台大轉學考微積分 107 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)
$\mathop {\lim }\limits_{x \to -\infty } {\rm{(3}}x + \sqrt {9x^2 – x} {\rm{)}} =?$

提示

極限雖然是很基本的題目,遇到趨近於 $-\infty$ 時可以先變換變數,即可繼續用平常的習慣作答。而函數本身是 $\infty-\infty$ 型,通常可以對這個部分做整理進而消除一些項,本題使用有理化。

解答

令 $y=-x$
$\begin{array}{cl} \mathop {\lim }\limits_{x \to -\infty } {\rm{(3}}x + \sqrt {9x^2 – x} {\rm{)}} &=\mathop {\lim }\limits_{y \to \infty } ( – 3y + \sqrt {9y^2 + y} ) \cdot \frac{{3y + \sqrt {9y^2 + y} }}{{3y + \sqrt {9y^2 + y} }} \\ &=\mathop {\lim }\limits_{y \to \infty } \frac{{ – 9y^2 + 9y^2 + y}}{{3y + \sqrt {9y^2 {\rm{ + }}y} }} \\ &=\mathop {\lim }\limits_{y \to \infty } \frac{{y/y}}{{3y/y + \sqrt {9y^2 {\rm{ + }}y} /y}} \\&=\mathop {\lim }\limits_{y \to \infty } \frac{1}{{3 + \sqrt {9{\rm{ + }}\frac{{\rm{1}}}{y}} }} = \frac{1}{{3 + 3}} = \frac{1}{6} ∎\end{array}$

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此題考點極限運算定理(四則運算篇)老大比較法(上)

第二題 (10 分)
${\rm{(}}1 – \frac{1}{4}{\rm{)(1}} – \frac{1}{9}{\rm{)(1}} – \frac{1}{{16}}{\rm{)}} \cdots {\rm{(1}} – \frac{1}{{n^2 }}{\rm{)}} \cdots$

提示

此題為無窮項數列的基本題,一定要拿分。

解答

$\begin{array}{cl} \text{Let }a_n &=(1^2 – (\frac{1}{2})^2 )(1^2 – (\frac{1}{3})^2 )(1^2 – (\frac{1}{4})^2 ) \cdots (1 – (\frac{1}{n})^2 ) \\ &=(1 – \frac{1}{2})(1 + \frac{1}{2})(1 – \frac{1}{3})(1 + \frac{1}{3}) \cdots (1 – \frac{1}{n})(1 + \frac{1}{n}) \\ &=\frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times \frac{5}{4} \times \cdots \frac{{n – 1}}{n} \times \frac{{n + 1}}{n}\\&=(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \frac{{n – 1}}{n})(\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \frac{{n + 1}}{n}) \\&=\frac{{n + 1}}{{2n}}\end{array}$

${\rm{(}}1 – \frac{1}{4}{\rm{)(1}} – \frac{1}{9}{\rm{)(1}} – \frac{1}{{16}}{\rm{)}} \cdots {\rm{(1}} – \frac{1}{{n^2 }}{\rm{)}} \cdots=\mathop {\lim }\limits_{n \to \infty } a_n = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{{2n}} = \frac{1}{2} ∎$

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此題考點數列與數列的極限

第三題 (10 分)
Cardiord $x = 2\sin \theta – \sin 2\theta$, $y = 2\cos \theta – \cos 2\theta$. Find its total length.

提示

此題要算參數型曲線長,公式一定要很熟悉,可以使用三角函數的性質簡化計算過程。

解答

◎ 弧長公式:
$d\ell = \sqrt {(dx)^2 + (dy)^2 } \cdot \frac{{dt}}{{dt}} = \sqrt {(\frac{{dx}}{{dt}})^2 + (\frac{{dy}}{{dt}})^2 } \cdot dt = \sqrt {\left[ {x'(t)} \right]^2 + \left[ {y'(t)} \right]^2 } dt$
$\ell = \int {d\ell } = \int_b^a {\sqrt {\left[ {x'(t)} \right]^2 + \left[ {y'(t)} \right]^2 } } dt$
◎ 降次公式:
$\left\{ {\begin{array}{c} {\sin ^2 \theta = \frac{{1 – \cos 2\theta }}{2}} \\ {\cos ^2 \theta = \frac{{1 + \cos 2\theta }}{2}} \end{array}} \right.$
$\sin ^2 \theta = \frac{{1 – \cos 2\theta }}{2} \Rightarrow \sin ^2 \frac{\theta }{2} = \frac{{1 – \cos \theta }}{2} \Rightarrow 2\sin ^2 \frac{\theta }{2} = 1 – \cos \theta$

