點二重積分座標變換

乙.計算、證明題

第一題 (12 分)
A heat-seeking particle is located at the point $(-1,2)$ on a metal plate whose temperature at $(x,y)$ is $T(x,y)=64-2x^2-y^2$. (1) In what direction from $(-1,2)$ does the temperature increase most rapidly? What is this rate of increase? (2) Find the path of the particle as it continuously moves in the direction of maximum temperature increase.

提示

本題討論一個鐵盤上的溫度函數。 第一題求某一點走哪個方向溫度上升最快,其實問的就是梯度的特性,只需算出來即可; 第二題則是要找出一條曲線沿著走,都是溫度上升最快的,這就要討論切向量跟梯度的關係了,於是會列出微分方程,並且要求出它的解。

解答

increase most rapidly:
$\nabla T( – 1,2) = \left. {( – 4x, – 2y)} \right|_{( – 1,2)} = (4, – 4)$
rate: $\sqrt {4^2 + ( – 4)^2 } = 4\sqrt 2 ∎$

Let the path be $\left\{ {\begin{array}{c} {x = x(t),} & {x(0) = – 1} \\ {y = y(t),} & {y(0) = 2}\end{array}} \right.$
$\Rightarrow (x'(t),y'(t)) = \lambda (t)( – 4x(t), – 2y(t))$
$\Rightarrow \left\{ {\begin{array}{l} {x'(t) = – 4\lambda (t) \cdot x(t)} \\ {y'(t) = – 2\lambda (t) \cdot y(t)}\end{array}} \right.$
$\Rightarrow \frac{{x'{\rm{(}}t{\rm{)}}}}{{y'(t)}} = 2 \cdot \frac{{x(t)}}{{y(t)}} \Rightarrow \frac{{x'(t)}}{{x(t)}} = 2 \cdot \frac{{y'(t)}}{{y(t)}}$
$\mathop \Rightarrow \limits^{\int {dt} } \ln \left| {x(t)} \right| = 2\ln \left| {y(t)} \right| + C$
$\mathop \Rightarrow \limits^{t = 0} \ln 1 = 2\ln 2 + C \Rightarrow C = – 2\ln 2 = – \ln 4$
$\Rightarrow \left| {x(t)} \right| = \frac{1}{4}y(t)^2 \Rightarrow y^2 = 4\left| x \right|$
$\Rightarrow$ 因要求沿溫度上升方向,故取 $y^2=-4x,\ y\ge 0 ∎$

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此題考點梯度、旋度、散度微分方程〔1〕〔2〕

第二題 (12 分)
Determine if the given series converges or diverges. Explain your reasoning.
〔1〕$\sum\limits_{n = 1}^\infty (\frac{3n+2}{n+3})^n$ (6 分)
〔2〕$\sum\limits_{n = 1}^\infty {\frac{{e^{\frac{2}{n}} }}{{n^2 }}}$ (6 分)

提示

級數的歛散性為必考題,第一小題可以直接觀察一般項的極限;第二小題可以用已知級數搭配極限比較審斂法。

解答

〔1〕
$\because\ \mathop {\lim }\limits_{n \to \infty } \left| {(\frac{{3n + 2}}{{n + 3}})^n } \right|^{\frac{1}{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3n + 2}}{{n + 3}} = 3 > 1$
$\therefore\ \sum\limits_{n = 1}^\infty {(\frac{{3n + 2}}{{n + 3}})^n }$ diverges (by root test)$. ∎$

〔2〕
$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{e^{\frac{2}{n}} }}{{n^2 }}}}{{\frac{1}{{n^2 }}}} = \mathop {\lim }\limits_{n \to \infty } e^{\frac{2}{n}} = 1 \\ & \text{and }\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}}\text{ converges (by }p\text{-series)}. \\ \therefore & \sum\limits_{n = 1}^\infty {\frac{{e^{\frac{2}{n}} }}{{n^2 }}}\text{ converges (by limit comparison test).} ∎  \end{array}$

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此題考點級數p-級數極限比較審斂法

第三題 (12 分)
Find the maximum value of $\int_C {y^3 dx} + (27x – x^3 )dy$, where $C$ is any circle in the $xy$-plane, oriented counterwise.

提示

此題要求線積分最大值,由於路徑已明確要求為一個封閉圓,且 $y^3$ 和 $27x-x^3$ 偏導函數連續,因此可以用格林定理轉為面積分討論。

解答

◎ Green theorem:
$\oint_C {Pdx} + Qdy =\iint_{\Omega}Q_x-P_ydxdy$

By Green’s theorem,
$\int_C {y^3 } dx + (27x – x^3 )dy =\iint_{\Omega}27-3x^2-3y^2dxdy$

為使積分最大,取 $\Omega =\left\{ \left. {(x,y)} \right|27 – 3x^2 – 3y^2 \ge 0 \right\}\Rightarrow \left\{ \left. {(x,y)} \right|x^2 +y^2 \leq 9 \right\}$

$\begin{array}{cl} \Rightarrow \int_C {y^3 dx} + (27x – x^3 )dy & =\iint_{\Omega}27-3x^2-3y^2dxdy \\ & =\int_0^{2\pi } {\int_0^3 {(27 – 3r^2 )r} dr} d\theta \\ & (\text{Let }u = 27 – 3r^2 \Rightarrow du = – 6rdr)\\&=\int_0^{2\pi } {\int_{27}^0 {u \cdot \frac{{du}}{{ – 6}}} } d\theta\\&=\frac{1}{6}\int_0^{2\pi } {\left. {\frac{{u^2 }}{2}} \right|_{u = 0}^{u = 27} } d\theta\\&=\frac{1}{{12}}\int_0^{2\pi } {729} d\theta\\&=\frac{{729}}{{12}} \cdot 2\pi = \frac{{243}}{2}\pi  ∎ \end{array}$

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此題考點格林定理

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