台綜大轉學考微積分 109 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〕 $\mathop {\lim }\limits_{x \to 1} \frac{{e^x – 4}}{x}$.
〔2〕 $\mathop {\lim }\limits_{x \to – \infty } (\sqrt {x^2 + x + 1} + x)$.

提示

第一小題雖是基本題,但並非不定型,不可使用羅必達法則;第二題注意 $x$ 是趨近 $-\infty$ 需要特別留意正負號。

解答

〔1〕$\mathop {\lim }\limits_{x \to 1} \frac{{e^x – 1}}{x} = \frac{{e^1 – 1}}{1} = e – 1 ∎$
〔2〕

Let $y=-x$
$\mathop {\lim }\limits_{x \to – \infty } (\sqrt {x^2 + x + 1} + x)$
$=\mathop {\lim }\limits_{y \to \infty } (\sqrt {y^2 – y + 1} – y) \times \frac{{\sqrt {y^2 – y + 1} + y}}{{\sqrt {y^2 – y + 1} + y}}$
$=\mathop {\lim }\limits_{y \to \infty } \frac{{y^2 – y + 1 – y^2 }}{{\sqrt {y^2 – y + 1} + y}}$
$=\mathop {\lim }\limits_{y \to \infty } \frac{{ – 1 + \frac{1}{y}}}{{\sqrt {1 – \frac{1}{y} + \frac{1}{{y^2 }}} + 1}} = – \frac{1}{2} ∎$

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此題考點老大比較法

第二題 (10 分)
Find the tangent line to the curve defined by the equation $4^{\frac{x}{y}}=x-y$ when $x=0$.

提示

要求在 $x=0$ 的切線方程式,因此再找出該點的微分作為斜率即可。

解答

At $x=0$, $4^{\frac{x}{y}}=0-y\Rightarrow y=-1$
$\because \ 4^{\frac{x}{y}}=x-y$
$\therefore \ 4^{\frac{x}{y}} \cdot \ln 4 \cdot \frac{{1 \cdot y – x \cdot y’}}{{y^2 }} = 1 – y’$

At $(x,y)=(0,-1)$
$\Rightarrow 4^{\frac{0}{{ – 1}}} \cdot \ln 4 \cdot \frac{{1 \cdot ( – 1) – 0 \cdot y’}}{{( – 1)^2 }} = 1 – y’$
$\Rightarrow \ln 4 \cdot ( – 1) = 1 – y’ \Rightarrow y’ = 1 + \ln 4$

the tangent line:
$y + 1 = (1 + \ln 4)(x – 0)$
$y = (1 + \ln 4)x – 1 ∎$

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此題考點切線專論

第三題 (10 分)
Calculate the definite integral $\int_0^{4\sqrt 2 } {\int_{x^{\frac{2}{5}} }^2 {\frac{{e^y }}{{\sqrt y }}} } dydx$.

提示

不易直接算的二重積分,可以嘗試極座標變換或交換積分次序( Fubini 定理),本題使用交換積分次序。

解答

$\int_0^{4\sqrt 2 } {\int_{x^{\frac{2}{5}} }^2 {\frac{{e^y }}{{\sqrt y }}} } dydx$
$=\int_0^2 {\int_0^{y^{\frac{5}{2}} } {\frac{{e^y }}{{\sqrt y }}} } dxdy$
$= \int_0^2 {y^{ – \frac{1}{2}} e^y } \underbrace {\int_0^{y^{\frac{5}{2}} } {dx} }_{y^{\frac{5}{2}} }dy$
$=\int_0^2 {y^2 \cdot e^y } dy$
$=\left. {y^2 e^y } \right|_0^2 – 2\left. {ye^y } \right|_0^2 + \int_0^2 {2e^y } dy$
$=\int_0^2 {y^2 \cdot e^y } dy = 4e^2 – 4e^2 + 2\left[ {e^y } \right]_{y = 0}^{y = 2} = 2(e^2 – 1) ∎$

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此題考點二變數函數的積分分部積分

第四題 (10 分)
Find the arc length of the polar curve $r=\theta$ with $0\leq\theta\leq \pi$.

