- (10 pts) Let $w=f\left(x,y\right)$ be a function of two variables, and let $x=u+v$ and $y=u-v$. Determine $a$ and $b$, $a,b\in R$, such that
$\displaystyle\frac{\partial^2w}{\partial u\partial v}=a\frac{\partial ^2w}{\partial x^2}+b\frac{\partial^2w}{\partial y^2}$.
- (15 pts) Let $A$ be the closed and bounded region defined by $-1\leq x,y\leq1$, and let $f:A\to R$ be given by
$f\left(x,y\right)=x^2y-2xy$.
Find the points in $A$ where $f$ attains a global maximum and a global minimum. - 當 $x=-1$ 時，函數為 $f\left(-1,y\right)=3y$，故在 $y=-1$ 時有極小值為 $f\left(-1,-1\right)=-3$，而在 $y=1$ 處有極大值 $f\left(-1,1\right)=3$；
- 當 $x=1$ 時有 $f\left(1,y\right)=-y$，那麼當 $y=1$ 時有極小值為 $f\left(1,1\right)=-1$，而當 $y=-1$ 時有極大值 $f\left(1,-1\right)=1$；
- 當 $y=-1$ 時有 $f\left(x,-1\right)=-x^2+2x=1-\left(x-1\right)^2$，故當 $x=1$ 時有極大值為 $f\left(1,-1\right)=1$，而當 $x=-1$ 時有極小值為 $-3$；
- 當 $y=1$ 時有 $f\left(x,1\right)=x^2-2x=\left(x-1\right)^2-1$，故當 $x=1$ 時有極小值為 $f\left(1,1\right)=-1$，而當 $x=-1$ 時有極大值為 $f\left(-1,1\right)=3$。
- (10 pts) Find the critical points for the function
$f\left(x,y\right)=3x^2+2xy+2x+y^2+y+4$
and determine whether they are local maxima, local minima, or saddle points. - (10 pts) Compute the length $\ell$ of the polar curve whose polar model is
$r\left(\theta\right)=2\cos\left(\theta\right),~0\leq\theta\leq\pi$.
- (15 pts) Calculate the integral $\displaystyle\int_{R^2}\exp\left(-3x^2+4xy-3y^2\right)dxdy$.
- (10 pts) Assume that $f$ is differentiable at $a$. Evaluate
$\displaystyle\lim_{x\to a}\frac{a^nf\left(x\right)-x^nf\left(a\right)}{x-a}$ where $n$ is a natural number.
- (10 pts) Determine whether or not the following limit exists, and find its value if it exists:
$\displaystyle\lim_{n\to+\infty}\left[n-\frac{n}e\left(1-\frac1n\right)^n\right]$.
- (20 pts) Let $C$ be the curve $x^2+y^2=1$ lying in the plane $z=1$. Let ${\bf F}=\left(z-y\right){\bf i}+y{\bf k}$.
- (5 pts) Calculate $\nabla\times{\bf F}$.
- (5 pts) Calculate $\displaystyle\int_C{\bf F}\cdot d{\bf s}$ using a parametrization of $C$ and a chosen orientation for $C$.
- (5 pts) Write $C=\partial S$ for a suitably chosen surface $S$ and, applying Stokes' theorem, verify your answer in (b).
- (5 pts) Consider the sphere with radius $\sqrt2$ and center the origin. Let $S'$ be the part of the sphere that is above the curve (i.e., lies in the region $z\geq1$), and has $C$ as boundary. Evaluate the surface integral of $\nabla\times{\bf F}$ over $S'$. Specify the orientation you are using for $S'$.
- 計算旋度可知
$\nabla\times{\bf F}=\begin{vmatrix}{\bf i}&{\bf j}&{\bf k}\\\partial_x&\partial_y&\partial_z\\z-y&0&y\end{vmatrix}={\bf i}+{\bf j}+{\bf k}$
- 將曲線參數化為 $\left\{\begin{aligned}&x=\cos\theta\\&y=\sin\theta\\&z=1\end{aligned}\right.$，如此所求的線積分可表達並計算如下
$\displaystyle\int_C{\bf F}\cdot d{\bf s}=\int_0^{2\pi}\left(1-\sin\theta,0,\sin\theta\right)\cdot\left(-\sin\theta,\cos\theta,0\right)d\theta=\int_0^{2\pi}\left(\sin^2\theta-\sin\theta\right)d\theta=\left.\left(\frac{2\theta-\sin2\theta}4+\cos\theta\right)\right|_0^{2\pi}=\pi$
- 取 $S=\left\{\left(x,y,z\right)\in\mathbb{R}^3:\,x^2+y^2\leq1,\,z=1\right\}$，那麼由 Stokes 定理可知
$\displaystyle\int_C{\bf F}\cdot d{\bf s}=\iint_S\left(\nabla\times{\bf F}\right)d{\bf S}=\iint_S\left(1,1,1\right)\cdot\left(0,0,1\right)dA=\iint_SdA=\pi$
- 同樣使用 Stokes 定理，其中在 $S'$ 上的曲面積分之法向量指向上方，從而有
$\displaystyle\iint_{S'}\nabla\times{\bf F}\cdot d{\bf S}=\int_C{\bf F}\cdot d{\bf s}=\pi$
