114年政治大學應數碩士班-微積分詳解

國立政治大學114學年度碩士班招生考試

考試科目:微積分
系所別:應用數學系

解答:

$$\text{Change the order of the integration}\\ \qquad I=\int_0^4 \int_{\sqrt x}^2 {x\over y^5+1}\,dy\,dx = \int_0^2 \int_0^{y^2} {x\over y^5+1}\,dx\,dy  = \int_0^2  {y^4\over 2(y^5+1)} \,dy \\ u=y^5+1 \Rightarrow du=5y^4\,dy \Rightarrow I=\int_1^{33} {1\over 10u}du =\left. \left[ {1\over 10}\ln u \right] \right|_1^{33} = \bbox[red, 2pt]{{1\over 10}\ln 33}$$

解答: $$\cases{x= r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_D y+ \sqrt{x^2 +y^2} \,dA =\int_{\pi/2}^\pi \int_0^3 \left( r\sin \theta+ r\right)r \,dr\,d\theta =\int_{\pi/2}^\pi (9\sin \theta+ 9)\,d\theta \\=\left. \left[ -9\cos\theta+9 \theta \right] \right|_{\pi/2}^\pi = \bbox[red, 2pt]{9+{9\over 2}\pi}$$

解答: $$\cases{u=\tan^{-1}x\\ dv={1\over 1+x^2}\,dx} \Rightarrow \cases{du={1\over 1+x^2}dx \\ v=\tan^{-1}x} \Rightarrow I_1 =\int{\tan^{-1}x\over 1+x^2}\,dx =(\tan^{-1}x)^2-I_1 \Rightarrow I_1={1\over 2} (\tan^{-1} x)^2 +C_1\\\cases{u= \tan^{-1}x \\ dv= (1-{1\over x^2+1})\,dx} \Rightarrow \cases{du={1\over 1+x^2} dx\\ v=x-\tan^{-1}x} \\\Rightarrow I=\int{ \tan^{-1}x \over 1+{1\over x^2}}\,dx = \int \left( 1-{1\over x^2+1}\right) \tan^{-1} \,dx =\tan^{-1}x (x-\tan^{-1}x) -\int {x-\tan^{-1}x\over 1+x^2}\,dx \\= x\tan^{-1}x-(\tan^{-1}x)^2- \int{x\over 1+x^2}\,dx + I_1 =x\tan^{-1} -(\tan^{-1}x)^2-{1\over 2}\ln(1+x^2) +{1\over 2}(\tan^{-1})^2+C_1 \\= \bbox[red, 2pt]{x\tan^{-1}x-{1\over 2}(\tan^{-1}x)^2 -{1\over 2}\ln(1+x^2) +C_1}$$

解答: $$\textbf{(a) }\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \lim_{(x,y) \to (0,0)} f(x,y)= \lim_{r\to 0} {2r^3\cos^2 \theta\sin \theta+ r^3 \sin^3\theta \over r^2} =\lim_{r\to 0}r(2\cos^2 \theta \sin \theta+ \sin^3\theta) \\\qquad = \bbox[red, 2pt]0 \\\textbf{(b) } \frac{\partial }{\partial x}\left({2x^2y+y^3 \over x^2+y^2} \right) ={4xy\over x^2+y^2} -{2x(2x^2y+y^3) \over (x^2+y^2)^2} ={2xy^3 \over (x^2+y^2)^2} \\\qquad \Rightarrow \bbox[red, 2pt] {\frac{\partial f}{\partial x}(x,y)=\begin{cases} \displaystyle {2xy^3 \over (x^2+y^2)^2}, & \text{if }(x,y)\ne (0,0) \\ 0, &\text{if }(x,y)=(0.0)\end{cases}} \\\qquad  \frac{\partial }{\partial y} \left({2x^2y+y^3 \over x^2+y^2} \right) ={2x^2+3y^2\over x^2+y^2} -{2y(2x^2y +y^3)\over (x^2+y^2)^2} ={2x^4+x^2y^2+y^4 \over (x^2+y^2)^2} \\\qquad \Rightarrow \bbox[red, 2pt] {\frac{\partial f}{\partial y}(x,y)=\begin{cases} \displaystyle {2x^4+ x^2y^2 +y^4 \over (x^2+y^2)^2}, & \text{if }(x,y)\ne (0,0) \\ 0, &\text{if }(x,y)=(0.0)\end{cases}} \\\textbf{(c) } \cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \lim_{(x,y)\to (0,0)} {2xy^3 \over (x^2+y^2)^2} =\lim_{r\to 0} {2r^4\cos \theta \sin^3\theta\over r^4} =2\cos \theta \sin^3\theta \text{ is dependent on }\theta \\ \qquad \Rightarrow {\partial f\over \partial x}(x,y) \text{ is not continuous at }(0,0) \Rightarrow f(x,y) \text{ is }\bbox[red, 2pt]{\text{ not differentiable at}(0,0)}$$