提示

此題是極座標求弧長的基本題。

解答

$\begin{array}{cl} l & =\int_a^b {\sqrt {r(\theta )^2 + r'(\theta )^2 } } d\theta \\ & =\int_0^\pi {\sqrt {\theta ^2 + 1^2 } } d\theta \end{array}$

$\left( \begin{array}{c} \text{Let }\theta=\tan u\Rightarrow d\theta=\sec^2 udu \\ \theta = 0 \Rightarrow u = 0 \\ \theta = \pi \Rightarrow u = \tan ^{ – 1} \pi \mathop = \limits^{{\rm{Let}}} \alpha \\ \end{array} \right)$

$\begin{array}{cl} \ & =\int_0^\alpha {\sec u \cdot \sec ^2 u} du \\ & =\int_0^\alpha {\sec ^3 u} du \\ & =\left. {\frac{{\sec u\tan u + \ln \left| {\sec u + \tan u} \right|}}{2}} \right|_{u = 0}^{u = \alpha } \\ &=\frac{1}{2}(\sec \alpha \tan \alpha + \ln \left| {\sec \alpha + \tan \alpha } \right|) \\ &=\frac{1}{2}(\pi \sqrt {1 + \pi ^2 } + \ln (\pi + \sqrt {1 + \pi ^2 } )) ∎ \end{array}$

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此題考點極座標統整與應用

第五題 (10 分)
Find the volume of the solid of revolution about the $x$-axis bounded by the curve $f(x) = \frac{1}{{x\sqrt {x^2 + 1} }}$ with $x \in [1,\infty )$.

提示

旋轉體積分可以使用圓盤法或剝殼法,本題使用圓盤法。

解答

$\begin{array}{cl} V & =\int_1^\infty {\pi (\frac{1}{{x^2 (x\sqrt {x^2 + 1} )}})^2 } dx \\ & =\pi \int_1^\infty {\frac{1}{{x^2 (x^2 + 1)}}} dx \\ & =\pi \int_1^\infty {\frac{a}{x} + \frac{b}{{x^2 }} + \frac{c}{{x^2 + 1}}} dx \\&=\pi \int_1^\infty {\frac{1}{{x^2 }} + \frac{{ – 1}}{{x^2 + 1}}} dx \\&=\pi (\int_1^\infty {\frac{1}{{x^2 }}} dx – \int_1^\infty {\frac{1}{{1 + x^2 }}} dx) \\&=\pi (\left[ { – \frac{1}{x}} \right]_{x = 1}^{x = \infty } – \left[ {\tan ^{ – 1} x}\right]_{x = 1}^{x = \infty } ) \\ &=\pi \left[ {0 – ( – 1) – (\frac{\pi }{2} – \frac{\pi }{4})} \right] = \pi (1 – \frac{\pi }{4}) ∎ \end{array}$

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此題考點旋轉體積分

第六題 (10 分)
Consider the power series $\sum\limits_{n = 1}^\infty {\frac{{( – 3)^n (x + 2)^2 }}{{\sqrt {n + 1} }}}$. Find its maximal domain of convergence.

提示

求收斂區間,可從八大審斂法擇其一找出收斂半徑後,再去驗證兩端點是否收斂。

解答

$\begin{array}{cl} \text{radius of convergence} & =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}{{a_n }}} \right|}} \\ & =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{( – 3)^{n + 1} }}{{\sqrt {n + 2} }}}}{{\frac{{( – 3)^n }}{{\sqrt {n + 1} }}}}} \right|}} \\&= \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{{3\sqrt {n + 1} }}{{\sqrt {n + 2} }}}} = \frac{1}{3} \end{array}$

At $x=-\frac{5}{3}$
$\sum\limits_{n = 1}^\infty {\frac{{( – 3)^n (x + 2)^2 }}{{\sqrt {n + 1} }}}=\sum\limits_{n = 1}^\infty {\frac{{( – 3)^n (\frac{1}{3})^n }}{{\sqrt {n + 1} }}} = \sum\limits_{n = 1}^\infty {( – 1)^n \frac{1}{{\sqrt {n + 1} }}}$
$\because \ \frac{1}{\sqrt{n+1}}$ 遞減至 $0$
$\therefore \sum\limits_{n = 1}^\infty {( – 1)^n \frac{1}{{\sqrt {n + 1} }}}$ convergences. (by alternating series test)

