台大轉學考微積分 109 C 卷解答

此極限為不定型，可以使用羅必達法則。

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt[x]{2} – 1}}{{{\rm{1/}}x}} \mathop = \limits^{{\rm{Let }}\ t = \frac{1}{x}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{2^t – 1}}{t}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{2^t \ln 2}}{1} = \ln 2　∎$

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：羅必達法則

積分的上下界有未知數，想到微積分基本定理。

$\begin{array}{cl} f'(x) & =e^{g(x)} g'(x) \\ & =e^{g(x)} \cdot \frac{d}{{dx}}\int_4^{x^2 } {\frac{t}{{1 + t^3 }}} dt \\ & =e^{g(x)} \cdot \frac{{x^2 }}{{1 + x^6 }} \cdot (2x) \end{array}$
$f'(2) = e^{\overbrace {g(2)}^0} \cdot \frac{4}{{1 + 64}} \cdot 4 = \frac{{16}}{{65}}　∎$

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此題考點：微積分基本定理

當函數給參數式的時候可以讓 $\frac{dy}{dx}$ 跟參數的微分 $\frac{dx}{dt}$, $\frac{dy}{dt}$ 連結起來

$\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{\sin \theta }}{{2 – \cos \theta }}$
$\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{2}} = \frac{{\sin \frac{\pi }{2}}}{{2 – \cos \frac{\pi }{2}}} = \frac{1}{2}$

At $\theta =\frac{\pi}{2}$, $(x,y)=(\pi -\sin \frac{\pi}{2},2-\cos \frac{\pi}{2})=(\pi -1, 2)$
所求: $x-2y=\pi-5　∎$

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此題考點：曲線分析(上)

這題是給參數式求面積，雖然不是習慣的變數，但可經由變數變換將面積公式變成題目的格式。

$\begin{array}{cl} \int_a^b {f(x)} dx & =\int_0^{2\pi }{\underbrace {(2 – \cos \theta )}_{y = f(x)}}\underbrace {(2 – \cos \theta )d\theta }_{dx} \\ & =\int_0^{2\pi } {4 – 4\cos \theta + \frac{1}{2}(1 + \cos 2\theta )} d\theta \\&=\left[ {4\theta – 4\underbrace {\sin \theta }_0\frac{\theta }{2} + \underbrace {\frac{{\sin 2\theta }}{4}}_0}\right]_{\theta = 0}^{\theta = 2\pi }\\&=8\pi+\pi=9\pi 　∎\end{array}$

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此題考點：曲線分析(上)

遇旋轉體積分可以使用剝殼法或圓盤法，本題使用剝殼法。

$\begin{array}{cl} V & =\int_{ – 1}^1 {2\pi (1 – x)(2 – {\rm{2}}x^2 )} dx \\ & =4\pi \int_{ – 1}^1 {(1 – x)(1 – x^2 )} dx \\ & =4\pi \int_{ – 1}^1 {{\rm{1}} – \underbrace x_0 – x^2 + \underbrace {x^3 }_0} dx \\&=8\pi \left[ {x – \frac{{x^3 }}{3}} \right]_{x = 0}^{x = 1} = 8\pi \times \frac{2}{3} = \frac{{16\pi }}{3} 　∎\end{array}$

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此題考點：旋轉體積分

作圖後發現圖形有對稱性，分段會好算很多。

令 $l$ 為 $(1,1)$ 至 $(2,0)$ 的曲線長。
$\begin{array}{cl} \text{arc length} & =4l \\ & =4\int_0^1 {\sqrt {1 + (2x)^2 } } dx \end{array}$

Let $2x = \tan \theta \Rightarrow 2dx = \sec ^2 \theta d\theta$
$\begin{array}{cl} \Rightarrow 4l & =4\int_0^\alpha {\sec \theta \cdot \frac{{\sec ^2 \theta d\theta }}{2}} \\ & =2\int_0^\alpha {\sec ^3 \theta } d\theta \\ & =2\left[ {\frac{{\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|}}{2}} \right]_{\theta = 0}^{\theta = \alpha } \\&=\sec \alpha \tan \alpha + \ln \left|{\sec \alpha + \tan \alpha } \right| \\&= 2\sqrt 5 + \ln (2 + \sqrt 5 )　∎ \end{array}$

