114年升官等薦任-微積分詳解

 114年公務人員升官等考試

等 級：薦任
類科（別）：物理
科 目：微積分
考試時間： 2 小時

解答:$$e^x = \sum_{n=0}^\infty {x^n\over n!} \Rightarrow f(x)=e^{x^2} = \sum_{n=0}^\infty {x^{2n}\over n!} \Rightarrow x^{100} 的係數為{1\over 50!} \\ \Rightarrow f^{(100)} 的常數項為{100! \over 50!} \Rightarrow f^{(100)}(0) = \bbox[red, 2pt]{100! \over 50!} $$

解答:$$\cases{x= r\cos \theta\\ y= r\sin \theta} \Rightarrow r^4=r^2-4r^2\cos \theta \sin \theta =r^2-2r^2 \sin 2\theta \Rightarrow r^2=1-2\sin 2\theta \ge 0 \\ \Rightarrow \sin 2\theta\le {1\over 2} \Rightarrow 2\theta \in [0,{\pi\over 6}] \cup [{5\over 6}\pi, 2\pi] \cup [2\pi, 2\pi+{\pi\over 6}] \cup [2\pi+{5\over 6}\pi, 4\pi] \\ \Rightarrow \theta \in \left[0,{\pi\over 12}  \right] \cup \left[{5\pi\over 12} ,{13\pi\over 12} \right]  \cup\left[{17\pi\over 12}, 2\pi  \right] \Rightarrow 所圍面積={1\over 2} \int r^2 \,d \theta \\={1\over 2} \left( \int_0^{\pi/12} (1-2\sin 2\theta)\,d\theta + \int_{5\pi/12}^{13\pi/12} (1-2\sin 2\theta)\,d\theta +\int_{17\pi/12}^{2\pi} (1-2\sin 2\theta)\,d\theta\right) \\ ={1\over 2} \left( ({\sqrt 3\over 2}-1+{\pi\over 12}) + (\sqrt 3+{2\pi\over 3}) + (1+{\sqrt 3\over 2}+{7\pi\over 12})\right) ={1\over 2} \left( 2\sqrt 3+{4\pi\over 3} \right) = \bbox[red, 2pt]{\sqrt 3+{2\pi\over 3}}$$

解答:$$\cases{P(x,y) =e^x \sin(2y) \\Q(x,y) =2e^x \cos(2y)} \Rightarrow P_y=2e^x \cos(2y) =Q_x \Rightarrow 存在 \Psi(x,y)=\int P\,dx =\int Q\,dy \\\Rightarrow \Psi(x,y)= \int e^x\sin(2y) \,dx= \int2e^x\cos(2y)\,dy \Rightarrow \Psi =e^x \sin(2y)+\phi(y) =e^x \sin(2y)+ \rho(x) \\ \Rightarrow \Phi(x,y)= e^x \sin(2y) \Rightarrow \int_C P\,dx+Q\,dy = \Psi(1,1)- \Psi(0,0)= \bbox[red, 2pt]{e \sin(2)}$$

解答:$$若y=x \ne 0 \Rightarrow \lim_{(x,y) \to (0,0)} f(x,y)=\lim_{(x,x) \to (0,0)} f(x,x)=\lim_{(x,x) \to (0,0)} {x^4\over 2x^4} ={1\over 2} \\ 若y= -x \ne 0 \Rightarrow \lim_{(x,y) \to (0,0)} f(x,y)=\lim_{(x,-x) \to (0,0)} f(x,-x)=\lim_{(x,-x) \to (0,0)} {-x^4\over 2x^4} =-{1\over 2}  \\ \Rightarrow {1\over 2} \ne -{1\over 2} \Rightarrow f(x,y)在(0,0)\bbox[red, 2pt]{不連續}$$

解答:$$xy^3+4x^2=5 \Rightarrow y^3+3xy^2y'+8x=0 \Rightarrow y'=-{y^3+8x\over 3xy^2} \Rightarrow y'(1,1)=-3 \\ \Rightarrow 切線方程式: y=-3(x-1)+1 \Rightarrow \bbox[red, 2pt]{3x+y=4} $$

解答:$$z= \sin^{-1}(\tan(xy)) \Rightarrow \sin z= \tan(xy) \Rightarrow \sec z={1\over \sqrt{1-\tan^2(xy)}}\\ 又\cases{\displaystyle {d\over dx} \sin^{-1}x= {1\over \sqrt{1-x^2}} \\ \displaystyle {d\over dx} \tan x= \sec^2 x} \Rightarrow \cases{\displaystyle {\partial z\over \partial x} ={1\over \sqrt{1-\tan^2(xy)}} \cdot \sec^2 (xy)\cdot y \\ \displaystyle {\partial z\over \partial y} ={1\over \sqrt{1-\tan^2(xy)}} \cdot \sec^2 (xy)\cdot x } \\ \Rightarrow {\partial z\over \partial x}+{\partial z\over \partial y} ={\sec^2(xy)\over \sqrt{1-\tan^2(xy)}}(x+y) =(x+y) \sec^2(xy) \sec(z) \; \bbox[red, 2pt]{故得證}$$

解答:

$$將底面為正方形的金字塔(四角錐)倒過來，尖頂位於原點，如上圖$$

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