- Calculate (a) $\displaystyle\frac{d}{dx}\left(x^2\right)^x$ (b) $\displaystyle\int x\sqrt{x+1}dx$. (10%)
- 運用換底公式改寫後使用連鎖律計算如下
$\displaystyle\frac{d}{dx}\left(x^2\right)^x=\frac{d}{dx}e^{2x\ln x}=e^{2x\ln x}\cdot\left(2\ln x+2\right)=2x^{2x}\left(1+\ln x\right)$
- 令 $u=\sqrt{x+1}$,那麼有 $x=u^2-1$,求導得 $dx=2udu$。如此所求的不定積分可改寫並計算如下
$\displaystyle\begin{aligned}\int x\sqrt{x+1}dx=\int\left(u^2-1\right)\cdot u\cdot2udu&=2\int\left(u^4-u^2\right)du\\&=2\left(\frac{u^5}5-\frac{u^3}3\right)+C=\frac{2\left(x+1\right)^{5/2}}5-\frac{2\left(x+1\right)^{3/2}}3+C\end{aligned}$
- Sketch the graph of $f\left(x\right)=3x^{5/3}/5-3x^{2/3}$, and indicate the extrema, inflection points, concavity, cusp (if any), and asymptotes (if any). (20%)
- Calculate the arc length of the cardioid $r=\left(1-\cos\theta\right)$, locate the centroid of the arc, and determine the area of the surface generated by revolving the curve about the $x$-axis. (20%)
- 當 $\theta=0$ 時有 $u=1$;
- 當 $\theta=\pi$ 時有 $u=-1$;
- 求導可知 $du=-\sin\theta d\theta$。
- Determine the series $\displaystyle\sum_{k=2}^{\infty}a_k$ converges or diverges. $\displaystyle a_k=\sum_{n=2}^{\infty}\frac1{k^n}$. Find the sum, if it converges. (10%)
- Find the Taylor series expansion of $e^{-2x}$ and give the radius of converges. (10%)
- Find the absolute extreme values of the function $f\left(x,y\right)=4xy-x^2-y^2-6x$ on the triangular region $D=\left\{\left(x,y\right):\,0\leq x\leq2,\,0\leq y\leq3x\right\}$. (10%)
- 當 $y=0$ 時有 $f\left(x,0\right)=-x^2-6x=9-\left(x+3\right)^2$,又 $x\in\left[0,2\right]$,故在 $x=0$ 處有最大值為 $f\left(0,0\right)=0$,而在 $x=2$ 有最小值為 $f\left(2,0\right)=-16$。
- 當 $x=2$ 時有 $f\left(2,y\right)=-\left(y-4\right)^2$,因此當 $y=0$ 時有最小值 $f\left(2,0\right)=-16$,而在 $y=4$ 時有最大值為 $f\left(2,4\right)=0$。
- 當 $y=3x$ 時有 $\displaystyle f\left(x,3x\right)=2x^2-6x=2\left(x-\frac32\right)^2-\frac92$,故當 $\displaystyle x=\frac32$ 時有最小值為 $\displaystyle f\left(\frac32,\frac92\right)=-\frac92$,而當 $x=0$ 時有大值為 $f\left(0,0\right)=0$。
- Evaluate the double integral $\displaystyle\int_0^1\int_{x^2}^1\frac{x^3}{\sqrt{x^4+y^2}}dydx$. (10%)
- Let $T$ be a solid with a piecewise-smooth boundary. Show that if $f$ and $g$ have continuous second partials, then the flux of $\nabla f\times\nabla g$ out of $T$ is zero. (10%)
訣竅