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此題考點：曲線分析(上)

二變數積分不好算時，可以使用極座標變換或變數變換，此題使用變數變換，不論使用哪種都需要記得乘上 Jocobian: $|J|$。

Let $\left\{ {\begin{array}{c}
{u = y + x} \\
{v = y – x} \\
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{c}
{\frac{{u + v}}{2} = y} \\
{\frac{{u – v}}{2} = x} \\
\end{array}} \right.$

$|J|=\frac{{dxdy}}{{dudv}} = {\rm{abs}}(\left|{\begin{array}{c}
{x_u } & {x_v } \\
{y_u } & {y_v } \\
\end{array}} \right|)$
$\Rightarrow (\frac{{dudv}}{{dxdy}})^{ – 1} = {\rm{abs}}(\left| {\begin{array}{c}
{u_x } & {u_y } \\
{v_x } & {v_y } \\
\end{array}} \right|)^{ – 1} = {\rm{abs}}(\left| {\begin{array}{c}
1 & 1 \\
{ – 1} & 1 \\
\end{array}} \right|)^{ – 1} = \frac{1}{2}$
$\begin{array}{cl} Area & =\int_1^2 {\int_{ – u}^u {\cos (\frac{v}{u})\underbrace {\frac{1}{2}}_{\left| {\rm{J}}\right|}} } dvdu \\ & =\frac{1}{2}\int_1^2 {\left[ {u\sin (\frac{v}{u})} \right]} _{v = – u}^{v = u} du \\ & =\frac{1}{2}\int_1^2 {u(\sin (1) – \sin ( – 1))} du \\&=\sin (1) \cdot \int_1^2 u du \\&=\sin (1)\left[ {\frac{{u^2 }}{2}} \right]_{u = 1}^{u = 2} = \sin (1) \cdot (\frac{4}{2} – \frac{1}{2}) = \frac{3}{2}\sin (1)　∎ \end{array}$

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此題考點：二重積分座標變換

本題沒有範圍限制，因此使用二次微分檢驗法。

$\nabla f(x,y) = (4x^3 – 4y,4y^3 – 4x)\mathop = \limits^{{\rm{Let}}} (0,0)$
$\Rightarrow$ critical point $(0,0)$, $(1,1)$ and $(-1,-1)$

$\text{H}f=\left[ {\begin{array}{c}
{12x^2 } & { – 4} \\
{ – 4} & {12y^2 } \\
\end{array}} \right]$
$\begin{array}{cl} (1) & {\rm{D}}f(0,0) = \left| {\begin{array}{c}
0 & { – 4} \\
{ – 4} & 0 \\
\end{array}} \right| = – 16 < 0 \Rightarrow \text{saddle.}\\ (2) & {\rm{D}}f(1,1) = \left| {\begin{array}{c}{12} & { – 4} \\
{ – 4} & {12} \\ \end{array}} \right| = 144 – 16 > 0 \Rightarrow \text{min.}\ f(1,1)=-1 \\(3) & {\rm{D}}f( – 1, – 1) = \left| {\begin{array}{c}{12} & { – 4} \\
{ – 4} & {12} \\ \end{array}} \right| > 0 \Rightarrow \text{min.}\ f(-1,-1)=-1　∎\end{array}$

