- Only answers will be graded. Work need not be shown.
- $6\%$ for each of (1)-(10) and $8\%$ for each of (11)-(15).
- Find the interval on which $y=x^3-3x^2-6x+1$ is both decreasing and concave upward. 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 $x^3+3x^2y+y^3=5$.
- Evaluate $\displaystyle\int_4^5\frac{dx}{x^2-3x+2}=$ (4) .
- Rotate the region $\left\{\left(x,y\right):0\leq x\leq\pi\right.$ and $\left.0\leq y\leq\sin x\right\}$ about the $x$-axis. Find the volume of the solid obtained. Answer. (5) .
- Let $g\left(x\right)$ be the inverse function of the strictly increasing function $f\left(x\right)=x^5+3x^3+1$. Then $\displaystyle\int_{f\left(0\right)}^{f\left(1\right)}g\left(x\right)dx=$ (6) .
- The coefficient of $x^9$ in the Taylor expansion of $\displaystyle\int_0^xe^{-t^2}\,dt$ is (7) .
- Find the direction derivative of $f\left(x,y\right)=\ln\left(x^2+y^2\right)$ at the point $\left(1,2\right)$ in the direction $\displaystyle{\bf u}=\left(\frac35,\frac45\right)$. Answer. (8) .
- Find the plane tangent to the surface $x^3+y+z^2-xy^3z^4=8$ at the point $\left(2,1,1\right)$. Answer. (9) .
- Evaluate $\displaystyle\int_{y=0}^{y=2}\int_{x=y}^{x=2}e^{x^2}\,dx\,dy=$ (10) .
- Let $\displaystyle f\left(x,y\right)=\frac1{\left(x^2y^2-4x+3y+1\right)^2}$ and $g\left(u,v\right)=f\left(x\left(u,v\right),y\left(u,v\right)\right)$, where $x\left(u,v\right)=u^2-3uv+2v^2$ and $y\left(u,v\right)=u^4+3uv^3-4v^4$. Find $\displaystyle\frac{\partial g}{\partial u}$ at the point $u=1$ and $v=1$. Answer. (11) .
- Find point(s) (12) on the curve $x^3-y^3=1$ farthest from the line $y=x$.
- Evaluate $\displaystyle\iint_\Omega\left(\frac{y-x}{y+x}\right)^4\,dx\,dy$, where $\Omega$ is the region in the first quadrant bounded by the lines $x+y=1$ and $x+y=2$. Answer. (13) .
- Solve the differential equation $\displaystyle\frac{dy}{dt}=2\left(1+t\right)\left(1+y^2\right)$ with the initial condition $y\left(-1\right)=0$. Answer. (14) .
- Solve the differential equation $\displaystyle t\frac{dy}{dt}+2y=e^t$ with the initial condition $y\left(1\right)=2$. Answer. (15) .
訣竅
遞減為 $y'<0$，凹口向上為 $y''>0$，求其聯立。解法
根據題目的要求聯立解不等式組：$\left\{\begin{aligned}&y'=3x^2-6x-6<0,\\&y''=6x-6>0.\end{aligned}\right.$
如此可解得$\left\{\begin{aligned}&1-\sqrt3<x\leq1+\sqrt3,\\&x>1.\end{aligned}\right.$
因此取共同區間為 $1<x<1+\sqrt3$。訣竅
根據隱函數微分法求解即可。解法
先對原式進行一次微分可得$3x^2+6xy+3x^2y'+3y^2y'=0.$
取 $x=1$ 與 $y=1$ 代入可得 $\displaystyle y'=-\frac32$。另一方面對微分的結果再微分一次可得
$6x+6y+12xy'+3x^2y''+6yy'^2+3y^2y''=0.$
取 $x=1$、$y=1$ 與 $\displaystyle y'=-\frac32$ 可得 $\displaystyle y''=-\frac54$。訣竅
因式分解後拆項再積分。解法
$\displaystyle\int_4^5\frac{dx}{\left(x-1\right)\left(x-2\right)}=\int_4^5\left(\frac1{x-2}-\frac1{x-1}\right)dx=\left.\ln\left(\frac{x-2}{x-1}\right)\right|_4^5=\ln\frac98$.
