109年台綜大轉學考-微積分A詳解

臺灣綜合大學系統109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： $$\mathbf{(a)}\;\lim_{n\to \infty}{3n+2\over 2n+1} =\lim_{n\to \infty}{3+2/n\over 2+1/n} = \bbox[red,2pt]{3\over 2} \\\mathbf{(b)}\; \lim_{x\to 0} {\cos(2x)-1\over x^2} = \lim_{x\to 0} {(\cos(2x)-1)' \over (x^2)'} = \lim_{x\to 0} {-\sin(2x) \over x} =\lim_{x\to 0} {(-\sin(2x))' \over (x)'} \\\quad =\lim_{x\to 0} {-2\cos(2x) \over 1} = \bbox[red, 2pt]{-2}$$

解答： $$\cases{x=t^2 +2t+3\\ y=t^4-3t^3} \Rightarrow \cases{dx/dt = 2t+2\\ dy/dt=4t^3-9t^2} \Rightarrow dy/dx ={4t^3-9t^2\over 2t+2}\\ 其中\cases{x=6\\ y=-2} \Rightarrow t=1 \Rightarrow dy/dx = {4-9\over 2+2} = \bbox[red,2pt]{-{5\over 4}}$$

解答： $${\overline{BP}\over \overline{AP}} ={\overline{CQ}\over \overline{BQ}} ={\overline{AR}\over \overline{CR}} = x \Rightarrow \cases{\overline{AP}= 3/(x+1)\\ \overline{BP} =3x/(x+1)\\ \overline{BQ}= 5/(x+1)\\ \overline{CQ} =5x/(x+1) \\ \overline{CR} =4/(x+1)\\ \overline{AR}= 4x/(x+1)} \\\Rightarrow \cases{\triangle APR/\triangle ABC ={\overline{AP}\cdot \overline{AR} \over \overline{AB}\cdot \overline{AC}} ={x\over (x+1)^2} \\\triangle BPQ/\triangle ABC ={\overline{BP}\cdot \overline{BQ} \over \overline{AB}\cdot \overline{BC}} ={x\over (x+1)^2} \\ \triangle CQR/\triangle ABC ={\overline{CQ}\cdot \overline{CR} \over \overline{AC}\cdot \overline{BC}} ={x\over (x+1)^2}}\\ \Rightarrow \triangle APR= \triangle BQP= \triangle CQR = {x\over (x+1)^2}\triangle ABC= {6x\over (x+1)^2} \\ \Rightarrow f(x)=\triangle PQR面積=6-{18x\over (x+1)^2} \Rightarrow f'(x)= {18(x-1)\over (x+1)^2} =0 \Rightarrow x=1 \\ \Rightarrow f(1)=6-{18\over 4} ={3\over 2}，因此\cases{\text{critical point at:}\bbox[red,2pt]{x=1} \\\text{minimum of   }f(x)= \bbox[red,2pt]{3/2}}$$

解答： $$令\sum_{n=0}^\infty {(2n)^n \over n!}x^n = \sum_{n=0}^\infty a_n \Rightarrow \left|{a_{n+1} \over a_n}  \right|=\left|{(2n+2)^{n+1} x^{n+1}\over (n+1)!}  \cdot {n!\over (2n)^n x^n}\right| =\left|{(2n+2) \over (n+1)}\cdot ({2n+2\over 2n})^n x \right| \\= \left| 2(1+{1\over n})^n x\right|，因此\lim_{n\to \infty}\left|{a_{n+1} \over a_n}  \right|\lt 1 \Rightarrow 2e|x|\lt 1 \Rightarrow |x|\lt {1\over 2e} \Rightarrow 收斂半徑為\bbox[red, 2pt]{1\over 2e}$$

