- A car is racing along the elliptical route $x^2/9+y^2/16=2500$ m$^2$. When the car is at the point $\left(120\mbox{m},120\mbox{m}\right)$, a man at the point $\left(30\mbox{m},0\right)$ determines with radar that the distance between him and the car is increasing at the rate $37\mbox{m}/\mbox{sec}$. Find the speed $(\mbox{m}/\mbox{sec})$ of the car at that moment.
- Find $\displaystyle\lim_{x\to0}\left(\frac{x}{\sin^3x}-\frac1{x^2}\right)$.
- Given $f\left(x\right)=\ln\left(1+x+x^2\right)$, find $f^{\left(3k\right)}\left(0\right)$, where $k\geq0$ is an integer.
- Find the arc length of the curve $y=x^2/2$ from $x=0$ to $x=1$.
- 當 $x=0$ 時有 $\theta=0$；
- 當 $x=1$ 時有 $\displaystyle\theta=\frac\pi4$；
- 求導有 $dx=\sec^2\theta d\theta$。
- Find the volume of the solid generated by revolving the region enclosed by the curve $\displaystyle y=\frac{2x}{4-x^2}$, the $x$-axis, $x=0$ and $x=1$.
- Let $g\left(x\right)$ be the inverse function of $f\left(x\right)=x\left(\ln x\right)^3$ for $x\geq1$. Find $\displaystyle\int_0^eg\left(x\right)dx$.
- 當 $x=0$ 時有 $x=1$；
- 當 $x=e$ 時有 $u=e$；
- 求導有 $dx=f'\left(u\right)du=\left[\left(\ln u\right)^3+3\left(\ln u\right)^2\right]du$。
- Find the tangent plane to the surface $x^3+y^3+z^3-3xy^2z^3=0$ at the point $x=1$, $y=1$ and $z=1$.
- Find the local minimum values of $f\left(x,y\right)=x^3-12xy+8y^3$.
- $D\left(0,0\right)=-144<0$，故 $\left(0,0\right)$ 為鞍點；
- $D\left(2,1\right)=432>0$ 且 $f_{xx}\left(2,1\right)=12>0$，故 $\left(2,1\right)$ 為局部極小點。
- Evaluate $\displaystyle\int_{y=0}^{y=2}\int_{x=y}^{x=2}e^{x^2}dxdy$.
- Solve the differential equation $y'\cos x+y\sin x=\cos^3x$, $y\left(\pi/4\right)=1$.
訣竅
運用相關變率的概念與連鎖律求解即可。解法
考慮觀察者與汽車的距離函數為$r\left(t\right)=\sqrt{\left(x\left(t\right)-30\right)^2+\left(y\left(t\right)\right)^2}$
求導可知$\displaystyle\frac{dr}{dt}=\frac{\left(x\left(t\right)-30\right)x'\left(t\right)+y\left(t\right)y'\left(t\right)}{\sqrt{\left(x\left(t\right)-30\right)^2+\left(y\left(t\right)\right)^2}}$
按題意可知在該時刻 $t=t_0$ 時有$\displaystyle37=\frac{90x'\left(t_0\right)+120y'\left(t_0\right)}{\sqrt{90^2+120^2}}=\frac{3x'\left(t_0\right)+4y'\left(t_0\right)}5$
即 $3x'\left(t_0\right)+4y'\left(t_0\right)=185$。另一方面，對給定的橢圓方程求導則有$\displaystyle\frac{2x\left(t\right)}9x'\left(t\right)+\frac{2y\left(t\right)}{16}y'\left(t\right)=0$
取 $t=t_0$ 則有 $\displaystyle\frac{80}3x'\left(t_0\right)+15y'\left(t_0\right)=0$，至此可解得 $\left(x'\left(t_0\right),y'\left(t_0\right)\right)=\left(-45,80\right)$。故該汽車在該時刻的速率為$v\left(t_0\right)=\sqrt{\left(x'\left(t_0\right)\right)^2+\left(y'\left(t_0\right)\right)^2}=\sqrt{45^2+80^2}=5\sqrt{337}$
訣竅
通分後應用泰勒展開式即可。解法