使用換底公式與連鎖律可處理第一小題;運用變數代換後計算不定積分。解法
訣竅
透過計算極限、一階與二階導函數等尋求極值、反曲點、遞增遞減區間、凹凸區間以及漸近線等繪圖資訊。解法
首先計算極限可知 $\displaystyle\lim_{x\to\pm\infty}f\left(x\right)=\pm\infty$ 且 $\displaystyle\lim_{x\to\pm\infty}\frac{f\left(x\right)}x=\pm\infty$,故無水平與斜漸近線。又 $f$ 為連續函數,故對任何 $a\in\mathbb{R}$ 恆有 $\displaystyle\lim_{x\to a}f\left(x\right)=f\left(a\right)\in\mathbb{R}$,因此也無鉛直漸近線。
又計算一階導函數有 $\displaystyle f'\left(x\right)=x^{2/3}-2x^{-1/3}=x^{-1/3}\left(x-2\right)$。為了找出極值可發現 $x\in\left(-\infty,0\right)\cup\left(2,\infty\right)$ 時有 $f'\left(x\right)>0$,而 $x\in\left(0,2\right)$ 時有 $f'\left(x\right)<0$,因此在 $x=2$ 處有極小值,而在 $x=0$ 處有極大值,且因函數 $f$ 在 $x=0$ 處不可導且有 $\displaystyle\lim_{x\to0^\pm}f'\left(x\right)=\mp\infty$,故在 $\left(0,0\right)$ 處也為尖點(cusp)。
再計算二階導函數 $\displaystyle f''\left(x\right)=\frac23x^{-4/3}\left(x+1\right)$。那麼當 $x\in\left(-1,0\right)\cup\left(0,\infty\right)$ 時有 $f''\left(x\right)>0$ 且 $x\in\left(-\infty,-1\right)$ 時有 $f''\left(x\right)<0$,因此 $x=-1$ 處有反曲點 $\left(-1,-18/5\right)$。
將以上資訊繪圖如下訣竅
運用曲線弧長公式、形心公式(可搭配對稱性)與旋轉體體積公式求解。解法
使用曲線弧長公式可知$\displaystyle s=\int_0^{2\pi}\sqrt{r^2\left(\theta\right)+r'^2\left(\theta\right)}d\theta=\int_0^{2\pi}\sqrt{\left(1-\cos\theta\right)^2+\sin^2\theta}d\theta=2\int_0^{2\pi}\sqrt{\frac{1+\cos\theta}2}d\theta=4\int_0^{\pi}\cos\frac\theta2d\theta=\left.8\sin\frac\theta2\right|_0^{\pi}=8$
使用形心的計算公式與對稱性可知 $\bar{y}=0$ 且$\displaystyle\begin{aligned}\bar{x}&=\frac{\displaystyle\int_0^{\pi}r\left(\theta\right)\cos\theta\sqrt{2+2\cos\theta}d\theta}4=\frac12\int_0^{\pi}\left(\cos\theta-\cos^2\theta\right)\cos\frac\theta2d\theta\\&=\frac14\int_0^{\pi}\left(2\cos\theta-1-\cos2\theta\right)\cos\frac{\theta}2d\theta\\&=\left.\frac18\int_0^{\pi}\left(\cos\frac{3\theta}2-\cos\frac{5\theta}2\right)d\theta=\frac18\left(\frac23\sin\frac{3\theta}2-\frac25\sin\frac{5\theta}2\right)\right|_0^{\pi}=\frac18(-\frac23-\frac25)=-\frac2{15}\end{aligned}$
運用旋轉體體積公式可知$\displaystyle V=\int_{\theta=\pi}^{\theta=0}\pi y^2\left(\theta\right)d x\left(\theta\right)=\pi\int_0^{\pi}r^2\left(\theta\right)\sin^2\theta\cdot\left(\sin\theta-2\cos\theta\sin\theta\right)d\theta=\pi\int_0^\pi\sin^3\theta\left(1-\cos\theta\right)^2\left(1-2\cos\theta\right)d\theta$
