台聯大轉學考微積分 107 A2 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。

甲. 簡答題

第一題 (8 分)
Determine the limits of integration where $a\leq b$ such that $\int_a^b(x^2-16)dx$ has minimal value.

提示

此題要求積分式的最小值,可以從積分的基本觀念著手,也可利用微分觀念處理。

解答

法一 被積函數為開口向下之拋物線,由面積觀念得積分最大值為 $x^2-16 \ge 0$ 處
$\begin{array}{cl} \int_a^b(x^2-16)dx & =\int_{{\rm{ – 4}}}^{\rm{4}} {x^2 – {\rm{16}}} dx \\ & =2\int_0^4 {x^2 – 16} dx \\ & =2\left[ {\frac{{x^3 }}{3} – 16x} \right]_{x = 0}^{x = 4}\\&=2(\frac{{64}}{3} – 64) = – \frac{{256}}{3} ∎ \end{array}$

法二 使用二次微分檢驗法

Let $f(a,b) = \int_a^b {x^2 – {\rm{16}}} dx$
$\Rightarrow \nabla f( – (a^2 – 16),b^2 – 16)\mathop = \limits^{{\text{Let }}} (0,0)$
$\Rightarrow \left\{ {\begin{array}{c} {a^2 – 16 = 0} \\ {b^2 – 16 = 0}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{c} {a = \pm 4} \\ {b = \pm 4}\end{array}} \right.$
$\because a\leq b$
$\therefore (a,b)=(-4,-4),\ (4,4),\ (-4,4)$

$Hf(a,b) = \left[ {\begin{array}{c} { – 2a} & 0 \\ 0 & {2b} \end{array}} \right]$
If $(a,b)=(-4,-4)$
$\Rightarrow Df(a,b) = \left| {\begin{array}{c} 8 & 0 \\ 0 & { – 8} \end{array}} \right| < 0 \Rightarrow$ saddle
If $(a,b)=(4,4)$
$\Rightarrow Df(a,b) = \left| {\begin{array}{c} { – 8} & 0 \\ 0 & 8\end{array}} \right| < 0 \Rightarrow$ saddle
If $(a,b)=(-4,4)$
$\Rightarrow Df(a,b) = \left| {\begin{array}{c} 8 & 0 \\ 0 & 8 \end{array}} \right| > 0 \Rightarrow$ 有極值
$\because 8>0\ \therefore f(-4,4)$ is a local minimum where $f( – 4,4) = – \frac{{256}}{3} ∎$

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此題考點定積分直觀觀念相對、絕對極值、鞍點

第二題 (8 分)
Evaluate the integral $\int_0^{\frac{\pi }{2}} {\sqrt {1 – \sin x} } dx$.

提示

除了四大積分法外,有時候也能用以前學過的技巧讓計算變容易,三角函數有很多恆等式可以使用。

解答

$\begin{array}{cl} \int_0^{\frac{\pi }{2}} {\sqrt {1 – \sin x} } dx & =\int_0^{\frac{\pi }{2}} {\sqrt {(\sin ^2 \frac{x}{2} + \cos ^2 \frac{x}{2}) – 2\sin \frac{x}{2}\cos \frac{x}{2}} } dx \\ & =\int_0^{\frac{\pi }{2}} {\sqrt {(\sin \frac{x}{2} – \cos \frac{x}{2})^2 } } dx \\ & =\int_0^{\frac{\pi }{2}} {\left| {\sin \frac{x}{2} – \cos \frac{x}{2}} \right|} dx\\&(\text{Let }u = \frac{x}{2} \Rightarrow du = \frac{1}{2}dx)\\&= \int_0^{\frac{\pi }{4}} {\left| {\sin u – \cos u} \right| \cdot 2} du\\&=2\int_0^{\frac{\pi }{4}} {\cos u – \sin u} du\\&=2\left[ {\sin u + \cos u} \right]_{u = 0}^{u = \frac{\pi }{4}}\\&=2\left[ {(\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}) – (0 + 1)} \right] = 2\sqrt 2 – 2 ∎\end{array}$

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此題考點變數變換法

第三題 (8 分)
Evaluate the integral $\iint_{R} \sqrt {3-x^2-y^2}dA$, where $R = { (x,y)|x^2 + y^2 \le 3}$.

