台綜大轉學考微積分 109 B 卷解答

第一小題是基本題；第二小題看到底數和指數有 $x$，要反應換底公式：$\Delta = e^{\ln \Delta}$

〔1〕$\mathop {\lim }\limits_{x \to 2} \frac{{x + 4}}{{x – 7}} = \frac{{2 + 4}}{{2 – 7}} = \frac{6}{{ – 5}}$.
〔2〕原式 $= \mathop {\lim }\limits_{x \to 0} e^{\ln (1 – 7x)^{\frac{4}{x}} }= e^{\mathop {\lim }\limits_{x \to 0} \frac{{4\ln (1 – 7x)}}{x}}\mathop = \limits^{\rm{L}} e^{\mathop {\lim }\limits_{x \to 0} \frac{{4 \cdot \frac{{ – 7}}{{1 – 7x}}}}{1}}= e^{ – 28}　∎$

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

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此題考點：極限的四則運算、羅必達法則

第一小題看到多項式 $x$ 乘三角函數，使用分部積分；第二小題分母根號內有二次函數，可以使用三角置換

〔1〕

$\begin{array}{clr} \int_0^{\frac{\pi }{4}} {x\sec ^2 x} dx & = \left. {x\tan x} \right|_0^{\frac{\pi }{4}} – \int_0^{\frac{\pi }{4}} {\tan x} dx \\ & = \frac{\pi }{4} – \int_0^{\frac{\pi }{4}} {\frac{{\sin x}}{{\cos x}}} dx\\ & \text{Let } u=\cos x\Rightarrow du=-\sin xdx \\  &=\frac{\pi }{4} + \int_1^{\frac{{\sqrt 2 }}{2}} {\frac{1}{u}} du\\  & =\frac{\pi }{4} + \left[ {\ln \left| u \right|} \right]_{u = 1}^{u = \frac{{\sqrt 2 }}{2}} \\ & =\frac{\pi }{4} + \ln \frac{{\sqrt 2 }}{2}　∎ \\ \end{array} $

〔2〕

$\begin{array}{clr} \int_4^5 {\frac{{x – 3}}{{\sqrt {x^2 – 6x + 12} }}} dx & =\int_4^5 {\frac{{x – 3}}{{\sqrt {(x – 3)^2 + 2^2 } }}} dx \\ & \text{Let } x – 3 = 2\tan u \Rightarrow dx = 2\sec ^2 udu \\ & =\int_\alpha ^{\frac{\pi }{4}} {\frac{{2\tan u}}{{2\sec u}} \cdot 2\sec ^2 u} du \\ &=2\int_\alpha ^{\frac{\pi }{4}} {\tan u} \cdot \sec udu\\ &=2\left[ {\sec u} \right]_{u = \alpha }^{u = \frac{\pi }{4}}\\&= 2(\sqrt 2 – \frac{{\sqrt 5 }}{2}) = 2\sqrt 2 – \sqrt 5　∎\\ \end{array}$

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此題考點：分部積分法、三角置換法

定積分之上下界有未知數，想到微積分基本定理

$\frac{{\partial u}}{{\partial x}} = (e^x )\int_{ – 2}^{2x} {\sqrt {4 – t^2 } } dt + (e^x – \frac{y}{6}) \cdot \sqrt {4 – (2x)^2 } \cdot 2 \\ \Rightarrow \left. {\frac{{\partial u}}{{\partial x}}} \right|_{(0,1)}=\int_{- 2}^0 {\sqrt {4 – t^2 } } dt + (1 – \frac{1}{6}) \cdot \sqrt 4 \cdot 2 = \int_{-2}^0 {\sqrt {4 – t^2 } } dt + \frac{{10}}{3}$

