- (10%) Find the limit $\displaystyle\lim_{x\to0}x\left(\frac1{x^3}-\frac1{\sin^3x}\right)$.
- (10%) Evaluate the integral $\displaystyle\int\frac{dx}{\sin x+8\cos x+4}$.
- (10%) Evaluate the integral
$\displaystyle\int_{-2}^2\int_0^{\sqrt{4-y^2}}\int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}y^2\sqrt{x^2+y^2+z^2}dzdxdy$
- (10%) Find all normal lines of $y=x^3+2x^2$ which pass through the point $\left(-2,0\right)$.
- 假若 $x_0=0$ 或 $\displaystyle x_0=-\frac43$，那麼在這些位置的法線為鉛直線，明顯不通過 $\left(-2,0\right)$。
- 假若 $x_0\neq0$ 且 $\displaystyle x_0=-\frac43$，那麼其法線斜率為 $\displaystyle-\frac1{3x_0^2+4x_0}$，故法線方程式為
$\displaystyle y-\left(x_0^3+2x_0^2\right)=-\frac{x-x_0}{3x_0^2+4x_0}$
因為要通過 $\left(-2,0\right)$，代入可知$\displaystyle x_0^3+2x_0^2=\frac{-2-x_0}{3x_0^2+4x_0}$
即 $3x_0^5+10x_0^4++8x_0^3+x_0+2=0$，可因式分解得 $\left(x_0+2\right)\left(x_0+1\right)^2\left(3x_0^2-2x_0+1\right)=0$，故得實根為 $x_0=-1$ 與 $x_0=-2$。 - (10%) Find the relative extreme values and the absolute extreme values of
$\displaystyle f\left(x\right)=\left(1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)e^{-x}$, $x\in\mathbb{R}$,
where $n$ is a positive integer. - 若 $n$ 為偶數，那麼 $f'\left(x\right)\leq0$ 且等號成立僅當 $x=0$，故 $x=0$ 不為相對極值，此函數無相對極值。且由 $\displaystyle\lim_{x\to\infty}f\left(x\right)=0$ 與 $\displaystyle\lim_{x\to-\infty}f\left(x\right)=\infty$，因此函數 $f$ 也無絕對極值。
- 若 $n$ 為奇數，那麼在 $\left(-\infty,0\right)$ 時有 $f'\left(x\right)>0$，而在 $f'\left(x\right)<0$，因此函數 $f$ 在 $x=0$ 處有絕對極大值（也是相對極大值）。但因 $\displaystyle\lim_{x\to\infty}f\left(x\right)=0$ 且 $\displaystyle\lim_{x\to-\infty}f\left(x\right)=-\infty$，故無絕對極小值。
- (10%) Find the sum of the series $\displaystyle\frac{x}{1\cdot2}+\frac{x^2}{2\cdot3}+\frac{x^3}{3\cdot4}+\cdots+\frac{x^n}{n\left(n+1\right)}+\cdots$.
- (10%) The plane $x+y+z=a$ cuts the unit sphere $x^2+y^2+z^2=1$ into two pieces $S_1$ and $S_2$, where $0\leq a\leq\sqrt3$. Compute $\left|\text{area}\left(S_1\right)-\text{area}\left(S_2\right)\right|$.
- (10%) Find the area of the region enclosed by the curve $x^3+y^3=3xy$.
- 當 $\theta=0$ 時有 $u=0$；
- 當 $\displaystyle\theta=\frac\pi2$ 時有 $u\to+\infty$；
- 由三角函數的特性可知 $\displaystyle\sin\theta=\frac{u}{\sqrt{1+u^2}}$ 與 $\displaystyle\cos\theta=\frac1{\sqrt{1+u^2}}$，求導可知 $d\theta=\frac{du}{1+u^2}$。
- (10%) An 80 kg man carries a 5 kg can of paint up a helical staircase that encircles a silo with a radius 10 m. Suppose there is a hole in the can of paint and 2 kg of paint leaks steadily out of the can during the man' ascent. If the silo is 40 m high and the man makes exactly four complete revolutions climbing to the top, how much work is done by the man against gravity?
