- (10 pts) Find the equation of the tangent line drawn to the graph of $-3x^2-16xy-2y^2+3y=178$ at the point $\left(-3,5\right)$.
- (10 pts) For what real values of $x$ does the following series converge?
$\displaystyle\sum_{n=0}^{\infty}\frac{e^n\left(x-\pi\right)^{2n}}{\sqrt{n^2+1}}$
Justify your claim. - (15 pts) Compute, with justification, the value of the following limit in terms of $\alpha$.
$\displaystyle\lim_{n\to\infty}n^\alpha\int_0^{\infty}\exp\left(-nx^2\right)dx$.
(Hint: It might be useful to use the fact that $\displaystyle\int_0^{\infty}t^b\exp\left(-at\right)dt=\Gamma\left(b+1\right)/a^{b+1}$.) - (15 pts) Determine the set of real numbers $x$ for which
$\displaystyle\sum_{n=1}^{\infty}\left(\frac1n\csc\frac1n-1\right)^x$
converges. - (10 pts) Find, with explanation, the maximum value of $f\left(x\right)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36\leq13x^2$.
- (10 pts) Evaluate the integral
$\displaystyle\iint_{\Omega}\left(2x-y\right)dA$
where $\Omega$ is bounded by the line $y=x$ and the parabola $x=2-y^2$. - (10 pts) Let $R$ be the parallelogram with vertices $\left(0,0\right)$, $\left(1,1\right)$, $\left(2,-1\right)$ and $\left(3,0\right)$. Evaluate the integral $\displaystyle\iint_R\left(x+2y\right)^2\exp\left(x-y\right)dA$.
- (10 pts) Captain Kurt of the starship Interprize has steered his starship toward a wormhole in space. The Interprize's current location is $\left(0,2,2\right)$. The wormhole is quite oddly shaped, and its surface is given by the equation $-x^2+2y^2+2z^2=1$. Captain Kurt is really eager to get to the wormhole as soon as possible. Find a point on the wormhole's surface that is nearest to the Interprize so that Kurt knows where to send the ship. Be sure to write up your solution to this problem carefully, explaining all the steps involved in getting to your answer.
- 若 $\lambda=1$，那麼第二式與第三式則得 $\displaystyle y=z=\frac23$，那麼由第四式知 $\displaystyle x=\pm\frac{\sqrt7}3$。
- 若 $x=0$，那麼第四式寫為 $2y^2+2z^2=1$。再者，第二式與第三式相減可知 $2\left(y-z\right)\left(1+2\lambda\right)=0$。若 $y=z$，那麼解得 $\displaystyle y=z=\pm\frac12$；若 $\displaystyle\lambda=-\frac12$，那麼容易發現第二式或第三式產生矛盾。
- (10 pts) Let ${\bf S}$ be the portion of the unit sphere centered at the origin that is cut out by the cone $z\geq\sqrt{x^2+y^2}$. Evaluate $\displaystyle\iint_{\bf S}{\bf F}\cdot d{\bf S}$ where
${\bf F}\left(x,y,z\right)=\left(xy+\cos z,-yx+x^2+z^3,2z^2+x\right)$.
