114年台聯大轉學考-微積分A3A4A6詳解

 台灣聯合大學系統114學年度學士班轉學生考試

科目:微積分 類組別:A3/A4/A6

甲、填充題：共10題，每題8分，共80分

解答:$$\text{Taylor series:} \cases{\sqrt{1+x} =1+{1\over 2}x-{1\over 8}x^2+\cdots\\ \sin x=1-{x^3\over 6}+{x^5\over 120}-\cdots\\ \tan x=x+{x^3\over 3}+{2x^5 \over 15} +\cdots} \Rightarrow \cases{\sqrt{1+ \tan(3x)}=1+{3\over 2}x+ {9\over 2}x^3+ \cdots\\ \sqrt{1+\sin(3x)} =1+{3\over 2}x-{9\over 4}x^3+ \cdots} \\ \Rightarrow \lim_{x\to 0}{\sqrt{1+\tan(3x)}-\sqrt{1+\sin(3x)} \over x^3} =\lim_{x\to 0} {{27\over 4}x^3+ \cdots\over x^3} = \bbox[red, 2pt]{27\over 4}$$

解答:$$g(x)=\int_0^{\cos(2x)} \left[1+\sin(t^2)\right]\,dt \Rightarrow g'(x)= -2\sin(2x)(1+\sin(\cos^2(2x))) \\ \Rightarrow g'(\pi/4)=-2(1+0) =-2 \text{ and }g(\pi/4)=0\\ f(x)=\int_0^{g(x)} {1\over \sqrt{1+t^3}}\,dt \Rightarrow f'(x)= {1\over \sqrt{1+g^3(x)}}\cdot g'(x) \Rightarrow f'(\pi/4) ={1\over \sqrt{1+g^3(\pi/4)}} \cdot g'(\pi/4) \\=g'(\pi/4)= \bbox[red, 2pt]{-2}$$

解答:$$V= \int y^2\pi\,dx = \int_0^2 {16\pi\over (x^2+4)^2}\,dx = \pi \left. \left[ {2x\over x^2+4}+ \tan^{-1}{x\over 2}\right] \right|_0^2 = \bbox[red, 2pt]{{\pi\over 2}+{\pi^2\over 4}}$$

解答:$$f(x,y)=e^{-x^2-y^2}(x^2+2y^2) \Rightarrow \cases{f_x= (2x-2x(x^2+ 2y^2)) e^{-x^2-y^2} \\ f_y =(4y-2y(x^2+2y^2)) e^{-x^2-y^2}} \\ \cases{f_x=0 \Rightarrow 2x(x^2+2y^2-1)=0 \Rightarrow  x=0, x^2+2y^2=1 \\f_y=0 \Rightarrow 2y(x^2+2y^2-2)=0 \Rightarrow y=0,x^2+2y^2=2} \Rightarrow (x,y)=(0,0),(0,\pm 1),(\pm 1,0) \\ \Rightarrow \cases{f(0,0)=0\\ f(0,1)=f(0,-1) =2e^{-1} \\f(1,0)= f(-1,0) =e^{-1}} \Rightarrow \text{local maximum: }2e^{-1} \\\text{Consider the boundary of }D: x^2+y^2=4 \Rightarrow f(x,y)=e^{-4}(4+y^2) \Rightarrow max:8e^{-4} \lt 2e^{-1} \\ \Rightarrow \text{ absolute max: }\bbox[red, 2pt]{2e^{-1}}$$

解答:$$\cases{s=1\\ t=2} \Rightarrow \cases{x=g(s=1,t=2) =3\\ y=h(s=1,t=2) =6}\\z=f(x,y) \Rightarrow {\partial z\over \partial s}(s=1,t=2) =f_x(x =3,y=6)g_s( s=1,t=2) + f_y(x=3,y=6)h_s(s=1,t=2) \\\qquad =7\cdot (-1)+8\cdot (-5) = \bbox[red, 2pt]{-47}$$

解答:$$f(x,y)={x\over x^2+y^2} \Rightarrow \nabla f=(f_x,f_y) = \left \langle {y^2-x^2\over (x^2+y^2)^2}, -{2xy \over (x^2+y^2)}\right \rangle  \Rightarrow \nabla f(1,2)= \left \langle {3\over 25}, -{4 \over 25}\right \rangle\\ \mathbf u={\mathbf v\over |\mathbf v|} = \left\langle {3\over 5}, {4\over 5} \right \rangle \Rightarrow D_{\mathbf u} f(1,2) =\nabla f(1,2) \cdot \mathbf u =\left \langle{3\over 25}, -{4 \over 25}\right \rangle \cdot \left\langle {3\over 5}, {4\over 5} \right \rangle = \bbox[red, 2pt]{-{7\over 125}}$$

解答:$$\cases{u=x+y\\ v=y/x} \Rightarrow \cases{x= u/(v+1)\\ y=uv/ (v+1)} \Rightarrow {\partial(x,y) \over \partial (u,v)} =\begin{vmatrix}\displaystyle {1\over v+1}& -\displaystyle {u\over (v+1)^2}\\ \displaystyle {v\over v+1}& \displaystyle {u\over (v+1)^2} \end{vmatrix} ={u\over (v+1)^2} \\ \Rightarrow \iint_R \left( 1+{y\over x}\right)\,dA= \int_{1/2}^2 \int_1^3 (1+v)\cdot {u \over (v+1)^2}\, dudv = \int_{1/2}^2 \int_1^3 {u\over v+1}\, dudv \\= \int_{1/2}^2   {4\over v+1 }\,  dv =4(\ln 3-\ln {3\over 2}) = \bbox[red, 2pt]{4\ln 2}$$

