台綜大轉學考微積分 108 C 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。

第一題 (10 分)
Find the following limits:
〔1〕$\mathop {\lim }\limits_{x \to 0} \frac{{2^x – 1}}{{4^x – 1}}$ (5分)
〔2〕$\mathop {\lim }\limits_{x \to 0} \sin ^{ – 1} (\frac{{1 – x}}{{1 – x^2 }})$, where $\sin^{-1}$ is the inverse function of sine. (5分)

提示

第一題為不定型,有觀察到平方差,可以把零因子去掉;也可以用羅必達法則。第二題用連續函數的極限性質即可。

解答

〔1〕
法一 使用去零因子法
$\mathop {\lim }\limits_{x \to 0} \frac{{2^x – 1}}{{4^x – 1}}=\mathop {\lim }\limits_{x \to 0} \frac{{2^x – 1}}{{(2^x – 1)(2^x + 1)}} = \frac{1}{2} ∎$
法二 使用羅必達法則
$\mathop {\lim }\limits_{x \to 0} \sin ^{ – 1} (\frac{{1 – x}}{{1 – x^2 }})=\mathop {\rm{ = }}\limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{2^x \ln 2}}{{4^x \ln 4}} = \frac{{\ln 2}}{{2\ln 2}} = \frac{1}{2} ∎$

附註:$\mathop = \limits^{\rm{L}}$ 表示此等號使用羅必達法則 (L’Hôpital’s rule)

〔2〕
$\mathop {\lim }\limits_{x \to 0} \sin ^{ – 1} (\frac{{1 – x}}{{1 – x^2 }})=\sin ^{ – 1} {\rm{(}}\mathop {\lim }\limits_{x \to 0} \frac{{{\rm{1}} – x}}{{1 – x^2 }}) = \sin ^{ – 1} 1 = \frac{\pi }{2} ∎$

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此題考點羅必達法則去零因子法極限和連續的聯手

第二題 (10 分)
Define $f(x) = \tan ^2 (x)$ for $x\in (0,\frac{\pi}{2})$ and let $f^{-1}$ be its inverse function. Find $(f^{-1})'(3)$.

提示

求反函數的微分使用反函數定理。

解答

◎ $(f^{ – 1} )'(y_0 ) = \left[ {f'(x_0 )} \right]^{ – 1}$

$\because\ \tan \frac{\pi }{3} = \sqrt 3$
∎∎$\Rightarrow \tan ^2 \underbrace {\frac{\pi }{3}}_{x_0 } =\underbrace { 3 }_{y_0 }$
$\begin{array}{cl} \therefore (f^{ – 1} )'{\rm{(}}3) & =\left[ {f'(\frac{\pi }{3})} \right]^{ – 1}= (2\tan x \cdot \left. {\sec ^2 x} \right|_{x = \frac{\pi }{3}} )^{ – 1} \\ & =\left[ {2 \cdot \sqrt 3 \cdot 4} \right]^{ – 1} = \frac{1}{{8\sqrt 3 }} ∎ \end{array}$

◎ Let $y_0=\tan^2 x_0=3\Rightarrow \tan x_0=\pm \sqrt{3}$
$\because x_0\in (0,\frac{\pi}{2})\ \therefore \tan x_0=\sqrt{3}$

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此題考點微分合成律(連鎖律)

第三題 (10 分)
Calculate the following integrals
〔1〕$\int_0^{\frac{\pi }{2}} {\frac{{\sin 2x}}{{2 + \cos x}}} dx$. (5分)
〔2〕$\int_1^2 {\frac{{(\ln x)^2 }}{{x^3 }}} dx$. (5分)

提示

四大積分法:變數變換、三角置換、分部積分、部分分式法為必考題,一定要練熟。

解答

〔1〕
Let $u = {\rm{2 + }}\cos x$. ($du = – \sin xdx$)
$\begin{array}{cl} \Rightarrow \int_0^{\frac{\pi }{2}} {\frac{{\sin 2x}}{{2 + \cos x}}} dx & = \int_3^2 {\frac{{2(u – 2)}}{u} \cdot } ( – du)\\ & =2\int_2^3 {(1 – \frac{2}{u})} du \\ & =2\left[ {u – 2\ln \left| u \right|} \right]_{u = 2}^{u = 3} \\&=2\left[ {(3 – 2\ln 3) – 2(2 – \ln 2)} \right]\\&=2(1 – \ln \frac{4}{9}) ∎\end{array}$