提示

此題為二重積分,不好直接算時,通常可以嘗試交換積分次序或極座標變換。由於積分區域與被積函數有 $x^2+y^2$ 因此使用極座標變換。

解答

$\begin{array}{cl} \iint_{R} \sqrt {3-x^2-y^2}dA & =\iint_{R} \sqrt {3-(x^2+y^2)}dA \\ & =\int_0^{2\pi } {\int_0^{\sqrt 3 } {\sqrt {3 – r^2 } } rdr} d\theta \\ & (\text{Let }u = 3 – r^2 \Rightarrow du = – 2rdr)\\&=\int_0^{2\pi } {\int_3^0 {u^{\frac{1}{2}} \cdot \frac{{du}}{{ – 2}}} } d\theta\\&=\frac{1}{2}\int_0^{2\pi } {\left[ {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right]} _{u = 0}^{u = 3} d\theta \\&=\frac{1}{3}\int_0^{2\pi } {3^{\frac{3}{2}} } d\theta = 2\sqrt 3 \pi ∎\end{array}$

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此題考點二重積分座標變換

第四題 (8 分)
Find the interval of convergence of the power series $\sum\limits_{n = 0}^\infty {\frac{{2n(x – 3)^n }}{{(n + 1)!}}}$.

提示

此題為求收斂區間的基本題。

解答

收斂半徑$=\frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{2n + 2}}{{(n + 2)!}}}}{{\frac{{2n}}{{(n + 1)!}}}}} \right|}} = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{{2n + 2}}{{2n(n + 2)}}}} = \infty$
所求$=\mathbb{R} ∎$

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此題考點冪級數

第五題 (8 分)
Find the volume of the solid bounded above by the surface $z = f(x,y)$ and below by the plane region $R$, where $f(x,y)=\ln x$ and $R$ is bounded by the graphs $y=2x$ and $y=0$ from $x=1$ to $x=3$.

提示

此題要算體積,給的條件有積分區域與對應的 $z$ 座標(可看作高),由體積觀念列好後即是一個二重積分,剩下只要使用以往二重積分常用的手法即可。

解答

$\begin{array}{cl} \iint_{R}\ln xdA & =\int_1^3 {\int_0^{2x} {\ln x} } dydx \\ & =\int_1^3 {(\ln x)(2x)} dx \\ & =\left. {x^2 \ln x} \right|_1^3 – \int_1^3 x dx\\&=\left. {9\ln 3 – \frac{{x^2 }}{2}} \right|_1^3 = 9\ln 3 – 4 ∎\end{array}$

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此題考點二變數函數的積分

第六題 (8 分)
Let $z = f(x,y)=\ln (xy)^{\frac{1}{2}}$. Find the approximate change in $z$ when the point changes from $(5,9)$ to $(5.03,9.96)$.

提示

此題要估計二變數函數於兩個點的間的誤差值,由於只需要估計值因此不用直接算出來相減,這時運用微分量的性質即可有效率的求出誤差。

解答

◎ 微分量
(1) $f(x) = f(a) + f'(a)(x – a) + \frac{{f”(a)}}{{2!}}(x – a)^2 + \cdots$
$\Rightarrow f(x) – f(a) = f'(a)(x – a) + \frac{{f”(a)}}{{2!}}(x – a)^2 + \cdots$, 當 $x$ 夠接近 $a$
$\Rightarrow f(x) – f(a) \approx f'(a)(x – a)$
(2) $f(x,y) – f(a,b) \approx \nabla f(a,b) \cdot (x – a,y – b)$

$\begin{array}{cl} f(5.03,9.96) – f(5,10) & \approx \nabla f(5,10) \cdot (5.03 – 5,9.96 – 10) \\ & =\left. {(\frac{1}{2} \cdot \frac{{10}}{{50}},\frac{1}{2} \cdot \frac{x}{{xy}})} \right|_{(x,y) = (5,10)} \cdot (0.03, – 0.04) \\ & =(\frac{1}{2} \cdot \frac{{10}}{{50}},\frac{1}{2} \cdot \frac{5}{{50}}) \cdot (\frac{3}{{100}}, – \frac{4}{{100}})\\&=\frac{1}{{10}} \cdot \frac{3}{{100}} – \frac{1}{{20}} \cdot \frac{4}{{100}} = \frac{1}{{1000}} = 0.001 ∎\end{array}$

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此題考點多變數函數的微分量

第七題 (8 分)
Consider a differetial equation $\frac{{dy}}{{dt}} = \frac{k}{v}(10 – y)$, $y(0)=y_0$, where $k$, $v$ and $y_0$, are positive constants with $y<0$. Find $\mathop {\lim }\limits_{t \to \infty } y$.

