112年臺綜大轉學考-微積分C詳解

臺灣綜合大學系統112學年度學士班轉學生聯合招生

科目名稱：微積分C

解答： $$\lim_{x\to 1} {\sqrt x-1 \over x-1} =\lim_{x\to 1} {\sqrt x-1 \over (\sqrt x-1)(\sqrt x+1)} = \lim_{x\to 1} {1 \over \sqrt x+1} = \bbox[red, 2pt]{1\over 2}$$

解答： $$\int_0^{\pi/2} \sin^2 x\cos x\,dx = \left. \left[ {1\over 3}\sin^3 x \right] \right|_0^{\pi/2} =\bbox[red,2pt]{1\over 3}$$

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

解答： $$\cases{x=r(\theta-\sin \theta)\\ y=r(1-\cos \theta)} \Rightarrow \cases{dx= r(1-\cos \theta)d\theta \\ dy = r\sin \theta \,d\theta} \Rightarrow 擺線長=\int_0^{\pi/2} \sqrt{({dx\over d\theta})^2+({dy\over d\theta})^2} \,d\theta \\=\int_0^{\pi/2} \sqrt{r^2(1-\cos\theta)^2 +r^2\sin^2 \theta}\,d\theta =\int_0^{\pi/2} r\sqrt{2(1-\cos \theta)}\,d\theta =\int_0^{\pi/2} r\sqrt{4\sin^2(\theta/2)}\,d\theta \\ =\int_0^{\pi/2}2r\sin(\theta/2)\,d\theta = \left. \left[ -4r\cos(\theta/2) \right] \right|_0^{\pi/2} =\bbox[red,2pt]{(4-2\sqrt 2)r}$$

解答： $$\cases{\ln(1+x)= \sum_{n=1}^\infty {1\over n}(-1)^{n+1}x^n \\[1ex] \tan^{-1}x =\sum_{m= 0}^\infty {1\over 2m+1}(-1)^m x^{2m+1}} \\\Rightarrow 考量\ln(1+x)\tan^{-1}x 的x^5係數:\cases{m=0,n=4: -{1\over 4}\cdot 1=-{1\over 4} \\ m=1,n=2:-{1\over 2}\cdot (-{1\over 3})={1\over 6} } \\ \Rightarrow f(x)=\ln(1+x)\tan^{-1}x 的x^5係數=(-{1\over 4}+{1\over 6}) =-{1\over 12} \\ \Rightarrow f^{[5]}(0)=-{1\over 12}\times 5!= \bbox[red, 2pt]{-10}$$

解答： $$x-y+z=0 \Rightarrow y=x+z \Rightarrow \cases{x+2y+3z=3x+5z\\ x^2+y^2=x^2+(x+z)^2}\\ 取\cases{g(x,z)=3x+5z\\ h(x,z)=x^2+(x+z)^2-29}，利用\text{Lagrange }算子求解\\ \cases{g_x= \lambda h_x\\ g_z=\lambda h_z\\ h=0} \Rightarrow \cases{3=\lambda(4x+2z) \\ 5= \lambda(2x+2z)} \Rightarrow {3\over 5}={4x+2z \over 2x+2z} \Rightarrow x=-{2\over 7}z \Rightarrow y=x+z={5\over 7}z\\ \Rightarrow h(-2z/7,z)=0 \Rightarrow {29\over 49}z^2 =29 \Rightarrow \cases{z=7 \Rightarrow x=-2,y=5\\ z=-7 \Rightarrow x=2,y=-5} \\ \Rightarrow \cases{f(-2,5,7)= 29\\f(2,-5,-7)= -29} \Rightarrow 最大值為\bbox[red, 2pt]{29}$$

解答： $$此題相當於求曲線y=\sqrt x繞x軸旋轉(x=1 \to 9)所得的表面積\\ 即\int_1^9 2\pi \sqrt x\,dx =\left. \left[ {4\over 3}\pi x^{3/2} \right] \right|_1^9 =\bbox[red, 2pt]{{104\over 3}\pi}$$

解答： $$令\cases{x=r\sin \phi \cos \theta\\ y= r\sin \phi \sin \theta\\ z=r \cos \phi} \Rightarrow \iiint_B e^{(x^2+y^2+z^2)^{3/2}}\, dV = \int_0^{2\pi} \int_0^\pi \int_0^1 e^{r^{3}} \cdot r^2 \sin \phi\,drd\phi d\theta \\= {1\over 3}(e-1)\int_0^{2\pi} \int_0^\pi \sin \phi \,d\phi d\theta= {2\over 3}(e-1)\int_0^{2\pi} 1\,d\theta= \bbox[red, 2pt]{{4\over 3}(e-1)\pi}$$

解答： $$利用\text{ Green theorem }求解，令\cases{P(x,y))= y^2+\sin x\\ Q(x,y)=3xy-e^y} \Rightarrow \cases{{\partial \over \partial y}P=2y\\ {\partial \over \partial x} Q=3y} \\ \Rightarrow \oint_C (y^2+\sin x)dx+(3xy-e^y)dy = \iint_D \left({\partial Q\over \partial x} -{\partial P\over \partial y}\right)dxdy = \iint_D y\,dxdy \\=\int_0^\pi \int_2^3 r^2\sin \theta \,drd\theta   =\int_0^\pi {19\over 3}\sin \theta\,d\theta =\bbox[red, 2pt]{38\over 3}$$

解答： $$\mathbf F(x,y,z)=xy\mathbf i+(y^2+e^{xz})\mathbf j+ \cos(xy)\mathbf k \Rightarrow \text{div }\mathbf F =F_x +F_y+F_z = y+2y+0=3y\\ \cases{z=1-x^2\\ z=0\\ y=0\\ y+z=2} \Rightarrow E\{(x,y,z)\mid -1\le x\le 1,0\le y\le 2-z, 0\le z\le 1-x^2\}\\利用發散定理: \\\iint_S \mathbf F\cdot d\mathbf S =\iiint_E \text{div }\mathbf F\,dV =\int_{-1}^1 \int_0^{1-x^2} \int_0^{2-z} 3y\,dydzdx=\int_{-1}^1 \int_0^{1-x^2} {3\over 2}(z-2)^2\,dzdx \\= \int_{-1}^1 4-{1\over 2}(x^2+1)^3\,dx =\bbox[red, 2pt]{184\over 35}$$

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