$\begin{array}{cl} \ell & =\int_{\rm{0}}^{{\rm{2}}\pi } {\sqrt {\left[ {x'(\theta )} \right]^2 + \left[ {y'(\theta )} \right]^2 } } d\theta \\ & =\int_0^{2\pi } {\sqrt {(2\cos \theta – 2\cos 2\theta )^2 + ( – 2\sin \theta + 2\sin 2\theta )^2 } } d\theta \\ & =\int_0^{2\pi } \sqrt{\begin{array}{l} 4\cos ^2 \theta – 8\cos \theta \cos 2\theta + 4\cos ^2 2\theta \\ + 4\sin ^2 \theta – 8\sin \theta \sin 2\theta + 4\sin ^2 2\theta \end{array} } d\theta\\&=\int_0^{2\pi } {\sqrt {4 + 4 – 8(\cos \theta \cos 2\theta + \sin \theta \sin 2\theta )} } d\theta\\&= 2\sqrt 2 \int_0^{2\pi } {\sqrt {1 – \cos \theta } } d\theta\\&=2\sqrt 2 \int_0^{2\pi } {\sqrt {2\sin ^2 \frac{\theta }{2}} } d\theta = 4\int_0^{2\pi } {\left| {\sin \frac{\theta }{2}} \right|} d\theta\\&=4\int_0^{2\pi } {\sin ^2 \frac{\theta }{2}} d\theta = 4( – 2\left. {\cos \frac{\theta }{2}} \right|_0^{2\pi } ) = 4 \times 4 = 16 ∎ \end{array}$

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此題考點曲線分析(上)

第四題 (10 分)
When $x = 2\sin \theta – \sin 2\theta$, $y = 2\cos \theta – \cos 2\theta$, determine $\frac{{d^2 y}}{{dx^2 }}$ at $\theta = \frac{\pi }{3}$.

提示

本題給定參數式曲線要求二次微分。 正常情況應該會想要把參數消掉,變成 $f(x,y)=0$ ,但也有不好做的時候。這裡的做法是對 $\frac{dy}{dx}$ 去處理,變成 $\frac{dy/dt}{dx/dt}$,二次微分的做法類似。

解答


$\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{ – 2\sin \theta + 2\sin 2\theta }}{{2\cos \theta – 2\cos 2\theta }} = \frac{{ – \sin \theta + \sin 2\theta }}{{\cos \theta – \cos 2\theta }}$

$\begin{array}{cl} \frac{{d^2 y}}{{dx^2 }} & =\frac{{d(\frac{{dy}}{{dx}})}}{{dx}} = \frac{{d(\frac{{dy}}{{dx}})/d\theta }}{{dx/d\theta }} \\ & =\frac{{( – \cos \theta + 2\cos 2\theta )(\cos \theta – \cos 2\theta ) – ( – \sin \theta + \sin 2\theta )( – \sin \theta + 2\sin 2\theta )}}{{2\cos \theta – 2\cos 2\theta }} \\ & =\frac{{( – \cos ^2 \theta + 3\cos 2\theta \cos \theta – 2\cos ^2 2\theta ) – ( – \sin ^2 \theta – 3\sin 2\theta \sin \theta + 2\sin ^2 2\theta )}}{{2(\cos \theta – \cos 2\theta )^3 }}\\&=\frac{{ – 1 – 2 + 3(\overbrace {\cos 2\theta \cos \theta + \sin 2\theta \sin \theta }^{\cos \theta })}}{{2(\cos \theta – \cos 2\theta )^3 }}\\&=\frac{{ – 3(1 – \cos \theta )}}{{2(\cos \theta – \cos 2\theta )^3 }}  \end{array}$

$\frac{{d^2 y}}{{dx^2 }}=\left. {\frac{{d^2 y}}{{dx^2 }}} \right|_{\theta = \frac{\pi }{3}} = \frac{{ – 3(1 – \frac{1}{2})}}{{2(\frac{1}{2} + \frac{1}{2})^3 }} = – \frac{3}{2} \times \frac{{\frac{1}{2}}}{1} = – \frac{3}{4} ∎$

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此題考點曲線分析(上)

第五題 (10 分)
Consider $y = \frac{x}{{(1 + kx)}}$ which is a family of hyperbolas. Find its orthogonal trajectories.

