台綜大轉學考微積分 109 A 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〕 $\mathop {\lim }\limits_{n \to 0} \frac{{3n + 2}}{{2n + 1}}$.
〔2〕 $\mathop {\lim }\limits_{x \to 0} \frac{{\cos (2x) – 1}}{{x^2 }}$.

提示

第一小題是基本題;第二小題為不定型,可以使用羅必達法則;或將 $\cos x$ 泰勒展開後再去零因子

解答

〔1〕$\mathop {\lim }\limits_{n \to \infty } \frac{{3n + 2}}{{2n + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3 + \frac{2}{n}}}{{2 + \frac{1}{n}}} = \frac{3}{2} ∎$.
〔2〕

法一 羅必達法則

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos (2x) – 1}}{{x^2 }}= \mathop {\lim }\limits_{x \to 0} e^{\ln (1 – 7x)^{\frac{4}{x}} }= e^{\mathop {\lim }\limits_{x \to 0} \frac{{4\ln (1 – 7x)}}{x}}\mathop = \limits^{\rm{L}} e^{\mathop {\lim }\limits_{x \to 0} \frac{{4 \cdot \frac{{ – 7}}{{1 – 7x}}}}{1}}= e^{ – 28}$

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

法二 泰勒展開式

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos (2x) – 1}}{{x^2 }}$

$=\mathop {\lim }\limits_{x \to 0} \frac{{\left[ {1 – \frac{{(2x)^2 }}{{2!}} + \frac{{(2x)^4 }}{{4!}} – \frac{{(2x)^6 }}{{6!}} + \cdots } \right] – 1}}{{x^2 }}$

${\rm{ = }}\mathop {\lim }\limits_{x \to 0} \frac{{ – 2x^2 }}{{x^2 }} = – 2 ∎$

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此題考點老大比較法羅必達法則泰勒展開式

第二題 (10 分)
A curve in $\mathbb{R}^2$ is given parametrically by $\left\{ \begin{array}{l}{x = t^2 + 2t + 3} \\ y = t^4 – 3t^3 \\ \end{array}\right.$ for all $t>0$. Find $\frac{dy}{dx}$ at the point $(6,-2)$.

提示

$x$ 和 $y$ 皆是 $t$ 的函數,$\frac{dy}{dx}$ 與 $\frac{dy/dt}{dx/dt}$ 等價。

解答

$\frac{{dy}}{{dx}}{\rm{ = }}\frac{{dy/dt}}{{dx/dt}} = \frac{{4t^3 – 9t^2 }}{{2t + 2}}$

Let $\left\{ \begin{array}{l} x=t^2+2t+3\Rightarrow t^2+2t-3=0\\ y=t^4-3t^3=-2 \end{array} \right.$

$\Rightarrow t=-3$ (負不合) or $=t=1$

$\frac{dy}{dx}=\left. {\frac{{dy}}{{dx}}} \right|_{(6, – 2)} = \left. {\frac{{4t^3 – 9t^2 }}{{2t + 2}}}\right|_{t = 1} = – \frac{5}{4} ∎$

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此題考點曲線分析

第三題 (10 分)
Let $x>0$ and $\Delta ABC$ be a triangle whose side lengths are $\overline {BC} = 5$, $\overline {AC} = 4$, $\overline {AB} = 3$. Choose a point $P$ on $\overline {AC}$, and a point on $\overline {BC}$, and a point on $\overline {AC}$ so that $\frac{{\overline {BP} }}{{\overline {AP} }} = \frac{{\overline {CQ} }}{{\overline {BQ} }} = \frac{{\overline {AR} }}{{\overline {CR} }} = x$
Let $f(x)$ be the area of $\Delta PQR$. Find the critical point and the minimum of $f(x)$

提示

作圖找出面積的表示法後,使用微分求極值。

解答

Let $\left\{ {\begin{array}{c}\overline {AB} =a(1 + x) = 3 \\ \overline {BC} =b(1 + x) = 5 \\ \overline {AC} =c(1 + x) = 4 \\ \end{array}} \right.$