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此題考點：微分求極值法

本題的限制條件不只是一條線，因此使用 Lagrange 乘數法與二次微分檢驗法。

On $x^2+y^2=1$, let $g(x,y)=x^2+y^2$
Let $\nabla f+\lambda \nabla g =0$
$\Rightarrow (4x^3 – 4y,4y^3 – 4x) + \lambda (2x,2y) = (0,0)$
$\begin{array}{cl} \Rightarrow & \left\{ {\begin{array}{c}
{2x^3 – 2y + \lambda x = 0 \cdots (1)} \\
{2y^3 – 2x + \lambda y = 0 \cdots (2)} \\
\end{array}} \right.\\ & x^2+y^2=1 \end{array}$
$\begin{array}{cl} (1)+(2) & \Rightarrow 2(x^3 + y^3 ) – 2(x + y) + \lambda (x + y) = 0\\ & \Rightarrow 2(x + y)(x^2 – xy + y^2 ) + (x + y)(\lambda – 2) = 0 \\ & \Rightarrow (x + y)\left[ {2(x^2 – xy + y^2 ) + \lambda – 2} \right] = 0 \\&\Rightarrow (x + y)(\lambda – 2xy) = 0 \\&\Rightarrow x=-y \text{ or }\lambda=2xy \end{array}$
If $x=-y$
$\because x^2+y^2=1$
$\therefore (x,y)=\pm (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$
If $\lambda=2xy$,
$\mathop \Rightarrow \limits^{(1)} 2x^3 – 2y + 2x^2 y = 0 \Rightarrow x^3 – y + x^2 y = 0 = y(x^2 – 1) – y^3$
$\because x^2+y^2=1$
$\therefore x^3-y^3=0$
$\Rightarrow x^3=y^3\Rightarrow x=y$
$\because x^2+y^2=1$
$\therefore\ (x,y)=\pm (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

For $x^2+y^2<1$
$\nabla f = (4x^3 – 4y,4y^3 – 4x)\mathop = \limits^{{\rm{Let}}} (0,0)$
$\Rightarrow \left\{ \begin{array}{cl} x^3-y=0 & \Rightarrow y=x^3 \\ y^3-x=0& \end{array} \right.\Rightarrow (x^3)^3-x=0$
$x(x^3-1)=0$
$x=0 \text{ or }x=\pm 1$
$\Rightarrow$ critical points: $(0,0)$, $(1,1)$ and $(-1,-1)$
${\rm{H}}f = \left[ {\begin{array}{c}
{12x^2 } & { – 4} \\
{ – 4} & {12y^2 } \\ \end{array}} \right]$
$\Rightarrow \text{D}f(0,0)=\left| {\begin{array}{c}
0 & { – 4} \\
{ – 4} & 0 \\
\end{array}}\right| = – 16 < 0$: saddle

$f(\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}) = \frac{1}{4} + \frac{1}{4} – 4 \cdot \frac{1}{2} + 1 = \frac{1}{2} – 1 = – \frac{1}{2}$
$f( – \frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }}) = – \frac{1}{2}$
$f(\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }}) = \frac{1}{4} + \frac{1}{4} + 4 \cdot \frac{1}{2} + 1 = \frac{7}{2}$
$f( – \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}) = \frac{7}{2}$
$\left\{ {\begin{array}{c}
{{\rm{absolute\ minima}}: – \frac{1}{2}} \\
{{\rm{absolute\ maxima}}:\text{ }\frac{7}{2}} \\
\end{array}} \right.　∎$

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此題考點：微分求極值法、Lagrange 乘數法

為一階微分方程常考的題型之一，本題使用公式解。

◎ 解 $y’+p(x)y=q(x)$ 型
$y’\underbrace { – \frac{1}{x}}_{p(x)}y = \underbrace {x\sin x}_{q(x)}$
$\Rightarrow I(x) = e^{\int { – \frac{1}{x}} dx} = e^{ – \ln x} = \frac{1}{x}$
$\Rightarrow \frac{1}{x}y’ – \frac{1}{{x^2 }}y = \sin x$
$\Rightarrow (\frac{1}{x}y)’ = \sin x$
$\Rightarrow \frac{y}{x} = – \cos x + C$,
$\because \ y(\pi)=2\pi$
$\therefore \ \frac{{2\pi }}{\pi } = – \cos \pi + C$
$\Rightarrow 2=1+C\Rightarrow C=1$
$\Rightarrow \frac{y}{x} = – \cos x + 1 \Rightarrow y = x – x\cos x　∎$

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此題考點：常見的一階微分方程題型及解法