Find the critical number of $y = 1 – \frac{4}{{\pi ^2 }}(\tan ^{ – 1} x)^2$.

$y’ = – \frac{4}{{\pi ^2 }} \cdot 2 \cdot (\tan ^{ – 1} x) \cdot \frac{1}{{1 + x^2 }}\mathop = \limits^{{\text{Let}}} 0$
$\Rightarrow \tan ^{ – 1} x = 0 \Rightarrow x = 0 ∎$

Determine the limits of integration where $a\leq b$ such that $\int_a^b {(x^2 – 16)} dx$ has minimal value.

$\begin{array}{cl} \int_a^b(x^2-16)dx & =\int_{{\rm{ – 4}}}^{\rm{4}} {x^2 – {\rm{16}}} dx \\ & =2\int_0^4 {x^2 – 16} dx \\ & =2\left[ {\frac{{x^3 }}{3} – 16x} \right]_{x = 0}^{x = 4}\\&=2(\frac{{64}}{3} – 64) = – \frac{{256}}{3} ∎ \end{array}$

Evaluate $\int_{ – \infty }^\infty {\frac{{e^x }}{{1 + e^{2x} }}} dx$.

Let $u = e^x \Rightarrow e^x dx = du$
$\begin{array}{cl} \int_{ – \infty }^\infty {\frac{{e^x }}{{1 + e^{2x} }}} dx=\int_0^\infty {\frac{1}{{1 + u^2 }}} du &= \left. {\tan ^{ – 1} u} \right|_{u = 0}^{u = \infty } \\ & =\frac{\pi }{2} – 0 = \frac{\pi }{2} ∎ \end{array}$

Find the slope of the surface $f{\rm{(}}x,y{\rm{) = (}}x^3 + y^3 )^{\frac{1}{3}}$ at the point $(0,0)$ in the $y$-direction.

Find the surface area of the portion of the plane $z=4-2x-2y$ that lies above the circle $x^2 + y^2 \le 1$ in the first quadrant.

◎ 曲面面積公式：
(一般型) $\text{A}=\iint_{\Omega}\sqrt {1 + f_x ^2 + f_y ^2 }dA$
(參數型) $\text{A}=\iint_{\Omega}\parallel \overset{\rightharpoonup}{u_x}\times \overset{\rightharpoonup}{u_y} \parallel dA$

Find an equation of the tangent plane to the paraboloid $\textbf{r}(u,v)=u\textbf{i}+v\textbf{j}+(u^2+v^2)\textbf{k}$ at the point $(1,2,5)$.

$\left\{ {\begin{array}{l} {\overset{\rightharpoonup}{\textbf{r}_u} = (1,0,2u)} \\ {\overset{\rightharpoonup}{\textbf{r}_v} = (0,1,2v)} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{l} {\overset{\rightharpoonup}{\textbf{r}_u} = (1,0,4)} \\ {\overset{\rightharpoonup}{\textbf{r}_v} (1,2) = (0,1,4)}\end{array}} \right. \Rightarrow$ 取 $\overset{\rightharpoonup}{\textbf{r}_u}\times \overset{\rightharpoonup}{\textbf{r}_v}=(2,4,1)$
$\Rightarrow$ 切平面方程式為：$2x+4y-z=5 ∎$

$\begin{array}{cl} \Rightarrow \overset{\rightharpoonup}{n} & =\nabla f(1,2,5) \\ & =\left. {(2x,2y, – 1)} \right|_{(1,2,5)} = (2,4, – 1) \end{array}$
$\Rightarrow$ 切平面方程式為：$2x+4y-z=5 ∎$

Evaluate the integral $\int_0^\infty {\int_0^\infty {\frac{1}{{(1 + x^2 + y^2 )^2 }}} } dxdy$.

$\begin{array}{cl} \int_0^\infty {\int_0^\infty {\frac{1}{{(1 + x^2 + y^2 )^2 }}} } dxdy & =\int_0^{\frac{\pi }{2}} {\int_0^\infty {\frac{1}{{(1 + r^2 )^2 }}} } \cdot rdrd\theta \\ & =\int_0^{\frac{\pi }{2}} {\int_0^\infty {\frac{1}{{u^2 }} \cdot \frac{{du}}{2}} } d\theta \\ & =\frac{1}{2}\int_0^{\frac{\pi }{2}} {\left. { – \frac{1}{u}} \right|} _{u = 1}^{u = \infty } d\theta\\&=\frac{1}{2}\int_0^{\frac{\pi }{2}} {0 – ( – 1)} d\theta = \frac{1}{2}\underbrace {\int_0^{\frac{\pi }{2}} {d\theta } }_{\frac{\pi }{2}} = \frac{\pi }{4} ∎ \end{array}$

Use a change of variable to find the volume of the solid region lying below the surface $z = \sqrt {(x + 4y)(x – y)}$ and above the plane region $R$: region bounded by the parallelogram with vertices $(0,0),\ (1,1),\ (5,0)$ and $(4,-1)$.

