※注意：請於試卷上「非選擇題作答區」標明題號並依序作答。
Correctly number each of your answer
to indicate which question is answered.
- Work need not be shown.
- Only answers will be graded.
- $5$ points each and $100$ points total.
- Find the interval on which $y=x^3-3x^2+2x+1$ is both decreasing and concave downward. Answer. (1) .
- Find $\displaystyle\frac{dy}{dx}=$ (2) and $\displaystyle\frac{d^2y}{dx^2}=$ (3) at the point $x=1$, $y=1$ of the curve $x^3+xy-2y^4=0$.
- Evaluate $\displaystyle\int_2^3\sqrt{3+2x-x^2}dx$. Answer: (4) .
- 當 $x=2$ 時則有 $\displaystyle\theta=\frac\pi6$；
- 當 $x=3$ 時則有 $\displaystyle\theta=\frac\pi2$；
- 求導有 $dx=2\cos\theta\,d\theta$。
- Find the volume of the solid generated by revolving the following region about the $y$-axis: $\left\{\left(x,y\right):0\leq x\leq\pi\mbox{ and }0\leq y\leq\sin x\right\}$. Answer: (5) .
- Find the arc length of the curve $\displaystyle y=\int_0^x\sqrt{\sin t}dt$ from $x=0$ to $x=\pi/3$. Answer: (6) .
- Find the solution of the differential equation $y'=\left(1+2x\right)\left(1+y^2\right)$ with the initial condition $y\left(0\right)=1$. Answer: (7) .
- Find the solution of the differential equation $y'+y\tan x=\sec x$ with the initial condition $y\left(0\right)=2$. Answer: (8) .
- Find the first three nonzero terms of the McLaurin series of $\tan x$. Answer: (9) .
- Let $u=u\left(x,y\right)$ be a function of $x$, $y$. Express $\displaystyle\frac{\partial u}{\partial x}$ in terms of polar coordinates $r$, $\theta$ together with $\displaystyle\frac{\partial u}{\partial r}$ and $\displaystyle\frac{\partial u}{\partial\theta}$. Answer: (10) .
- Find the directional derivative of $f\left(x,y,z\right)=x^yz^2$ at the point $\left(e,e,1\right)$ in the direction $\displaystyle{\bf u}=\frac3{13}{\bf i}+\frac{12}{13}{\bf j}+\frac{4}{13}{\bf k}$. Answer: (11) .
- Find the critical points of $f\left(x,y\right)=x^2+y^4+3xy^2-5x$ which are saddle points. Answer: (12) .
- 若 $y=0$，則 $\displaystyle x=\frac52$；
- 若 $2y^2+3x=0$，則代回第一式可得 $x=-2$，從而有 $y=\pm\sqrt3$。
- Find the minimum of $x^2+y^2+z^2$ for $\left(x,y,z\right)$ on the intersection curve of the two surfaces $y+2z=1$ and $3x^2+y^2-z^2=1$. Answer: (13) .
- 若 $x=0$，則代入第三式可得 $z=0$ 或 $\displaystyle z=\frac43$；
- 若 $\displaystyle\lambda=-\frac13$，則代入第二式有 $\displaystyle z=\frac13$，因此 $\displaystyle x=\pm\frac{\sqrt3}3$。
- 若 $x=0$，則第五式可寫為 $y^2-z^2=1$，而第四式表明 $y=1-2z$，代入則有 $3z^2-4z=0$，因此 $z=0$ 或 $\displaystyle\frac43$，從而 $\displaystyle y=1$ 或 $\displaystyle y=-\frac53$。
- 若 $\displaystyle\lambda_2=-\frac13$，則第二式與第三式可分別寫為 $4y+3\lambda_1=0$ 及 $8z+6\lambda_1=0$。將前者乘以兩倍後減去後者，即有 $y=z$。並由第四式可得 $\displaystyle y=z=\frac13$，再代入最後一式則能解得 $\displaystyle x=\pm\frac{\sqrt3}3$。
- Evaluate $\displaystyle\int_0^3\int_{\sqrt{x/3}}^1e^{y^3}dydx$. Answer: (14) .