At $x=-\frac{7}{3}$
$\sum\limits_{n = 1}^\infty {\frac{{( – 3)^n (x + 2)^2 }}{{\sqrt {n + 1} }}}=\sum\limits_{n = 1}^\infty {\frac{{( – 3)^n ( – \frac{1}{3})^n }}{{\sqrt {n + 1} }}} = \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}}$
$\because \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{\sqrt {n + 1} }}}}{{\frac{1}{{\sqrt n }}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt {n + 1} }} = 1$ and $\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt n }}}$ divergence (by p-series)
$\therefore \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}}$ diverges. (by limit comparison test)

interval of convergence: $( – \frac{7}{3}, – \frac{5}{3}] ∎$

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此題考點冪級數

第七題 (10 分)
Recall that $\tanh x = \frac{{e^x – e^{ – x} }}{{e^x + e^{ – x} }}$, $x\in \mathbb{R}$ is one to one and has the inverse function $f(x)=\tanh ^{-1}x$. Find the Maclaurin series of $f(x)=\tanh^{-1}x$.

提示

可以先將 $\tanh x$ 的反函數推出,再配合已知的馬克勞林級數得到答案。

解答

$\begin{array}{cl} ◎\ \ln (1 + x) & =\int_0^x {\frac{1}{{1 + t}}} dt,\ \left| t \right| < 1,\ \left| x \right| < 1 \\ & =\int_0^x {(1 – t + t^2 – t^3 + t^4 – \cdots )} dt \\ & =\left. {t – \frac{{t^2 }}{2} + \frac{{t^3 }}{3} – \frac{{t^4 }}{4} + \cdots } \right|_{t = 0}^{t = x} \\&=x – \frac{{x^2 }}{2} + \frac{{x^3 }}{3} – \frac{{x^4 }}{4} + \cdots \end{array}$

Let $y=\tanh^{-1}x$
$\Rightarrow x = \tanh y = \frac{{e^y – e^{ – y} }}{{e^y + e^{ – y} }}$
$\Rightarrow xe^y + xe^{ – y} = e^y – e^{ – y}$
$\Rightarrow xe^{2y} + x = e^{2y} – 1$
$\Rightarrow (e^y )^2 (x – 1) = – 1 – x$
$ \begin{array}{cl} \Rightarrow e^y & =\sqrt {\frac{{1 + x}}{{1 – x}}} \end{array}$
$\begin{array}{cl} \Rightarrow y & =\ln \sqrt {\frac{{1 + x}}{{1 – x}}} = \frac{1}{2}\ln (\frac{{1 + x}}{{1 – x}}) \\ & =\frac{1}{2}(\ln (1 + x) – \ln (1 – x)) \\ & =\frac{1}{2}\left[ {(x – \frac{{x^2 }}{2} + \frac{{x^3 }}{3} – \frac{{x^4 }}{4} + \cdots ) – (x + \frac{{x^2 }}{2} + \frac{{x^3 }}{3} + \frac{{x^4 }}{4} + \cdots )} \right] \\&=\frac{1}{2}\left[ {2(x + \frac{{x^3 }}{3} + \frac{{x^5 }}{5} + \cdots )} \right] = x + \frac{{x^3 }}{{\rm{3}}}{\rm{ + }}\frac{{x^5 }}{{\rm{5}}} + \cdots = \sum\limits_{k = 1}^\infty {\frac{{x^{2k – 1} }}{{2k – 1}}} ∎ \end{array}$

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此題考點泰勒級數與泰勒定理雙曲函數

第八題 (10 分)
Find the absolute maximum of $f(x,y)=e^{-xy}$ on the region $x^2+4y^2\leq 1$.

