107年台綜大轉學考-微積分A詳解

臺灣綜合大學系統107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：$$\mathbf{(a)}\;\lim_{n \to \infty} {n\over n+1}= \lim_{n \to \infty} {1\over 1+1/n}= \bbox[red, 2pt]1 \\ \mathbf{(b)}\; \lim_{n \to \infty} e^{-1/x^2}\left( {1\over x^5} -{3\over x^3}\right) =\lim_{n \to \infty} e^{-1/x^2} \cdot {1-3x^3\over x^5}   = 1\cdot 0=\bbox[red, 2pt] 0$$

解答：$$假設\cases{P在\;y=x^{3/2}上 \Rightarrow P(t,t^{3/2}),t\in \mathbb R \\ Q(5/2,0)} \Rightarrow  \overline{PQ} = \sqrt{(t-5/2)^2 + t^3} =\sqrt{t^3+t^2-5t+25/4}\\ 令f(t)= t^3+t^2-5t+25/4 \Rightarrow f'(t)=  3t^2+2t-5 \Rightarrow f''(t)=6t+2\\ 因此f'(t)=0 \Rightarrow (3t+5)(t-1)=0 \Rightarrow t=1,-5/3 \Rightarrow \cases{f''(1) \gt 0 \\ f''(-5/3) \lt 0} \\ \Rightarrow 當t=1時,\overline{PQ}有最小值,此時P＝ \bbox[red,2pt]{(1,1)}$$

解答：$$\cases{x=3t^2\\ y=2t^3} \Rightarrow \cases{x'=6t\\ y'= 6t^2} \Rightarrow 曲線長＝ \int_0^2\sqrt{ (x')^2+ (y')^2}\,dt =\int_0^2\sqrt{ 36t^2 +36t^4}\,dt\\ = \int_0^1 6t\sqrt{1+t^2}\,dt = \int_1^2 3\sqrt u\,du\;( u=1+t^2) =\left.\left[2 u^{3/2} \right]\right|_1^2 = \bbox[red, 2pt]{2^{5/2}-2}$$

解答：$$\mathbf{(a)}\;\lim_{n\to \infty}\left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty} \left|{ (n+1)^2 x^{n+1} \over n^2x^n} \right| = |x|\lt 1 \Rightarrow 收斂半徑＝\bbox[red, 2pt]1 \\\mathbf{(b)}\; 令f(x)={1\over 1-x} =\sum_{n=0}^\infty x^n \Rightarrow f'(x)={1\over (1-x)^2} = \sum_{n=1}^\infty nx^{n-1} \Rightarrow xf'(x) = {x\over (1-x)^2} =\sum_{n=1}^\infty nx^{n} \\ \Rightarrow (xf'(x))' = {1+x\over (1-x)^3} =\sum_{n=1}^\infty n^2x^{n-1} \Rightarrow g(x)={x(1+x)\over (1-x)^3} =\sum_{n=1}^\infty n^2x^{n} \\ \Rightarrow g(1/2)=6= \sum_{n=1}^\infty {n^2\over 2^n} \Rightarrow \sum_{n=1}^\infty {n^2\over 2^n}=\bbox[red, 2pt]6$$

解答：$$x^2y +e^{xyz}-2\cos(xz)=0 \Rightarrow \vec n|_{(1,1,0)} \\=(2xy+yze^{xyz} +2z\sin(xz), x^2+xze^{xyz}, xye^{xyz}+2 x\sin(xz) )|_{(1,1,0)}  =(2,1,1) \\ \Rightarrow 切平面方程式:(2,1,1)\cdot (x-1,y-1,z)=0 \Rightarrow 2(x-1)+(y-1)+z=0\\ \Rightarrow \bbox[red, 2pt]{2x+y+z =3}$$

