國立臺灣大學一百零三學年度研究所碩士班入學考試試題：微積分(B)

1. A cycloid can be parametrized by the functions $x\left(\theta\right)=\left(\theta-\sin\theta\right)$, $y\left(\theta\right)=\left(1-\cos\theta\right)$.
   1. Find $dy/dx$ and $d^2y/dx^2$. (10%)
   2. Plot the cycloid curve. $0\leq\theta\leq2\pi$. (10%)
   3. Find the length of the cycloid arch. (10%)
   4. Find the surface area of the solid generated by revolving the cycloid arch about the $x$-axis. (10%)
訣竅
運用連鎖律求參數曲線的斜率與凹性並據此繪圖，隨後透過曲線弧長公式以及旋轉體表面積計算。
解法

1. 利用連鎖律可知

   $\displaystyle\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{\sin\theta}{1-\cos\theta}$,　$\displaystyle\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{\displaystyle\frac{d}{d\theta}\left(\frac{\sin\theta}{1-\cos\theta}\right)}{\displaystyle\frac{dx}{d\theta}}=\frac{\displaystyle\frac{\cos\theta}{1-\cos\theta}-\frac{\sin^2\theta}{\left(1-\cos\theta\right)^2}}{1-\cos\theta}=\frac{\cos\theta-1}{\left(1-\cos\theta\right)^3}=-\frac1{\left(1-\cos\theta\right)^2}$

2. 描點繪圖如下
   
3. 由曲線弧長公式可計算如下

   $\displaystyle\begin{aligned}s&=\int_0^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta=\int_0^{2\pi}\sqrt{\left(1-\cos\theta\right)^2+\left(\sin\theta\right)^2}d\theta\\&=2\int_0^{2\pi}\sqrt{\frac{1+\cos\theta}2}d\theta=2\int_0^{2\pi}\left|\cos\frac\theta2\right|d\theta=4\int_0^{\pi}\cos\frac\theta2d\theta=\left.8\sin\frac\theta2\right|_0^{\pi}=8\end{aligned}$

4. 由旋轉體體積公式可知

   $\displaystyle\begin{aligned}V&=\int_0^{2\pi}2\pi y\left(\theta\right)\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta=4\pi\int_0^{2\pi}\left(1-\cos\theta\right)\sqrt{\frac{1+\cos\theta}2}d\theta\\&=8\pi\int_0^{\pi}\left(1-\cos\theta\right)\cos\frac\theta2d\theta=4\pi\int_0^{\pi}\left(\cos\frac\theta2-\cos\frac{3\theta}2\right)d\theta=\left.8\pi\left(\sin\frac\theta2-\frac13\sin\frac{3\theta}2\right)\right|_0^{\pi}=\frac{32\pi}3\end{aligned}$



2. Calculate $\displaystyle\int\sin^5xdx$. (10%)
訣竅
運用變數代換的概念求解即可。
解法

利用三角恆等式可改寫並計算所求的不定積分如下

$\displaystyle\int\sin^5xdx=\int\left(1-\cos^2x\right)^2\sin xdx=\int\left(-1+2\cos^2x-\cos^4x\right)d\cos x=-\cos x+\frac{2\cos^3x}3-\frac{\cos^5x}5+C$



3. Test the convergence of the series $\displaystyle\sum\frac1k\left(\frac1{\ln k}\right)^{3/2}$. (10%)
訣竅
由積分審歛法判斷即可。
解法
設 $\displaystyle f\left(x\right)=\frac1x\left(\frac1{\ln x}\right)^{3/2}$，容易注意到 $f$ 恆正且遞減。又有
$\displaystyle\int_2^{\infty}f\left(x\right)dx=-2\left(\ln x\right)^{-1/2}\Big|_2^{\infty}=2\left(\ln 2\right)^{-1/2}<\infty$
因此由積分審歛法知其收斂。