改寫後使用羅必達法則如下$\displaystyle\begin{aligned}\lim_{x\to0}\left(\frac{x}{\sin^3x}-\frac1{x^2}\right)&=\lim_{x\to0}\frac{x^3-\sin^3x}{x^2\sin^3x}=\lim_{x\to0}\frac{\displaystyle x^3-\left(x-\frac{x^3}6+\cdots\right)^3}{x^2\cdot\left(x+\cdots\right)^3}=\lim_{x\to0}\frac{\displaystyle x^3-\left(x^3-\frac12x^5+\cdots\right)}{x^5+\cdots}=\frac12\end{aligned}$
訣竅
留意到 $1-x^3=\left(1-x\right)\left(1+x+x^2\right)$，據此考慮輔助的函數處理。解法
注意到如下的等式$f\left(x\right)=\ln\left(1-x^3\right)-\ln\left(1-x\right)$
那麼由無窮等比級數公式可注意到$\displaystyle\frac1{1-x}=\sum_{k=0}^{\infty}x^k$
兩邊同取積分可知$\displaystyle-\ln\left|1-x\right|=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}$
接著用 $x^3$ 取代 $x$ 可得$\displaystyle\ln\left|1-x^3\right|=-\sum_{k=0}^{\infty}\frac{x^{3k+3}}{k+1}$
故有$\displaystyle f\left(x\right)=-\sum_{k=0}^{\infty}\frac{x^{3k+3}}{k+1}+\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}$
那麼 $\displaystyle f^{\left(3k\right)}\left(0\right)=\left(3k\right)!\cdot-\frac1k-\left(3k\right)!\cdot-\frac1{3k}=-2\left(3k-1\right)!$。訣竅
運用曲線弧長公式求解。解法
運用曲線弧長公式可知$\displaystyle s=\int_0^1\sqrt{1+y'^2}dx=\int_0^1\sqrt{1+x^2}dx$
令 $x=\tan\theta$，那麼$\displaystyle s=\int_0^{\frac\pi4}\sec\theta\cdot\sec^2\theta d\theta=\int_0^{\frac\pi4}\sec^3\theta d\theta=\left.\frac{\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|}2\right|_0^{\frac\pi4}=\frac{\sqrt2+\ln\left(1+\sqrt2\right)}2$
其中我們所使用的不定積分公式推導如下$\displaystyle\int\sec^3\theta d\theta=\int\left(\sec\theta+\sec\theta\tan^2\theta\right)d\theta=\int\sec\theta d\theta+\int\tan\theta d\sec\theta\int\sec\theta d\theta+\sec\theta\tan\theta-\int\sec^3\theta d\theta$
因此所求為$\displaystyle\int\sec^3\theta d\theta=\frac{\sec\theta\tan\theta}2+\frac12\int\sec\theta d\theta=\frac{\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|}2+C$
訣竅
使用旋轉體體積公式計算即可。解法
使用旋轉體體積公式如下$\displaystyle V=\int_0^1\pi y^2dx=\pi\int_0^1\frac{4x^2}{\left(4-x^2\right)^2}dx=4\pi\int_0^1\frac{x^2}{\left(2-x\right)^2\left(2+x\right)^2}dx$
考慮待定的係數 $A,B,C,D$ 滿足$\displaystyle\frac{x^2}{\left(2-x\right)^2\left(2+x\right)^2}=\frac{A}{2-x}+\frac{B}{2+x}+\frac{C}{\left(2-x\right)^2}+\frac{D}{\left(2+x\right)^2}$
同乘以 $\left(4-x^2\right)^2$ 可得$\left(-A+B\right)x^3+\left(-2A-2B+C+D\right)x^2+\left(4A-4B+4C-4D\right)x+\left(8A+8B+4C+4D\right)=x^2$
藉由比較係數可知 $\displaystyle A=B=-\frac18$、$\displaystyle C=D=\frac14$。如此所求可改寫並計算如下$\displaystyle V=-\frac\pi2\int_0^1\left(\frac1{2-x}+\frac1{2+x}-\frac2{\left(2-x\right)^2}-\frac2{\left(2+x\right)^2}\right)dx=\left.-\frac\pi2\left(\ln\frac{2+x}{2-x}-\frac2{2-x}+\frac2{2+x}\right)\right|_0^1=\frac\pi2\left(\frac43-\ln3\right)$
訣竅
應用變數變換處理該積分即可。解法