提示

限制條件的邊界(等式)使用拉格朗日重數法,區域內部使用二次微分檢驗法。

解答

On $x^2+4y^2<1$

$\nabla f = ( – ye^{ – xy} , – xe^{ – xy} )\mathop = \limits^{{\rm{Let}}} (0,0)$

$\Rightarrow \left\{ {\begin{array}{l}
– ye^{ – xy} = 0 \\ – xe^{ – xy} = 0 \\ \end{array}} \right. \Rightarrow (x,y) = (0,0)$
$\Rightarrow {\rm{D}}f(0,0) = \left| {\begin{array}{cc}
0 & { – 1} \\
{ – 1} & 0 \\
\end{array}} \right| = – 1 < 0$
$\Rightarrow (0,0)$ is a saddle point.

On $x^2+4y^2=1$, let $g(x,y)=x^2+4y^2=1$
Let $\nabla f + \lambda \nabla g = 0$
$\Rightarrow ( – ye^{ – xy} , – xe^{ – xy} ) + \lambda (2x,8y) = (0,0)$
$\begin{array}{cl} \Rightarrow & \left\{ {\begin{array}{c}
{ – ye^{ – xy} + 2\lambda x = 0 \cdots (5)} \\
{ – xe^{ – xy} + 8\lambda y = 0 \cdots (6)} \\
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{c}
{ – xye^{ – xy} + 2\lambda x^2 = 0 \cdots (1)} \\
{ – xye^{ – xy} + 8\lambda y^2 = 0 \cdots (2)} \\
\end{array}} \right.\\ & x^2+4y^2=1 \end{array}$

$\begin{array}{cl} (1)+(2) & \Rightarrow – 2xye^{ – xy} + 2\lambda \overbrace {(x^2 + 4y^2 )}^1 = 0 \\ & \Rightarrow xye^{ – xy} = \lambda \Rightarrow \left\{ {\begin{array}{l}
{ye^{ – xy} = \frac{\lambda }{x} \cdots (3)} \\
{xe^{ – xy} = \frac{\lambda }{y} \cdots (4)} \\
\end{array}} \right.\end{array}$
$(3)$ 代入 $(5)\ \Rightarrow – \frac{\lambda }{x} + 2\lambda x = 0 \Rightarrow \lambda (2x – \frac{1}{x}) = 0$
$(4)$ 代入 $(6)\ \Rightarrow – \frac{\lambda }{y} + 2\lambda y = 0 \Rightarrow \lambda (8y – \frac{1}{y}) = 0$
If $\lambda =0$
代入 $(5),\ (6) \Rightarrow \left\{ {\begin{array}{c}
{ – ye^{ – xy} = 0} \\
{ – xe^{ – xy} = 0} \\
\end{array} \Rightarrow \left\{ {\begin{array}{c}
{x = 0} \\{y = 0} \\ \end{array}} \right.{\rm{ }} \to \leftarrow } \right. \ (x^2+4y^2=1)$

Hence $\lambda \ne 0$
$\Rightarrow (x,y) = \pm (\frac{1}{{\sqrt 2 }},\frac{1}{{2\sqrt 2 }})$ or $\pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{2\sqrt 2 }})$
$\begin{array}{cl} \because & f( \pm (\frac{1}{{\sqrt 2 }},\frac{1}{{2\sqrt 2 }})) = e^{ – \frac{1}{{\sqrt 2 }} \cdot \frac{1}{{2\sqrt 2 }}} = e^{ – \frac{1}{4}} \\ & f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{2\sqrt 2 }})) = e^{\frac{1}{{\sqrt 2 }} \cdot \frac{1}{{2\sqrt 2 }}} = e^{\frac{1}{4}} \\ \therefore & \text{absolutely maximum}=e^{\frac{1}{4}} ∎ \end{array}$

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此題考點拉格朗日乘數法相對、絕對極值、鞍點

第九題 (10 分)
Evaluate the integral $\int_C {y^3 dx} – x^3 dy$ where $C$ is the circle $x^2+y^2=8$ with positive orientation.