訣竅
運用多變量函數的連鎖律計算即可。解法
計算一階偏導函數可知$\displaystyle\frac{\partial w}{\partial v}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial v}=\frac{\partial w}{\partial x}-\frac{\partial w}{\partial y}$
接著計算二階偏導函數則有$\displaystyle\frac{\partial^2w}{\partial u\partial v}=\frac{\partial}{\partial u}\frac{\partial w}{\partial x}-\frac{\partial}{\partial u}\frac{\partial w}{\partial y}=\frac{\partial^2w}{\partial x^2}+\frac{\partial^2w}{\partial y\partial x}-\frac{\partial^2w}{\partial x\partial y}-\frac{\partial^2w}{\partial y^2}=\frac{\partial^2w}{\partial x^2}-\frac{\partial^2w}{\partial y^2}$
因此 $a=1$ 且 $b=-1$。訣竅
在內部找出偏導為零的位置，而在邊界則化為單變函數求極值的位置處理。解法
為了求出內部的極值，我們解下列的聯立方程組$\left\{\begin{aligned}&f_x\left(x,y\right)=2xy-2y=0\\&f_y\left(x,y\right)=x^2-2x=0\end{aligned}\right.$
由第二式可知 $x=0$ 或 $x=2$，從而由第一式可知 $y=0$，但由於 $\left(2,0\right)\notin A$，故僅得極值候選點 $\left(0,0\right)$。現分別針對矩形區域的四條邊界考察如下訣竅
首先透過解偏導為零的位置以獲得極值候選點，隨後計算二階判別式以確定極值候選點的特性。解法
為了求出極值候選點，我們解下列的聯立方程組$\left\{\begin{aligned}&f_x\left(x,y\right)=6x+2y+2=0\\&f_y\left(x,y\right)=2x+2y+1=0\end{aligned}\right.$
如此可解得 $\displaystyle\left(x,y\right)=\left(-\frac14,-\frac14\right)$。進一步地，我們計算二階判別式如下$D\left(x,y\right)=\begin{vmatrix}f_{xx}\left(x,y\right)&f_{xy}\left(x,y\right)\\f_{yx}\left(x,y\right)&f_{yy}\left(x,y\right)\end{vmatrix}=\begin{vmatrix}6&2\\2&2\end{vmatrix}=8>0$
且因 $f_{xx}\left(x,y\right)=6>0$，故 $\displaystyle\left(-\frac14,-\frac14\right)$ 為局部極小值（事實上也是全域極小值）。訣竅
運用極座標下的曲線弧長公式計算即可。解法
運用曲線弧長公式可知$\displaystyle s=\int_0^{\pi}\sqrt{r^2\left(\theta\right)+r'^2\left(\theta\right)}d\theta=\int_0^{\pi}\sqrt{4\cos^2\theta+4\sin^2\theta}d\theta=\int_0^{\pi}2d\theta=2\pi$
訣竅
運用配方法與極座標變換處理之。解法
首先可以注意到指數之次方可配方如下$\displaystyle-3x^2+4xy-3y^2=-3\left(x-\frac{2y}3\right)^2-\frac{8y^2}3$
據此，令 $\displaystyle u=\sqrt3x-\frac{2y}{\sqrt3}$、$\displaystyle v=\frac{2\sqrt2y}{\sqrt3}$，又其 Jacobian 行列式能計算如下$\displaystyle\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}=\Big|\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}\Big|^{-1}=\Big|\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\[3mm]\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{vmatrix}\Big|^{-1}=\Big|\begin{vmatrix}\sqrt3&\displaystyle-\frac2{\sqrt3}\\[3mm]0&\displaystyle\frac{2\sqrt2}{\sqrt3}\end{vmatrix}\Big|^{-1}=\frac1{2\sqrt2}=\frac{\sqrt2}4$
因此所求的二重積分可改寫如下$\displaystyle\int_{R^2}\exp\left(-3x^2+4xy-3y^2\right)dxdy=\frac{\sqrt2}4\iint_{\mathbb{R}^2}\exp\left(-u^2-v^2\right)dudv=\frac{\sqrt2}4\int_0^{2\pi}\int_0^{\infty}e^{-r^2}rdrd\theta=\frac{\sqrt2}4\cdot2\pi\cdot\frac12=\frac{\pi\sqrt2}4$
訣竅
改寫極限式並化為對函數求導的定義即可。解法
容易注意到極限式可改寫為$\displaystyle\lim_{x\to a}\frac{a^nf\left(x\right)-x^nf\left(a\right)}{x-a}=a^n\lim_{x\to a}\frac{f\left(x\right)-f\left(a\right)}{x-a}+f\left(a\right)\lim_{x\to a}\frac{a^n-x^n}{x-a}=a^nf'\left(a\right)-na^{n-1}f\left(a\right)$
註：本題不可使用羅畢達法則，因為題目敘述僅提及函數 $f$ 在 $a$ 處可求導，但未能清楚知道 $f$ 在他處是否也能求導，或者求導後是否能取極限。
訣竅
運用自然指數的定義觀察即可知其極限發散。解法
首先注意到$\displaystyle\lim_{n\to\infty}\left(1-\frac1n\right)^n=\left[\lim_{n\to\infty}\left(1+\frac1{-n}\right)^{-n}\right]^{-1}=e^{-1}$
假若給定的極限收斂至 $L\in\mathbb{R}$，那麼由極限的四則運算定理可知$\displaystyle\frac{L}{1-e^{-2}}=\frac{\displaystyle\lim_{n\to\infty}\left[n-\frac{n}e\left(1-\frac1n\right)^n\right]}{\displaystyle\lim_{n\to\infty}\left(1-\frac1e\left(1-\frac1n\right)^n\right)}=\lim_{n\to\infty}n=\infty$
矛盾，故給定之極限不存在。
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