〔2〕
Let $u = \ln x$. ($du = \frac{1}{x}dx$)
$\begin{array}{cll} \Rightarrow \int_1^2 {\frac{{(\ln x)^2 }}{{x^3 }}} dx & =\int_0^{\ln 2} {\frac{{u^2 }}{{e^{2u} }}} du & \\ & =\int_0^{\ln 2} {u^2 \cdot e^{ – 2u} } du&=\left. {\frac{{u^2 e^{ – 2u} }}{{ – 2}}} \right|_0^{\ln 2} -\left. {\frac{{ue^{ – 2u} }}{2}} \right|_0^{\ln 2} +\int_0^{\ln 2} {\frac{{e^{ – 2u} }}{2}} du \\ && =\frac{{(\ln 2)^2 e^{ – 2\ln 2} }}{{ – 2}} – \frac{{(\ln 2)e^{ – 2\ln 2} }}{2} + \frac{1}{2} \cdot \left. {\frac{{e^{ – 2u} }}{{ – 2}}} \right|_0^{\ln 2} \\&&=\frac{{(\ln 2)^2 \cdot \frac{1}{4}}}{{ – 2}} – \frac{{(\ln 2) \cdot \frac{1}{4}}}{2} – \frac{1}{4}(\frac{1}{4} – 1)\\ &&=\frac{1}{8}((\ln 2)^2 – \ln 2 – \frac{3}{2}) ∎ \end{array}$

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此題考點三角置換變數變換分部積分

第四題 (10 分)
Find the arc length of the curve with equation $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$ within the region $\left\{ {\left. {(x,y)} \right|x \ge 0{\text{ and }}y \ge 0} \right\}$.

提示

弧長也是常考的範圍,除了公式要熟悉之外,也要注意積分範圍有無多取。

解答

◎ Arc length:
$\begin{array}{cl} \ell & =\int_b^a {\sqrt {1 + \left[ {f'(x)} \right]^2 } } dx \\ & =\int_b^a {\sqrt {\left[ {x'(t)} \right]^2 + \left[ {y'(t)} \right]^2 } } dt \end{array}$

Let $\left\{ {\begin{array}{c}{x = \cos ^3 t} \\{y = \sin ^3 t}\end{array}} \right.$, $0 \le t \le \frac{\pi }{2}$
$\begin{array}{cl} \ell & =\int_0^{\frac{\pi }{2}} {\sqrt {\left[ {3\cos ^2 t \cdot ( – \sin t)} \right]^2 + \left[ {3\sin ^2 t \cdot \cos t} \right]^2 } } dt \\ & =\int_0^{\frac{\pi }{2}} {\sqrt {9\cos ^4 t \cdot \sin ^2 t + 9\sin ^4 t \cdot \cos ^2 t} } dt \\ & =\int_0^{\frac{\pi }{2}} {\sqrt {9\cos ^2 t \cdot \sin ^2 t \cdot (\underbrace {\cos ^2 t + \sin ^2 t}_1)} } dt \\&=\int_0^{\frac{\pi }{2}} {3\cos t \cdot \sin t} dt\\& (\text{Let }u=\sin t\Rightarrow du=\cos tdt)\\&=3\int_0^1 u du\\&=3 \cdot \left. {\frac{{u^2 }}{2}} \right|_{u = 0}^{u = 1} = \frac{3}{2} ∎ \end{array}$

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此題考點曲線分析(上)

第五題 (10 分)
Find the slope of the tangent line to the polar curve $r = 1 + \sin (2\theta )$ at the point specified by $\theta =\frac{\pi}{3}$.