- (10%) A particle travelling in a straight line with constant velocity $-{\bf i}-3{\bf j}+5{\bf k}$ passes through the point $\displaystyle\left(\frac52,4,-\frac74\right)$ and hit the surface $z=x^2+y^2$. The particle ricochets off the surface, the angle of reflection being equal to the angle of incidence. Assuming no loss of speed, what is the velocity of the particle after the ricochet?
訣竅
通分整理後使用泰勒展開式則即可。解法
通分整理後使用羅必達法則可知$\displaystyle\lim_{x\to0}x\left(\frac1{x^3}-\frac1{\sin^3x}\right)=\lim_{x\to0}\frac{\sin^3x-x^3}{x^2\sin^3x}=\lim_{x\to0}\frac{\displaystyle\left(x-\frac{x^3}6+\cdots\right)^3-x^3}{x^2\cdot\left(x^3-\cdots\right)}=\lim_{x\to0}\frac{\displaystyle-\frac{x^5}2+\cdots}{x^5-\cdots}=-\frac12$
訣竅
使用正切函數的半角代換。解法
設 $\displaystyle t=\tan\frac{x}2$，那麼整理可知 $x=2\tan^{-1}\left(t\right)$，使用倍角公式可知$\displaystyle\sin x=2\sin\frac{x}2\cos\frac{x}2=2\cdot\frac{t}{\sqrt{1+t^2}}\cdot\frac1{\sqrt{1+t^2}}=\frac{2t}{1+t^2},\qquad\cos x=2\cos^2\frac{x}2-1=\frac2{1+t^2}-1=\frac{1-t^2}{1+t^2}$
再者求導有 $\displaystyle dx=\frac{2dt}{1+t^2}$，如此所求的不定積分可改寫並計算如下$\displaystyle\begin{aligned}\int\frac{dx}{\sin x+8\cos x+4}&=\int\frac{\displaystyle\frac{2dt}{1+t^2}}{\displaystyle\frac{2t}{1+t^2}+8\cdot\frac{1-t^2}{1+t^2}+4}\\&=\int\frac{dt}{-2t^2+t+6}=-\int\frac{dt}{\left(2t+3\right)\left(t-2\right)}=\frac17\int\left(\frac2{2t+3}-\frac1{t-2}\right)dt\\&=\frac17\left(\ln\left|2t+3\right|-\ln\left|t-2\right|\right)+C=\frac17\ln\left|\frac{2t+3}{t-2}\right|+C\\&=\frac17\ln\left|\frac{\displaystyle2\tan\frac{x}2+3}{\displaystyle\tan\frac{x}2-2}\right|+C\end{aligned}$
訣竅
由球面座標變換改寫三重積分即可求解。解法
令 $\left\{\begin{aligned}&x=\rho\cos\theta\sin\phi\\&y=\rho\sin\theta\sin\phi\\&z=\rho\cos\phi\end{aligned}\right.$，那麼按三重積分之區域可得變數範圍為 $\left\{\begin{aligned}&0\leq\rho\leq2\\&-\frac\pi2\leq\theta\leq\frac\pi2\\&0\leq\phi\leq\pi\end{aligned}\right.$，如此所求之三重積分可改寫並計算如下$\displaystyle\begin{aligned}\int_{-2}^2\int_0^{\sqrt{4-y^2}}\int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}y^2\sqrt{x^2+y^2+z^2}dzdxdy&=\int_{-\frac\pi2}^{\frac\pi2}\int_0^\pi\int_0^2\rho^2\sin^2\theta\sin^2\phi\cdot\rho\cdot\rho^2\sin\phi\,d\rho d\phi d\theta\\&=\left(\int_0^2\rho^5d\rho\right)\left(\int_0^{\pi}\sin^3\phi\,d\phi\right)\left(\int_{-\frac\pi2}^{\frac\pi2}\sin^2\theta d\theta\right)\\&=\left.