台綜大轉學考微積分 108 B 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)
Evaluate
〔1〕$\mathop {\lim }\limits_{x \to {\rm{0}}} \frac{1}{{\cos x}}$ (5分)
〔2〕$\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{e^x – 1 – x}}$. (5分)

提示

第一小題沒有陷阱,連續函數的極限值等於函數值。第二小題是不定型,可用羅必達法則,也可用泰勒展開式(須注意有無在收斂半徑內)。

解答

〔1〕
$\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x}} = 1 ∎$

〔2〕
法一 使用羅必達法則
$\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{e^x – 1 – x}}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{e^x – 1}}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{e^x }} = 1 ∎$
法二 使用泰勒展開式
$\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{e^x – 1 – x}}{\rm{ = }}\mathop {\lim }\limits_{x \to 0} \frac{{1 – (1 – \frac{{x^2 }}{{2!}} + \frac{{x^4 }}{{4!}} – \frac{{x^6 }}{{6!}} + \cdots )}}{{(1 + x + \frac{{x^2 }}{{2!}} + \frac{{x^3 }}{{3!}} + \cdots ) – 1 – x}} = \frac{{\frac{1}{2}}}{{{\rm{ }}\frac{1}{2}{\rm{ }}}} = 1 ∎$

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泰勒定理

第二題 (10 分)
Water is pumped into a spherical balloon so that its volume increases at a rate 5 $cm^3/s$. How fast is the radius of the balloon increasing when the diameter is 4 $cm$? Here the volume of a ball of radius $r$ is $\frac{4}{3}\pi r^3$.

提示

本題要求半徑增加的速率,將體積與時間的關係列式後微分即可。

解答

$V(t){\rm{ = }}\frac{4}{3}\pi \left[ {r(t)} \right]^3$
$\Rightarrow \underbrace {V'(t)}_{5cm^3 /s} = 4\pi \underbrace {\left[ {r(t)} \right]^2 }_{(2cm)^2 } \cdot r'(t)$

Let $r(t) = 2$
$\Rightarrow 5 = 16\pi \cdot r'(t) \Rightarrow r'(t) = \frac{5}{{16\pi }} ∎$

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此題考點微分求極值

第三題 (10 分)
Let $f:(-\delta ,\delta)\to \mathbb{R}$ be a function so that $f(0)=\frac{1}{2}$ for $\delta >0$. Assume that $f(x)$ satisfies the equation $x^2 + (f(x))^2 = (2x^2 + 2(f(x)^2 ) – x)^2$ for all $x \in ( – \delta ,\delta )$. Suppose we know that $f$ is differentiable at $0$. Compute the tangent line to the curve $y=f(x)$ at the point $(0,\frac{1}{2})$.

提示

本題要求 $f(x)$ 的切線,已經有切點了,因此只要找出 $f'(x)$ 即可。

解答

$\because x^2+(f(x))^2={\rm{(2}}x^2 {\rm{ + 2(}}f(x))^2 – x{\rm{)}}^2$
$\therefore 2x + 2f(x) \cdot f'(x) = 2(2x^2 + 2(f(x))^2 – x) \cdot (4x + 4f(x) \cdot f'(x) – 1)$

At $x=0$. $(x,f(x)) = (0,\frac{1}{2})$
$\Rightarrow 2 \cdot \frac{1}{2} \cdot f'(0) = 2 \cdot (2 \cdot \frac{1}{4})(4 \cdot \frac{1}{2} \cdot f'(0) – 1)$
$\Rightarrow f'(0) = 2f'(0) – 1$
$\Rightarrow f'(0) = 1$

Tangent line: $x-y=-\frac{1}{2} ∎$

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此題考點切線專論

第四題 (10 分)
Find the minimum of the function $f(x) = \int_1^{\sqrt x } {\frac{{e^t }}{t}} dt$ on $[1,\infty)$. Explain how you obtain the minimum and find the point where the minimum of $f$ occurs.

