114年台綜大轉學考-微積分C詳解

臺灣綜合大學系統114學年度學士班轉學生聯合招生考試

科目名稱:微積分C

解答:$$\textbf{(a) }\lim_{h\to 1}{\int_0^{h^2-1} e^{x^2-2}\,dx \over h-1} =\lim_{h\to 1}{ 2he^{{(h^2-1)}^2-2}  \over 1} = \bbox[red, 2pt]{2e^{-2}} \\\textbf{(b) }\sin(3x)\gt x, x\in (0,\pi/6) \Rightarrow \lim_{x\to 0^+}{|x-\sin(3x)| \over 7x} =\lim_{x\to 0^+}{\sin(3x)-x \over 7x} =\lim_{x\to 0^+}{3\cos(3x)-1 \over 7}\\\quad =\bbox[red, 2pt]{2\over 7}$$

解答:$$\sqrt x+\sqrt y=12 \Rightarrow {1\over 2\sqrt x}+{y'\over 2\sqrt y}=0 \Rightarrow y'=-{\sqrt y\over \sqrt x} =-{12-\sqrt x\over \sqrt x} =1-{12\over \sqrt x}=-3 \\ \Rightarrow \sqrt x=3 \Rightarrow x=9 \Rightarrow 3+\sqrt y=12 \Rightarrow \sqrt y=9\Rightarrow y=81 \Rightarrow (x_0,y_0) = \bbox[red, 2pt]{(9,81)}$$

解答:$$e^x = \sum_{k=0}^\infty {1\over k!}x^k \Rightarrow f(x)=xe^x =\sum_{k=0}^\infty {1\over k!}x^{k+1} \Rightarrow f'(x)=(x+1)e^x =\sum_{k=0}^\infty {(k+1)\over k!}x^{k}  \\ \Rightarrow f'(-1)=0= \sum_{k=0}^\infty {(k+1)\over k!}(-1)^{k} =1-2+{3\over 2}+ \sum_{k=3}^\infty {(k+1)\over k!}(-1)^{k} \\ \Rightarrow \sum_{k=3}^\infty {(k+1)\over k!}(-1)^{k}= \bbox[red, 2pt]{-{1\over 2}}$$

解答:$$\cases{A(2t,0) \\ B(6t \cos 30^\circ, 6t\sin 30^\circ) =(3\sqrt 3 t, 3t)} \Rightarrow f(t)=\overline{AB} =\sqrt{(3\sqrt 3-2)^2t^2+ 9t^2} = \sqrt{40-12\sqrt 3}t \\ \Rightarrow f'(t)=\sqrt{40-12\sqrt 3} \Rightarrow f'({1\over 2}) = \bbox[red, 2pt]{\sqrt{40-12\sqrt 3}}$$

解答:$$f(x,y,z) =2xe^y-x \sin z \Rightarrow \nabla f=(2e^y-\sin z, 2xe^y,-x\cos z) \Rightarrow \nabla f(1,0,0) =(2,2,-1) \\ \Rightarrow \nabla f(1,0,0) \cdot \mathbf u=0 \Rightarrow 2a-b=0 \Rightarrow b=2a \text{ and } a^2+b^2=1 \Rightarrow 5a^2=1 \Rightarrow a={1\over \sqrt 5} \\ \Rightarrow b={2\over \sqrt 5} \Rightarrow \bbox[red, 2pt]{\cases{a=\sqrt 5/5\\ b=2\sqrt 5/5}}$$

解答:$$\cases{V=f(x,y,z)=8xyz\\ g(x,y,z) =x^2/9+y^2/4+z^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y \\ f_z= \lambda g_z\\ g=0} \Rightarrow \cases{8yz= \lambda \cdot {2x\over 9} \\ 8xz =\lambda\cdot {y\over 2} \\ 8xy = \lambda\cdot 2z \\x^2/9+y^2/4+z^2=1} \\ \Rightarrow \cases{9y^2=4x^2\\ y^2=4z^2 } \Rightarrow {y^2\over 4}+{y^2\over 4}+{y^2\over 4}=1 \Rightarrow y^2={4\over 3} \Rightarrow \cases{x^2=3\\ z^2=1/3} \Rightarrow \cases{x=\sqrt 3\\ y=2/\sqrt 3\\ z=1/\sqrt 3} \\ \Rightarrow V=f(\sqrt 3, {2 \over \sqrt 3},{1\over \sqrt 3}) ={16\over \sqrt 3} = \bbox[red, 2pt]{16\sqrt 3\over 3}$$