$\begin{array}{cll} \int_{ – 2}^0 {\sqrt {4 – t^2 } } dt & =\int_{ – \frac{\pi }{2}}^0 {2\cos u \cdot \cos u} du \\ & \text{Let } t=2\sin u \Rightarrow dt=2\cos udu \\ & =4\int_{ – \frac{\pi }{2}}^0 {\frac{{1 + \cos 2u}}{2}} du \\&=2\left[ {u + \frac{{\sin 2u}}{2}} \right]_{u = – \frac{\pi }{2}}^{u = 0} \\ &=2\left[ {0 – ( – \frac{\pi }{2} + 0)} \right] = \pi　∎ \end{array}$

$\Rightarrow \left. \frac{\partial u}{\partial x} \right|_{(0,1)}=\pi+\frac{10}{3}　∎$

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此題考點：微積分基本定理 II：先積再微型

基本的偏微分題目，如有需要可以使用樹狀圖表示變數間的關係。

$\left\{ \begin{array}{c} s = x^2 + y^2 \mathop = \limits^{(x,y) = (2,0)} 4 \\ t = 3x – 4y\mathop = \limits^{(x,y) = (2,0)} 6 \\ \end{array} \right.$

$\begin{array}{cll} \frac{{\partial u}}{{\partial y}} = \frac{{\partial h}}{{\partial y}} & =\frac{{\partial h}}{{\partial s}} \cdot \frac{{\partial s}}{{\partial y}} + \frac{{\partial h}}{{\partial t}} \cdot \frac{{\partial t}}{{\partial y}} \\ & =(\frac{t}{4} \cdot \frac{{ – 1}}{{s^2 }} + \frac{{(\sec ^2 s)\ln s – (\tan s) \cdot \frac{1}{s}}}{{(\ln s)^2 }})(2y) + (\frac{1}{{4s}} – 2t) \cdot ( – 4) \\ \end{array}$

$\begin{array}{cl} \Rightarrow \left. {\frac{{\partial u}}{{\partial y}}} \right|_{(2,0)} & =(\frac{6}{4} \cdot \frac{{ – 1}}{{4^2 }} + \frac{{(\sec ^2 4) \cdot \ln 4 – (\tan 4) \cdot \frac{1}{4}}}{{(\ln 4)^2 }}) \cdot 0 + (\frac{1}{{16}} – 12) \cdot ( – 4) \\ & = 48 – \frac{1}{4} = \frac{{191}}{4}　∎\\ \end{array}$

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此題考點：偏微分運算律

基本旋轉體積分題，使用剝殼法或圓盤法。此題使用剝殼法。

$\begin{array}{cl} V &=\int_{0}^{1} 2\pi x\sin (x^2)dx \\ &=2\pi \int_0^1 {\sin \overbrace {(x^2 )}^u} \underbrace {xdx}_{\frac{{du}}{2}} \\ &=2\pi \int_0^1 {\sin u} \frac{{du}}{2} \\&=\pi \left[ { – \cos u} \right]_{u = 0}^{u = 1} = \pi (1 – \cos 1)　∎ \end{array}$

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此題考點：旋轉體積分

題目已提示使用泰勒展開式來計算所求級數和。

$\begin{array}{cl} \int_0^x {f(t)} dt = \int_0^x {te^t } dt &=\left. {te^t } \right|_{t = 0}^{t = x} – \int_0^x {e^t } dt = xe^x – \left[ {e^t } \right] {t = 0}^{t = x} \\ &=xe^x – e^x + 1 \cdots (1) \\ \end{array}$

$\begin{array}{clr} f(x) = xe^x & =x(1 + \frac{x}{{1!}} + \frac{{x^2 }}{{2!}} + \frac{{x^3 }}{{3!}} + \cdots ) \\ & =x + \frac{{x^2 }}{{1!}} + \frac{{x^3 }}{{2!}} + \frac{{x^4 }}{{3!}} + \cdots \\ \end{array}$