訣竅
按旋轉體體積公式計算體積。解法
$\displaystyle V=\int_0^\pi\pi\sin^2x\,dx=\pi\int_0^\pi\frac{1-\cos2x}2\,dx=\left.\frac{\pi}2\left(x-\frac{\sin2x}2\right)\right|_0^\pi=\frac{\pi^2}2$.
訣竅
做變數代換後使用分部積分。解法
令 $x:=f\left(x\right)$，因此所求計算如下$\begin{aligned}\int_{f\left(0\right)}^{f\left(1\right)}g\left(x\right)dx&=\int_0^1g\left(f\left(x\right)\right)df\left(x\right)\\&=\int_0^1xf'\left(x\right)dx\\&=xf\left(x\right)\Big|_0^1-\int_0^1f\left(x\right)dx\\&=5-\int_0^1\left(x^5+3x^3+1\right)dx\\&=5-\left.\left(\frac{x^6}6+\frac{3x^4}4+x\right)\right|_0^1\\&=\frac{37}{12}.\end{aligned}$
訣竅
避免對此函數做九次微分，先考慮其 Taylor 展開式後積分。解法
由於 $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$，因此 $\displaystyle e^{-t^2}=\sum_{k=0}^\infty\frac{\left(-1\right)^k}{k!}t^{2k}$。據此同取積分（積分範圍為 $\left[0,x\right]$）可得$\displaystyle\int_0^xe^{-t^2}\,dt=\sum_{k=0}^\infty\frac{\left(-1\right)^k}{k!\left(2k+1\right)}x^{2k+1}$.
因此其中當 $k=4$ 時可得 $x^9$ 的係數為 $\displaystyle\frac1{4!\cdot9}=\frac1{216}$。訣竅
根據方向導數的計算公式有 $D_{\bf u}f\left(x,y\right)=\nabla f\left(x,y\right)\cdot{\bf u}$ 即可。解法
$\displaystyle D_{\bf u}f\left(1,2\right)=\left.\left(\frac{2x}{x^2+y^2},\frac{2y}{x^2+y^2}\right)\right|_{\left(x,y\right)=\left(1,2\right)}\cdot\left(\frac35,\frac45\right)=\left(\frac25,\frac45\right)\cdot\left(\frac35,\frac45\right)=\frac{22}{25}.$
訣竅
利用梯度找出法線方向，根據點法式求解。解法
$\nabla\left(x^3+y+z^2-xy^3x^4-8\right)=\left(3x^2-y^3z^4,1-3xy^2z^4,2z-4xy^3z^3\right).$
而在座標 $\left(2,1,1\right)$ 時梯度為 $\left(11,-5,-6\right)$，因此由點法式可得切平面為 $11\left(x-2\right)-5\left(y-1\right)-6\left(z-1\right)=0$，即 $11x-5y-6z=11$。訣竅
原函數無法直接積分，因此使用 Fubini 定理改變積分次序以利積分。解法
根據訣竅改寫原重積分並計算如下$\displaystyle\int_{y=0}^{y=2}\int_{x=y}^{x=2}e^{x^2}\,dx\,dy=\int_{x=0}^{x=2}\int_{y=0}^{y=x}e^{x^2}\,dy\,dx=\int_0^2ye^{x^2}\Big|_{y=0}^{y=x}dx=\int_0^2xe^{x^2}\,dx=\left.\frac12e^{x^2}\right|_0^2=\frac{e^4-1}2.$
訣竅
根據多變數函數微分的連鎖律即可。解法
$\begin{aligned}\displaystyle\frac{\partial g}{\partial u}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\\&=\left[\frac{-2\left(2xy^2-4\right)}{\left(x^2y^2-4x+3y+1\right)^3}\right]\left(2u-3v\right)+\left[\frac{-2\left(2x^2y+3\right)}{\left(x^2y^2-4x+3y+1\right)^3}\right]\left(4u^3+3v^3\right).\end{aligned}$
取 $u=1$、$v=1$ 可得 $x=0$、$y=0$，由此代入上式可得 $\displaystyle\frac{\partial g}{\partial u}\left(1,1\right)=-50$。訣竅
先由點到直線的距離列出目標函數，再由 Lagrange 乘子法求其極值。解法