提示

本題為基本線積分,由於路徑和圓有關,可利用極座標變換來積分;亦滿足格林定理的條件,也可以使用。

解答

法一 參數法
$C:x^2 + y^2 = 8 \Rightarrow \left\{ {\begin{array}{c}
{x = 2\sqrt 2 \cos t} \\
{y = 2\sqrt 2 \sin t} \\
\end{array}} \right. ,\ 0\leq t\leq 2\pi$
$\Rightarrow \int_C {y^3 dx} – x^3 dy$
$= \int_0^{2\pi } {(2\sqrt 2 \sin t)^3 ( – 2\sqrt 2 \sin t)dt} – (2\sqrt 2 \cos t)^3 (2\sqrt 2 \cos t)dt$
$= – 64\int_0^{2\pi } {(\sin ^4 t + \cos ^4 t)} dt$
$= – 64\int_0^{2\pi } {(1 – 2\sin ^2 t\cos ^2 t)} dt$
$= – 64\int_0^{2\pi } {(1 – \frac{1}{2}\sin ^2 2t)} dt$
$= – 64\int_0^{2\pi } {(1 – \frac{1}{2} \cdot \frac{{1 – \cos 4t}}{2})} dt$
$= – 64\int_0^{2\pi } {(1 – \frac{1}{4} + \underbrace {\frac{{\cos 4t}}{4}}_{ {\rm{0}}})} dt$
$= – 64 \times \frac{{\rm{3}}}{{\rm{4}}} \times \underbrace {\int_0^{{\rm{2}}\pi } {dt} }_{2\pi } = – 96\pi ∎ $

法二 Green’s theorem

$\oint_C {\overbrace {y^3 }^P} dx – \overbrace {x^3 }^Qdy$
$=\iint_{\Omega} -3x^2-3y^2 dA$
$=-3\iint_{\Omega}x^2+y^2 dA$
$=-3\int_0^2\int_0^{2\sqrt{2}}r^2\cdot rdrd\theta$
$=3\int_0^{2\pi } {\left. {\frac{{r^4 }}{4}} \right|_{r = 0}^{r = 2\sqrt 2 } } d\theta$
$=3 \times \frac{1}{4} \times (2\sqrt 2 )^4 \times \int_0^{2\pi } {d\theta } = – 96\pi ∎$

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此題考點線積分格林定理

第十題 (10 分)
Let $\textbf{F}(x,y,z)=x\textbf{i}+y\textbf{j}-2z\textbf{k}$. Suppose that the closed surface $S$ is defined by the paraboloid $y=x^2+z^2$ with $0\leq y\leq 1$. Find the flux of $\textbf{F}$ outward across $S$.

提示

通量之積分範圍是拋物面的一部份,直接用定義不好做時,通常會套用高斯散度定理,但條件是封閉曲面,所以必須先確保曲面是封閉的,必要時可以自行補足。

解答

第一種可能
$S$ 為封閉曲面,令 $\Omega$ 為 $S$ 內部的體積
$S:\left\{ {(x,y,z) \in \mathbb{R}^3 :y = x^2 + z^2 ,0 \le y \le 1} \right\} \cup \left\{ {(x,y,z) \in \mathbb{R}^3 :x^2 + z^2 = 1} \right\}$
By Gauss’s theorem
$\Rightarrow \iint_S \textbf{F}\cdot\hat{n}dS=\iiint_{\Omega}\text{div}\cdot\textbf{F}dV=\iiint_{\Omega}\underbrace {(1 + 1 – 2)}_0dV=0 ∎$

第二種可能 題目多打 closed

By Gauss’s theorem
$\Rightarrow \iint_{S\cup D} \textbf{F}\cdot\hat{n}dS=\iiint_{\Omega}\text{div}\cdot\textbf{F}dV=0$
$\begin{array}{cl} \because & \iint_{S\cup D} \textbf{F}\cdot\hat{n}dS=\iint_{S} \textbf{F}\cdot\hat{n}dS+\iint_{D} \textbf{F}\cdot\hat{n}dS \\ &\text{and }\iint_{D} \textbf{F}\cdot\hat{n}dS =\iint_{D} (x,y,-2z)\cdot (0,1,0)dS=\iint_{D} ydS=\pi \\ \therefore & \iint_{S} \textbf{F}\cdot\hat{n}dS=\iint_{S\cup D} \textbf{F}\cdot\hat{n}dS-\pi=0-\pi=-\pi  ∎\\ \end{array}$

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此題考點梯度、旋度、散度散度定理

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