解答:$$I={1\over 2}\int_0^2 f(x)\,dx = {1\over 2}\int_0^2 \int_x^2 \cos(t^2) \,dt \,dx ={1\over 2} \int_0^2 \int_0^t \cos(t^2)\, dxdt= {1\over 2}\int_0^2 t\cos (t^2)\,dt \\={1\over 2}\left. \left[ {1\over 2}\sin(t^2) \right] \right|_0^2 = \bbox[red, 2pt]{{1\over 4}\sin 4}$$

解答:$$I=\int_{-2}^2 \int_0^\sqrt{4-y^2} \int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} y^2 \sqrt{x^2+y^2+z^2} \,dzdxdy = \iiint_E y^2\,dV, \\\qquad \text{where }E=\{(x,y,z) \mid x^2+y^2+z^2\le 4,x\ge 0\}.  \text{Then let }\cases{x=\rho \sin \phi \cos \theta\\ y=\rho \sin \phi \sin \theta\\ z=\rho \cos \phi}, \\ \text{we have }I =\int_{-\pi/2}^{\pi/2} \int_0^\pi \int_0^2 \rho^2\sin^2\phi \sin^2 \theta\cdot \rho^2\sin \phi\,d\rho\,d\phi\,d\theta =\int_{-\pi/2}^{\pi/2} \int_0^\pi \int_0^2 \rho^4 \sin^3\phi \sin^2 \theta  \,d\rho\,d\phi\,d\theta \\={32\over 5}\int_{-\pi/2}^{\pi/2} \int_0^\pi \sin^3 \phi \sin^2 \theta\,d\phi\,d\theta={32\over 5}\cdot {4\over 3} \int_{-\pi/2}^{\pi/2}\sin^2 \theta\,d\theta ={32\over 5}\cdot {4\over 3}\cdot {\pi\over 2} = \bbox[red, 2pt]{{64\over 15}\pi}$$

解答:$$\text{By Green Theorem, }\int_C \mathbf F\cdot d\mathbf r =\int_D \left( {\partial \over \partial x}\tan^{-1}x -{\partial \over \partial y} \sqrt{x^2+1} \right)\,dydx =\int_0^1 \int_x^1 {1\over 1+x^2}\,dydx \\=\int_0^1 {1-x\over 1+x^2}\,dx =\int_0^1 \left({1\over 1+x^2}-{x\over 1+x^2} \right) dx =\left. \left[ \tan^{-1}x-{1\over 2}\ln(1+x^2) \right] \right|_0^1 = \bbox[red, 2pt]{{\pi\over 4}-{1\over 2}\ln 2}$$

乙、計算、證明題：共2題，每題10分，共20分

解答:$$\textbf{(a) }u=\ln x \Rightarrow du={dx\over x} \Rightarrow \int_2^\infty {1\over x\sqrt{\ln x}}\,dx =\lim_{b\to \infty} \int_{\ln 2}^b {1\over \sqrt u}\,du =\lim_{b\to \infty} (2\sqrt b-2\sqrt{\ln 2})= \infty \\\qquad \Rightarrow \bbox[red, 2pt] {\text{divergent}} \\\textbf{(b) }a_n={(x+2)^2\over n4^n} \Rightarrow \lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| = \lim_{n\to \infty}  \left|{(x+2)^{n+1}\over (n+1)4^{n+1}} \cdot { n4^n\over (x+2)^n} \right| =\lim_{n\to \infty}  \left|{(x+2)n\over 4(n+1)} \right|  \\\qquad =\left| {x+2\over 4}\right| \lt 1 \Rightarrow -4\lt x+2\lt 4 \Rightarrow -6\lt x\lt 2 \\ x=-6 \Rightarrow \sum_{n=1}^\infty{(x+2)^n\over n4^n} = \sum_{n=1}^\infty{(-4)^n\over n4^n} = \sum_{n=1}^\infty{(-1)^n\over n } \text{ convergent by alternating series test} \\x=2 \Rightarrow \sum_{n= 1}^\infty{(x+2)^n\over n4^n} = \sum_{n=1}^\infty{ 4^n\over n4^n} =   \sum_{n =1}^\infty{1\over n } \text{ is a harmonic series} \Rightarrow \text{divergent} \\ \text{Finally, we have the interval of convergence: }\bbox[red, 2pt]{[-6,2)}$$

解答:$$\textbf{(a) }\int_1^\infty{1\over x+x^3} \,dx = \int_1^\infty{1\over x(1+x^2)} \,dx = \int_1^\infty \left({1\over x} -{x\over 1+x^2} \right) \,dx = \left. \left[ \ln x-{1\over 2}\ln(1+x^2) \right] \right|_1^\infty \\ = \left. \left[ \ln {x\over \sqrt{1+x^2}} \right] \right|_1^\infty =\ln 1-\ln {1\over \sqrt 2} =\bbox[red, 2pt]{{1\over 2}\ln 2} \\\textbf{(b) }\cases{u=\tan^{-1}x \\ dv=dx/x^2} \Rightarrow \cases{du =dx/(1+x^2) \\ v=-1/x} \Rightarrow \int_1^\infty {\tan^{-1}x\over x^2}dx = \left. -{\tan^{-1}x\over x} \right|_1^\infty +\int_1^\infty {1\over x(1+x^2)}\,dx \\\qquad = \bbox[red, 2pt]{{\pi\over 4}+{1\over 2}\ln 2}$$

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