令 $x=f\left(u\right)$，那麼$\displaystyle\begin{aligned}\int_0^eg\left(x\right)dx&=\int_1^eu\cdot\left[\left(\ln u\right)^3+3\left(\ln u\right)^2\right]du=\frac12\int_1^e\left[\left(\ln u\right)^3+3\left(\ln u\right)^2\right]du^2\\&=\left.\frac{u^2\left[\left(\ln u\right)^3+3\left(\ln u\right)^2\right]}2\right|_1^e-\frac32\int_1^eu\cdot\left[\left(\ln u\right)^2+2\ln u\right]du\\&=2e^2-\frac34\int_1^e\left[\left(\ln u\right)^2+2\ln u\right]du^2\\&=2e^2-\left.\frac34u^2\left[\left(\ln u\right)^2+2\ln u\right]\right|_1^e+\frac32\int_1^e\left(u\ln u+u\right)du\\&=-\frac14e^2+\frac34\int_1^e\left(\ln u+1\right)du^2\\&=-\frac14e^2+\left.\frac34u^2\left(\ln u+1\right)\right|_1^e-\frac34\int_1^eudu\\&=\frac54e^2-\frac34-\frac{3e^2-3}8=\frac{7e^2-3}8\end{aligned}$
訣竅
運用梯度求出曲面在該處的法向量，隨後使用點法式寫出切平面方程式。解法
設 $F\left(x,y,z\right)=x^3+y^3+z^3-3xy^2z^3$，則曲面由方程式 $F\left(x,y,z\right)=0$ 所定義。計算其梯度有$\nabla F\left(x,y,z\right)=\left(F_x\left(x,y,z\right),F_y\left(x,y,z\right),F_z\left(x,y,z\right)\right)=\left(3x^2-3y^2z^3,3y^2-6xyz^3,3z^2-9xy^2z^2\right)$
那麼曲面在 $\left(x,y,z\right)=\left(1,1,1\right)$ 處的法向量為$\nabla F\left(1,1,1\right)=\left(0,-3,-6\right)$
如此使用點法式寫出切平面方程式$0\cdot\left(x-1\right)-3\left(y-1\right)-6\left(z-1\right)=0$
或 $y+2z=3$。訣竅
為了找出極值的位置，我們解一階偏導為零的位置，隨後使用二階行列式判別其性質。解法
為了找出極值，我們解下列的聯立方程組
$\left\{\begin{aligned}&f_x\left(x,y\right)=3x^2-12y=0\\&f_y\left(x,y\right)=-12x+24y^2=0\end{aligned}\right.$
第一式給出 $\displaystyle y=\frac{x^2}4$，代入第二式則有$\displaystyle x=2y^2=2\cdot\frac{x^4}{16}$
故有 $x^4-8x=0$，因此 $x=0$ 或 $x=2$，從而 $y=0$ 或 $y=1$。至此我們得到兩候選點 $\left(0,0\right)$ 與 $\left(2,1\right)$。現在計算二階行列式如下
$D\left(x,y\right)=\left|\begin{matrix}f_{xx}\left(x,y\right)&f_{xy}\left(x,y\right)\\f_{yx}\left(x,y\right)&f_{yy}\left(x,y\right)\end{matrix}\right|=\left|\begin{matrix}6x&-12\\-12&48y\end{matrix}\right|=144\left(2xy-1\right)$
那麼訣竅
交換積分次序求解。解法
原積分範圍 $\left\{\begin{aligned}&y\leq x\leq2\\&0\leq y\leq2\end{aligned}\right.$ 可改寫為 $\left\{\begin{aligned}&0\leq x\leq2\\&0\leq y\leq x\end{aligned}\right.$，如此所求的重積分可改寫並計算如下$\displaystyle\int_{y=0}^{y=2}\int_{x=y}^{x=2}e^{x^2}dxdy=\int_0^2\int_0^xe^{x^2}dydx=\int_0^2xe^{x^2}dx=\left.\frac{e^{x^2}}2\right|_0^2=\frac{e^4-1}2$
訣竅
運用積分因子求解。解法
先同除以 $\cos x$ 化為標準的一階線性常微分方程：$y'+y\tan x=\cos^2x$
如此取積分因子為 $e^{\int\tan xdx}=e^{\ln\sec x}=\sec x$，故同乘以 $\sec x$ 可得$\left(y\sec x\right)'=y'\sec x+y\sec x\tan x=\cos^2x\sec x=\cos x$
據此在區間 $\displaystyle\left[\frac\pi4,x\right]$ 上取定積分可得$\displaystyle y\left(x\right)\sec x-y\left(\frac\pi4\right)\sec\frac\pi4=\sin x-\sin\frac\pi4$
如此所求為$\displaystyle y\left(x\right)=\left(\sin x+\frac{\sqrt2}2\right)\cos x$
- Evaluate $\displaystyle\iint_{\Omega}\sqrt{x^2+xy+y^2}dxdy$, where $\Omega=\left\{\left(x,y\right):x^2+xy+y^2\leq1\right\}$, by first making the change of variables $\displaystyle u=x+\frac{y}2$, $\displaystyle v=\frac{\sqrt{3}y}2$ and then by polar coordinates.