提示

這題給極座標方程式要求切線斜率,若轉回 $xy$ 關係式會很難計算,因此要會在極座標的情況下求斜率。

解答

◎ $\left\{ {\begin{array}{c}{x = r\cos \theta = r(\theta )\cos \theta }\\{y = r\sin \theta = r(\theta )\sin \theta } \end{array}} \right.$
$\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{r'(\theta )\sin \theta + r(\theta )\cos \theta }}{{r'(\theta )\cos \theta – r(\theta )\sin \theta }}$

$r(\theta ) = 1 + \sin (2\theta )$
$\Rightarrow r'(\theta ) = 2\cos (2\theta )$
$\Rightarrow \left\{ {\begin{array}{l}{r(\frac{\pi }{3}) = 1 + \sin \frac{{2\pi }}{3} = 1 + \frac{{\sqrt 3 }}{2} = – 1} \\ {r'(\frac{\pi }{3}) = 2\cos \frac{{2\pi }}{3} = 2 \times ( – \frac{1}{2}) = – 1} \\ \end{array}} \right.$

$\begin{array}{cl} \text{slope} & =\frac{{r'(\frac{\pi }{3})\sin \frac{\pi }{3} + r(\frac{\pi }{3})\cos \frac{\pi }{3}}}{{r'(\frac{\pi }{3})\cos \frac{\pi }{3} – r(\frac{\pi }{3})\sin \frac{\pi }{3}}} \\ & =\frac{{( – 1) \cdot \frac{{\sqrt 3 }}{2} + (1 + \frac{{\sqrt 3 }}{2}) \cdot \frac{1}{2}}}{{( – 1) \cdot \frac{1}{2} – (1 + \frac{{\sqrt 3 }}{2}) \cdot \frac{{\sqrt 3 }}{2}}} \\ & =\frac{{ – 2\sqrt 3 + 2 + \sqrt 3 }}{{ – 2 – \sqrt 3 (2 + \sqrt 3 )}} \\&=\frac{{2 – \sqrt 3 }}{{ – 2 – 3 – 2\sqrt 3 }} = \frac{{\sqrt 3 – 2}}{{5 + 2\sqrt 3 }} ∎\end{array}$

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此題考點極座標統整應用

第六題 (10 分)
Find the radius of convergence of the series $\sum\limits_{n = 1}^\infty {\frac{{n(x + 3)^n }}{{4^{n + 1} }}}$.

提示

此題求收斂半徑,帶公式即可。若要求收斂區間,記得多考慮兩邊端點是否收斂。

解答

r收斂半徑:
$\begin{array}{c}r &=\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left|{\frac{{a_{n + 1} }}{{a_n }}} \right|}}\\&=\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {a_n } \right|^{\frac{1}{n}} }} \end{array}$

$a_n = \frac{n}{{4^{n + 1} }}$

$r =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{n + 1}}{{4^{n + 2} }}}}{{\frac{n}{{4^{n + 1} }}}}} \right|}} =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{{4n}}}} = \frac{{{\rm{ }}1{\rm{ }}}}{{\frac{1}{4}}} = 4 ∎$

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此題考點比較審斂法根值審斂法

第七題 (10 分)
Find the Maclaurin series of order 4 for the function $f(x) = e^{ – x^2 } \cdot \cos x$, i.e., approximate $f(x)$ by a polynomial $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$.

提示

求馬克勞林級數首先要觀察題目,有看到熟悉的部分可以用已知的泰勒展開式快速推得。建議除了定義之外,常用的泰勒展開式要很熟悉。

解答

$e^x = 1 + x + \frac{{x^2 }}{{2!}} + \frac{{x^3 }}{{3!}} + \frac{{x^4 }}{{4!}} + \cdots$
$\begin{array}{cl} e^{ – x^2 } & =1 + ( – x^2 ) + \frac{{( – x^2 )^2 }}{{2!}} + \cdots \\ & =1 – x^2 + \frac{{x^4 }}{{2!}} – \cdots \end{array}$
$\begin{array}{cl} \cos x & =1 – \frac{{x^2 }}{{2!}} + \frac{{x^4 }}{{4!}} – \cdots \\ & =1 – \frac{{x^2 }}{2} + \frac{{x^4 }}{{24}} – \cdots \end{array}$
$\Rightarrow e^{ – x^2 } \cos x = 1 + ( – \frac{{x^2 }}{2} – x^2 ) + (\frac{{x^4 }}{{24}} + \frac{{x^4 }}{2} + \frac{{x^4 }}{2}) + \cdots$

所求 $=1 – \frac{3}{2}x^2 + \frac{{25}}{{24}}x^4 ∎$

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此題考點泰勒定理

第八題 (10 分)
Find the Maximum value of the function $f(x,y,z) = x{\rm{ + 2}}y + 3z$ on the curve of intersection of the plane $x-y+z=1$ and the cylinder $x^2+y^2=1$.