$\begin{array}{clr} \Rightarrow \int_0^x {f(t)} dt & =\int_0^x {(t + \frac{{t^2 }}{{1!}}{\rm{ + }}\frac{{t^3 }}{{2!}} + \cdots )} dt \\ & =\left. {\frac{{t^2 }}{2} + \frac{{t^3 }}{3}{\rm{ + }}\frac{{t^4 }}{{4 \cdot 2!}} + \frac{{t^5 }}{{5 \cdot 3!}} + \cdots } \right|_0^x \\ & =(\frac{{t^2 }}{2} + \frac{{t^3 }}{3}{\rm{ + }}\frac{{t^4 }}{{4 \cdot 2!}} + \frac{{t^5 }}{{5 \cdot 3!}} + \cdots ) \cdots (2)\\ \end{array}$

Since $(1)=(2)$, let $x=1$

$\begin{array}{clr} \Rightarrow \sum\limits_{k = 0}^\infty {\frac{1}{{(k + 2)k!}}} & =\frac{1}{2} + \frac{1}{3} + \frac{1}{{4 \cdot 2!}} + \frac{1}{{5 \cdot 3!}} + \cdots \\ & =e^1 – e^1 + 1 = 1　∎ \\ \end{array}$

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此題考點：泰勒級數與泰勒定理

將位置與時間的關係列式，經由微分得到瞬時速率。

1 分鐘走 0.1 km ，$t$ 分鐘走 $\frac{t}{10}$ km

$\Rightarrow h = \sqrt {1^2 + 1^2 – 2 \cdot 1 \cdot {\rm{1}} \cdot \cos \frac{t}{{10}}} = \sqrt {2 – 2\cos \frac{t}{{10}}} \sqrt 2$

$\begin{array}{clr} \Rightarrow h'(t) & =\frac{1}{2} \cdot \frac{1}{{\sqrt {2 – 2\cos \frac{t}{{10}}} }} \cdot 2 \cdot \frac{1}{{10}} \cdot \sin \frac{t}{{10}} \\ & =\frac{1}{{10}} \cdot \frac{{\sin \frac{t}{{10}}}}{{\sqrt {2 – 2\cos \frac{t}{{10}}} }} \\ \end{array}$

When dog is at D, $\frac{t}{10}=\frac{3}{4}\pi$

$\Rightarrow h'(t) = \frac{1}{{10}} \cdot \frac{{\sin \frac{3}{4}\pi }}{{\sqrt {2 – 2\cos \frac{3}{4}\pi } }}$

$\Rightarrow h'(t) = \frac{1}{{10}} \cdot \frac{{\frac{{\sqrt 2 }}{2}}}{{\sqrt {2 – 2( – \frac{{\sqrt 2 }}{2})} }} = \frac{{\sqrt 2 }}{{20}} \cdot \frac{1}{{\sqrt {2 + \sqrt 2 } }}　∎$

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此題考點：導數與微分的概念

反函數定理：若 $f(x_0)=y_0$ ，則 $(f^{ – 1} )'(y_0 ) = \left[ {f'(x_0 )} \right]^{ – 1}$。

Let $5=x^5+2x^3+2x$

$\Rightarrow x=1$

$\Rightarrow f(1)=5$

$\Rightarrow f'(5)=1$

$(f^{ – 1} )(5) = \left[ {f'(1)} \right]^{ – 1} = (\left. {5x^4 + 6x^2 + 2} \right|_{x = 1} )^{ – 1} = 13^{ – 1} = \frac{1}{{13}}　∎$

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此題考點：微分合成律

遇到二重積分可以使用極座標變換或交換積分次序，本題使用及座標變換。

$\int_0^1 {\int_{\sqrt {1 – x^2 } }^{\sqrt {9 – x^2 } } {e^{x^2 + y^2 } } dydx} + \int_1^3 {\int_0^{\sqrt {9 – x^2 } } {e^{x^2 + y^2 } } } dydx$