Let $\left\{ {\begin{array}{c} {u = x + 4y} \\ {v = x – y} \end{array}} \right. \Rightarrow \frac{{dudv}}{{dxdy}} = \left| {\begin{array}{l} 1 & 4 \\ 1 & { – 1}\end{array}} \right| = – 5$
$\Rightarrow \left| {\frac{{dudv}}{{dxdy}}} \right| = 5 \Rightarrow \left| {\frac{{dxdy}}{{dudy}}} \right| = \frac{1}{5}$

$\begin{array}{cl} V & =\iint_{R}\sqrt{(x+4y)(x-y)}dxdy \\ & =\iint_{R^{\prime}}\sqrt {uv} \cdot \left| {\frac{{dxdy}}{{dudv}}} \right| dudv \\ & =\frac{1}{5}\int_0^5 {\int_0^5 {\sqrt {uv} } } dudv\\&=\frac{1}{5}\int_0^5 {\sqrt v } (\int_0^5 {\sqrt u } du)dv = \frac{1}{5}(\int_0^5 {\sqrt u } du)(\int_0^5 {\sqrt v } dv)\\&=\frac{1}{5}(\int_0^5 {\sqrt u } du)^2 = \frac{1}{5}(\left. {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{u = 0}^{u = 5} )^2\\&=\frac{1}{5} \cdot \frac{4}{9} \cdot (5\sqrt 5 )^2 = \frac{{100}}{9} ∎ \end{array}$

A heat-seeking particle is located at the point $(-1,2)$ on a metal plate whose temperature at $(x,y)$ is $T(x,y)=64-2x^2-y^2$. (1) In what direction from $(-1,2)$ does the temperature increase most rapidly? What is this rate of increase? (2) Find the path of the particle as it continuously moves in the direction of maximum temperature increase.

increase most rapidly:
$\nabla T( – 1,2) = \left. {( – 4x, – 2y)} \right|_{( – 1,2)} = (4, – 4)$
rate: $\sqrt {4^2 + ( – 4)^2 } = 4\sqrt 2 ∎$

Let the path be $\left\{ {\begin{array}{c} {x = x(t),} & {x(0) = – 1} \\ {y = y(t),} & {y(0) = 2}\end{array}} \right.$
$\Rightarrow (x'(t),y'(t)) = \lambda (t)( – 4x(t), – 2y(t))$
$\Rightarrow \left\{ {\begin{array}{l} {x'(t) = – 4\lambda (t) \cdot x(t)} \\ {y'(t) = – 2\lambda (t) \cdot y(t)}\end{array}} \right.$
$\Rightarrow \frac{{x'{\rm{(}}t{\rm{)}}}}{{y'(t)}} = 2 \cdot \frac{{x(t)}}{{y(t)}} \Rightarrow \frac{{x'(t)}}{{x(t)}} = 2 \cdot \frac{{y'(t)}}{{y(t)}}$
$\mathop \Rightarrow \limits^{\int {dt} } \ln \left| {x(t)} \right| = 2\ln \left| {y(t)} \right| + C$
$\mathop \Rightarrow \limits^{t = 0} \ln 1 = 2\ln 2 + C \Rightarrow C = – 2\ln 2 = – \ln 4$
$\Rightarrow \left| {x(t)} \right| = \frac{1}{4}y(t)^2 \Rightarrow y^2 = 4\left| x \right|$
$\Rightarrow$ 因要求沿溫度上升方向，故取 $y^2=-4x,\ y\ge 0 ∎$

Determine if the given series converges or diverges. Explain your reasoning.
〔1〕$\sum\limits_{n = 1}^\infty (\frac{3n+2}{n+3})^n$ (6 分)
〔2〕$\sum\limits_{n = 1}^\infty {\frac{{e^{\frac{2}{n}} }}{{n^2 }}}$ (6 分)

〔1〕
$\because\ \mathop {\lim }\limits_{n \to \infty } \left| {(\frac{{3n + 2}}{{n + 3}})^n } \right|^{\frac{1}{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3n + 2}}{{n + 3}} = 3 > 1$
$\therefore\ \sum\limits_{n = 1}^\infty {(\frac{{3n + 2}}{{n + 3}})^n }$ diverges (by root test)$. ∎$

〔2〕
$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{e^{\frac{2}{n}} }}{{n^2 }}}}{{\frac{1}{{n^2 }}}} = \mathop {\lim }\limits_{n \to \infty } e^{\frac{2}{n}} = 1 \\ & \text{and }\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}}\text{ converges (by }p\text{-series)}. \\ \therefore & \sum\limits_{n = 1}^\infty {\frac{{e^{\frac{2}{n}} }}{{n^2 }}}\text{ converges (by limit comparison test).} ∎ \end{array}$

Find the maximum value of $\int_C {y^3 dx} + (27x – x^3 )dy$, where $C$ is any circle in the $xy$-plane, oriented counterwise.

◎ Green theorem:
$\oint_C {Pdx} + Qdy =\iint_{\Omega}Q_x-P_ydxdy$

By Green’s theorem,
$\int_C {y^3 } dx + (27x – x^3 )dy =\iint_{\Omega}27-3x^2-3y^2dxdy$

$\begin{array}{cl} \Rightarrow \int_C {y^3 dx} + (27x – x^3 )dy & =\iint_{\Omega}27-3x^2-3y^2dxdy \\ & =\int_0^{2\pi } {\int_0^3 {(27 – 3r^2 )r} dr} d\theta \\ & (\text{Let }u = 27 – 3r^2 \Rightarrow du = – 6rdr)\\&=\int_0^{2\pi } {\int_{27}^0 {u \cdot \frac{{du}}{{ – 6}}} } d\theta\\&=\frac{1}{6}\int_0^{2\pi } {\left. {\frac{{u^2 }}{2}} \right|_{u = 0}^{u = 27} } d\theta\\&=\frac{1}{{12}}\int_0^{2\pi } {729} d\theta\\&=\frac{{729}}{{12}} \cdot 2\pi = \frac{{243}}{2}\pi ∎ \end{array}$