109年普考-微積分詳解

109年公務人員普通考試試題

類科:氣象
科目:微積分

解：

$$h(x)=f(g(x))+{f(2x)\over f(2x)+g(x)} \Rightarrow h'(x)=f'(g(x))g'(x) +{2f'(2x)\over f(2x)+g(x)}-{f(2x) (2f'(2x)+g'(x)) \over (f(2x)+g(x))^2}\\ \Rightarrow h(1) = f(g(1))+{f(2)\over f(2)+g(1)} =f(2)+ {f(2)\over f(2)+g(1)} =2+ {2\over 2+2}={5\over 2}\\ \Rightarrow h'(1)= f'(2)g'(1) +{2f'(2) \over f(2)+g(1)}-{f(2)(2f'(2)+g'(1)) \over (f(2)+g(1))^2} =9+{6\over 4}-{2(6+3) \over 4^2} ={75\over 8}\\ 過(1,h(1))=(1,5/2)的直線斜率為75/8 \Rightarrow 方程式為:y-{5\over 2}={75\over 8}(x-1) \Rightarrow \bbox[red, 2pt]{75x-8y=55}$$

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解：

$$先計算左半部: \int e^{5x} \cos 2x\;dx\\ \cos 2x +i\sin 2x=e^{2xi} \Rightarrow \int e^{5x} \cdot e^{2xi}\;dx = \int e^{(5+2i)x}\; dx = {1\over 5+2i}e^{(5+2i)x} ={1\over 5+2i}e^{5x}(\cos 2x+i\sin 2x) \\ = {5-2i \over 29}e^{5x}(\cos 2x+i\sin 2x),實數部份為{5\over 29}e^{5x}\cos 2x+ {2\over 29}e^{5x}\sin 2x ={e^{5x} \over 29}(5\cos 2x+2\sin 2x)\\ 再計算右半部: \int x\sec x^2\;dx\\ 令u=x^2 \Rightarrow du=2xdx \Rightarrow \int x\sec x^2\;dx ={1\over 2}\int \sec u\;du ={1\over 2}\ln |\sec u+ \tan u| ={1\over 2}\ln |\sec x^2+ \tan x^2| \\ \Rightarrow \int e^{5x} \cos 2x+x\sec x^2 \;dx = \bbox[red, 2pt]{{e^{5x} \over 29}(5\cos 2x+2\sin 2x)+ {1\over 2}\ln |\sec x^2+ \tan x^2| +C}$$

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解：

$$f(x,y,z)={2\over \sqrt{x^2+y^2+z^2}} =2(x^2 +y^2+z^2)^{-1/2}\Rightarrow \cases{{\partial f\over \partial x} = -2x(x^2+y^2+z^2)^{-3/2} \\ {\partial f\over \partial y} = -2y(x^2+y^2+z^2)^{-3/2} \\{\partial f\over \partial z} = -2z(x^2+y^2+z^2)^{-3/2} } \\ \Rightarrow \cases{{\partial^2 f\over \partial x^2} = -2(x^2+y^2+z^2)^{-3/2}+ 6x^2(x^2+y^2+z^2)^{-5/2} \\ {\partial^2 f\over \partial y^2} = -2(x^2+y^2+z^2)^{-3/2} +6y^2(x^2+y^2+z^2)^{-5/2} \\{\partial^2 f\over \partial z^2} = -2(x^2+y^2+z^2)^{-3/2} +6z^2(x^2+y^2+z^2)^{-5/2}}  \\ \Rightarrow {\partial^2 f\over \partial x^2}+{\partial^2 f\over \partial y^2} +{\partial^2 f\over \partial z^2} = -6(x^2+y^2+z^2)^{-3/2}+ 6(x^2+y^2+z^2)^{-3/2}= \bbox[red, 2pt]{0}$$

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解：
$$\cases{F_x=\lambda G_x+ \mu H_x \\ F_y=\lambda G_y+ \mu H_y \\F_z=\lambda G_z+ \mu H_z \\G=2\\ H=1} \Rightarrow  \cases{1=2\lambda x+ \mu  \\ 3=18\lambda y  \\5= 5 \mu  \\x^2+9y^2 =2\\ x+5z=1} \Rightarrow \cases{\lambda x=0  \\ \lambda y=1/6  \\\mu=1  \\x^2+9y^2 =2\\ x+5z=1} \\ \lambda x=0 \Rightarrow \lambda=0\; 或\;x=0,若\lambda=0 \Rightarrow \lambda y\ne 1/6,因此\lambda\ne 0\Rightarrow x=0 \Rightarrow \cases{9y^2=2\\ 5z=1}\\ \Rightarrow (0,\pm \sqrt 2/3,1/5) 為臨界點 \Rightarrow \cases{F(0,\sqrt 2/3,1/5) =0 + \sqrt 2+1= \sqrt 2+1\\ F(0,-\sqrt 2/3,1/5) =0 - \sqrt 2+1= 1-\sqrt 2}\\ 最大值:\bbox[red, 2pt]{1+\sqrt 2},最小值:\bbox[red, 2pt]{1-\sqrt 2}$$

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解：

$$\iiint_V \left(x+ {y \over 3}+ {z\over 5} \right)^2\;dxdydz =\int_0^5 \int_0^{(15-3z)/5} \int_0^{(15-5y-3z)/15} \left(x+ {y \over 3}+ {z\over 5} \right)^2\;dxdydz \\ =\int_0^5 \int_0^{(15-3z)/5}\left .\left[ {1\over 3}\left(x+ {y \over 3}+ {z\over 5} \right)^3 \right] \right|_0^{1-y/3-z/5}dydz = \int_0^5 \int_0^{(15-3z)/5} {1\over 3}(1-({1\over 3}y+{1\over 5}z)^3)\;dydz \\ ={1\over 3}\int_0^5 \left. \left[ y-{3\over 4}({1\over 3}y+{1\over 5}z)^4\right] \right|_0^{(15-3z)/5} =\int_0^5 {3\over 4}-{1\over 5}z+{1\over 4}({1\over 5}z)^4\;dz =\left. \left[{3\over 4}z-{1\over 10}z^2+{1\over 4}({1\over 5}z)^5 \right] \right|_0^5 \\= \bbox[red, 2pt]{3\over 2}$$

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