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台聯大系統轉學考107微積分A2詳解

 

甲、填充

 

1. Determine the limits of integration where a\le b ( a>b ) such that \displaystyle\int_a^b(x^2-16)\dx has minimal value.

 

2. Evaluate \displaystyle \int_0^{\frac{\pi}2}\sqrt{1-\sin x}\dx .

 

3. Evaluate the integral \displaystyle \iint_R\sqrt{3-x^2-y^2}\dA , where R=\bigl\{(x,y)\bigm\vert x^2+y^2\le 3\bigr\} .

4. Find the interval of convergence of the power series \displaystyle \sum_{n=0}^{\infty}\frac{2n(x-3)^n}{(n+1)!} .

 

5. Find the volume of the solid bounded 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.

 

6. Let z=f(x,y)=\ln(xy)^{\frac 12}. Find the approximate change in z when the point changes from (5,10) to (5.03,9.96).

 

7. Consider a differential equation \mfrac{\dy}{\dt} =\mfrac k v(10-y), y(0)=y_0, where k,v and y_0 are positive constants with y<10. Find \mlim{t\to\infty} y.

 

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

 

乙、計算

 

1. 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?

 

2. Determine if the given series converges or diverges. Explain your reasoning.
a. (6分) \displaystyle \sum_{n=1}^{\infty}\frac{e^{\frac 2 n}}{n^2}          b. (6分) \displaystyle\sum_{n=1}^{\infty} \frac n{\sqrt{3n^2+5}}

 

3. Consider the function f(x,y)= \mycases{3ex}{-6pt}\begin{array}{ll} ke^{-\frac{x+y}{a}}&,\;\text{if}\,x\ge 0, y\ge0\\ 0&,\;\text{elsewhere} \end{array}
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.

 

 

 

甲、填充

1.

 解1

被積分函數 x^2-16(-\infty,-4)\bigcup(4,\infty) 為正,在 (-4,4) 為負。易知取 a=-4, b=4 可使 \displaystyle\int_a^b(x^2-16)\dx 最小。

 

 解2

視為雙變數函數 \displaystyle f(a,b)=\int_a^b (x^2-16)\dx,則問題等同於求雙變數函數 f(a,b) 在限定範圍 b>a 的最小值。由微積分基本定理,偏導數

    \begin{align*} f_a(a,b)=&\,-(a^2-16)=16-a^2\\ f_b(a,b)=&\,b^2-16 \end{align*}

f_a(a,b)=f_b(a,b)=0 可解得 a=\pm 4, b=\pm4 。又 b>a 故取 b=4, a=-4
二階偏導數

(1)   \begin{align*} f_{aa}(a,b)=&\,-2a &\hspace{-15mm} f_{ab}(a,b)=&\,0\\ f_{ba}(a,b)=&\,0 &\hspace{-15mm} f_{bb}(a,b)=&\,2b \end{align*}

黑塞行列式

(2)   \begin{align*} H(-4,4)=\begin{vmatrix}f_{aa}(-4,4) & f_{ab}(-4,4)\\ f_{ba}(-4,4) & f_{bb}(-4,4) \end{vmatrix}=\begin{vmatrix}8 & 0\\ 0 & 8 \end{vmatrix}>0 \end{align*}

(a,b)=(-4,4) 處為極值。又 f_{aa}(-4,4)=8>0 ,故其為極小值。

 

2.

    \begin{align*} &\,\int_0^{\frac{\pi}2}\sqrt{1-\sin x}\dx\\[1mm] =&\,\int_0^{\frac{\pi}2}\sqrt{1-\cos x}\dx& \eqnote{\(\int_0^{\frac{\pi}2}f(\sin(\theta))\dtheta =\int_0^{\frac{\pi}2}f(\cos(\theta))\dtheta\)}\\[1mm] =&\,\int_0^{\frac{\pi}2}\sqrt{2\sin^2(\frac x 2)}\dx &\eqnote{half angle formula}\\[1mm] =&\,\sqrt{2}\int_0^{\frac{\pi}2}\abs{\sin(\frac x 2)}\dx &\eqnote{\(\sqrt{x^2}=\abs{x}\)}\\[1mm] =&\,\sqrt{2}\int_0^{\frac{\pi}2}\sin(\frac x 2)\dx\\[1mm] =&\,\Big. -2\sqrt{2}\cos(\frac x 2)\Big\vert_0^{\frac{\pi}2} =2\sqrt{2}-2 \end{align*}

 

3. 使用極坐標代換

    \begin{align*} &\iint_R\sqrt{3-x^2-y^2}\dA\\ =&\int_0^{2\pi}\int_0^{\sqrt{3}}\sqrt{3-r^2}r\dr\dtheta\\ =&2\pi\cdot\left[-\frac13(3-r^2)^{\frac32}\right]_0^{\sqrt{3}}\\ =&2\sqrt{3}\pi \end{align*}

 

4.