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

解答： $$h(x,y,z)= g(\sqrt{x^2+y^2+z^2}) \Rightarrow \cases{{\partial h \over \partial x } ={x \over \sqrt{x^2+y^2+z^2}} g'(\sqrt{x^2+ y^2+z^2}) \\{\partial h \over \partial y } ={y \over \sqrt{x^2+y^2+z^2}} g'(\sqrt{x^2+ y^2+z^2}) \\{\partial h \over \partial z } ={z \over \sqrt{x^2+y^2+z^2}} g'(\sqrt{x^2+ y^2+z^2}) } \\ \Rightarrow \cases{{\partial^2 h \over \partial x^2 } =\left({1 \over \sqrt{x^2+y^2+z^2}}-{x^2 \over (x^2+y^2+z^2)^{3/2}} \right) g'(\sqrt{x^2+ y^2+z^2}) +{x^2 \over x^2+y^2+z^2}g''(\sqrt{x^2 +y^2 +z^2}) \\ {\partial^2 h \over \partial y^2 } =\left({1 \over \sqrt{x^2+y^2+z^2}}-{y^2 \over (x^2+y^2+z^2)^{3/2}} \right) g'(\sqrt{x^2+ y^2+z^2}) +{y^2 \over x^2+y^2+z^2}g''(\sqrt{x^2 +y^2 +z^2}) \\ {\partial^2 h \over \partial z^2 } =\left({1 \over \sqrt{x^2+y^2+z^2}}-{z^2 \over (x^2+y^2+z^2)^{3/2}} \right) g'(\sqrt{x^2+ y^2+z^2}) +{z^2 \over x^2+y^2+z^2}g''(\sqrt{x^2 +y^2 +z^2}) } \\ \Rightarrow \cases{{\partial^2 h\over \partial x^2}(P)= {5\over 9}g'(1)+{4\over 9}g''(1) = -{1\over 9} \\{\partial^2 h\over \partial y^2}(P)= {5\over 9}g'(1)+{4\over 9}g''(1) = -{1\over 9} \\{\partial^2 h\over \partial z^2}(P)= {8\over 9}g'(1)+{1\over 9}g''(1) = {20\over 9} } \Rightarrow {\partial^2 h\over \partial x^2}(P) +{\partial^2 h\over \partial y^2}(P)+ {\partial^2 h\over \partial z^2}(P) =\bbox[red, 2pt]2$$

解答： $$假設w=x\cos(xy) +z^2y^4-7xz \\\Rightarrow \nabla w=(\cos(xy)-xy\sin(xy) -7z,-x^2\sin(xy) +4z^2y^3, 2zy^4-7x)\\ \Rightarrow \nabla w|_{P}=(-6,4,2) \Rightarrow 過P(0,1,1)且法向量為(-6,4,2)的切平面方程式:-6x+4(y-1) +2(z-1)=0\\ \Rightarrow \bbox[red,2pt]{3x-2y-z+3=0};過P且方向向量為(-6,4,2)的直線方程式:{x\over -6}={y-1\over 4}={z-1\over 2}\\，即\bbox[red, 2pt]{\cases{x=-3t\\ y=2t+1\\ z=t+1},t\in \mathbb{R}}$$

解答： $$\cases{x=r\cos \theta \\ y=r\sin \theta} \Rightarrow R=\{(x,y)\mid 1\le x^2+y^2\le 4, x\ge 0\} \Rightarrow S=\{(r,\theta)\mid 1\le r\le 2,\theta \in[-\pi/2,\pi/2 ]\} \\ \Rightarrow \iint_R (y-x)\,dA = \int_{-\pi/2}^{\pi/2} \int_1^2  r^2(\sin\theta -\cos\theta)\,drd\theta = {7\over 3}\int_{-\pi/2}^{\pi/2}(\sin\theta -\cos\theta)\,d\theta\\ ={7\over 3} \left.\left[ -\cos\theta-\sin\theta \right]\right|_{-\pi/2}^{\pi/2} ={7\over 3}\cdot (-2)= \bbox[red, 2pt]{-{14\over 3}}$$

解答： $$\cases{x(t)= \cos(t) \\ y(t)=\sin(t)\\ z(t)=t} \Rightarrow \cases{x'(t)= -\sin(t)\\ y'(t)= \cos(t)\\ z'(t)=1 \\ \mathbf{F}= (t\cos(t), t\sin(t), \cos^3(t))} \Rightarrow 曲線C長度=\int_0^a \sqrt{x'(t)^2 +y'(t)^2 +z'(t)^2}\,dt\\ =\int_0^a \sqrt{2}\,dt = \sqrt 2a = {\sqrt 2\over 4}\pi  \Rightarrow a={\pi\over 4}\\ \\\Rightarrow \int_C \mathbf{F}\cdot dr = \int_0^a \left(-t\sin(t)\cos(t)+ t\sin(t)\cos(t) +\cos^3(t) \right)\,dt =\int_0^a \cos^3(t)\,dt \\ ={1\over 9}\left(9\sin(t)+ \sin(3t) \right)|_0^{\pi/4} ={1\over 9}(9\cdot {\sqrt 2\over 2}+ {\sqrt 2\over 2})= \bbox[red,2pt]{5\sqrt 2/9}$$

解答： $$F=(3x,2y,5z) \Rightarrow \nabla\cdot F=3+2+5=10 \Rightarrow \iiint \nabla\cdot F\,dV = \iiint 10\,dV =10\times {4\over 3}\pi = \bbox[red, 2pt]{{40\over 3}\pi}$$

========================== END ============================