$=\underbrace {\int_0^1 {\int_{\sqrt {1 – x^2 } }^{\sqrt {9 – x^2 } } {e^{x^2 + y^2 } } } dydx}_{\rm{I}} + \underbrace {\int_1^3 {\int_0^{\sqrt {9 – x^2 } } {e^{x^2 + y^2 } } } dydx}_{{\rm{II}}}$

$=\int_0^{\frac{\pi }{2}} {\int_1^3 {e^{r^2 } } } rdrd\theta$

$=\frac{1}{2}\int_0^{\frac{\pi }{2}} {\left. {e^u } \right|_{u = 1}^{u = 9} } d\theta$

$=\frac{{e^9 – e}}{2}\int_0^{\frac{\pi }{2}} {d\theta } = \frac{\pi }{4}(e^9 – e)　∎$

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此題考點：二重積分座標變換

依題意使用拉格朗日乘數法。

Let $\nabla f + \lambda \nabla g = 0$
$\Rightarrow (yze^{xyz} ,xze^{xyz} ,xye^{xyz} ,xye^{xyz} ) + \lambda (3x^2 , – 3y^2 ,3z^2 ) = 0$
$\Rightarrow \left\{ {\begin{array}{l} {yze^{xyz} + 3\lambda x^2 = 0} \\ {xze^{xyz} – 3\lambda y^2 = 0} \\ {xye^{xyz} + 3\lambda z^2 = 0} \end{array}} \right.$ $\Rightarrow \left\{ {\begin{array}{l} {xyze^{xyz} + 3\lambda x^3 = 0} \\ {xyze^{xyz} – 3\lambda y^3 = 0} \\ {xyze^{xyz} + 3\lambda z^3 = 0} \end{array}} \right.$
$\Rightarrow 3xyze^{xyz} + 3\lambda \underbrace {(x^3 – y^3 + z^3 )}_{24} = 0$
$\Rightarrow xyze^{xyz} = – 24\lambda$
$\Rightarrow \left\{ {\begin{array}{l} { – 24\lambda + 3\lambda x^3 = 0} \\{ – 24\lambda – 3\lambda y^3 = 0} \\ { – 24\lambda + 3\lambda z^3 = 0} \\ \end{array}} \right.$ $\Rightarrow \left\{ {\begin{array}{l}
{\lambda (x^3 – 8) = 0} \\ { – \lambda (y^3 + 8) = 0} \\ {\lambda (z^3 – 8) = 0} \\ \end{array}} \right.$
$\Rightarrow \left\{ {\begin{array}{l} {\lambda = 0{\text{ or }}x = 2} \\ {\lambda = 0{\text{ or }}y = – 2} \\ {\lambda = 0{\text{ or }}z = 2} \\ \end{array}} \right.$
If $\lambda =0$
$\left\{ {\begin{array}{l} {yze^{xyz} = 0 \Rightarrow y = 0{\text{ or }}z = 0} \\ {xze^{xyz} = 0 \Rightarrow x = 0{\text{ or }}z = 0} \\ {xye^{xyz} = 0 \Rightarrow x = 0{\text{ or }}y = 0} \\ \end{array}} \right.$
$\because x^3 – y^3 + z^3 = 24$
$\therefore (x,y,z) = (0,0,\sqrt[3]{{24}})\text{ or }(\sqrt[3]{{24}},0,0)\text{ or }(0, – \sqrt[3]{{24}},0)$
If $\lambda \ne 0$
$\Rightarrow (x,y,z) = (2, – 2,2)$
$f(0,0,\sqrt[3]{{24}}) = e^0 = 1$
$f(2, – 2,2) = e^{ – 8} < 1$
$f(0, – \sqrt[3]{{24}},0) = e^0 = 1$
$f(\sqrt[3]{{24}},0,0) = e^0 = 1$
$\Rightarrow \left\{ {\begin{array}{l} {{\rm{Max}}{\rm{.}} = 1} \\ {{\rm{min}}{\rm{. = }}e^{ – 8} } \\ \end{array}} \right.　∎$

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此題考點：Lagrange 乘數法