提示

求 $f(x)$ 的極值需要微分,而 $f(x)$ 裡有積分式且上下界包含變數,想到微積分基本定理。

解答

$\begin{array}{cl} f'(x) & =\frac{d}{{dx}}\int_1^{\sqrt x } {\frac{{e^t }}{t}} dt \\ & =\frac{{e^{\sqrt x } }}{{\sqrt x }} \cdot \frac{1}{2} \cdot x^{ – \frac{1}{2}} \\ & =\frac{{e^{\sqrt x } }}{{2x}} > 0\ \text{on }[1,\infty) \end{array}$

$\because \ f'(x)>0$ on $[1,\infty)$
$\therefore f'(x)$ increases on $[1,\infty)$
$\Rightarrow \mathop {\min }\limits_{x \in \left[ {1,\infty } \right)} f(x) = f(1) = \int_1^1 {\frac{{e^t }}{t}} dt = 0 ∎$

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此題考點微積分基本定理 I微分求極值

第五 (10 分)
Find the volume of the solid $S$ where the base of $S$ is a circular disk of radius $1$ and parallel cross sections perpendicular to its base are squares.

提示

本題並非旋轉體積分,因此回到積分的觀念下手;積分時,可以多多利用對稱的性質讓計算變簡單。

解答

V $=2\int_0^1 {2(1 – x^2 )} dx$ (利用對稱性質算 $x \ge 0$ 的部分再乘 2)
V $=8\left[ {x – \frac{{x^3 }}{3}} \right]_{x = 0}^{x = 1}$
V $=8(1 – \frac{1}{3}) = \frac{{16}}{3} ∎$

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此題考點定積分直觀觀念

第六 (10 分)
Evaluate the improper integral $\int_0^\infty {\frac{{x\tan ^{ – 1} x}}{{(1 + x^2 )^2 }}} dx$ if it exists.
Here $\tan ^{ – 1} x = \arctan x$.

提示

四大積分法:變數變換、三角置換、分部積分和部分分式 是一定會考的內容,一定要熟練。

解答

Let $u = \tan ^{ – 1} x$
$\begin{array}{cl} \int_0^\infty {\frac{{x\tan ^{ – 1} x}}{{(1 + x^2 )^2 }}} dx & =\int_0^{\frac{\pi }{2}} {\frac{{(\tan u) \cdot u}}{{(\underbrace {1 + \tan ^2 u}_{\sec ^2 u})}}} du \\ & =\int_0^{\frac{\pi }{2}} {\frac{{\sin u}}{{\cos u}} \cdot \cos ^2 u \cdot u} du \\ & =\frac{1}{2}\int_0^{\frac{\pi }{2}} {u \cdot \sin (2u)} du \\&=\frac{1}{2}\left[ {\left. {\frac{{ – u \cdot \cos (2u)}}{2}} \right|_{u = 0}^{u = \frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\frac{{\cos (2u)}}{2}} du} \right]\\&=\frac{1}{2}\left[ {\frac{{ – \frac{\pi }{2} \times ( – 1)}}{2} + \frac{1}{2} \cdot \left. {\frac{{\sin (2u)}}{2}} \right|_{u = 0}^{u = \frac{\pi }{2}} } \right]\\&=\frac{1}{2} \times \frac{\pi }{4} = \frac{\pi }{8} ∎ \end{array}$

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此題考點三角置換法分部積分

第七 (10 分)
Find the radius of convergence and the interval of convergence of the power series $\sum\limits_{n = 1}^\infty {\frac{{( – 2)^n }}{{n^2 }}x^n }$.