提示

這題是基本的微分方程題目,為可分離變數型,但最後要求的是 $\mathop {\lim }\limits_{t \to \infty } y$ ,因此運算中於適合的地方取極限即可。

解答

$\frac{{dy}}{{dt}} = \frac{k}{v}(10 – y)$
$\Rightarrow \int {\frac{1}{{10 – y}}} dy = \int {\frac{k}{v}} dt$
$\Rightarrow – \ln \left| {10 – y} \right| = \frac{k}{v}t + C$
$\Rightarrow \ln \left| {10 – y} \right| = – \frac{k}{v}t + C$
$\Rightarrow \left| {10 – y} \right| = Ce^{ – \frac{k}{v}t}$
$\Rightarrow \mathop {\lim }\limits_{t \to \infty } \left| {10 – y} \right| = \mathop {\lim }\limits_{t \to \infty } Ce^{ – \frac{k}{v}t}$
$\Rightarrow \left| {10 – \mathop {\lim }\limits_{t \to \infty } y} \right| = 0$
$\Rightarrow \mathop {\lim }\limits_{t \to \infty } y = 10 ∎$

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此題考點微分方程〔1〕〔2〕

第八題 (8 分)
Find the minimum of the function $f(x,y,z)=xy+2yz+2xz$ subject to the constraint $xyz=108$.

提示

有條件限制的多變數極值問題,要想到拉格朗日乘數法。

解答

$\nabla f + \lambda \nabla g = (y + 2z,x + 2z,2x + 2y) + \lambda (yz,xz,xy) = (0,0,0)$
$\Rightarrow \left\{ {\begin{array}{c} {y + 2z + \lambda yz = 0\cdots (1)} \\ {x + 2z + \lambda xz = 0\cdots (2)} \\ {2x + 2y + \lambda xy = 0} \\ {xyz = 108} \end{array}} \right.$
由 (1), (2) $\Rightarrow (x – y) + \lambda z(x – y) = 0\Rightarrow (x – y)(1 + \lambda z) = 0\Rightarrow x=y\text{ or }\lambda z = – 1$

If $x=y$:
$\Rightarrow \left\{ \begin{array}{l}x+2z+\lambda xz=0\\4x+\lambda x^2=0\\x^2z=108\end{array} \right.$
$\Rightarrow x(4+\lambda x)=0 \Rightarrow \lambda x=-4\text{ or }x=0(\to\leftarrow \text{by }x^2z=108)$
$\Rightarrow \left\{ {\begin{array}{c} {x + 2z – 4z = 0 \Rightarrow x = 2z} \\ {x^2 z = 108 \Rightarrow x = 2}\end{array}} \right. \Rightarrow 4z^3 = 108 \Rightarrow z = 3$
$\mathop \Rightarrow \limits^{x = 2z} x = 6\mathop \Rightarrow \limits^{x = y} y = 6 \Rightarrow (x,y,z) = (6,6,3)$

If $\lambda z=-1$
$\Rightarrow \left\{ \begin{array}{c} y+2z-y=0 \\ x+2z-x=0 \\ 2x+2y+\lambda xy=0\\ xyz=108  \end{array} \right.$
$\Rightarrow 2z=0\Rightarrow z=0\ (\to\leftarrow \because xyz=108)$

所求$= f(6,6,3) = 6^2 + 2 \times 6 \times 3 + 2 \times 6 \times 3 = 36 \times 3 = 108$
$f(x,y,z)=xy+2yz+2xz$
考慮 $(x,y,z) = (1,t,\frac{{108}}{t}),\ t \in\mathbb{R},\ t \ne 0$
$\Rightarrow f(x,y,z) = t + 216 + \frac{{108}}{t}$
$\because \mathop {\lim }\limits_{t \to \infty } f(1,t,\frac{{108}}{t}) = \infty\text{ and }\mathop {\lim }\limits_{t \to – \infty } f(1,t,\frac{{108}}{t}) =-\infty$
∴$\therefore$ the minimum of $f(x,y,z)$ does not exist$. ∎$

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此題考點Lagrange 乘數法

乙.計算、證明題

第一題 (12 分)
An airplane is flying on a flight path that will take it directly over a radar tracking station. The distance $s$ is decreasing at a rate of 640 kilometers per hour when $s=16$ km. What is the speed of the plane ?