解答: $$S= \sum_{n=1}^\infty{(n!+1)^2 \over [(n+1)!]^2} =\sum_{n=1}^\infty{(n!)^2 \over [(n+1)!]^2} + \sum_{n=1}^\infty{2(n!) \over [(n+1)!]^2} + \sum_{n=1}^\infty{1 \over [(n+1)!]^2} \\ S_1= \sum_{n =1}^\infty{(n!)^2 \over [(n+1)!]^2}=\sum_{n=1}^\infty{(n!)^2 \over (n+1)^2(n!)^2} = \sum_{n=1}^\infty{1\over (n+1)^2} \lt \sum_{n=1}^\infty{1\over n^2} ={\pi\over 6} \Rightarrow S_1\text{ converges} \\ S_2= \sum_{n =1}^\infty{2(n!) \over [(n+1)!]^2} = \sum_{n=1}^\infty{2 \over (n+1)(n+1)! } \lt \sum_{n=1}^\infty{2 \over  (n+2)! } \Rightarrow \lim_{n\to \infty} \left({ 2\over (n+3)!}\cdot {(n+2)! \over 2} \right) =0\\\qquad \Rightarrow \sum_{n =1}^\infty{2 \over  (n+2)! } \text{ converges} \Rightarrow S_2 \text{ converges} \\ S_3= \sum_{n=1}^\infty{1 \over [(n+1)!]^2}  \Rightarrow \lim_{n\to \infty}\left({1\over [(n+2)!]^2} \cdot {[(n+1)!]^2 \over 1} \right) =0 \Rightarrow S_3 \text{ converges}\\ \Rightarrow S=S_1+S_2 +S_3 \Rightarrow S \text{ is convergent}. \quad \bbox[red, 2pt]{QED}$$

解答: $$\cases{f(x,y)=x^2y+5\\ g(x,y)=x^2+y^2-9} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ g=0} \Rightarrow \cases{2xy=\lambda(2x) \\ x^2= \lambda(2y) \\ x^2+y^2=9} \Rightarrow {2xy\over x^2} ={\lambda(2x) \over \lambda(2y)} \Rightarrow {2y\over x}={x\over y} \\ \Rightarrow x^2=2y^2 \Rightarrow x^2+y^2=2y^2+y^2=3y^2=9 \Rightarrow y=\sqrt 3 (-\sqrt 3\not \ge 0) \Rightarrow x=-\sqrt 6 \\ \Rightarrow \text{absolute maximum =}f(-\sqrt 6, \sqrt 3) = \bbox[red, 2pt]{6\sqrt 3+5}\\ (x,y) \in  R \Rightarrow f(x,y) = x^2+y+5 \ge 5 \Rightarrow \text{absolute minimum =}\bbox[red, 2pt]5$$