- Let $R$ be the region in the first quadrant of the $xy$-plane bounded by $xy=1$, $xy=2$, $y=x$ and $y=2x$. Evaluate $\displaystyle\iint_Re^{xy}dxdy$. Answer: (15) .
- Evaluate $\displaystyle\iiint_{\Omega}\frac{\cos\sqrt{x^2+y^2+z^2}}{x^2+y^2+z^2}dxdydz$, where $\Omega$ is given by $\Omega=\left\{\left(x,y,z\right):1\leq x^2+y^2+z^2\leq2\right\}$. Answer: (16) .
- Evaluate $\displaystyle\iiint_{\Omega}ze^{x^2+y^2+3z^2}dxdydz$, where $\Omega$ is the cylinder defined by $\Omega=\left\{\left(x,y,z\right):x^2+y^2\leq1\mbox{ and }0\leq z\leq1\right\}$. Answer: (17) .
- Let $S$ be the surface described by $z=x^2+\frac{y^2}2$ with $4x^2+y^2\leq1$ oriented with normals with positive ${\bf k}$-components. ${\bf F}\left(x,y,z\right)=x{\bf i}-y{\bf j}+{\bf k}$. Evaluate $\displaystyle\iint_S{\bf F}\cdot d{\bf S}$. Answer: (18) .Also, evaluate $\displaystyle\iint_SdS$. Answer: (19) .
- Let $C$ be the counterclockwise oriented boundary of the region in the $xy$-plane enclosed by $x^2+y^2-2x=0$ and $x^2+y^2-2y=0$. Evaluate the line integral $\displaystyle\oint_C\left(y+e^{-x^2}\right)dx+\left(3x+\sin\left(y^2\right)\right)dy$. Answer: (20) .
訣竅
函數遞減表明 $y'<0$；而凹口向下表明 $y''<0$。解法
為了找出所求的區間，我們解下列的聯立不等式$\left\{\begin{aligned}&y'=3x^2-6x+2<0\\&y''=6x-6<0\end{aligned}\right.$
因此可得$\displaystyle\left\{\begin{aligned}1-\frac1{\sqrt3}<x&<1+\frac1{\sqrt3}\\x&<1\end{aligned}\right.$
即$\displaystyle1-\frac1{\sqrt3}< x<1$
訣竅
利用隱函數微分即可。解法
隱函數微分一次可得 $3x^2+y+xy'-8y^3y'=0$，取 $\left(x,y\right)=\left(1,1\right)$ 代入後可得$\displaystyle3+1+\left.\frac{dy}{dx}\right|_{(x,y)=(1,1)}-8\left.\frac{dy}{dx}\right|_{(x,y)=(1,1)}=0$
整理即有 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(1,1\right)}=\frac47$。接著對已經隱函數微分一次後的方程再微分一次可得$6x+2y'+\left(x-8y^3\right)y''-24y^2y'^2=0$
取 $x=1$、$y=1$、$\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(1,1\right)}=\frac47$ 代入可得$\displaystyle6+2\cdot\frac47+\left(1-8\right)\left.\frac{d^2y}{dx^2}\right|_{(x,y)=(1,1)}-24\cdot1^2\cdot\left(\frac47\right)^2=0$
整理求得 $\displaystyle\left.\frac{d^2y}{dx^2}\right|_{\left(x,y\right)=\left(1,1\right)}=-\frac{34}{343}$。訣竅
配方法後三角代換。解法