提示

所謂的正交曲線族,它的任何一條曲線在行經原本的各個曲線時,兩者的切向量是彼此垂直的。 因此,把各個點的切線斜率計算好,只要利用兩個斜率相乘等於 -1 即可。接著會出現微分方程式,把它解出來就是答案。

解答


$\frac{{dy}}{{dx}} = \frac{{1 \cdot (1 + kx) – x(k)}}{{(1 + kx)^2 }} = \frac{1}{{(1 + kx)^2 }} = (\frac{y}{x})^2$

令 $g(x,y,c)$ 為 $y = \frac{x}{{1 + kx}}$ 之正交曲線族
則 $\frac{{dy}}{{dx}} \cdot (\frac{y}{x})^2 = – 1 \Rightarrow \frac{{dy}}{{dx}} = – \frac{{x^2 }}{{y^2 }}$
$\Rightarrow \int {y^2 } dy = \int { – x^2 } dx$
$\Rightarrow \frac{{y^3 }}{3} + C_1 = – \frac{{x^3 }}{3} + C_2$
$\Rightarrow y^3 = – x^3 + C$
$\Rightarrow x^3 + y^3 = C ∎$

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此題考點導數與微分的概念微分方程[1][2]

第六題 (10 分)
When $y = \frac{x}{{(1 + kx)}}$, derive $\frac{{\partial z}}{{\partial x}}$ and $\frac{{\partial z}}{{\partial y}}$.

提示

此題為多變數函數偏微分的基本題。

解答


$\because\ x^3+y^3+z^3-3xyz=1$
$\therefore\ 3x^2 + 3z^2 \frac{{\partial z}}{{\partial x}} – (3yz + 3xy\frac{{\partial z}}{{\partial x}}) = 0$
$\frac{{\partial z}}{{\partial x}}(z^2 – xy) + x^2 – yz = 0$
$\Rightarrow \frac{{\partial z}}{{\partial x}} = \frac{{yz – x^2 }}{{z^2 – xy}}$

(直覺: $\frac{{\partial z}}{{\partial y}} = \frac{{xz – y^2 }}{{z^2 – xy}}$ )
$\because\ x^3+y^3+z^3-3xyz=1$
$\therefore\ 3y^2 + 3z^2 \frac{{\partial z}}{{\partial y}} – (3xz + 3xy\frac{{\partial z}}{{\partial y}}) = 0$
$\Rightarrow \frac{{\partial z}}{{\partial y}}(z^2 – xy) + y^2 – xz = 0 \Rightarrow \frac{{\partial z}}{{\partial y}} = \frac{{xz – y^2 }}{{z^2 – xy}} ∎$

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此題考點高階偏微分

第七題 (10 分)
When $f(x,y) = 2xy – \frac{1}{2}(x^4 + y^4 ) + 1$, find local maxima, local minima and saddle points.

提示

沒有範圍限制的多變數函數求極值使用二次微分檢驗法。

解答

◎ 二次微分檢驗法
i. $\nabla f(x,y) = (f_x ,f_y )\mathop = \limits^{{\rm{Let}}} 0 \Rightarrow$ 解 critical point
ii. $\text{H}f(x,y) = \left[ {\begin{array}{c} {f_{xx} } & {f_{yx} } \ {f_{xy} } & {f_{yy} } \end{array}} \right]$
iii. $\text{D}f(x_0 ,y_0 ) = \text{detH}f(x_0 ,y_0 ) = \left| \begin{array}{c} f_{xx} (x_0 ,y_0 ) & f_{yx} (x_0 ,y_0 ) \\ f_{xy} (x_0 ,y_0 ) & f_{yy} (x_0 ,y_0 ) \end{array} \right|$
(1) $\text{D}>0\Rightarrow$ 有極值 $\left\{ \begin{array}{c} a > 0:\text{min} \\ a < 0:\text{Min} \end{array}\right.$, where $a = f_{xx} (x_0 ,y_0 )$
(2) $\text{D}<0\Rightarrow$ saddle.
(3) $\text{D}=0\Rightarrow$ 無結論.