提示

計算題遇到有限制條件求極值時,要想到拉格朗日乘數法,平時可以多練習解聯立的技巧。

解答

Let $g_1 (x,y,z) = x – y + z – 1$
Let $g_2 (x,y,z) = x^2 + y^2 – 1$
Let $\nabla f + \lambda _1 \nabla g_1 + \lambda _2 \nabla g_2 = 0$
$\Rightarrow (1,2,3) + \lambda _1 (1, – 1,1) + \lambda _2 (2x,2y,0) = 0$
$\Rightarrow \left\{ {\begin{array}{l}{1 + \lambda _1 + 2\lambda _2 x = 0} \\ {2 – \lambda _1 + 2\lambda _2 y = 0} \\ {3 + \lambda _1 = 0} \\ \end{array}} \right.\Rightarrow \left\{ {\begin{array}{l}{ – 2 + 2\lambda _2 x = 0} \\ {5 + 2\lambda _2 y = 0{\rm{ }}} \\ \end{array}} \right.\Rightarrow \left\{ {\begin{array}{l}{x = \frac{1}{{\lambda _2 }}{\rm{ }}} \\ {y = – \frac{5}{{2\lambda _2 }}} \\ \end{array}} \right.$
代入 $g_2 (x,y,z) = x^2 + y^2 – 1$
$\Rightarrow \frac{1}{{\lambda _2 ^2 }} + \frac{{25}}{{4\lambda _2 ^2 }} = 1 \Rightarrow \lambda _2 ^2 = \frac{{29}}{4} \Rightarrow \lambda _2 = \pm \frac{{\sqrt {29} }}{2}$
$\Rightarrow (x,y) = (\frac{2}{{\sqrt {29} }}, – \frac{5}{{\sqrt {29} }})$ 或 $( – \frac{2}{{\sqrt {29} }},\frac{5}{{\sqrt {29} }})$
代入 $g_1 (x,y,z) = x – y + z – 1$
$\Rightarrow (x,y,z) = (\frac{2}{{\sqrt {29} }}, – \frac{5}{{\sqrt {29} }},1 – \frac{7}{{\sqrt {29} }})$ 或 $( – \frac{2}{{\sqrt {29} }},\frac{5}{{\sqrt {29} }},1 + \frac{7}{{\sqrt {29} }})$

$f(\frac{2}{{\sqrt {29} }},\frac{{ – 5}}{{\sqrt {29} }},1 – \frac{7}{{\sqrt {29} }}) = \frac{2}{{\sqrt {29} }} + 2 \cdot \frac{{( – 5)}}{{\sqrt {29} }} + 3(1 – \frac{7}{{\sqrt {29} }}) = 3 – \sqrt {29}$ (minimum)
$f(\frac{{ – 2}}{{\sqrt {29} }},\frac{5}{{\sqrt {29} }},1 + \frac{7}{{\sqrt {29} }}) = 3 + \sqrt {29}$ (maximum)
The maximum of $f(x,y,z)$ is $3+\sqrt{29} ∎$

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此題考點Lagrange 乘數法

第九題 (10 分)
Let $\textbf{F}(x,y) = \frac{{ – 2}}{{x^2 + y^2 }}\textbf{i}{\rm{ + }}\frac{{{\rm{2}}x}}{{x^2 + y^2 }}\textbf{j}$ and let $\text{D} = \left\{ {\left. {(x,y)} \right|x^2 + y^2 \le 9} \right\}$. Find $\int_{\partial D} {\textbf{F}\cdot \textbf{T}} ds$, where we traverse the boundary $\partial D$ in the counterclockwise direction and $\textbf{T}$ is the unit tangent vector.