$\Rightarrow a=\frac{3}{1+x}$, $b=\frac{5}{1+x}$, $c=\frac{4}{1+x}$

$\Delta ABC = \frac{{4 \times 3}}{2} = 6$

$\Delta APR = \frac{{a \times cx}}{2} = \frac{{acx}}{2}$

$\Delta BPQ = \frac{{ax \times 4 \times \frac{1}{{1 + x}}}}{2} = \frac{{2ax}}{{1 + x}}$

$\Delta CRQ = \frac{{c \times 3 \times \frac{1}{{1 + x}}}}{2} = \frac{{3cx}}{{2(1 + x)}}$

$\begin{array}{cl} \Rightarrow \Delta PQR & =6 – \frac{{acx}}{2} – \frac{{2ax}}{{1 + x}} – \frac{{3cx}}{{2(1 + x)}} \\ & =6 – \frac{x}{2} \cdot \frac{3}{{1 + x}} \cdot \frac{4}{{1 + x}} – \frac{{2x}}{{1 + x}} \cdot \frac{3}{{1 + x}} – \frac{{3x}}{{2(1 + x)}} \cdot \frac{4}{{1 + x}} \\ & =6 – \frac{{6x}}{{(1 + x)^2 }} – \frac{{6x}}{{(1 + x)^2 }} – \frac{{6x}}{{(1 + x)^2 }} \\&=6 – \frac{{18x}}{{(1 + x)^2 }} = f(x) \end{array}$

$f'(x) = – 18 \cdot \frac{{1 \cdot (1 + x)^2 – x \cdot 2 \cdot (1 + x)}}{{(1 + x)^4 }}\mathop = \limits^{{\rm{Let}}} 0$

$\Rightarrow 1 + x – 2x = 0 \Rightarrow x = 1$ (critical point)

$f'(x) = – 18 \cdot \frac{{1 – x}}{{(1 + x)^3 }}\left\{ \begin{array}{cl} >0, & x>1 \\ =0, & x=1 \\ <0, & 0<x<1 \end{array} \right.$

$\Rightarrow $minima$=f(1) = 6 – \frac{{18}}{4} = \frac{3}{2} ∎$

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此題考點微分求極值法

第四題 (10 分)
Find the radius of the convergence of the power series $\sum\limits_{n = 0}^\infty {\frac{{(2n)^n }}{{n!}}x^n }$ .

提示

本題要求冪級數的收斂半徑,一般求收斂半徑的公式有兩個,只要極限存在即可套用,一個是需要計算開 $n$ 次方的極限,另一個是需要計算相鄰兩項比例的極限。

解答

$\begin{array}{cl} r & =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{(2(n + 1))^{n + 1} }}{{(n + 1)!}}}}{{\frac{{(2n)^n }}{{n!}}}}} \right|}} \\ & =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{{2n + 2}}{{n + 1}} \cdot (\frac{{2n + 2}}{{2n}})^n }} \\ & =\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \underbrace {\frac{{2n + 2}}{{n + 1}}}_2 \cdot \underbrace {(1 + \frac{1}{n})^n }_e}} \\ &=\frac{1}{{2e}} ∎ \end{array}$

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此題考點根值審斂法

第五題 (10 分)
Evaluate the improper integral $\int_0^{\frac{\pi }{2}} {\cos x \cdot (\ln \cos \frac{x}{2} + \ln \sin \frac{x}{2})} dx$ .

提示

三角函數的積分可嘗試使用分部積分。

解答

$\int_0^{\frac{\pi }{2}} {\cos x \cdot (\ln \cos \frac{x}{2} + \ln \sin \frac{x}{2})} dx$

$=\int_0^{\frac{\pi }{2}} {\cos x \cdot \ln \frac{{\sin x}}{2}} dx$

$=\left. {\sin x \cdot \ln \frac{{\sin x}}{2}} \right|_0^{\frac{\pi }{2}} – \int_0^{\frac{\pi }{2}} {\cos } xdx$

$=\ln \frac{1}{2} – \underbrace {\mathop {\lim }\limits_{x \to 0} \sin x \cdot \ln \frac{{\sin x}}{2}}_0 – \underbrace {\left[ {\sin x}\right]_{x = 0}^{x = \frac{\pi }{2}} }_{\rm{1}}={\rm{ln}}\frac{{\rm{1}}}{{\rm{2}}} – 1 ∎$

