111年成大環工碩士班-微積分詳解

 國立成功大學111學年度碩士班招生考試

系所:環境工程學系
科目:微積分

解答: $$\textbf{(1)}\; f(x)=5x(4^{-3x}) =5x e^{-6x\ln 2} \Rightarrow f'(x) =5e^{-6x\ln 2}-30\ln 2xe^{-3x\ln 4}\\ \quad=\bbox[red, 2pt]{5(4^{-3x}) -30\ln 2  (4^{-3x})x} \\\textbf{(2)}\; g(x)=\log_5 {4\over x^2\sqrt{1-x}} =\log_5 4-\log_5 x^2\sqrt{1-x} = 2\log_5 2-2\log_5x -{1\over 2}\log_5(1-x) \\\quad =2\log_5 2-{2\over \ln 5}\ln x -{1\over 2\ln 5}\ln(1-x) \Rightarrow g'(x)=\bbox[red, 2pt]{-{2\over x\ln 5}+{1\over 2(1-x)\ln 5}} \\ \textbf{(3)}\; h(t)= \sin(\arccos t) =\sin(\arcsin \sqrt{1-t^2}) =\sqrt{1-t^2} \Rightarrow h'(t)= \bbox[red, 2pt]{-{t\over \sqrt{1-t^2}}}$$

解答: $$1+\ln(xy)=e^{x-y} \Rightarrow {y+xy'\over xy} =(1-y')e^{x-y} \Rightarrow {y(1)+1\cdot y'(1)\over 1\cdot y(1)} =(1-y'(1))e^{1-y(1)} \\ \Rightarrow 1+y'(1)  =1-y'(1) \Rightarrow y'(1)=0 \Rightarrow \text{ tangent line:}\bbox[red,2pt]{y=1}$$

解答: $$V=圓錐體體積={1\over 3}\times A(底面圓面積) \times h(高)\\ 底面圓周長=R\theta = 2\pi r (r是底面圓半徑) \Rightarrow r={R\theta\over 2\pi} \\ 又R^2= r^2+h^2 \Rightarrow h=\sqrt{R^2-r^2} \Rightarrow V={1\over 3}r^2\pi \times \sqrt{R^2-r^2} \\\Rightarrow V(\theta)={\pi\over 3} \left({R\theta \over 2\pi}\right)^2\sqrt{R^2- \left( {R\theta \over 2\pi}\right)^2} ={R^2\theta^2 \over 12\pi} \sqrt{R^2-{R^2 \theta^2 \over 4\pi^2}} \\\Rightarrow V'(\theta)={ R^2 \theta\over 6\pi } \sqrt{R^2- {R^2 \theta^2 \over 4\pi^2}}-{R^4 \theta^3 \over 48 \pi^3} \cdot {1\over \sqrt{R^2-{R^2\theta^2 \over 4\pi^2}}} \\因此V'(\theta)=0 \Rightarrow R^2-{R^2\theta^2 \over 4\pi^2} ={R^2 \theta^2 \over 8\pi^2} \Rightarrow {3\theta^2 \over 8\pi^2}=1 \Rightarrow \bbox[red, 2pt]{\theta={2\sqrt 6\pi \over 3 }}$$

解答:

$$此題相當於求藍色面積繞y軸旋轉所得體積的兩倍\\ 假設\overline{OA}=a \Rightarrow r^2 = a^2 +({h\over 2})^2 \Rightarrow a^2=r^2-(h/2)^2\\ V_1=圓柱體體積 =a^2 h\pi= (r^2-{h^2\over 4})h\pi \\ V_2=\stackrel{\frown}{PB}繞y軸旋轉體積= \int_0^{h/2} (r^2-y^2) \pi\,dy =\left({r^2\over 2}h-{1\over 24} h^3\right) \pi\\ 欲求之體積=2V_2-V_1= \left(r^2h-{1\over 12}h^3\right)\pi -(r^2-{h^2\over 4})h\pi = \bbox[red, 2pt]{{1\over 6}h^3\pi}$$

解答: $${x^2\over a^2} +{y^2\over b^2}=1 \Rightarrow y^2=b^2-{b^2\over a^2}x^2 \Rightarrow y=\pm \sqrt{b^2-{b^2\over a^2}x^2} \\ \Rightarrow \text{area of the ellipse: }4\int_0^a \int_0^{ \sqrt{b^2-{b^2\over a^2}x^2}}1\,dydx =4\int_0^a \sqrt{b^2-{b^2\over a^2}x^2}\,dx =4b\int_0^a \sqrt{1-{1\over a^2}x^2}\,dx \\= 4b\int_{\pi/2}^0 \sqrt{1-\cos^2\theta}(-a\sin \theta)d\theta= -4ab \int_{\pi/2}^0 \sin^2\theta \,d\theta = -2ab \int_{\pi/2}^0 1-\cos(2\theta),d\theta \\=-2ab\times (-{\pi \over 2}) =ab\pi$$