設 $P\left(x,y\right)$、直線 $L$ 為 $y=x$，則目標函數為 $\displaystyle d\left(P,L\right)=\frac{\left|x-y\right|}{\sqrt2}$，因此我們可以先考慮 $x-y$ 的極大值與極小值。首先設 Lagrange 乘子函數 $F\left(x,y,\lambda\right)=x-y+\lambda\left(x^3-y^3-1\right)$，接著解以下聯立方程：
$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=1+3\lambda x^2=0,\\&F_y\left(x,y,\lambda\right)=-1-3\lambda y^2=0,\\&F_\lambda\left(x,y,\lambda\right)=x^3-y^3-1=0.\end{aligned}\right.$
將前兩式相加可得 $\left(x-y\right)\left(x+y\right)\lambda=0$，由第三式可知 $x\neq y$，再由前兩式可知 $\lambda\neq0$，因此 $x+y=0$。如此可解得 $\displaystyle\left(x,y,\lambda\right)=\left(\frac{\sqrt[3]4}2,-\frac{\sqrt[3]4}2,-\frac{\sqrt[3]4}2\right)$。因此我們知道最遠的距離為 $\displaystyle\frac{\left|x-y\right|}{\sqrt2}=\sqrt[6]2$。訣竅
做適當的代換以詮釋此積分區域。解法
令 $u=y+x$、$v=y-x$，由邊界條件可以容易發現 $u$ 介於 $1$ 與 $2$ 之間。又 $\Omega$ 落於第一象限，因此 $u+v=2y\geq0$、$u-v=2x\geq0$，因此 $-u\leq v\leq u$。故 $u,v$ 應滿足 $\left\{\begin{aligned}&1\leq u\leq2\\&-u\leq v\leq u\end{aligned}\right.$。如此原重積分可改寫並計算如下$\begin{aligned}\iint_\Omega\left(\frac{y-x}{y+x}\right)^4\,dx\,dy&=\int_1^2\int_{-u}^u\left(\frac vu\right)^4|\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right||\,dv\,du\\&=\frac12\int_1^2\int_{-u}^u\frac{v^4}{u^4}\,dv\,du\\&=\frac1{10}\int_1^2\left.\frac{v^5}{u^4}\right|_{-u}^u\,du\\&=\frac15\int_1^2u\,du\\&=\left.\frac{u^2}{10}\right|_1^2\\&=\frac3{10}.\end{aligned}$
訣竅
使用分離變數法即可求解。解法
首先移項後有$\displaystyle\frac{dy}{1+y^2}=2\left(1+t\right)dt$.
同取積分可得$\arctan y=2t+t^2+C$,
其中 $C$ 為積分常數。透過初始條件可確定出 $C=1$，因此 $y=\tan\left[\left(t+1\right)^2\right]$。訣竅
本題可以先考慮齊次解，再使用常數變異法求出一般解後用初始條件確定積分常數。另外本題亦可直接使用積分因子法。解法一
先考慮齊次微分方程：$\displaystyle t\frac{dy}{dt}+2y=0$.
如此透過移項後取積分即可得到$\displaystyle y\left(t\right)=\frac{C}{t^2}$,
其中 $C$ 為積分常數。接著我們運用常數變異法，設$\displaystyle y\left(t\right)=\frac{K\left(t\right)}{t^2}$.
代回題目的方程中可得$\displaystyle\frac{K'\left(t\right)}t=e^t$.
如此可得 $K\left(t\right)=te^t-e^t+C$，其中 $C$ 為積分常數。因此我們得到通解為$\displaystyle y\left(t\right)=\frac{te^t-e^t+C}{t^2}$.
透過初始條件可得 $C=2$，於是滿足本題條件的函數為$\displaystyle y\left(t\right)=\frac{te^t-e^t+2}{t^2}$.
解法二
注意到 $t$ 為此方程的積分因子，即同乘 $t$ 可得$\displaystyle t^2\frac{dy}{dt}+2ty=te^t$,
亦可寫為$\displaystyle\frac d{dt}\left(t^2y\right)=te^t$.
因此同取積分可得$t^2y=te^t-e^t+C$,
其中 $C$ 為積分常數，透過初始條件可確定 $C=2$。因此滿足題目條件的函數為$\displaystyle y\left(y\right)=\frac{te^t-e^t+2}{t^2}$.
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