$\begin{array}{cl} ◎\ \mathop {\lim }\limits_{x \to 0} \sin x \cdot \ln \frac{{\sin x}}{2} & =\mathop {\lim }\limits_{x \to 0} \frac{{\ln \frac{{\sin x}}{2}}}{{\csc x}}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\cos x}}{2} \cdot \frac{2}{{\sin x}}}}{{ – \csc x \cdot \cot x}} \\ & = \mathop {\lim }\limits_{x \to 0} ( – \sin x) = 0 \end{array}$

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此題考點分部積分法

第六題 (10 分)
Let $g:(0,\infty)\to \mathbb{R}$ be twice differentiable function. Assume that $g(1)=1$, $g'(1)=3$, $g^{\prime \prime }(1)=-4$.
Define a real value function $h$ on $\mathbb{R}^3 \backslash \{ (0,0,0)\}$ by $h(x,y,z) = g(\sqrt {x^2 + y^2 + z^2 } )$.
Calculate $\frac{{\partial ^2 h}}{{\partial x^2 }}(P) + \frac{{\partial ^2 h}}{{\partial y^2 }}(P) + \frac{{\partial ^2 h}}{{\partial z^2 }}(P)$ where $P = (\frac{2}{3},\frac{2}{3},\frac{{ – 1}}{3})$.

提示

本題為多變數函數偏微分的基本題,要注意連鎖律。

解答

$\begin{array}{cl} \frac{{\partial h}}{{\partial x}} & =g'(\sqrt {x^2 + y^2 + z^2 } ) \cdot \frac{1}{2} \cdot \frac{{2x}}{{\sqrt {x^2 + y^2 + z^2 } }} \\ & =g'(\sqrt {x^2 + y^2 + z^2 } ) \cdot \frac{x}{{\sqrt {x^2 + y^2 + z^2 } }} \end{array}$

$\frac{{\partial h}}{{\partial y}} = g'(\sqrt {x^2 + y^2 + z^2 } ) \cdot \frac{y}{{\sqrt {x^2 + y^2 + z^2 } }}$

$\frac{{\partial h}}{{\partial z}} = g'(\sqrt {x^2 + y^2 + z^2 } ) \cdot \frac{z}{{\sqrt {x^2 + y^2 + z^2 } }}$

$\begin{array}{rl} \frac{{\partial ^2 h}}{{\partial x^2 }}= & g^{\prime \prime }(\sqrt {x^2 + y^2 + z^2 } ) \cdot (\frac{x}{{\sqrt {x^2 + y^2 + z^2 } }})^2 \\ & +g'(\sqrt {x^2 + y^2 + z^2 } )\frac{{\sqrt {x^2 + y^2 + z^2 } – x \cdot \frac{x}{{\sqrt {x^2 + y^2 + z^2 } }}}}{{x^2 + y^2 + z^2 }}\\ =&{g^{\prime \prime}(\sqrt {x^2 + y^2 + z^2 } )\frac{{x^2 }}{{x^2 + y^2 + z^2 }} + g'(\sqrt {x^2 + y^2 + z^2 } )\frac{{y^2 + z^2 }}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}}\cdots (1) \end{array}$

$\begin{array}{rl} {\frac{{\partial ^2 h}}{{\partial y^2 }}} = & {g^{\prime \prime}(\sqrt {x^2 + y^2 + z^2 } )\frac{{y^2 }}{{x^2 + y^2 + z^2 }} + g'(\sqrt {x^2 + y^2 + z^2 } )\frac{{x^2 + z^2 }}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}} \cdots (2) \\ {\frac{{\partial ^2 h}}{{\partial z^2 }}} =& {g^{\prime \prime}(\sqrt {x^2 + y^2 + z^2 } )\frac{{z^2 }}{{x^2 + y^2 + z^2 }} + g'(\sqrt {x^2 + y^2 + z^2 } )\frac{{x^2 + y^2 }}{(x^2 + y^2 + z^2 )^{\frac{3}{2}}}} \cdots (3) \end{array}$