那麼令 $u=\cos\theta$,則$\displaystyle\begin{aligned}V&=\pi\int_{-1}^1\left(1-u^2\right)\left(1-u\right)^2\left(1-2u\right)du=\pi\int_{-1}^1\left(1-4u+4u^2+2u^3-5u^4+2u^5\right)du\\&=2\pi\int_0^1\left(1+4u^2-5u^4\right)du=\left.\pi\left(2u+\frac{8u^3}3-2u^5\right)\right|_0^1=\frac{8\pi}3\end{aligned}$
訣竅
先釐清其一般項之形式,隨後計算其部分和並取極限即可。解法
注意到 $a_k$ 為無窮等比級數,故由無窮等比級數和公式可知$\displaystyle a_k=\frac{\displaystyle\frac1{k^2}}{\displaystyle1-\frac1k}=\frac1{k^2-k}=\frac1{k-1}-\frac1k$
如此所求之級數為$\displaystyle\sum_{k=2}^{\infty}a_k=\lim_{n\to\infty}\sum_{k=2}^n\left(\frac1{k-1}-\frac1k\right)=\lim_{n\to\infty}\left(\frac11-\frac1n\right)=1$
訣竅
按照泰勒級數的計算規則處理,隨後由比值審歛法求其收斂半徑。解法
求導可知 $f^{\left(n\right)}\left(x\right)=\left(-1\right)^n2^ne^{-2x}$,如此可知所求的泰勒級數為$\displaystyle\sum_{n=0}^{\infty}\frac{f^{\left(n\right)}\left(0\right)}{n!}x^n=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n2^n}{n!}x^n$
又由比值審歛法的概念可知$\displaystyle R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\left|\frac{\left(-1\right)^n2^n}{n!}\div\frac{\left(-1\right)^{n+1}2^{n+1}}{\left(n+1\right)!}\right|=\infty$
故收斂半徑為 $\infty$。訣竅
在內部由一階偏導函數求極值,而在邊界則逐段化為單變數函數求極值。解法
首先先解聯立方程組$\left\{\begin{aligned}&f_x\left(x,y\right)=4y-2x-6=0\\&f_y\left(x,y\right)=4x-2y=0\end{aligned}\right.$
如此解得 $\left(x,y\right)=\left(1,2\right)\in D$,此時 $f\left(1,2\right)=-3$。又對於邊界上的狀況可分為三段討論如下:訣竅
改變積分次序後直接計算求解即可。解法
原積分範圍 $\left\{\begin{aligned}&0\leq x\leq1\\&x^2\leq y\leq1\end{aligned}\right.$ 可改寫為 $\left\{\begin{aligned}&0\leq x\leq\sqrt{y}\\&0\leq y\leq1\end{aligned}\right.$,如此所求的重積分可改寫並計算如下$\displaystyle\int_0^1\int_0^{\sqrt{y}}\frac{x^3}{\sqrt{x^4+y^2}}dxdy=\int_0^1\left.\frac{\sqrt{x^4+y^2}}2\right|_0^{\sqrt{y}}dy=\frac{\sqrt2-1}2\int_0^1ydy=\frac{\sqrt2-1}4$
訣竅
利用高斯散度定理計算即可。解法
先計算向量函數 $\nabla f\times\nabla g$ 如下$\nabla f\times\nabla g=\left(f_x,f_y,f_z\right)\times\left(g_x,g_y,g_z\right)=\left(f_yg_z-f_zg_y,f_zg_x-f_xg_z,f_xg_y-f_yg_x\right)$
故其散度為$\displaystyle\begin{aligned}\text{div}\left(\nabla f\times\nabla g\right)&=\frac{\partial}{\partial x}\left(f_yg_z-f_zg_y\right)+\frac{\partial}{\partial y}\left(f_zg_x-f_xg_z\right)+\frac{\partial}{\partial z}\left(f_xg_y-f_yg_x\right)\\&=f_{xy}g_z+f_yg_{xz}-f_{xz}g_y-f_zg_{xy}+f_{yz}g_x+f_zg_{xy}-f_{xy}g_z-f_xg_{yz}+f_{xz}g_y+f_xg_{yz}-f_{yz}g_x-f_yg_{xz}\\&=0\end{aligned}$
因此運用散度定理可知 $\nabla f\times\nabla g$ 流出 $T$ 的通量為 $0$。
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