因為 $3+2x-x^2=4-\left(x-1\right)^2$，令 $x=1+2\sin\theta$，則$\begin{aligned}\displaystyle\int_2^3\sqrt{3+2x-x^2}dx&=\int_{\frac\pi6}^{\frac\pi2}2\cos\theta\cdot2\cos\theta\,d\theta\\&=2\int_{\frac\pi6}^{\frac\pi2}\left(1+\cos2\theta\right)d\theta\\&=\left(2\theta+\sin2\theta\right)\Big|_{\frac\pi6}^{\frac\pi2}\\&=\frac{2\pi}3-\frac{\sqrt3}2.\end{aligned}$
訣竅
根據旋轉體體積公式即可。解法
利用旋轉體體積公式可得$\displaystyle\begin{aligned}V&=\int_0^{\pi}2\pi xf\left(x\right)dx\\&=2\pi\int_0^{\pi}x\sin x\,dx\\&=\left(-2\pi x\cos x\right)\Big|_0^\pi+2\pi\int_0^{\pi}\cos x\,dx\\&=2\pi^2+\left(2\pi\sin x\right)\Big|_0^\pi\\&=2\pi^2.\end{aligned}$
訣竅
利用曲線弧長的公式即可。解法
運用弧長的計算公式列式並計算如下：$\displaystyle\begin{aligned}s&=\int_0^{\frac\pi3}\sqrt{1+f'^2\left(x\right)}dx\\&=\int_0^{\frac\pi3}\sqrt{1+\sin x}\,dx\\&=\int_0^{\frac\pi3}\left(\sin\frac{x}2+\cos\frac{x}2\right)dx\\&=\left.-2\cos\frac{x}2+2\sin\frac{x}2\right|_0^{\frac\pi3}\\&=3-\sqrt3.\end{aligned}$
訣竅
利用分離變數法即可。解法
移項整理有$\displaystyle\frac{dy}{1+y^2}=\left(1+2x\right)dx$
同取範圍為 $\left[0,x\right]$ 的定積分可得$\arctan y\left(x\right)-\arctan y\left(0\right)=x+x^2$
運用初始條件並整理之可得$\displaystyle y\left(x\right)=\tan\left(x^2+x+\frac{\pi}{4}\right)$
訣竅
利用積分因子即可。解法
對該微分方程同乘以 $\sec x$ 如此可得$\left(y\sec x\right)'=y'\sec x+y\sec x\tan x=\sec^2x$
同取範圍為 $\left[0,x\right]$ 的定積分，於是有$y\left(x\right)\sec x-y\left(0\right)\sec\left(0\right)=\tan x$
使用初始條件並整理之，可解得$y\left(x\right)=\left(2+\tan x\right)\cos x$
訣竅
直接計算 Taylor 展開式即可。解法
設 $f\left(x\right)=\tan x$，則$\begin{aligned}f'\left(x\right)&=\sec^2x\\f''\left(x\right)&=2\sec^2x\tan x\\f'''\left(x\right)&=2\sec^4+4\sec^2x\tan^2x\\f^{\left(4\right)}\left(x\right)&=16\sec^4x\tan x+8\sec^2x\tan^2x\\f^{\left(5\right)}\left(x\right)&=16\sec^6x+32\sec^4x\tan^2x+16\sec^4x\tan x+16\sec^2x\tan^3x\end{aligned}$
如此我們有$f''\left(0\right)=0=f^{\left(4\right)}\left(0\right)$ ； $f'\left(0\right)=1$ ； $f'''\left(0\right)=2$ ； $f^{\left(5\right)}\left(0\right)=16$
因此 Taylor 展開後的前三個非零項為 $\displaystyle x+\frac{x^3}3+\frac{2x^5}{15}$。訣竅
用多變數微分的連鎖律。解法
直接計算有$\displaystyle\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial\theta}\frac{\partial\theta}{\partial x}=\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}r\frac{\partial u}{\partial\theta}$