$(1)+(2)+(3)$

$\Rightarrow \frac{{\partial ^2 h}}{{\partial x^2 }} + \frac{{\partial ^2 h}}{{\partial y^2 }} + \frac{{\partial ^2 h}}{{\partial z^2 }} = g^{\prime \prime}(\sqrt {x^2 + y^2 + z^2 } ) + g'(\sqrt {x^2 + y^2 + z^2 } )\frac{2}{{\sqrt {x^2 + y^2 + z^2 } }}$

$\Rightarrow \frac{{\partial ^2 h}}{{\partial x^2 }}(P) + \frac{{\partial ^2 h}}{{\partial y^2 }}(P) + \frac{{\partial ^2 h}}{{\partial z^2 }}(P) = g^{\prime \prime}(1) + g'(1) \cdot \frac{2}{1} = – 4 + 3 \times 2 = 2 ∎$

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此題考點偏微分運算律

第七題 (10 分)
Let $S$ be the surface defined by the equation $g(1)=1$, $g'(1)=3$, $x\cos (xy)+z^2y^4-7xz=1$ and $P(0,1,1)$ be a point on $S$. Find an equation that defines the tangent plane to $S$ at $P$ and a parametric equation of the normal line to $S$ at $P$.

提示

本題為曲面上求切平面與法線的基本題。

解答

Let $f(x,y,z) = x\cos (xy) + z^2 y^4 – 7xz$
$\nabla f(x,y,z) = (\cos (xy) + x \cdot ( – \sin (xy)) \cdot y – 7z,x( – \sin (xy)) \cdot x + z^2 \cdot 4y^3 ,2zy^4 – 7x)$
$\Rightarrow \overset{\rightharpoonup}{n} \mathbin{/\mkern-5mu/} \nabla f(0,1,1) = (1 – 7,4,2) = ( – 6,4,2)$
取 $\overset{\rightharpoonup}{n}=(3,-2,-1)$
tangent plane: $3x-2y-z=-3$
normal line: $\left\{ \begin{array}{l} {x = 0 + 3t} \\ {y = 1 – 2t} \\ {z = 1 – t} \end{array} \right.$, $t\in \mathbb{R} ∎$

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此題考點等值面與切平面

第八題 (10 分)
Evaluate the double integral $\iint_{R}(y-x)dA$ where $R = \left\{ {(x,y) \in \mathbb{R}^2 :{\rm{ }}1 \le x^2 + y^2 \le 4,x \ge 0} \right\}$.

提示

遇到二重積分,可以嘗試使用極座標變換或交換積分次序。本題積分區域和圓有關,因此使用極座標變換。

解答

$\iint_{R}(y-x)dA$

$=\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\int_1^2 {(r\sin \theta – r\cos \theta )} } \cdot rdrd\theta$

$=\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {(\underbrace {\sin \theta }_0 – \cos \theta )}\underbrace {\int_1^2 {r^2 } dr}_{\frac{7}{3}}d\theta$

$=\frac{{14}}{3}\int_0^{\frac{\pi }{2}} {( – \cos \theta )} d\theta$

$=\frac{{14}}{3}\left[ { – \sin \theta } \right]_{\theta = 0}^{\theta = \frac{\pi }{2}} = \frac{{14}}{3} \times ( – 1) = – \frac{{14}}{3} ∎$

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此題考點二重積分座標變換

第九題 (10 分)
Let $C$ be the curve in $\mathbb{R}^3$ define by the parametric equation $x(t)=\cos (t)$, $y(t)=\sin (t)$, $z(t)=t$ for $0\leq t\leq a$. Suppose that the arc length of $C$ is $\frac{\sqrt{2}}{4}\pi$. Evaluate the line integral $\int_{C} \textbf{F}\cdot d\textbf{r}$ of the vector $\textbf{F}=xz\textbf{i}+yz\textbf{j}+x^3\textbf{k}$ on $\mathbb{R}^3$