訣竅
運用隱函數微分求得切線斜率，隨後按點斜式寫出切線方程式。解法
使用隱函數微分有$\displaystyle-6x-16\left(y+x\frac{dy}{dx}\right)-4y\frac{dy}{dx}+3\frac{dy}{dx}=0$
如此取 $\left(x,y\right)=\left(-3,5\right)$ 可得$\displaystyle18-16\left(5-3\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(-3,5\right)}\right)-\left.20\frac{dy}{dx}\right|_{\left(x,y\right)=\left(-3,5\right)}+3\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(-3,5\right)}=0$
因此 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(-3,5\right)}=2$，從而切線方程式為$y-5=2\left(x+3\right)$
或寫為 $y=2x+11$。訣竅
先運用比值審歛法求得收斂半徑，隨後確認端點是否收斂。解法
運用比值審歛法可知
$\displaystyle e\left|x-\pi\right|^2=\lim_{n\to\infty}\left|\frac{e^{n+1}\left(x-\pi\right)^{2n+2}/\sqrt{\left(n+1\right)^2+1}}{e^n\left(x-\pi\right)^{2n}/\sqrt{n^2+1}}\right|<1$
故得 $\displaystyle\pi-e^{-1/2}<x<\pi+e^{-1/2}$。現檢查端點可知當 $x=\pi\pm e^{-1/2}$ 時，級數可寫為 $\displaystyle\sum_{n=0}^{\infty}\frac1{\sqrt{n^2+1}}$。由於 $\displaystyle\lim_{n\to\infty}\frac{1/n}{1/\sqrt{n^2+1}}=1$，故兩級數 $\displaystyle\sum_{n=1}^{\infty}\frac1n$ 與 $\displaystyle\sum_{n=1}^{\infty}\frac1{\sqrt{n^2+1}}$ 有相同的歛散性，而前者為調和級數，故發散。如此我們知道給定的冪級數恰僅在 $\left(\pi-e^{-1/2},\pi+e^{-1/2}\right)$ 上收斂。
訣竅
先求出該瑕積分之值，隨後再求其極限；亦可利用題目所給之提示使用變數代換處理。解法一
令 $s=\sqrt{n}x$，那麼瑕積分可以改寫並計算如下$\displaystyle\int_0^{\infty}\exp\left(-nx^2\right)dx=\frac1{\sqrt{n}}\int_0^{\infty}\exp\left(-s^2\right)ds=\frac{\sqrt\pi}{2\sqrt{n}}$
故所求之極限為 $\displaystyle\lim_{n\to\infty}n^\alpha\cdot\frac{\sqrt\pi}{2\sqrt{n}}=\frac{\sqrt\pi}2\lim_{n\to\infty}n^{\alpha-1/2}$，如此有$\displaystyle\lim_{n\to\infty}n^\alpha\int_0^{\infty}\exp\left(-nx^2\right)dx=\begin{cases}\infty,&\mbox{if}~\alpha>1/2,\\\sqrt\pi/2,&\mbox{if}~\alpha=1/2,\\0,&\mbox{if}~\alpha<1/2.\end{cases}$
解法二
為了使用提示，我們考慮變數代換 $t=x^2$，那麼有 $x=\sqrt{t}$，求導則 $\displaystyle dx=\frac{t^{-1/2}}2dt$。如此該瑕積分可轉為$\displaystyle\int_0^{\infty}\exp\left(-nx^2\right)dx=\frac12\int_0^{\infty}t^{-1/2}\exp\left(-nt\right)dt=\frac{\Gamma\left(1/2\right)}{2n^{1/2}}$
因此所求的極限為 $\displaystyle\lim_{n\\to\infty}n^{\alpha}\cdot\frac{\Gamma\left(1/2\right)}{2n^{1/2}}=\frac{\Gamma\left(1/2\right)}2\lim_{n\to\infty}n^{\alpha-1/2}$。據此得到$\displaystyle\lim_{n\to\infty}n^\alpha\int_0^{\infty}\exp\left(-nx^2\right)dx=\begin{cases}\infty,&\mbox{if}~\alpha>1/2,\\\Gamma\left(1/2\right)/2,&\mbox{if}~\alpha=1/2,\\0,&\mbox{if}~\alpha<1/2.\end{cases}$
【註】 事實上 $\Gamma\left(1/2\right)=\sqrt\pi$。訣竅
針對各種可能的 $x$ 進行討論分析即可。解法
