106年台綜大轉學考-微積分A詳解

臺灣綜合大學系統106學年度學士班轉學生聯合招生考試

科目名稱：微積分A

解答：$$\ln(1-x^3+y^3)-4=0 \Rightarrow {-3x^2+3y^2y' \over 1-x^3+y^3} =0 \Rightarrow y'={x^2 \over y^2} \Rightarrow 當y=0時,y'\to \infty \\ y=0代入原式 \Rightarrow \ln(1-x^3)-4=0 \Rightarrow x=\sqrt[3]{1-e^4} \Rightarrow 在位置\bbox[red,2pt]{(\sqrt[3]{1-e^4}, 0)}有垂直切線$$

解答：$$\lim_{x\to 0} (3x) =\lim_{x\to 0}(\sin(4x)+a-2b) \Rightarrow a-2b=0 \Rightarrow a=2b \cdots(1)\\ 又\lim_{x\to 0}{(\sin(4x)+a-2b)' \over (3x)'} =\lim_{x\to 0}{ 4\cos(4x)\over 3} ={4\over 3}= 2a+b \cdots(2)\\ 將(1)代入(2) \Rightarrow 5b={4\over 3} \Rightarrow b={4\over 15} \Rightarrow a={8\over 15} \Rightarrow (a,b)=\bbox[red,2pt]{({8\over 15},{4\over 15})}$$

解答：$$f(x)=\tan^{-1}(2x) \Rightarrow f'(x)={2\over 1+4x^2} =2(1-4x^2+16x^4-64x^6+\cdots)=\\ \Rightarrow f''(x)=2(-8x+64x^3-384x^5+\cdots) \Rightarrow f'''(x)=2(-8+192x^2-1920x^4+\cdots)\\ \Rightarrow \cases{f(0)=0 \\ f'(0)=2 \\ f''(0)=0\\ f'''(0)= -16} \Rightarrow f(x)的泰勒級數=f(0)+ f'(0)x +{1\over 2}f''(0)x^2+ {1\over 6}f'''(0)x^3+\cdots \\= 2x-{8\over 3}x^3+\cdots\\ 又g(x)={1\over 1-x}   \Rightarrow {f(x)/g(x)} = {2x-{8\over 3}x^3+\cdots \over 1-x} =\bbox[red,2pt]{2x+2x^2-{2\over 3}x^3}+\cdots $$

解答：

$$積分區域如上圖，改變積分順序，則\int_0^1 \int_{\sqrt x}^1 x\cos(y^5+2)\,dy\,dx = \int_0^1 \int_0^{y^2}  x\cos(y^5+2)\,dx\,dy \\=\int_0^1 \left.\left[ {1\over 2}x^2 \cos(y^5+2)\right] \right|_0^{y^2} \;dy =\int_0^1 {1\over 2}y^4\cos(y^5+2) \,dy \\ 令u=y^5+2 \Rightarrow du =5y^4dy，則上式變為\int_2^3{1\over 10}\cos u\,du ={1\over 10}\left. \left[ \sin u \right]\right|_2^3 =\bbox[red,2pt]{{1\over 10}(\sin 3-\sin 2)}$$

解答：$$e^{x^2} +y^2\sin(2x)= 4y \Rightarrow 2xe^{x^2} +2 y^2\cos(2x) +2yy'\sin(2x) =4y' \\\Rightarrow (4-2y\sin(2x)) y'=2xe^{x^2} +2 y^2\cos(2x) \Rightarrow y'= \bbox[red,2pt]{{dy\over dx} ={2xe^{x^2} +2 y^2\cos(2x)  \over 4-2y\sin(2x)}}$$

解答：$$3a-1 \le 1 \Rightarrow \bbox[red, 2pt]{a\le {2\over 3}}$$

解答：$$I=\int_0^\infty e^{-4x^2}\,dx \Rightarrow 2I=\int_{-\infty}^\infty e^{-4x^2}\,dx \Rightarrow (2I)^2 =4I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-4(x^2+y^2)}\,dydx \\ 令\cases{x=r\cos\theta \\ y=r \sin\theta} \Rightarrow  \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-4(x^2+y^2)}\,dydx =\int_0^{2\pi} \int_0^\infty re^{-4r^2}\,drd\theta =\int_0^{2\pi} \left. \left[ -{1\over 8}e^{-4r^2} \right]\right|_0^\infty \,d\theta\\=\int_0^{2\pi} {1\over 8}\,d\theta = {\pi\over 4} \Rightarrow 4I^2 ={\pi \over 4} \Rightarrow I= \bbox[red, 2pt]{\sqrt \pi\over 4}$$

解答：$$令\cases{A(-2,0)\\ B(4,0)\\ C(3,3)\\ D(-1,3)},則ABCD為一梯形，面積為{(6+4)\times 3\over 2} =15\\ 依格林定理\oint_C \vec F\cdot d\vec r = \oint_C (4y+6ye^{2x})dx +(6x+3e^{2x})dy \\=\iint_R {\partial \over \partial x}(6x+3e^{2x})- {\partial \over \partial y}(4y+6ye^{2x})\,dA\\ =\iint_R ((6+6e^{2x})-(4+6e^{2x}))dA =\iint_R 2\,dA = 2\times 15=\bbox[red, 2pt]{30}$$

解答：$$F=e^{x+y^2+ \cos z} \Rightarrow \nabla F=(F_x,F_y,F_z) =(e^{x+y^2+ \cos z},2ye^{x+y^2+ \cos z}, -\sin ze^{x+y^2+ \cos z})\\ \Rightarrow \nabla F(0,0,0)=(e,0,0) \Rightarrow 遞減最快的方向=-{\nabla F(0,0,0) \over |\nabla F(0,0,0)|} =\bbox[red,2pt]{(-1,0,0)} \\遞減率為-|\nabla F|=\bbox[red,2pt]{-e}$$

解答：$$折疊後的盒子體積f(x)=(3-2x)^2 x \Rightarrow f'(x)= 3(2x-1)(2x-3) =0 \Rightarrow x=1/2,3/2 \\ \Rightarrow \cases{f(1/2)= 2 \\ f(3/2)=0} \Rightarrow 最大積體為\bbox[red,2pt]{2}$$

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