提示

本題為基本線積分。

解答

$\begin{array}{cl} \int_C \textbf{F} \cdot d\textbf{r} & =\int_C {(xz,yz,x^3 )} \cdot \overbrace {(dx,dy,dz)}^{(x'(t),y'(t),z'(t))dt} \\ & =\int_0^a {(t\cos t,t\sin t,\cos ^3 t) \cdot ( – \sin t,\cos t,1)} dt \\ & =\int_0^a {( – t\sin t\cos t + t\sin t\cos t + \cos ^3 t)} dt \\&=\int_0^a {\cos ^3 t} dt \end{array}$

$\because$ the arc length is $\frac{\sqrt{2}}{4}\pi$

$\therefore \int_0^a {\sqrt {\underbrace {( – \sin t)^2 + (\cos t)^2 }_1 + 1} } dt$

$\Rightarrow \int_0^a {\sqrt 2 } dt = \frac{{\sqrt 2 }}{4}\pi \Rightarrow a = \frac{\pi }{4}$

$\begin{array}{cl} \int_{C} \textbf{F}\cdot d\textbf{r} & =\int_0^{\frac{\pi }{4}} {\cos ^3 t} dt \\ & =\int_0^{\frac{\pi }{4}} {\underbrace {\cos ^2 t}_{1 – \sin ^2 t}\cos t} dt\ (\text{Let }u=\sin t,\ du=\cos tdt) \\ & =\int_0^{\frac{{\sqrt 2 }}{2}} {(1 – u^2 } )du \\&=\left. {u – \frac{{u^3 }}{3}} \right|_{u = 0}^{u = \frac{{\sqrt 2 }}{2}}\\&=\frac{{\sqrt 2 }}{2} – \frac{1}{3} \cdot \frac{{\sqrt 2 }}{4} = \frac{{5\sqrt 2 }}{{12}} ∎ \end{array}$

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此題考點線積分

第十題 (10 分)
Find the flux of the vector field $\textbf{F}$ on $\mathbb{R}^3$ defined by $\textbf{F}=3x\textbf{i}+2y\textbf{j}+5z\textbf{k}$ through the surface $S = \left\{ {(x,y,\sqrt {1 – x^2 – y^2 } ) \in \mathbb{R}^3 :x^2 + y^2 \le 1} \right\}$ oriented with upward pointing normal vector field.

提示

若區域內向量場處處存在,計算散度可使用高斯散度定理。

解答

By Gauss’s divergence theorem

$\begin{array}{cl} \iint_{S\cup E}\textbf{F}\cdot \hat{n}ds & =\iiint_{\Omega} \text{div}\cdot \textbf{F}dV \\ & =\iint_{S}\textbf{F}\cdot \hat{n}ds + \underbrace {\iint_{E}\textbf{F}\cdot\hat{n}ds}_0 \end{array}$
$\begin{array}{cl} ◎\ \iint_{E} \textbf{F}\cdot \hat{n}ds & =\iint_{E}(3x,2y,5z)\cdot (0,0,-1)ds \\ & =\iint_{x^2+y^2\leq 1,z=0}-5zds=0 \end{array}$

$\begin{array}{cl} \iiint_{\Omega} \text{div}\cdot \textbf{F}dV & =\iiint_{\Omega} (3+2+5)dV \\ & =10\iiint_{\Omega}dV=\frac{20}{3}\pi \end{array}$
$\begin{array}{cl} ◎ \iiint_{\Omega}dV & =\int_0^{2\pi } {\int_0^{\frac{\pi }{2}} {\int_0^1 {r^2 \sin \rho } } } drd\rho d\theta \\ & =\int_0^{2\pi } {\int_0^{\frac{\pi }{2}} {\sin \rho \left[ {\frac{1}{3}r^3 } \right]_{r = 0}^{r = 1} } } d\rho d\theta \\ & =\frac{1}{3}\int_0^{2\pi } {\left[ { – \cos \rho } \right]_{\rho = 0}^{\rho = \frac{\pi }{2}} } d\theta \\&=\frac{1}{3} \cdot 1 \cdot \int_0^{2\pi } {d\theta } = \frac{{20}}{3}\pi \end{array}$

$\Rightarrow \iint_{S}\textbf{F}\cdot \hat{n}ds=\frac{20}{3}\pi ∎$

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此題考點散度定理

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