首先可以注意到函數 $\displaystyle\frac{\sin x}x$ 在 $\left(0,\infty\right)$ 上隨 $x$ 遞減且 $\displaystyle\lim_{x\to0^+}\frac{\sin x}x=1$，故對於正整數 $n$ 有 $\displaystyle\frac{\sin n}n<1$，即 $\displaystyle\frac1n\csc\frac1n-1>0$。再者，因 $\displaystyle\lim_{x\to\infty}\frac{\sin x}x=0$，故 $\displaystyle\lim_{n\to\infty}\frac1n\csc\frac1n-1=0$。
由於當 $x\leq0$ 時，$\displaystyle\lim_{n\to\infty}\left(\frac1n\csc\frac1n-1\right)^x$ 發散或不為零，故給定的級數發散。
現在觀察底數發現
$\displaystyle\frac1{n\sin\frac1n}-1=\frac1{n\cdot\left(\frac1n-\frac1{6n^3}+\cdots\right)}-1=\frac1{1-\frac1{6n^2}+\cdots}-1=\left(1+\frac1{6n^2}+\cdots\right)-1=\frac1{6n^2}+\cdots$
因此當 $x\leq\frac12$ 時級數發散，而當 $x>\frac12$ 時級數收斂。訣竅
首先分析定義域，隨後在這些區域與邊界上尋求極值的候選點並計算即可。解法
由於 $x$ 滿足 $x^4+36\leq13x^2$，即 $\left(x^2-4\right)\left(x^2-9\right)\leq0$，故有 $4\leq x^2\leq9$，即 $x\in\left[-3,-2\right]\cup\left[2,3\right]$。而為了找出極值候選點，我們解方程式 $f'\left(x\right)=0$，即 $3x^2-3=0$，可得 $x=\pm1$，但這兩點都不落在討論範圍中，因此我們只需要考慮所考慮範圍中的端點即可。容易驗算如下$f\left(\pm2\right)=\pm2$, $f\left(\pm3\right)=\pm18$
故最大值為 $18$。訣竅
將邊界條件明確列出後可以將重積分寫為迭代積分並計算之。解法
首先可以得到 $y=x$ 與 $x=2-y^2$ 的交點為 $\left(1,1\right)$ 與 $\left(-2,-2\right)$，如此 $\Omega$ 可描述為$\left\{\left(x,y\right)\in\mathbb{R}^2:y\leq x\leq2-y^2,\,-2\leq y\leq1\right\}$
因此所求的重積分可以表達並計算如下$\displaystyle\begin{aligned}\iint_{\Omega}\left(2x-y\right)dA&=\int_{-2}^1\int_y^{2-y^2}\left(2x-y\right)dxdy=\int_{-2}^1\left(x^2-xy\right)\Big|_{x=y}^{x=2-y^2}dy\\&=\int_{-2}^1\left[\left(2-y^2\right)^2-\left(2-y^2\right)y\right]dy=\int_{-2}^1\left(y^4+y^3-4y^2-2y+4\right)dy\\&=\left.\left(\frac{y^5}5+\frac{y^4}4-\frac{4y^3}3-y^2+4y\right)\right|_{-2}^1=\frac{117}{20}\end{aligned}$
訣竅
考慮到被積分函數與積分區域的形狀，考慮變數變換來處理之。解法
令 $u=x+2y$ 與 $v=x-y$，那麼由頂點可知邊界可寫為 $u=0$、$u=3$、$v=0$ 與 $v=3$。而這樣的變數變換所對應的 Jacobian 行列式為$\displaystyle\Big|\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|\Big|=\Big|\left|\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}\right|\Big|^{-1}=\Big|\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial v}{\partial x}\\\displaystyle\frac{\partial u}{\partial y}&\displaystyle\frac{\partial v}{\partial y}\end{vmatrix}\Big|^{-1}=\Big|\begin{vmatrix}1&1\\2&-1\end{vmatrix}\Big|^{-1}=\frac13$
因此所求的重積分可改寫並計算如下$\displaystyle\int_0^3\int_0^3u^2e^v\cdot\frac13dvdu=\frac13\left(\int_0^3u^2du\right)\left(\int_0^3e^vdv\right)=\frac13\cdot\frac{3^3}3\cdot\frac{e^3-1}3=e^3-1$
訣竅
按照題意列式並使用初等不等式即可；亦可運用拉格朗日乘子法求條件極值。解法一