\frac{\rho^6}6\right|_0^2\cdot\left(\int_0^{\pi}\left(1-\cos^2\phi\right)\sin\phi\,d\phi\right)\left(\int_{-\frac\pi2}^{\frac\pi2}\frac{1-\cos2\theta}2d\theta\right)\\&=\frac{32}3\cdot\frac43\cdot\frac{\pi}2=\frac{64\pi}9\end{aligned}$
訣竅
直接求導可得切線斜率進一步得法線斜率，隨後由點斜式列式，最後由所過之點求得該法線。解法
直接微分可得 $y'=3x^2+4x$，故在 $\left(x_0,x_0^3+2x_0^2\right)$ 處的切線斜率為 $3x_0^2+4x_0$。訣竅
微分求極值，其中在求導的過程中應細心計算。解法
首先對函數 $f$ 求導可得$\displaystyle f'\left(x\right)=\left(1+x+\frac{x^2}{2!}+\cdots+\frac{x^{n-1}}{\left(n-1\right)!}\right)e^{-x}-\left(1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)e^{-x}=-\frac{x^n}{n!}e^{-x}$
為了找出相對極值，我們應解方程式 $f'\left(x\right)=0$，容易獲得 $x=0$。訣竅
由經典的無窮等比級數觀察改寫即可。解法
回憶起經典的無窮等比級數和為$\displaystyle\frac1{1-x}=\sum_{n=0}^{\infty}x^n=1+x+x^2+\cdots+x^n+\cdots$
取定積分可得$\displaystyle-\ln\left(1-x\right)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=x+\frac{x^2}2+\frac{x^3}3+\cdots+\frac{x^{n+1}}{n+1}+\cdots$
再取一次定積分得$\displaystyle x+\left(1-x\right)\ln\left(1-x\right)=-x\ln\left(1-x\right)+\ln\left(1-x\right)+x=\frac{x^2}{1\cdot2}+\frac{x^3}{2\cdot3}+\cdots+\frac{x^{n+2}}{\left(n+1\right)\cdot\left(n+2\right)}+\cdots$
兩邊同除以 $x$ 可得$\displaystyle\frac{x}{1\cdot2}+\frac{x^2}{2\cdot3}+\cdots+\frac{x^n}{n\left(n+1\right)}+\cdots=1+\frac{1-x}x\ln\left(1-x\right)$
訣竅
運用旋轉體表面積的概念求解，其中適當地改變轉軸可較容易地表達該旋轉曲面進而獲得表面積。解法
首先注意到 $x+y+z=a$ 到原點 $\left(0,0,0\right)$ 的距離為 $\displaystyle\frac{a}{\sqrt3}$，因此可以將被分割的兩塊曲面分別視為 $y=\sqrt{1-x^2}$ 在 $\displaystyle\left[-1,\frac{a}{\sqrt3}\right]$ 與 $\displaystyle\left[\frac{a}{\sqrt3},1\right]$ 上對 $x$ 軸旋轉後的曲面，並且分別記為 $S_1$ 與 $S_2$。那麼$\displaystyle\text{area}\left(S_1\right)=\int_{-1}^{a/\sqrt3}2\pi y\cdot\sqrt{1+y'^2}dx=2\pi\int_{-1}^{a/\sqrt3}\sqrt{1-x^2}\cdot\sqrt{1+\frac{x^2}{1-x^2}}dx=2\pi\left(\frac{a}{\sqrt3}+1\right)$
由類似的計算或由球面表面積為 $4\pi$ 可知$\displaystyle\text{area}\left(S_2\right)=2\pi\left(1-\frac{a}{\sqrt3}\right)$
因此所求為$\displaystyle\left|\text{area}\left(S_1\right)-\text{area}\left(S_2\right)\right|=\frac{4\pi a}{\sqrt3}$
訣竅