提示

本題為應用問題,要從雷達測出的距離變化來推算飛機飛行的速率。主要的做法就是把飛機與雷達站的位置關係畫出來,然後利用微分求出所要的答案。

解答

依題意:
$\left\{ {\begin{array}{l} {\left. {\frac{{ds}}{{dt}}} \right|_{s = 16} = – 640} \\ {\left[ {s(t)} \right]^2 = h^2 + [x(t)]^2 }\end{array}}\right.$
$\Rightarrow 2s\frac{{ds}}{{dt}} = 2x\frac{{dx}}{{dt}}$
$\mathop \Rightarrow \limits^{s = 16} 16 \cdot ( – 640) = \left. {\sqrt {16^2 – h^2 } \frac{{dx}}{{dt}}} \right|_{s = 16}$
$\Rightarrow \left. {\frac{{dx}}{{dt}}} \right|_{s = 16} = \frac{{ – 10240}}{{\sqrt {256 – h^2 } }}\ \text{km/hrs}. ∎$

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此題考點微分求極值

第二題 (12 分)
Determine if the given series converges or diverges. Explain your reasoning.
〔1〕$\sum\limits_{n = 1}^\infty {\frac{{e^{\frac{2}{n}} }}{{n^2 }}}$ (6 分)
〔2〕$\sum\limits_{n = 1}^\infty {\frac{n}{{\sqrt {3n^2 + 5} }}}$ (6 分)

提示

級數的歛散性為必考題,第一小題可以搭配已知級數用極限比較審斂法;第二小題可以直接觀察一般項的極限。

解答

〔1〕
$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{e^{\frac{2}{n}} }}{{n^2 }}}}{{\frac{1}{{n^2 }}}} = \mathop {\lim }\limits_{n \to \infty } e^{\frac{2}{n}} = 1 \\&\text{and }\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}}\text{ converges (by }p-\text{series.)}\\ \therefore &\sum\limits_{n = 1}^\infty {\frac{{e^{\frac{2}{n}} }}{{n^2 }}}\text{ converges (by limit comparison test).} ∎ \end{array}$

〔2〕
$\because\ \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\sqrt {3n^2 + 5} }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt {3 + \frac{5}{{n^2 }}} }} = \frac{1}{{\sqrt 3 }} \ne 0$
$\therefore\ \sum\limits_{n = 1}^\infty {\frac{n}{{\sqrt {3n^2 + 5} }}}\text{ diverges.} ∎$

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此題考點級數p-級數極限比較審斂法

第三題 (12 分)
Consider the function $f(x,y) = \left\{ {\begin{array}{c} {ke^{ – \frac{{(x + y)}}{a}} ,} & {{\text{if }}x \ge 0,y \ge 0} \\ {0,} & {{\text{elsewhere}}}\end{array}} \right.$. Find the relationship between the positive constants $a$ and $k$ such that $f$ is a joint probability density function of the continuous random variables $x$ and $y$.

提示

這一題用機率的應用來考瑕積分,一個函數 $f(x,y)$ 是機率密度函數,代表它在積分之後要是 1,因此以這個條件作出發點列式。

解答

$\int_{ – \infty }^\infty {\int_{ – \infty }^\infty {f(x,y)} } dA = 1$ (joint probability density function)
$\begin{array}{cl} \Rightarrow 1 & =\int_0^\infty {\int_0^\infty {ke^{ – \frac{{x + y}}{a}} } } dxdy \\ & =k\int_0^\infty {\int_{\rm{0}}^\infty {e^{ – \frac{x}{a}} e^{ – \frac{y}{a}} } } dxdy \\ & =k(\int_0^\infty {e^{ – \frac{x}{a}} dx} )(\int_{\rm{0}}^\infty {e^{ – \frac{y}{a}} } dy)\\&=k(\int_0^\infty {e^{ – \frac{x}{a}} dx} )^2\\&=k(\left. { – ae^{ – \frac{x}{a}} } \right|_0^\infty )^2 = k(0 – ( – a))^2 = ka^2  \end{array}$
$\Rightarrow a^2k=1 ∎$

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此題考點二變數函數的積分

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