其中$\displaystyle\begin{aligned} &\frac{\partial r}{\partial x}=\frac{\partial}{\partial x}\sqrt{x^2+y^2}=\frac1{2\sqrt{x^2+y^2}}\cdot2x=\frac{r\cos\theta}{r}=\cos\theta\\&\frac{\partial\theta}{\partial x}=\frac{1}{\displaystyle1+\left(\frac yx\right)^2}\cdot-\frac y{x^2}=-\frac{r\sin\theta}{r^2}=-\frac{\sin\theta}r.\end{aligned}$
訣竅
方向導數可利用梯度與方向的內積求解即可。解法
首先計算梯度為 $\nabla f=\left(yx^{y-1}z^2,x^yz^2\ln x,2x^yz\right)$，於是在 $\left(e,e,1\right)$ 時的梯度為$\nabla f\left(e,e,1\right)=\left(e^e,e^e,2e^e\right)$
因此方向導數為$\displaystyle D_{\bf u}f\left(e,e,1\right)=\left(e^e,e^e,2e^e\right)\cdot\left(\frac3{13},\frac{12}{13},\frac4{13}\right)=\frac{23e^e}{13}$
訣竅
利用一階偏導為零求出臨界點，再用二階判別式找出鞍點。解法
為了找出臨界點，我們解以下的聯立方程組：
$\left\{\begin{aligned} &f_x=2x+3y^2-5=0\\&f_y=4y^3+6xy=0\end{aligned}\right.$
由第二式可知 $y=0$ 或 $2y^2+3x=0$。再者，我們計算二階偏導函數可得
$f_{xx}=2$ ；$f_{xy}=6y=f_{yx}$ ； $f_{yy}=12y^2+6x$
如此由 Hessian 行列式有$H\left(x,y\right)=\left|\begin{array}{cc}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{array}\right|=\left|\begin{array}{cc}2&6y\\6y&12y^2+6x\end{array}\right|=12x-12y^2$
因此代入座標檢查可知$\displaystyle H\left(\frac{5}{2},0\right)=30>0$，因此不為鞍點；
$H\left(-2,\pm\sqrt{3}\right)=-48<0$，故皆為鞍點。
訣竅
本題使用限制條件並搭配基本的不等式即可求解；亦可先化簡後再利用 Lagrange 乘子法來求解；亦可直接使用多條件的 Lagrange 乘子法直接計算。解法一
由 $3x^2+y^2-z^2=1$ 來消去 $x^2+y^2+z^2$ 中的 $x$ 可得$\displaystyle x^2+y^2+z^2=\frac{1-y^2+z^2}3+y^2+z^2=\frac{1+2y^2}3$
接著運用 Cauchy 不等式與另一個限制條件可知$\left(y^2+\left(\sqrt2z\right)^2\right)\left(1^2+\sqrt2^2\right)\geq\left(y+2z\right)^2=1$
這表明$\displaystyle x^2+y^2+z^2=\frac13+\frac23\left(y^2+2z^2\right)\geq\frac13+\frac23\cdot\frac13=\frac59$
由不等式的等號成立條件容易知道 $y=z$，故得 $\displaystyle y=z=\frac13$、$\displaystyle x=\pm\frac{\sqrt3}3$。解法二
欲求極小值的函數可先透過第一個條件改寫為 $f\left(x,z\right)=x^2+5z^2-4z+1$，而第二個條件則改寫為 $3x^2+3z^2-4z=0$，如此我們設 Lagrange 乘子函數為$F\left(x,z,\lambda\right)=\left(x^2+5z^2-4z+1\right)+\lambda\left(3x^2+3z^2-4z\right)$
如此解下列聯立方程組：$\left\{\begin{aligned}&F_x(x,z,\lambda)=2x+6x\lambda=0,\\&F_z(x,z,\lambda)=10z-4+\lambda\left(6z-4\right)=0,\\&F_{\lambda}(x,z,\lambda)=3x^2+3z^2-4z=0.\end{aligned}\right.$
由第一條式子可知 $x=0$ 或 $\displaystyle\lambda=-\frac13$。$f\left(0,0\right)=1$ ； $\displaystyle f\left(0,\frac43\right)=\frac{41}9$ ； $\displaystyle f\left(\pm\frac{\sqrt3}3,\frac13\right)=\frac59$
因此最小值為 $\displaystyle\frac59$。解法三
設 Lagrange 乘子函數如下$F\left(x,y,z,\lambda_1,\lambda_2\right)=x^2+y^2+z^2+\lambda_1\left(y+2z-1\right)+\lambda_2\left(3x^2+y^2-z^2-1\right)$