4. Minimize $x^2+y^2$ on the curve $x^4+7x^2y^2+y^4=1$. (10%)
訣竅
運用拉格朗日乘子法求條件極值。本題與100學年度碩士班微積分(B)第五題雷同，惟本題改求最小值。
解法
設定拉格朗日乘子函數如下
$F\left(x,y,\lambda\right)=x^2+y^2+\lambda\left(x^4+7x^2y^2+y^4-1\right)$
據此解聯立方程組
$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=2x+\lambda\left(4x^3+14xy^2\right)=0\\&F_y\left(x,y,\lambda\right)=2y+\lambda\left(14x^2y+4y^3\right)=0\\&F_{\lambda}\left(x,y,\lambda\right)=x^4+7x^2y^2+y^4-1=0\end{aligned}\right.$
第一式乘以 $x$ 後減去第二式乘以 $y$ 可知 $x^2-y^2+\lambda\left(2x^4-2y^4\right)=0$，或因式分解為 $\left(x^2-y^2\right)\left[1+2\lambda\left(x^2+y^2\right)\right]=0$。
• 假若 $x^2-y^2=0$，則第三式可寫為 $9x^4=1$，即得 $\displaystyle x=\pm\frac{\sqrt3}3$，而 $\displaystyle y=\pm\frac{\sqrt3}3$，故有四點座標為 $\displaystyle\pm\left(\frac{\sqrt3}3,\frac{\sqrt3}3\right)$ 與 $\displaystyle\pm\left(\frac{\sqrt3}3,-\frac{\sqrt3}3\right)$。
• 假若 $2\lambda\left(x^2+y^2\right)=-1$，則 $\lambda\neq0$，故有 $\displaystyle x^2+y^2=-\frac1{2\lambda}$。又第一式乘以 $x$ 加上第二式乘以 $y$ 並搭配第三式便有 $\displaystyle-\frac1{\lambda}+7\lambda=0$，故得 $\displaystyle\lambda=\pm\frac{\sqrt7}7$。
綜上考慮各種情形可以知道函數 $x^2+y^2$ 的取值可能有 $\displaystyle\frac{\sqrt7}2$ 或 $\displaystyle\frac23$，因此最大值為 $\displaystyle\frac{\sqrt7}2$，而最小值為 $\displaystyle\frac23$。


5. Integrate $\vec{h}\left(x,y\right)=\left(x+2\right)y\vec{i}+\left(2x+y\right)\vec{j}$ over the indicated path $y=x^2$ from $\left(0,0\right)$ to $\left(2,4\right)$. (10%)
訣竅
利用參數化計算線積分即可。本題與102學年度碩士班微積分(B)第八題相同。
解法

利用參數化計算線積分可知

$\displaystyle\int_L\vec{h}\cdot d\vec{s}=\int_0^2\left[\left(t+2\right)t^2\cdot1+\left(2t+t^2\right)\cdot2t\right]dt=\int_0^2\left(3t^3+6t^2\right)dt=\left.\frac{3t^4}4+2t^3\right|_0^2=28$



6. Take $\Omega$ as the parallelogram bounded by $x-y=0$, $x-y=\pi$, $x+2y=0$, $x+2y=\pi/2$. Evaluate $\displaystyle\iint_{\Omega}\sin\left(x-y\right)\cos\left(x+2y\right)dxdy$. (10%)
訣竅
運用變數變換求解，其中需留意 Jacobian 行列式的計算。
解法

令 $\left\{\begin{aligned}&u=x-y\\&v=x+2y\end{aligned}\right.$，那麼 $\Omega$ 的四條邊界可表示為 $u=0$、$u=\pi$、$v=0$ 與 $v=\pi/2$。又其對應的 Jacobian 行列式為

$\displaystyle\left|J\right|=\Big|\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|\Big|=\Big|\left|\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}\right|\Big|^{-1}=\Big|\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{vmatrix}\Big|^{-1}=\Big|\begin{vmatrix}1&-1\\1&2\end{vmatrix}\Big|^{-1}=\frac13$

如此所求之重積分可改寫並計算如下

$\displaystyle\iint_{\Omega}\sin\left(x-y\right)\cos\left(x+2y\right)dxdy=\int_0^{\pi}\int_0^{\pi/2}\sin u\cos v\cdot\frac13dvdu=\frac13\left(\int_0^{\pi}\sin udu\right)\left(\int_0^{\pi/2}\cos vdv\right)=\frac23$



7. Evaluate $\displaystyle\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}\frac1{x^2+y^2+z^2}dzdydx$. (10%)
訣竅
運用球面座標變換計算即可。
解法

令 $\left\{\begin{aligned}&x=\rho\cos\theta\sin\phi\\&y=\rho\sin\theta\sin\phi\\&z=\rho\cos\phi\end{aligned}\right.$，那麼積分區域給出變數範圍為 $\left\{\begin{aligned}&0\leq\rho\leq1\\&0\leq\theta\leq\frac\pi2\\&0\leq\phi\leq\frac\pi2\end{aligned}\right.$。如此所求的三重積分可改寫並計算如下

$\displaystyle\begin{aligned}\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}\frac1{x^2+y^2+z^2}dzdydx&=\int_0^{\frac\pi2}\int_0^{\frac\pi2}\int_0^1\frac{\rho^2\sin\phi}{\rho^2}d\rho d\theta d\phi\\&=\left(\int_0^{\frac\pi2}\sin\phi d\phi\right)\left(\int_0^{\frac\pi2}d\theta\right)\left(\int_0^1d\rho\right)=\frac\pi2\end{aligned}$