(3)   \begin{align*} \rho=\lim_{n\to\infty}\frac{\;\mfrac{2n}{(n+1)!}\;}{\;\mfrac{2(n+1)}{(n+2)!}\;} =\lim_{n\to\infty}\frac{2n\cdot(n+2)}{2(n+1)}=\infty \end{align*}

則收斂半徑 R=0,收斂區間為 x=3

 

5.

    \begin{align*} &\int_1^3\int_0^{2x}\ln(x)\dy\dx\\ =&\int_1^32x\ln(x)\dx\\ =&\left.x^2\ln(x)\right\vert_1^3 -\int_1^3x\dx\\ =&9\ln 3-4 \end{align*}

 

6. z=f(x,y)=\frac12\ln(xy),先寫出全微分

    \begin{align*} \dz=&\,\frac{y}{2x}\dx+\frac{x}{2y}\dy\end{align*}

於是可估計

    \begin{align*} \Rightarrow\Delta z\doteqdot &\,\frac{10}{2\cdot5}\cdot0.03 +\frac{5}{2\cdot10}\cdot(-0.04)\\ =&\,0.03-0.01=0.02 \end{align*}

 

7. 首先看出答案明顯是10是一階線性常微分方程,於是按正常流程

    \begin{align*} y'(t)+\frac{k}{v}y(t)=&\frac{10k}{v}\\ e^{\frac{k}{v}t}y'(t)+\frac{k}{v}e^{\frac kvt} =&\frac{10k}{v}e^{\frac kvt}\\ \Rightarrow e^{\frac kvt}y(t) =&10e^{\frac kvt}+C\\ y(t)=&10+Ce^{-\frac kvt}\\ \Rightarrow \lim_{t\to\infty} y(t)=&\,\lim_{t\to\infty} 10+Ce^{-\frac kvt}=10 \end{align*}

 

8.利用拉格朗日乘子法

    \begin{align*} \mycases{5ex}{-6pt}\begin{array}{l} y+2z=yz\\ x+2z=xz\\ 2y+2x=xy\\ xyz=108 \end{array} \end{align*}

由前兩式知 x=y,代入第三式得x=y=4,再代入限制條件 xyz=108z=\frac{27}4。於是極小值為 f(4,4,\frac{27}4) =16+54+54=124

 

 

乙、計算

1. 此題為出題老師在抄 Larson 的教科書例題時沒抄好。原題在圖中標示高度,而考卷上的題目只抄了文字改數字,忘了補上高度。

 

2.
(a) 對於 n\in\INe^{\frac{2}{n}}\leq e^2 ,故 \frac{e^{\frac{2}{n}}}{n^2}\le\frac{e^2}{n^2} 對於所有正整數 n 恆成立。而 \medop\sum\mfrac{e^2}{n^2}=e^2\medop\sum\mfrac{1}{n^2} 是收斂的,故由比較審斂法知原級數收斂。
(b) 一般項的極限 \mlim{n\to\infty}\mfrac n{\sqrt{3n^2+5}}=\mfrac 1{\sqrt{3}}\ne 0 ,故級數發散。

 

3. 若 f 要是個聯合機率密度函數,須滿足 \iint_R f(x,y) \dx\dy=1,故寫下

    \begin{align*} \int_0^{\infty}\int_0^{\infty}ke^{-\frac{x+y}{a}}\dx\dy=&\,1\\ \int_0^{\infty}\left[-ae^{-\frac{x+y}{a}}\right]_0^{\infty}\dx\dy=&\,\frac{1}{k}\\ a\int_0^{\infty}e^{-\frac{y}{a}}\dy=&\,\frac{1}{k}\\ a\left[-ae^{-\frac{y}{a}}\right]_0^{\infty}=&\,\frac{1}{k}\\ \Rightarrow\;a^2=&\,\frac{1}{k} \end{align*}

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