- Let $\Gamma$ be the intersection curve of the two surfaces $x^3+2y^3+3z^3-2xyz=4$ and $x^2+y^2+z^2=3$. Find the unit tangent vector to $\Gamma$ at the point $\left(1,1,1\right)$.
訣竅
按題意的指示使用變數變換，其中應留意積分範圍的變化以及 Jacobian 行列式的計算。解法
容易看出 $u^2+v^2=x^2+xy+y^2\leq1$，由此設定$D=\left\{\left(u,v\right)\in\mathbb{R}^2:~u^2+v^2\leq1\right\}$
再者，其對應的 Jacobian 行列式為$\displaystyle\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}=\left|\begin{matrix}\displaystyle\frac{\partial x}{\partial u}&\displaystyle\frac{\partial x}{\partial v}\\\displaystyle\frac{\partial y}{\partial u}&\displaystyle\frac{\partial y}{\partial v}\end{matrix}\right|=\left|\begin{matrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{matrix}\right|^{-1}=\left|\begin{matrix}1&\displaystyle\frac12\\0&\displaystyle\frac{\sqrt3}2\end{matrix}\right|^{-1}=\frac2{\sqrt3}=\frac{2\sqrt3}3$
如此所求的重積分可先改寫如下$\displaystyle\iint_{\Omega}\sqrt{x^2+xy+y^2}dxdy=\iint_D\sqrt{u^2+v^2}\cdot\frac{2\sqrt3}3dudv$
接著應用極座標變換，即令 $\left\{\begin{aligned}&u=r\cos\theta\\&v=r\sin\theta\end{aligned}\right.$，其中 $\left\{\begin{aligned}&0\leq r\leq1\\&0\leq\theta\leq2\pi\end{aligned}\right.$，如此所求的重積分可改寫並計算如下$\displaystyle\iint_{\Omega}\sqrt{x^2+xy+y^2}dxdy=\frac{2\sqrt3}3\int_0^{2\pi}\int_0^1r\cdot rdrd\theta=\frac{2\sqrt3}3\left(\int_0^{2\pi}d\theta\right)\left(\int_0^1r^2dr\right)=\frac{4\pi\sqrt3}9$
訣竅
先求出這些曲面在該處的法向量，應用外積可得曲線的切向量。解法
設 $F_1\left(x,y,z\right)=x^3+2y^3+3z^3-2xyz-4$、$F_2\left(x,y,z\right)=x^2+y^2+z^2-3$，那麼先計算兩函數 $F_1\left(x,y,z\right)$ 與 $F_2\left(x,y,z\right)$ 的梯度如下$\displaystyle\nabla F_1\left(x,y,z\right)=\left(\frac{\partial F_1}{\partial x}\left(x,y,z\right),\frac{\partial F_1}{\partial y}\left(x,y,z\right),\frac{\partial F_1}{\partial z}\left(x,y,z\right)\right)=\left(3x^2-2yz,6y^2-2xz,9z^2-2xy\right)$
與$\displaystyle\nabla F_2\left(x,y,z\right)=\left(\frac{\partial F_2}{\partial x}\left(x,y,z\right),\frac{\partial F_2}{\partial y}\left(x,y,z\right),\frac{\partial F_2}{\partial z}\left(x,y,z\right)\right)=\left(2x,2y,2z\right)$
因此兩曲面 $F_1\left(x,y,z\right)=0$ 與 $F_2\left(x,y,z\right)=0$ 在 $\left(1,1,1\right)$ 處的法向量為$\nabla F_1\left(1,1,1\right)=\left(1,4,7\right)$, $\nabla F_2\left(1,1,1\right)=\left(2,2,2\right)\parallel\left(1,1,1\right)$
故所求的切向量為$\left(1,4,7\right)\times\left(1,1,1\right)=\left(-3,6,-3\right)\parallel\left(1,-2,1\right)$
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