解答: $$\textbf{(1)}\; y'={10x^2\over \sqrt{1+x^3}} \Rightarrow y=\int {10x^2\over \sqrt{1+x^3}}\,dx =\int {10\over \sqrt{u}}\cdot {1\over 3}du ={20\over 3}\sqrt u+c_1\\\quad \Rightarrow y=\bbox[red, 2pt]{{20\over 3}\sqrt{1+x^3} +c_1} \\\textbf{(2)}\; e^xy'+4e^x y=1 \Rightarrow y'+4y=e^{-x} \Rightarrow \text{ integration factor }I(x)=e^{\int 4\,dx} =e^{4x} \\\quad \Rightarrow e^{4x}y'+ 4e^{4x}y=e^{3x} \Rightarrow (e^{4x}y )'=e^{3x} \Rightarrow e^{4x}y= \int e^{3x}\,dx ={1\over 3}e^{3x}+c_1 \\\quad \Rightarrow \bbox[red, 2pt]{y={1\over 3}e^{-x}+ c_1e^{-4x}} \\\textbf{(3)}\; \cases{P(x,y)=x^2+y^2 \\ Q(x,y)=-2xy} \Rightarrow \cases{P_y=2y \\ Q_x=-2y} \Rightarrow P_x\ne Q_y \Rightarrow \text{Not Exact} \\\quad {P_y-Q_x\over Q} ={4y\over -2xy}=-{2\over x} \Rightarrow u'=-{2\over x}u \Rightarrow \text{integration factor }u(x)={1\over x^2} \\ \quad \Rightarrow \cases{uP=1+(y/x)^2\\ uQ=-2y/x} \Rightarrow (uP)_y= {2y\over x^2} =(uQ)_x\\ \quad \Rightarrow \text{potential function }\Phi(x,y)= \int uP\,dx =\int uQ\,dy \Rightarrow \int 1+{y^2\over x^2}\,dx = \int -{2y\over x}\,dy \\ \quad \Rightarrow x-{y^2\over x}+ \phi(y)= -{y^2\over x}+ \rho(x) \Rightarrow \Phi=x-{y^2\over x}+c_1=0 \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{x^2+c_1x}} \\\textbf{(4)}\;v(t)={1\over y} \Rightarrow v'=-{y'\over y^2} \Rightarrow y'=-y^2v' \Rightarrow 原式:y'=-y^2v'={3\over 20}y-{y^2\over 1600}\\\quad \Rightarrow v'=-{3\over 20y}+{1\over 1600} =-{3\over 20}v+{1\over 1600} \Rightarrow v'+{3\over 20}v={1\over 1600}\\\quad \Rightarrow e^{3t\over 20}v' +{3\over 20}e^{3t\over 20}v={1\over 1600}e^{3t\over 20}  \Rightarrow (e^{3t/20}v)'={1\over 1600}e^{3t\over 20} \\\quad \Rightarrow e^{3t/20}v={1\over 240}e^{3t\over 20}+c_1 \Rightarrow v={1\over y}=c_1e^{-3t\over 30} +{1\over 240} ={1+c_2e^{-3t/20}\over 240} \\\quad \Rightarrow y={240\over 1+c_2e^{-3t/20}} \Rightarrow y(0)=15={240\over 1+c_2} \Rightarrow c_2=15 \Rightarrow \bbox[red, 2pt] {y={240\over 1+ 15e^{-3t/20}}}$$

解答: $$\textbf{(1)}\; \int_0^\pi \tan{\theta\over 3}\,d\theta =\left. \left[ -3\ln(\cos(\theta /3)\right] \right|_0^\pi =-3\ln {1\over 2}+3\ln 1= \bbox[red, 2pt]{3\ln 2} \\\textbf{(2)}\; \int_{-2}^0 x^2 e^{x^3/2} \,dx = \left. \left[ {2\over 3} e^{x^3/2}\right] \right|_{-2}^0 = \bbox[red, 2pt]{{2\over 3}\left( 1-{1\over e^4}\right)} \\\textbf{(3)}\; x^2=5\tan^2 u \Rightarrow \cases{x^2+5=5(1+\tan^2 u)=5\sec^2 u \\ dx= \sqrt 5 \sec^2 u\,du}\\\quad  \Rightarrow \int {1\over (x^2+5)^{3/2}} \,dx = \int {\sqrt 5 \sec^2 u\over (5\sec^2 u)^{3/2}}\,du = \int {\sqrt 5 \sec^2 u\over 5\sqrt 5\sec^3 u }\,du =\int {1\over 5\sec u}\,du \\\quad = {1\over 5} \int \cos u\,du ={1\over 5}\sin u+c_1=\bbox[red, 2pt]{{x\over 5\sqrt{x^2+5}} +c_1} \\\textbf{(4)}\; \cases{u=\ln x\\ dv=x^5\,dx} \Rightarrow \cases{du=dx/x\\ v= x^6/6} \Rightarrow \int x^5\ln x\,dx = {1 \over 6}x^6\ln x-\int {1\over 6}x^5\,dx \\\quad = \bbox[red, 2pt]{{1\over 6}x^6 \ln x-{1\over 36}x^6+c_1} \\\textbf{(5)}\; \int_0^2 {dx\over x^2-2x+2} = \int_0^2 {dx\over (x-1)^2+1} = \left. \left[\tan^{-1}(x-1) \right] \right|_0^{2} =\tan^{-1} 1-\tan^{-1}(-1) \\\quad = {\pi\over 4}-(-{\pi \over 4})= \bbox[red, 2pt]{\pi \over 2}$$

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解題僅供參考,其他歷年試題及詳解