111年中央地質所碩士班-微積分詳解

 國立中央大學111學年度碩士班考試入學

所別:應用地質研究所
科目:微積分

解答: $$f'(x)=12x^2+4x-3 =0 \Rightarrow x={-1 \pm \sqrt{10} \over 6} \Rightarrow \cases{f((-1+\sqrt{10})/6 = {299-20\sqrt{10}\over 54}\\ f((-1-\sqrt{10})/6 ={299+20\sqrt{10}\over 54}} \\ \Rightarrow 斜率為零坐標: \bbox[red, 2pt]{ \left({-1+\sqrt{10} \over 6} ,{299-20\sqrt{10} \over 54}\right), \left({-1-\sqrt{10} \over 6} ,{299+20\sqrt{10} \over 54}\right)}$$

解答: $$x^2+xy-y^2=5 \Rightarrow 2x+y+xy'-2yy'=0 \Rightarrow 2+y'+y'+xy''-2y'y'-2yy'' =0 \\ \Rightarrow \cases{y'={2x+y\over 2y-x} \\ y''={2+2y'-2(y')^2\over 2y-x}} \Rightarrow y'(3)={6-1\over -2-3}=-1 \Rightarrow y''(3)={2-2-2\over -2-3} ={2\over 5} \Rightarrow \bbox[red, 2pt]{ \cases{y'=-1\\ y''=2/5}}$$

解答: $$x^2+y^2+z^2=22 為一圓球,且球心位於原點(0,0,0)，因此切平面法向量\vec n=(2,-3,3)\\ \Rightarrow 切平面方程式: 2(x-2)-3(y+3)+3(z-3)=0 \Rightarrow \bbox[red, 2pt]{2x-3y+3z=22}$$

解答: $$\textbf{(1)}\; a(t)= 6t^3-4t^2+t \Rightarrow v(t)= \int a(t)\,dt ={3\over 2}t^4-{4\over 3}t^3+{1\over 2}t^2 \\ \qquad \Rightarrow s=\int_0^5 v(t)\,dt = \left. \left[ {3\over 10}t^5-{1\over 3}t^4+{1\over 6}t^3 \right] \right|_0^5 =750 \Rightarrow \bbox[red, 2pt]{在東方750公分處} \\ \textbf{(2)}\; e=\int_0^5 \left( {3\over 2}t^4-{4\over 3}t^3+{1\over 2}t^2 \right)^2\,dt =\left. \left[ {1\over 4}t^9-{1\over 2}t^8+{59\over 126}t^7 -{2\over 9}t^6+{ 1\over 20}t^5\right] \right|_0^5 \\ \qquad = \bbox[red, 2pt]{13701875 \over 42} $$

解答: $$V=\pi \int_a^b y^2\,dx = \pi \int_{-3}^3 (x^2+1)^2\,dx = 2\pi \int_{0}^3 (x^4+2x^2+1) \,dx =  2\pi \left.\left[ {1\over 5}x^5+ {2\over 3}x^3+ x \right] \right|_{0}^3 \\\qquad = \bbox[red, 2pt]{{696\over 5}\pi}$$

解答: $$\textbf{(1)}\; f(x)=(1+2x)(2x+x^2) =2x^3+5x^2+2x \\\qquad \Rightarrow f'(x)= \bbox[red, 2pt]{6x^2+10x+2} \\\textbf{(2)}\; f(x)=\cos^3(2x) \sin(2x) ={1\over 2}\cos^2(2x)\cdot 2\sin(2x) \cos(2x) ={1\over 2}\left( {1\over 2} (\cos 4x+1) \sin(4x)\right) \\\qquad ={1\over 4}\cos(4x)\sin(4x)+ {1\over 4}\sin(4x)  ={1\over 8}\sin(8x) +{1\over 4}\sin(4x)\\ \qquad \Rightarrow f'(x)= \bbox[red, 2pt]{\cos(8x)  +\cos(4x)}$$

解答: $$\textbf{(1)}\; \int{2x^4\over (1+2x)^3}\,dx = \int {1\over 4}x-{3\over 8}+{3/4\over 1+2x}+{-1/2\over (1+2x)^2} +{1/8\over (1+2x)^3}\,dx \\\qquad = \bbox[red, 2pt]{{1\over 8}x^2-{3\over 8}x+ {3\over 8}\ln |2x+1|+{1\over 4}\cdot {1\over 1+2x}-{1\over 32}\cdot {1\over (1+2x)^2}+c} \\\textbf{(2)}\; \cases{u=\cot^{-1}(2x)\\ dv=dx} \Rightarrow \cases{du=-{2\over 1+4x^2}dx \\ v=x} \Rightarrow \int \cot^{-1}(2x)\,dx =x\cot^{-1}(2x) +\int{2x\over 1+4x^2}\, dx \\\qquad =\bbox[red, 2pt]{x\cot^{-1}(2x)+ {1\over 4}\ln(1+4x^2) +c}$$

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解題僅供參考,其他歷年試題及詳解