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 \(y\).
解
甲、填充
1.
| 解1 |
被積分函數 \(x^2-16\) 在 \((-\infty,-4)\bigcup(4,\infty)
\) 為正，在 \((-4,4)\) 為負。由積分的幾何意義易知取 \(a=-4, b=4\) 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*}
設 \(f_a(a,b)=f_b(a,b)=0\) 可解得 \(a=\pm 4, b=\pm4\) 。又 \(b>a\) 故取 \(b=4, a=-4\) 。
二階偏導數\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\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}
故 \((a,b)=(-4,4)\) 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.\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 \(xyz=108\) 得 \(
z=\frac{27}4\)。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 \(n\in\IN\) ， \(e^{\frac{2}{n}}\leq e^2\) ，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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