$\begin{array}{cl} \nabla f & =(2y – \frac{1}{2} \cdot 4x^3 ,2x – \frac{1}{2} \cdot 4y^3 ) \\ & =(2y – 2x^3 ,2x – 2y^3 )\mathop {\rm{ = }}\limits^{{\text{Let}}} {\rm{(0,0)}} \\ & \Rightarrow \left\{ {\begin{array}{c} {y – x^3 = 0} \\ {x – y^3 = 0} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{c} {y = x^3 } \\ {x = y^3 } \end{array}} \right. \Rightarrow x = x^9 \\&\Rightarrow x^9 – x = 0 \Rightarrow x(x^8 – 1) = 0 \Rightarrow x = 0\text{ or }\pm 1\\&\Rightarrow (x,y){\rm{ = (0,0),(1,1),(}} – 1, – 1{\rm{)}} \end{array}$

$\text{D}f = \left| {\begin{array}{c} { – 6x^2 } & 2 \ 2 & { – 6y^2 } \end{array}} \right|$
For $(x,y)=(0,0)$: $\text{D}f(0,0) = \left| {\begin{array}{c} 0 & 2 \\ 2 & 0 \end{array}} \right| < 0 \Rightarrow$ saddle.
For $f(x,y)=\pm (1,1)$
$\text{D}f(\pm (1,1))=\left| {\begin{array}{c} { – 6} & 2 \\ & { – 6} \end{array}} \right| = 32 > 0$
$\because\ a=-6<0$
$\therefore\ f(\pm (1,1)) \text{ are local maxima.} ∎$

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此題考點相對、絕對極值、鞍點

第八題 (10 分)
Find the volume determined by $x^2 + y^2+z^2\leq 1$ and $x^2+y^2\leq y$.

提示

本題要計算圓柱與單位球交出的體積,可以看成積分區域為一圓且曲面為 $z=\pm \sqrt{1-x^2-y^2}$ 的體積分。利用對稱性質及座標變換可讓計算變容易。

解答

$\begin{array}{cl} V & =2\iint_{x^2+(y-\frac{1}{2})^2\leq (\frac{1}{2})^2}\sqrt{1-(x^2+y^2)}dxdy \\ & =2\int_0^\pi {\int_0^{\sin \theta } {\sqrt {1 – r^2 } } } rdrd\theta \\ & \left( {\begin{array}{c} {{\rm{Let }}u = 1 – r^2 \Rightarrow du = – 2rdr} \\ {r = 0 \Rightarrow u = 1} \\ {r = \sin \theta \Rightarrow u = \cos ^2 \theta } \end{array}} \right)\\&=2\int_0^\pi {\int_1^{\cos ^2 \theta } {u^{\frac{1}{2}} } \frac{{du}}{{ – 2}}} d\theta\\&=\int_0^\pi {\left. {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{u{\rm{ = 1}}}^{u = \cos ^2 \theta } } d\theta\\&=\frac{2}{3}(\int_0^\pi {\left| {\cos ^3 \theta } \right|} d\theta – \pi )  \end{array}$

$\int_0^\pi {\left| {\cos ^3 \theta } \right|} d\theta$
$= 2\int_0^{\frac{\pi }{2}} {\cos ^3 \theta } d\theta$
$= 2\int_0^{\frac{\pi }{2}} {\cos ^2 \theta \cos \theta } d\theta$
$= 2\int_0^{\frac{\pi }{2}} {(1 – \sin ^2 \theta )} \cdot \cos \theta d\theta$
$\left( {\begin{array}{c} {{\text{Let }}u = \sin \theta \Rightarrow du = \cos \theta d\theta } \\ {\theta = 0 \Rightarrow u = 0} \\ {\theta = \frac{\pi }{2} \Rightarrow u = 1} \end{array}} \right)$
$= 2\int_0^{\frac{\pi }{2}} {1 – u^2 } du = 2\left[ {u – \frac{{u^3 }}{3}} \right]_0^1 = 2(1 – \frac{1}{3}) = \frac{4}{3}$

$V=\frac{2}{3}(\frac{4}{3} – \pi ) = \frac{{2\pi }}{3} – \frac{8}{9} ∎$

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此題考點二變數函數的積分二重積分座標變換

第九題 (10 分)
When $x^2 + y^2=z^2$ and $x+2z=4$, determine the maximum value of $z$.