據此解下列聯立方程組：$\left\{\begin{aligned}&F_x(x,y,z,\lambda_1,\lambda_2)=2x+6\lambda_2x=0,\\&F_y(x,y,z,\lambda_1,\lambda_2)=2y+\lambda_1+2\lambda_2y=0,\\&F_z(x,y,z,\lambda_1,\lambda_2)=2z+2\lambda_1-2\lambda_2z=0,\\&F_{\lambda_1}(x,y,z,\lambda_1,\lambda_2)=y+2z-1=0,\\&F_{\lambda_2}(x,y,z,\lambda_1,\lambda_2)=3x^2+y^2-z^2-1=0.\end{aligned}\right.$
由第一式可知 $x=0$ 或 $\displaystyle\lambda_2=-\frac13$。$f\left(0,1,0\right)=1$ ； $\displaystyle f\left(0,-\frac53,\frac43\right)=\frac{41}9$ ； $\displaystyle f\left(\pm\frac{\sqrt3}3,\frac13,\frac13\right)=\frac59$
因此最小值為 $\displaystyle\frac59$。訣竅
交換積分順序計算即可解法
原先的積分範圍為 $\left\{\begin{aligned}0\leq x&\leq3\\\sqrt{x/3}\leq y&\leq1\end{aligned}\right.$，可改寫為 $\left\{\begin{aligned}0\leq x&\leq3y^2\\0\leq y&\leq1\end{aligned}\right.$，如此原重積分可改寫並計算如下：$\displaystyle\int_0^1\int_0^{3y^2}e^{y^3}dxdy=\int_0^1xe^{y^3}\Big|_{x=0}^{x=3y^2}dy=\int_0^13y^2e^{y^3}dy=e^{y^3}\Big|_0^1=e-1$.
訣竅
利用變數代換後改寫積分範圍。解法
令 $u=xy$、$\displaystyle v=\frac{y}{x}$，如此有$\begin{aligned}\displaystyle&\int_1^2\int_1^2e^u\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|dvdu=\int_1^2\int_1^2e^u\cdot\frac1{2v}dvdu\\=&\left(\int_1^2e^udu\right)\left(\int_1^2\frac{dv}{2v}\right)=\frac{e\left(e-1\right)\ln2}2.\end{aligned}$
其中 $\displaystyle\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|$ 計算如下：$\displaystyle\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}=\left|\begin{array}{cc}\displaystyle\frac{\partial x}{\partial u}&\displaystyle\frac{\partial x}{\partial v}\\\displaystyle\frac{\partial y}{\partial u}&\displaystyle\frac{\partial y}{\partial v}\end{array}\right|=\left|\begin{array}{cc}\displaystyle\frac{1}{2\sqrt{uv}}&\displaystyle-\frac{\sqrt{u}}{2\sqrt{v^3}}\\\displaystyle\frac{\sqrt{v}}{2\sqrt{u}}&\displaystyle\frac{\sqrt{u}}{2\sqrt{v}}\end{array}\right|=\frac{1}{2v}$
訣竅
利用球極座標代換即可。解法
運用球極座標代換，令 $\left\{\begin{aligned} &x=\rho\sin\phi\cos\theta\\&y=\rho\sin\phi\sin\theta\\&z=\rho\cos\phi\end{aligned}\right.$，其中範圍為 $\left\{\begin{aligned}&1\leq\rho\leq\sqrt{2}\\&0\leq\theta\leq2\pi\\&0\leq\phi\leq\pi\end{aligned}\right.$，如此原三重積分可改寫並計算如下：$\begin{aligned}\displaystyle&\int_1^{\sqrt2}\int_0^{2\pi}\int_0^\pi\frac{\cos\left(\rho\right)}{\rho^2}\rho^2\sin\phi\,d\phi\,d\theta\,d\rho\\=&2\pi\left(\int_1^{\sqrt2}\cos\rho\,d\rho\right)\left(\int_0^{\pi}\sin\phi\,d\phi\right)\\=&4\pi\left(\sin\left(\sqrt2\right)-\sin\left(1\right)\right).\end{aligned}$