解答：$$依題意\cases{x=u^2-v^2\\ y=2uv} \Rightarrow \cases{{\partial x\over \partial u}=2u \\{\partial x\over \partial v}= -2v \\ {\partial y\over \partial u}=2v \\{\partial y\over \partial v}=2u } \Rightarrow \cases{g_u = {\partial f\over \partial x}{\partial x\over \partial u} +{\partial f\over \partial y}{\partial y\over \partial u} =2uf_x+ 2vf_y \\g_v = {\partial f\over \partial x}{\partial x\over \partial v} +{\partial f\over \partial y}{\partial y\over \partial v} = -2vf_x+ 2uf_y } \\\Rightarrow \cases{g_{uu} = 2f_x +2u({\partial \over \partial x}f_x\cdot {\partial x\over \partial u} +{\partial \over \partial y}f_x\cdot {\partial y\over \partial u}) +2v ({\partial \over \partial x}f_y\cdot {\partial x\over \partial u} +{\partial \over \partial y}f_y\cdot {\partial y\over \partial u})\\\qquad =2f_x +4u^2f_{xx} +8uvf_{xy} +4v^2f_{yy} \\g_{vv} =-2f_x -2v({\partial \over \partial x}f_x \cdot {\partial x\over \partial v} +{\partial \over \partial y}f_x \cdot {\partial y\over \partial v}) +2u({\partial \over \partial x}f_y \cdot {\partial x\over \partial v} +{\partial \over \partial y}f_y \cdot {\partial y\over \partial v})\\\qquad  =-2f_x +4v^2 f_{xx} -8uvf_{xy} +4u^2f_{yy}} \\ \Rightarrow g_{uu}+g_{vv}=4(u^2+v^2)f_{xx}+4(u^2+v^2)f_{yy} =4(u^2+v^2)(f_{xx}+f_{yy})=\bbox[red, 2pt]0$$

解答：$$取\cases{u=x+y \\ v= x-y} \Rightarrow \cases{x=(u+v)/2\\ y=(u-v)/2} \Rightarrow J={\partial(x,y)\over \partial (u,v)} =\begin{vmatrix} x_u & x_v\\ y_u & y_v\end{vmatrix}=\begin{vmatrix} 1/2 & 1/2\\ 1/2 & -1/2\end{vmatrix}= -{1\over 2}\\ \Rightarrow \iint_D \sin(x+y)\,dA = \int_\pi^0  \int_0^\pi -{1\over 2}\sin u\,dudv =\int_\pi^0 -1\,dv = \bbox[red, 2pt]\pi$$

解答：$$取\cases{x=r \cos\theta\\ y=r \sin \theta\\ z=z} \Rightarrow \iiint_E \sqrt{x^2+y^2}\,dV = \int_0^{2\pi} \int_0^1 \int_1^{r^2} r^2\,dzdrd\theta = \int_0^{2\pi} \int_0^1  r^4-r^2\, drd\theta \\ =\int_0^{2\pi} {7\over 12}\,d\theta= \bbox[red, 2pt]{{7\over 6}\pi}$$

解答：$$\cases{M(x,y)=ye^x +\sin y\\ N(x,y) = e^x+ x\cos y} \Rightarrow \cases{M_y= e^x+\cos y\\ N_x= e^x+\cos y} \Rightarrow M_y = N_x \\\Rightarrow 存在\text{potential function }f 滿足\cases{f_x= M\\ f_y= N} \Rightarrow f=\int M\,dx = \int N,dy\\ 由於\cases{\int ye^x +\sin y\,dx = ye^x + x\sin y+ \phi(y)\\ \int e^x+ x\cos y\,dy= ye^x +x\sin y +\rho(x)} \Rightarrow f(x,y) = ye^x+ x\sin y+ C\\ 又\cases{p_0=r(t=0)=(1,-1)\\ p_1=r(t=2)=(5,3)} \Rightarrow 該線積分=f(p_1)-f(p_0) = 3e^5+ 5\sin 3-(-e+\sin(-1))\\ = \bbox[red, 2pt]{3e^5+ 5\sin 3+e+\sin 1}$$

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