提示

本題要求收斂半徑,這裡示範比較審斂法。

解答

$r = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{( – 2)^{n + 1} }}{{(n + 1)^2 }}}}{{\frac{{( – 2)^n }}{{n^2 }}}}} \right|}} = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{2n^2 }}{{(n + 1)^2 }}} \right|}} = \frac{1}{2}$

$\Rightarrow $ the series converges on $( – \frac{1}{2},\frac{1}{2})$
For $x=\frac{1}{2}$
$\sum\limits_{n = 1}^\infty {\frac{{( – 2)^n }}{{n^2 }}x^n }=\sum\limits_{n{\rm{ = 1}}}^\infty {\frac{{( – 2)^n }}{{n^2 }} \cdot (\frac{1}{2})^n } = \sum\limits_{n = 1}^\infty {\frac{{( – 1)^n }}{{n^2 }}}$ converges. (by alternating series test)

For $x=-\frac{1}{2}$
$\sum\limits_{n = 1}^\infty {\frac{{( – 2)^n }}{{n^2 }} \cdot ( – \frac{1}{2})^n } = \sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}}$

$\Rightarrow $ interval of convergence: $\left[ { – \frac{1}{2},\frac{1}{2}} \right] ∎$

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此題考點比較審斂法根值審斂法

第八 (10 分)
Let $f$ be the function $\left\{ {\begin{array}{lc}{\frac{{xy^2 + \frac{1}{2}y^3 }}{{x^2 + y^2 }}} & {{\rm{if }}(x,y) \ne (0,0)} \\0 & {{\rm{if }}(x,y) = (0,0)} \\ \end{array}} \right.$.
Evaluate $f_x (0,0)$ and $f_y (0,0)$ if they exist.

提示

求(偏)微分遇到分段函數,有很高的機率需要使用定義處理。

解答

$\begin{array}{cl} f_x (0,0) & =\mathop {\lim }\limits_{h \to 0} \frac{{f(h,0) – f(0,0)}}{h} \\ & =\mathop {\lim }\limits_{h \to 0} \frac{{\frac{0}{{h^2 }} – 0}}{h} = 0 ∎\end{array}$

$\begin{array}{cl} f_y (0,0) & =\mathop {\lim }\limits_{k \to 0} \frac{{f(0,k) – f(0,0)}}{k} \\ & =\mathop {\lim }\limits_{k \to 0} \frac{{\frac{{\frac{1}{2}k^3 – 0}}{{k^2 }}}}{k} = \frac{1}{2} ∎ \end{array}$

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此題考點二變數函數的偏微分

第九 (10 分)
Use Lagrange multipliers to find the extremum of the function $f(x,y,z)=xyz$ subject to the constraint $xy+2xz+2yz=12$.

提示

題目已有要求使用拉格朗日乘數法找極值,而這個方法很需要熟練的計算技巧,平時要多練習。

解答

Let $g(x,y,z)=xy+2xz+2yz-12$
Let $\nabla f+\lambda \nabla g=0$
$\Rightarrow (yz,xz,xy) + \lambda (y + 2z,x + 2z,2x + 2y) = 0$
$\Rightarrow \left\{ {\begin{array}{cl}{yz + \lambda y + 2\lambda z = 0} & (1) \\{xz + \lambda x + 2\lambda z = 0} & (2) \\{xy + 2\lambda x + 2\lambda y = 0} & (3) \\ \end{array}} \right.$
◎◎ $\begin{array}{cl} xy+2xz+2yz=12 & (4) \end{array}$

$\begin{array}{cl} (1)-(2) & \Rightarrow z(y – x) + \lambda (y – x) = 0 \\ & \Rightarrow (y – x) \cdot (z + \lambda ) = 0 \Rightarrow x = y \text{ or }z=-\lambda \end{array}$
If $z=-\lambda$
$\mathop \Rightarrow \limits^{(1)} – \lambda y + \lambda y – 2\lambda ^2 = 0 \Rightarrow \lambda = 0 \Rightarrow z = 0$
$\mathop \Rightarrow \limits^{(2),(4)} \left\{ \begin{array}{l} xy = 0 \\{xy = 12}\end{array} ( \to \leftarrow ) \right. \Rightarrow z \ne \lambda$
$\Rightarrow x = y$
$\mathop \Rightarrow \limits^{(2),(3),(4)}\left\{ \begin{array}{ll} xz + \lambda x + 2\lambda z = 0 & (6) \\x^2 + 2\lambda x + 2\lambda x = 0 \Rightarrow x^2 + 4\lambda x = 0 & (5) \\ x^2 + 2xz + 2xz = 12 \Rightarrow x^2 + 4xz = 12 & (7)\end{array} \right.$