解答: $$e^x=1+x+{x^2\over 2}+{x^3\over 6}+{x^4\over 24}+ \cdots \Rightarrow e^{x^2} =1+x^2 +{x^4\over 2}+{x^6\over 6}+{x^8\over 24}+ \cdots \\ \Rightarrow xe^{x^2}  =x+x^3 +{x^5 \over 2}+{x^7\over 6}+{x^9\over 24}+ \cdots \\ {1\over 4+x^2} ={1\over 4}\cdot {1\over 1+{x^2 \over4}}={1\over 4}(1-{x^2\over 4} +{x^4\over 16} -{x^6\over 64} +{x^8\over 256}-\cdots)\\= {1\over 4}-{x^2\over 16} +{x^4\over 64} -{x^6\over 1024} +{x^8\over 4096}-\cdots \\ \Rightarrow xe^{x^2}-{1\over 4+x^2}= \bbox[red, 2pt]{-{1\over 4}+x+{x^2\over 16}+x^3}-{x^4\over 64}+{x^5\over 2}+\cdots$$

解答: $$\textbf{(a) }\cases{P(x,y)=ye^x \\Q(x,y)=e^x+3y^2} \Rightarrow P_y=e^x =Q_x \Rightarrow \bbox[red, 2pt]{\text{Yes, it is conservative}} \\\qquad \Phi(x,y)=\int P\,dx =\int Q\,dy \Rightarrow \Phi=\int ye^x\,dx =\int (e^x +3y^2)\,dy \\\qquad\Rightarrow  \Phi(x,y) =ye^x+\phi(y) =ye^x+y^3+\rho(x) \Rightarrow \text{potential function }\bbox[red, 2pt]{\Phi(x,y) =ye^x+y^3+C} \\\textbf{(b) }\int_C \vec F\cdot d\vec r =\Phi(x(t=3),y(t=3))-\Phi(x(t= 0), y(t=0)) =\Phi(e^6-1,e^{24}-1) -\Phi(0,0) \\\qquad = \bbox[red, 2pt]{(e^{24} -1)e^{e^{6}-1}+(e^{24}-1)^3 }$$

解答: $$x^2+y^2+z^2=5 \Rightarrow x^2=5-y^2-z^2=4y^2+4z^2 \Rightarrow y^2+z^2=1 \Rightarrow x^2=4 \\ \Rightarrow S_1\cap S_2 =\{(x,y,z) \mid x^2=4,y^2+z^2=1\} \\ \cases{\nabla (x^2+y^2+z^2-5) =(2x,2y,2z) \\ \nabla (x^2-4y^2-4z^2) =(2x,-8y,-8z)} \\\Rightarrow (2x,2y,2z) \cdot (2x,-8y,-8z) =4x^2-16y^2-16z^2 =0 ,\forall(x,y,z) \in S_1 \cap S_2 \\ \bbox[red, 2pt]{QED}$$

解答: $$\textbf{(a) } M_n= \sup\{a_n,a_{n+1},a_{n+2}, \dots\} \Rightarrow M_n \text{ is decreasing and bounded from below}\\ \Rightarrow  \text{By Monotone Convergence Theorem, its limit exists} \Rightarrow \lim_{n\to \infty} M_n \text{ exists}\\ \qquad \text{Similarly, }m_n \text{ is increasing and bounded from above. } \\\Rightarrow  \text{By Monotone Convergence Theorem, its limit exists} \Rightarrow \lim_{n\to \infty} m_n \text{ exists}\\ \qquad \bbox[red, 2pt]{\text{QED}} \\\textbf{(b) } \text{Suppose }\lim_{n\to \infty} M_n= \lim_{n\to \infty} m_n = L , \text{ and let }\epsilon \gt 0 \\\qquad \text{Since } \lim_{N\to \infty} M_N= L \Rightarrow M_N \lt L+ \epsilon, \text{ for some }N\in \mathbb N \Rightarrow a_n\lt L+\epsilon \text{ for }n \ge N \\ \qquad \text{Since }\lim_{K\to \infty} m_K= L \Rightarrow  L- \epsilon\lt m_K, \text{ for some }K\in \mathbb N \Rightarrow L-\epsilon\lt a_n  \text{ for }n \ge K \\ P=\max\{N,K\} \Rightarrow L-\epsilon \lt a_n \lt L+\epsilon \text{ for }n\ge P \Rightarrow \lim_{n\to \infty} a_n =L \Rightarrow \left.\{a_n \right\}_{n=1}^\infty \text{ converges} \\ \bbox[red, 2pt] {\text{QED}}$$

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解題僅供參考，碩士班歷年試題及詳解