使用極座標來表達曲線方程式，隨後使用極座標下的面積公式求解即可。解法
令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\end{aligned}\right.$，那麼代入曲線中有 $r^3\cos^3\theta+r^3\sin^3\theta=3r^2\cos\theta\sin\theta$，如此有$\displaystyle r=\frac{3\cos\theta\sin\theta}{\cos^3\theta+\sin^3\theta}$
而當 $\theta=0$ 與 $\displaystyle\theta=\frac\pi2$ 曲線自交於原點，故所求面積為$\displaystyle A=\frac12\int_0^{\frac\pi2}r^2\left(\theta\right)d\theta=\frac92\int_0^{\frac\pi2}\frac{\cos^2\theta\sin^2\theta}{\left(\cos^3\theta+\sin^3\theta\right)^2}d\theta$
令 $u=\tan\theta$，那麼$\displaystyle A=\frac92\int_0^{\infty}\frac{\displaystyle\frac{u^2}{\left(1+u^2\right)^2}}{\displaystyle\left(\frac1{\left(1+u^2\right)^{3/2}}+\frac{u^3}{\left(1+u^2\right)^{3/2}}\right)^2}\cdot\frac{du}{1+u^2}=\frac92\int_0^{\infty}\frac{u^2}{\left(1+u^3\right)^2}du=\left.-\frac32\left(1+u^3\right)^{-1}\right|_0^{\infty}=\frac32$
訣竅
按照題意參數化並列出作功所對應的定積分來計算。解法
由於繞四圈且四層樓高，故設路徑為 $\displaystyle\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)=\left(10\cos t,10\sin t,\frac{5t}\pi\right)$，其中 $0\leq t\leq8\pi$。那麼每增加一單位的 $t$，油漆就流出 $\displaystyle\frac{t}{4\pi}$ 公斤，從而所作之功為$\displaystyle W=\int_0^{8\pi}\left(80+5-\frac{t}{4\pi}\right)\cdot g\cdot\frac{5t}{\pi}dt=\frac{5g}{4\pi^2}\int_0^{8\pi}\left(340\pi t-t^2\right)dt=\left.\frac{5g}{4\pi^2}\left(170\pi t^2-\frac{t^3}3\right)\right|_0^{8\pi}=\frac{40160g\pi}3$
其中 $g$ 為重力加速度，近似值約為 $10$ 或 $9.8$，單位為 $\text{m}/\text{s}^2$。訣竅
利用梯度求出法向量，據此依序找出正射影與反射後的向量，從而得到碰撞後的速度。解法
設 $F\left(x,y,z\right)=x^2+y^2-z$，那麼曲面由方程式 $F\left(x,y,z\right)=0$ 所定義。此時這樣的曲面的法向量為 $\nabla F\left(x,y,z\right)=\left(F_x\left(x,y,z\right),F_y\left(x,y,z\right),F_z\left(x,y,z\right)\right)=\left(2x,2y,-1\right)$，故在 $\displaystyle\left(\frac52,4,-\frac74\right)$ 處的法向量為 $\displaystyle\nabla F\left(\frac52,4,-\frac74\right)=\left(5,8,-1\right)$。
設入射向量為 $v=\left(-1,-3,5\right)$，而法向量為 $n=\left(5,8,-1\right)$。可以知道 $v$ 投影到 $n$ 上的正射影為
$\displaystyle p=\frac{v\cdot n}{n\cdot n}n=-\frac{17}{45}\left(5,8,-1\right)$
那麼反射向量為$\displaystyle w=v-2p=\left(-1,-3,5\right)+\frac{34}{45}\left(5,8,-1\right)=\left(\frac{25}9,\frac{137}{45},\frac{191}{45}\right)$
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