訣竅
利用圓柱座標代換即可。解法
利用圓柱座標代換 $\left\{\begin{aligned} &x=r\cos\theta\\&y=r\sin\theta\\&z=z\end{aligned}\right.$，則 $\left\{\begin{aligned} &0\leq r\leq1\\&0\leq\theta\leq2\pi\\&0\leq z\leq1\end{aligned}\right.$，如此原三重積分可改寫並計算如下：$\begin{aligned}\iiint_\Omega ze^{x^2+y^2+z^2}dxdydz&=\int_0^1\int_0^{2\pi}\int_0^1ze^{r^2+3z^2}r\,dr\,d\theta\,dz\\=&2\pi\left(\int_0^1ze^{3z^2}\,dz\right)\left(\int_0^1re^{r^2}\,dr\right)\\&=2\pi\cdot\left.\frac{e^{3z^2}}6\right|_0^1\cdot\left.\frac{e^{r^2}}2\right|_0^1\\&=\frac{\pi\left(e^3-1\right)\left(e-1\right)}6.\end{aligned}$
訣竅
運用曲面積分的定義即可。解法
首先由 $z$ 分量為正，求出 $d{\bf S}=\left(-2x,-y,1\right)dA$，因此有$\displaystyle\iint_S{\bf F}\cdot d{\bf S}=\iint_S\left(-2x^2+y^2+1\right)dA$
運用橢圓座標變換，令 $\left\{\begin{aligned} &x=r\cos\theta/2\\&y=r\sin\theta\end{aligned}\right.$，其中範圍為 $\left\{\begin{aligned} &0\leq r\leq1\\&0\leq \theta\leq2\pi\end{aligned}\right.$，如此上式可改寫並計算如下：$\begin{aligned}\displaystyle\iint_S{\bf F}\cdot d{\bf S}&=\int_0^{2\pi}\int_0^1\left(-\frac{2r^2\cos^2\theta}4+r^2\sin^2\theta+1\right)\frac r2dr\,d\theta\\&=\int_0^{2\pi}\left.-\frac{r^4\cos^2\theta}{16}+\frac{r^4\sin^2\theta}8+\frac{r^2}4\right|_0^1d\theta\\&=\int_0^{2\pi}\left(\frac14+\frac{\sin^2\theta}8-\frac{\cos^2\theta}{16}\right)d\theta\\&=\int_0^{2\pi}\left(\frac9{32}-\frac{3\cos2\theta}{32}\right)d\theta\\&=\left.\frac{9\theta}{32}-\frac{3\sin2\theta}{64}\right|_0^{2\pi}\\&=\frac{9\pi}{16}.\end{aligned}$
另一方面，我們有$\begin{aligned}\displaystyle\iint_SdS&=\iint_S\sqrt{4x^2+y^2+1}\,dA\\&=\int_0^{2\pi}\int_0^1\sqrt{r^2+1}\frac r2\,dr\,d\theta\\&=2\pi\left.\frac{\left(r^2+1\right)^{\frac32}}6\right|_0^1\\=&\frac{\left(2\sqrt2-1\right)\pi}3.\end{aligned}$
訣竅
利用 Green 定理改寫為重積分計算即可。解法
設 $D$ 為 $x^2+y^2-2x\leq0$ 與 $x^2+y^2-2y\leq0$ 所圍成的共同區域，如下圖：
$\displaystyle\iint_D\left(3-1\right)dA=2\iint_DdA=2\cdot\left(\frac\pi2-1\right)=\pi-2$
其中 $\displaystyle\iint_DdA$ 代表 $D$ 區域的面積，可以看出此為兩個四分之一圓相加後扣去一個單位正方形的面積可得。
L題有解法3：
回覆刪除利用題目給定之兩平面，可求出兩平面之交線參數式x=f(t),y=g(t),z=h(t)，接著將其代入至x^2+y^2+z^2，可得到一串t的函數，接著微分=0可得出t值，再代回x^2+y^2+z^2即可得出最小值為5/9
給定的兩平面？
刪除