113年普考-微積分詳解

 113年公務人員普通考試試題

類 科：天文
科 目：微積分

解答:$$\textbf{(一)}\; \lim_{n\to \infty} \sum_{k=1}^n \left({2k\over n^2}\sin{k\pi \over 2n} \right) = \lim_{n\to \infty} \sum_{k=1}^n \left(2\cdot {1\over n} \cdot{k\over n}\sin({k \over n} \cdot{\pi \over 2}) \right) = \int_0^1 2x \sin{\pi x\over 2} \,dx \\\quad =\left. \left[ -{4x\over \pi} \cos{\pi x\over 2} +{8\over \pi^2} \sin{\pi x\over 2}\right] \right|_0^1 = \bbox[red, 2pt]{8\over \pi^2} \\\textbf{(二)}\; \lim_{x \to \infty}  {\int_0^{x^2} \ln(t^2+1)\,dt \over x^{2.01}} = \lim_{x \to \infty}  {\frac{d }{dx}\int_0^{x^2} \ln(t^2+1)\,dt \over \frac{d }{dx} x^{2.01}} = \lim_{x \to \infty}  {  2x\ln(x^4+1)  \over 2.01 x^{1.01}} = \lim_{x \to \infty}  {  2\ln(x^4+1)  \over 2.01 x^{0.01}} \\\quad =\lim_{x \to \infty}  {  \frac{d }{dx}2\ln(x^4+1)  \over \frac{d }{dx} 2.01 x^{0.01}}  =\lim_{x \to \infty} {{8x^3\over x^4+1} \over {201\over 10000}x^{-0.99}} =\lim_{x \to \infty} {80000 x^{3.99} \over 201(x^4+1)} = \bbox[red, 2pt]0 $$

解答:$$\int_0^{x^2+1} f(t)\,dt ={x+1\over e^x} \Rightarrow 2xf(x^2+1) =-{x\over e^x} \Rightarrow f(x^2+1) =-{1\over 2e^x} \Rightarrow f(2)=-{1\over 2e} \\ 又2xf'(x^2+1)= {1\over 2e^x} \Rightarrow 2f'(1+1)={1\over 2e} \Rightarrow f'(2)={1\over 4e} \\ \Rightarrow (2,f(2))=(2,-{1\over 2e})的切線斜率=f'(2)={1\over 4e} \Rightarrow 切線方程式:y+{1\over 2e}={1\over 4e}(x-2) \\ \Rightarrow \bbox[red, 2pt]{x-4ey=4}$$

解答:$$F(r)={e^r\over r} \Rightarrow F'(r) =e^r\left( {1\over r}-{1\over r^2}\right)  \Rightarrow F''(r) =e^r\left( {1\over r}-{2\over r^2} +{2\over r^3}\right)\\ G(x,y,z) =\sqrt{x^2+y^2+z^2} \Rightarrow \cases{G_x=x/\sqrt{x^2+y^2+z^2} \\ G_y=y/\sqrt{x^2+y^2+z^2} \\ G_z=z/\sqrt{x^2+y^2+z^2}} \Rightarrow \cases{G_{xx} =(y^2+z^2) /(x^2+y^2+z^2)^{3/2} \\G_{yy} =(x^2+z^2) /(x^2+y^2+z^2)^{3/2} \\G_{zz} =(x^2+ y^2) /(x^2+y^2+z^2)^{3/2} } \\ \frac{\partial }{\partial x} H= \frac{\partial }{\partial x}F(G) =F'(G)G_x \Rightarrow \frac{\partial^2 }{\partial x^2}H=\frac{\partial }{\partial x} F'(G)G_x =F''(G)(G_x)^2+ F'(G)G_{xx}\\ 同理\cases{\frac{\partial^2 }{\partial y^2}H =F''(G)(G_y)^2+ F'(G)G_{yy} \\\frac{\partial^2 }{\partial z^2}H =F''(G)(G_z)^2+ F'(G)G_{zz}} \Rightarrow \frac{\partial^2 }{\partial x^2}H +\frac{\partial^2 }{\partial y^2}H +\frac{\partial^2 }{\partial z^2}H \\ =F''(G)((G_x)^2+ (G_y)^2+ (G_z)^2) +F'(G)(G_{xx} +G_{yy} +G_{zz}) =F''(G)+F'(G)\cdot {2\over \sqrt{x^2+y^2+z^2}} \\=e^{\sqrt{x^2+y^2+z^2}}  \left({1\over \sqrt{x^2+y^2 +z^2}} -{2\over x^2+y^2 +z^2} +{2\over (x^2+y^2 +z^2)^{3/2}}\right) \\\qquad +e^{\sqrt{x^2+y^2+z^2}} \left({1\over \sqrt{x^2+y^2 +z^2}}-{1\over x^2+y^2+z^2} \right){2\over \sqrt{x^2+y^2+z^2}} \\=e^{\sqrt{x^2+y^2+z^2}} \left( {1\over \sqrt{x^2+y^2+ z^2}}  \right)= \bbox[red, 2pt] {   {e^{\sqrt{x^2+y^2+z^2}}\over (x^2+y^2 +z^2)^{3/2}} }$$

解答:$$\cases{u=3x+4y \\v=3x-4y} \Rightarrow \cases{x=(u+v)/6\\ y=(u-v)/8}\Rightarrow {\partial(x,y) \over \partial (u,v)} =\begin{Vmatrix} 1/6& 1/6\\ 1/8& -1/8 \end{Vmatrix} = {1\over 24} \\ \Rightarrow \iint_D (9x^2+16y^2)\,dxdy = {1\over 24}\int_0^2 \int_0^2 \left({1\over 4}(u+v)^2 +{1\over 4}(u-v)^2 \right)\,dudv \\ = {1\over 48}\int_0^2 \int_0^2 \left( u^2+v^2 \right)\,dudv ={1\over 48}\int_0^2 \left({8\over 3}+2v^2 \right)\,dv ={1\over 48} \cdot {32\over 3} = \bbox[red, 2pt]{2\over 9}$$

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