提示

因為向量場在原點不存在,因此不能使用格林定理,直接計算線積分即可。

解答

$\partial D:\ \left\{ {\begin{array}{c}{x = 3\cos t} \\ {y = 3\sin t} \end{array}} \right.$, $0 \le t \le 2\pi$

$\begin{array}{cl} \int_{\partial D} {\textbf{F}\cdot \textbf{T}} ds & =\int_{\partial D} {(\frac{{ – 2y}}{{x^2 + y^2 }},\frac{{2x}}{{x^2 + y^2 }}) \cdot (dx,dy)} \\ & =\int_0^{2\pi } {(\frac{{ – 2 \cdot 3\sin t}}{9},\frac{{2 \cdot 3\cos t}}{9}) \cdot ( – 3\sin t,3\cos t)} dt \\ & =\int_0^{2\pi } {(2\sin ^2 t + 2\cos ^2 t)} dt \\&=2\int_0^{2\pi } {dt} = 4\pi ∎ \end{array}$

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此題考點線積分

第十題 (10 分)
Let $\textbf{F}(x,y,z) =\frac{{x\textbf{i} + y\textbf{j} + z\textbf{k}}}{{(x^2 + y^2 + z^2 )^{\frac{3}{2}} }} $ and let $\text{D} = \left\{ \left. {(x,y,z)} \right|\frac{{x^2 }}{4} + \frac{{y^2 }}{4} + \frac{{z^2 }}{9} \le 1 \right\}$. Find the flux of $\textbf{F}$ outward across the boundary of $D$.

提示

因為向量場在原點不存在,因此不能直接使用高斯散度定理,但我們可以透過一些技巧,間接地使用散度定理來計算通量。

解答

令 $S_1:\ \frac{{x^2 }}{4} + \frac{{y^2 }}{4} + \frac{{z^2 }}{9} = 1,\ S_2:\ x^2+y^2+z^2=1$

先簡單計算 $\nabla \cdot \textbf{F}$
$\begin{array}{rl} \nabla \cdot \textbf{F}= &\frac{{(x^2 + y^2 + z^2 )^{\frac{3}{2}} – x \cdot \frac{3}{2}(x^2 + y^2 + z^2 )^{\frac{1}{2}} \cdot 2x}}{{(x^2 + y^2 + z^2 )^3 }} \\ & +\frac{{(x^2 + y^2 + z^2 )^{\frac{3}{2}} – y \cdot \frac{3}{2}(x^2 + y^2 + z^2 )^{\frac{1}{2}} \cdot 2y}}{{(x^2 + y^2 + z^2 )^3 }} \\ & +\frac{{(x^2 + y^2 + z^2 )^{\frac{3}{2}} – z \cdot \frac{3}{2}(x^2 + y^2 + z^2 )^{\frac{1}{2}} \cdot 2z}}{{(x^2 + y^2 + z^2 )^3 }} \\=&\frac{{3(x^2 + y^2 + z^2 )^{\frac{3}{2}} – (x^2 + y^2 + z^2 )^{\frac{1}{2}} (3x^2 + 3y^2 + 3z^2 )}}{{(x^2 + y^2 + z^2 )^3 }} = 0 \end{array}$

挖掉 $S_2$ (包含內部)後,向量場處處可微,因此我們可以用高斯散度定理來計算通量
$\begin{array}{cl} \iint_{S_1\cup S_2}\textbf{F}\cdot \hat{n}ds & =\iiint_{V}\overbrace {\nabla \cdot F}^0dV=0 \\ & =\iint_{S_1}\textbf{F}\cdot \hat{n}dA+\iint_{S_2}\textbf{F}\cdot\hat{n}dA \end{array}$
$\begin{array}{cl} \Rightarrow \iint_{s_1}\textbf{F}\cdot\hat{n}dA & =-\iint_{s_2}\textbf{F}\cdot\hat{n}dA \\ & =\iint_{s_2}\textbf{F}\cdot(-\hat{n})dA,\ S_2 = \left\{ (x,y,z):x^2 + y^2 + z^2 = 1 \right\} \end{array}$

$\begin{array}{cl} \text{flux} & =\iint_{s_2}\textbf{F}\cdot(-\hat{n})dA \\ & =\iint_{S_2}(x,y,z) \cdot \frac{{(x,y,z)}}{{\sqrt {x^2 + y^2 + z^2 } }}dA \\ & =\iint_{S_2}(\underbrace {x^2 + y^2 {\rm{ + }}z^2 }_1)^{\frac{1}{2}}dA \\&=\iint_{S_2}dA=4\pi\times 1^2=4\pi  ∎ \end{array}$

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此題考點散度定理

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