按照題意，我們要找 $\left(x,y,z\right)$ 滿足 $-x^2+2y^2+2z^2=1$ 且使 $f\left(x,y,z\right)=x^2+\left(y-2\right)^2+\left(z-2\right)^2$ 盡可能小，如此這樣的 $\left(x,y,z\right)$ 便是在曲面上離 $\left(0,2,2\right)$ 最近的點。首先我們可以發現利用這樣的條件便有$\displaystyle\begin{aligned}f\left(x,y,z\right)&=x^2+\left(y-2\right)^2+\left(z-2\right)^2=\left(2y^2+2z^2-1\right)+\left(y^2-4y+4\right)+\left(z^2-4z+4\right)\\&=3y^2-4y+3z^2-4z+7=3\left(y-\frac23\right)^2+3\left(z-\frac23\right)^2+\frac{13}3\geq\frac{13}3\end{aligned}$
此時等號成立條件為 $\displaystyle y=z=\frac23$，而 $\displaystyle x=\pm\frac{\sqrt7}3$。故曲面上與給定座標最近的點為 $\displaystyle\left(\pm\frac{\sqrt7}3,\frac23,\frac23\right)$。解法二
承解法一的假定，考慮拉格朗日乘子函數如下$F\left(x,y,z,\lambda\right)=x^2+\left(y-2\right)^2+\left(z-2\right)^2+\lambda\left(-x^2+2y^2+2z^2-1\right)$
據此解聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=2x-2\lambda x=0\\&F_y\left(x,y,z,\lambda\right)=2\left(y-2\right)+4\lambda y=0\\&F_z\left(x,y,z,\lambda\right)=2\left(z-2\right)+4\lambda z=0\\&F_{\lambda}\left(x,y,z,\lambda\right)=-x^2+2y^2+2z^2-1=0\end{aligned}\right.$
第一式給出 $2x\left(1-\lambda\right)=0$。訣竅
添加曲面化為封閉曲面後使用散度定理計算之。解法
考慮 $\displaystyle {\bf S}'=\left\{\left(x,y,z\right)\in\mathbb{R}^3:\,x^2+y^2\leq\frac12,\,z=\frac1{\sqrt2}\right\}$，且設 $D$ 為 ${\bf S}\cup{\bf S}'$ 所圍成之區域，那麼運用散度定理可知$\displaystyle\iint_{{\bf S}\cup{\bf S}'}{\bf F}\cdot d{\bf S}=\iiint_D\mbox{div}\,{\bf F}\,dV=\iiint_D\left(y-x+4z\right)dV$
為了計算該三重積分，令 $\left\{\begin{aligned}&x=r\cos\theta\\&y=r\sin\theta\\&z=z\end{aligned}\right.$，那麼容易注意到區域 $D$ 之變數範圍為 $\left\{\begin{aligned}&0\leq r\leq\frac1{\sqrt2}\\&0\leq\theta\leq2\pi\\&\frac1{\sqrt2}\leq z\leq\sqrt{1-r^2}\end{aligned}\right.$，從而所求的三重積分可計算如下$\displaystyle\begin{aligned}\iiint_D\left(y-x+4z\right)dV&=\int_0^{2\pi}\int_0^{\frac1{\sqrt2}}\int_{\frac1{\sqrt2}}^{\sqrt{1-r^2}}\left(r\sin\theta-r\cos\theta+4z\right)rdzdrd\theta=\\&=\int_0^{\frac1{\sqrt2}}\int_{\frac1{\sqrt2}}^{\sqrt{1-r^2}}4rzdzdr=\int_0^{\frac1{\sqrt2}}2rz^2\Big|_{\frac1{\sqrt2}}^{\sqrt{1-r^2}}dr\\&=\int_0^{\frac1{\sqrt2}}\left(r-2r^3\right)dr=\left.\frac{r^2-r^4}2\right|_0^{\frac1{\sqrt2}}=\frac18\end{aligned}$
因此所求的曲面積分可寫為$\displaystyle\iint_{\bf S}{\bf F}\cdot d{\bf S}=\frac18-\iint_{{\bf S}'}{\bf F}\cdot d{\bf S}$
而後者之曲面積分容易計算如下$\displaystyle\begin{aligned}\iint_{{\bf S}'}{\bf F}\cdot d{\bf S}&=\int_0^{2\pi}\int_0^{\frac1{\sqrt2}}\left(r^2\cos\theta\sin\theta+\cos\frac1{\sqrt2},-r^2\cos\theta\sin\theta+r^2\cos^2\theta+\frac18,\frac12+r\cos\theta\right)\cdot\left(0,0,-1\right)rdrd\theta\\&=-\int_0^{2\pi}\int_0^{\frac1{\sqrt2}}\left(\frac{r}2+r^2\cos\theta\right)drd\theta=-\frac12\int_0^{\frac1{\sqrt2}}rdr=-\frac18\end{aligned}$
故所求之值為 $\displaystyle\iint_{\bf S}{\bf F}\cdot d{\bf S}=\frac14$。
沒有留言:
張貼留言