解答:$$f(x,y)=e^y(y^2-x^2) \Rightarrow \cases{f_x=-2xe^y\\ f_y=(y^2-x^2+2y)e^y} \Rightarrow \cases{f_{xx}=-2e^y\\ f_{xy} =-2xe^y \\ f_{yy}=(y^2-x^2+4y+2)e^y} \\ \Rightarrow D(x,y)=f_{xx}f_{yy}-f_{xy}^2=-2(x^2+y^2+4y+2)e^{2y} \\ \cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(0,0),(0,-2) \Rightarrow \cases{D(0,-2)=4/e^4\gt 0\\ D(0,0)=-2/e^2 \lt 0} \Rightarrow \bbox[red, 2pt]{(0,0) \text{ is a saddle point}}$$

解答:$$y=\sqrt{x-x^2} \Rightarrow x^2-x+y^2=0 \Rightarrow (x-{1\over 2})^2+y^2={1\over 4} \Rightarrow 積分區域為一半圓\\ \cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow (r\cos \theta-{1\over 2})^2+ r^2\sin ^2\theta ={1\over 4 }\Rightarrow r^2-r\cos \theta =0 \Rightarrow r(r-\cos \theta)=0 \Rightarrow r=\cos \theta \\ \Rightarrow \int_0^1 \int_0^\sqrt{x-x^2} \sqrt{x^2+y^2} \,dydx =\int_0^{\pi/2} \int_0^{\cos \theta} r^2\,drd\theta= {1\over 3} \int_0^{\pi/2}\cos^3 \theta\,d\theta \\={1\over 12} \int_0^{\pi/2}(\cos 3\theta+3\cos \theta)\,d\theta ={1\over 12} \left. \left[ {1\over 3}\sin 3\theta+3\sin \theta\right] \right|_0^{\pi/2} =\bbox[red, 2pt]{2\over 9}$$

解答:

$$假設\cases{A(2,0) \\B(1,\sqrt 3)\\ O(0,0)} \Rightarrow \cases{C_1= \stackrel{\Large \frown}{AB} = \{(2\cos t, 2\sin t) \mid 0\le t\le \pi/3\}\\C_2=\overline{BO}=\{(1-t,\sqrt 3(1-t)) \mid 0\le t\le 1\} \\ C_2= \overline{OA}:\{(2t,0)  \mid 0\le t\le 1\}} \\ \text{By Green theorem, } \int_{C_1+C_2 +C_3} \mathbf F\cdot d\mathbf r =\iint_D \left[{\partial \over \partial x}\left({2xy\over y^2+1}+x \right)-{\partial \over \partial y}\left(\ln(y^2+1)-4y \right) \right]\,dA \\=\iint_D\left[ \left({2y\over y^2+1}+1 \right) -\left({2y\over y^2+1}-4 \right)\right]\,dA =\iint_D 5\,dA =5\cdot {1\over 6}\cdot 2^2\pi={10\over 3}\pi \\ \int_{C_3} \mathbf F\cdot d\mathbf r =\int_0^1 (e^{2t},2t)\cdot (2,0)\,dt = \int_0^1 2e^{2t}\,dt =\left.\left[ e^{2t}\right] \right|_0^1 =e^2-1 \\ \Rightarrow \int_C \mathbf F\cdot d\mathbf r =\int_{C_1+C_2+C_3} \mathbf F\cdot d\mathbf r -\int_{C_3} \mathbf F\cdot d\mathbf r = \bbox[red, 2pt]{{10\pi\over 3}-e^2+1}$$

解答:$$\text{By Divergence Theorem of Gauss, }\iint_S \mathbf F\cdot d\mathbf S =\iiint_V (\nabla \cdot \mathbf F)dV \\=\iiint_V (5-6+1)dV= \bbox[red, 2pt]0$$

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