提示

有限制條件求極值想到拉格朗日乘數法。

解答

◎ Lagrange multipliers:
目標函數 $f(x,y,z)$
限制條件 $g_1(x,y,z)=c_1$, $g_2(x,y,z)=c_2$
i. 解 $\nabla f + \lambda _1 \nabla g_1 + \lambda _2 g_2 = 0$
i. $\Rightarrow$極值候選人 $(x_1,y_1,z_1),\ (x_2,y_2,z_2)$
ii. 將 $(x_k,y_k,z_k)$ 代入 $f(x,y,z)$ 判斷即可

$\begin{array}{cl} \text{Let} & \nabla f + \lambda _1 g_1 + \lambda _2 g_2 = 0 \\ & \Rightarrow (0,0,1) + \lambda _1 (2x,2y, – 2z) + \lambda _2 (1,0,2) = (0,0,0) \\ & \Rightarrow \left\{ {\begin{array}{c} {2\lambda _1 x + \lambda _2 = 0} \\ {2\lambda _1 y = 0} \\ {1 – 2\lambda _1 z + 2\lambda _2 = 0} \end{array} \Rightarrow \lambda _1 = 0{\text{ or }}y = 0} \right.  \end{array}$

If $\lambda_1=0$:
$\lambda_1\ne 0$ and $y=0\Rightarrow \left\{ \begin{array}{c} 2\lambda _1 x + \lambda _2 = 0,x + 2z = 4 \\ 1 – 2\lambda _1 z + 2\lambda _2 = 0,x + 2z = 4 \\ x^2 + y^2 – z^2 = 0,x + 2z = 4 \end{array} \right.$
$\left\{ \begin{array}{l} (x + z)(x – z) = 0 \Rightarrow x = \pm z \\ x+2z=4  \end{array} \right.$
If $x=z$, then $z+2z=4$
$\Rightarrow z=\frac{4}{3}\Rightarrow (x,y,z)=(\frac{4}{3},0,\frac{4}{3})$
If $x=-z$, then $-z+2z=4$
$\Rightarrow z=4\Rightarrow (x,y,z)=(-4,0,4)$

$\because\ f(\frac{4}{3},0,\frac{4}{3}) = \frac{4}{3},\ f(-4,0,4)=4$
$\therefore\ \text{Max.}=4 ∎$

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此題考點Lagrange 乘數法

第十題 (10 分)
$\Omega = \left\{ \left. {(x,y)} \right|x \ge 0,y \ge 0,x + y \ge 3 \right\}$. Find $\iint_{\Omega}e^{-x}e^{-y}dxdy$

提示

因積分區域複雜時,可以分割成好算的區域,最後合併結果。

解答

$\begin{array}{cl} \int_0^\infty {\int_0^\infty {e^{ – x} e^{ – y} } } dxdy & =\int_0^\infty {e^{ – y} } \int_0^\infty {e^{ – x} } dxdy \\ & =(\int_0^\infty {e^{ – x} } dx)(\int_0^\infty {e^{ – y} } dy) \\ & =(\int_0^\infty {e^{ – x} } dx)^2\\&=( – \left. {e^{ – x} } \right|_0^\infty )^2 = (0 – ( – 1))^2 = 1  \end{array}$

$\begin{array}{cl} \int_0^3 {\int_0^{3 – y} {e^{ – x} e^{ – y} } } dxdy & =\int_0^3 {e^{ – y} } \int_0^{3 – y} {e^{ – x} } dxdy \\ & =\int_0^3 {e^{ – y} } \left[ { – e^{ – x} } \right]_{x = 0}^{x = 3 – y} dy \\ & =\int_0^3 {e^{ – y} } ( – e^{y – 3} + 1)dy\\&=\int_0^3 { – e^{ – 3} + e^{ – y} } dy\\&=\left. { – e^{ – 3} y – e^{ – y} } \right|_{y = 0}^{y = 3}\\&=( – 3e^{ – 3} – e^{ – 3} ) – ( – 1) = 1 – 4e^{ – 3}  \end{array}$

$\begin{array}{cl} \iint_{\Omega}e^{-x}e^{-y}dxdy & =\int_0^\infty {\int_0^\infty {e^{ – x} e^{ – y} } } dxdy – \int_0^3 {\int_0^{3 – y} {e^{ – x} e^{ – y} } } dxdy \\ & =1 – (1 – 4e^{ – 3} ) = 4e^{ – 3} ∎\end{array}$

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此題考點二變數函數的積分

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