$\begin{array}{cl} (5) & \Rightarrow x(x + 4\lambda ) = 0 \\ & \Rightarrow x = 0 \text{ or }x=-4\lambda \end{array}$
$\because x^2+4xz=12$
$\therefore x \ne 0$
$\Rightarrow x=-4\lambda$
$\because x=y$
$\therefore x=4y=-4\lambda$
$\mathop \Rightarrow \limits^{(6),(7)}\left\{ \begin{array}{l} -4\lambda z – 4\lambda ^2 + 2\lambda z = 0\Rightarrow \lambda (2\lambda + z) = 0\\ \begin{array}{ll} 16\lambda^2-16\lambda z=12 & (8)\end{array} \end{array}\right.$
$\Rightarrow \lambda = 0\text{ or }z=-2\lambda$
If $\lambda = 0\mathop \Rightarrow \limits^{(8)} 0 = 12{\rm{ }}( \to \leftarrow ) \Rightarrow \lambda \ne 0$
$\Rightarrow z=-2\lambda$
$\mathop \Rightarrow \limits^{(8)} (4\lambda )^2 + 32\lambda ^2 = 12 \Rightarrow \lambda = \pm \frac{1}{2}$
$\Rightarrow (x,y,z) = ( – 4\lambda , – 4\lambda , – 2\lambda ) = ( – 2, – 2, – 1)$ or $(2,2,1)$

當 $(x,y,z) = ( – 2, – 2, – 1)$ 時,$f(x,y,z)=-4$
當 $(x,y,z)=(2,2,1)$ 時,$f(x,y,z)=4$
極值 $\left\{ {\begin{array}{l} {{\rm{Max}} = 4} \\ {{\rm{min}} = – 4} \\ \end{array}} \right. ∎$

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此題考點Lagrange 乘數法

第十 (10 分)
Let $a>0$. Evaluate the double integral $\int_{ – a}^a {\int_{ – \sqrt {a^2 – x^2 } }^{\sqrt {a^2 – x^2 } } {\cos (x^2 + y^2 )} } dydx$.

提示

不易算的二重積分可嘗試極座標變換或交換積分次序(Fubini 定理),本題被積函數有 $x^2+y^2$ 因此使用極座標變換。

解答

Using polar coordinates:
Let $x = r\cos \theta$, $y = r\sin \theta$
$\Rightarrow x^2+y^2=r^2$, $dxdy=rdrd\theta$
$\int_{ – a}^a {\int_{ – \sqrt {a^2 – x^2 } }^{\sqrt {a^2 – x^2 } } {\cos (x^2 + y^2 )} } dydx=\int_0^{2\pi } {\int_0^a {\cos r^2 } } rdrd\theta$

Let $u=r^2\Rightarrow du=2rdr$
$\begin{array}{cl} \int_0^{2\pi } {\int_0^a {\cos r^2 } } rdrd\theta & =\int_0^{2\pi } {\int_0^{a^2 } {\cos u} \frac{{du}}{2}} d\theta \\ & =\frac{1}{2}\int_0^{2\pi } {\left. {\sin u} \right|_{u = 0}^{u = a^2 } } d\theta \\ & =\frac{1}{2}\int_0^{2\pi } {\sin a^2 } d\theta \\&=\pi \sin a^2 ∎ \end{array}$

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此題考點二重積分座標變換

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