112年清華生命科學碩士班-微積分詳解

國立清華大學112學年度碩士班考試入學

系所班組別: 生命科學暨醫學院(丙組)
考試科目: 微積分

解答: $$\textbf{(A)}\; \cases{f(x)=\ln x\\ g(x)=\ln |x|} \Rightarrow f'(x)={1\over x}=g'(x)\\\qquad 也就是說f'(x)=g'(x)不代表f(x)=g(x)+C \Rightarrow \bbox[red, 2pt]{不正確}\\\textbf{(B)} \; g(x)=x^3 \Rightarrow g^{-1}(x)=x^{1/3} \Rightarrow {d\over dx}g^{-1}(x)={1\over 3x^{2/3}} \Rightarrow {d\over dx}g^{-1}(0)不存在\\ \qquad\Rightarrow g^{-1}(x)在x＝0不可微 \Rightarrow \bbox[red, 2pt]{不正確}$$

解答: $$\lim_{x\to 0}{\tan^{-1} x\over x} =\lim_{x\to 0}{(\tan^{-1}x)' \over (x)'} = \lim_{x\to 0}{1\over 1+x^2} = \bbox[red, 2pt]1$$

解答: $$\textbf{(A)}\; xy+y^3=1 \Rightarrow y^3(0)=1 \Rightarrow y(0)=1\\\quad (xy+y^3)'=(1)' \Rightarrow y+xy'+3y^2y'=0 \Rightarrow (x+3y^2)y'=-y \Rightarrow y'=-{y\over x+3y^2} \\\quad \Rightarrow y'(0)=-{1\over 3y(0)} =\bbox[red, 2pt]{-{1\over 3}} \\\textbf{(B)}\; y-1=y'(0)(x-0) \Rightarrow \bbox[red, 2pt]{x+3y=3}$$

解答:

$$等腰\triangle ABC, 其中\cases{\overline{AB}= \overline{AC} \\ \angle ABC=\angle ACB= \theta} \Rightarrow \angle BAC=\pi-2\theta\\ 假設\overline{OP}\bot \overline{AC} \Rightarrow \overline{AP} = \overline{OA}\cos \angle OAP= r\cos({\pi\over 2}-\theta) = r\sin \theta \Rightarrow \overline{AC}=2r\sin \theta \\ \Rightarrow \triangle ABC面積=f(\theta)={1\over 2} \overline{AB}\overline{AC} \sin \angle BAC= 2r^2 \sin^2\theta \sin(\pi-2\theta) =2r^2\sin^2 \theta \sin(2\theta) \\ \Rightarrow f'(\theta)=4r^2\sin \theta \cos\theta \sin(2\theta)+4r^2\sin^2 \theta \cos(2\theta)  =4r^2\sin\theta (\cos\theta \sin(2\theta)+\sin \theta \cos(2\theta) \\=4r^2\sin \theta \sin(3\theta)= 0 \Rightarrow 3\theta=\pi \Rightarrow \theta=\pi/3 \Rightarrow f(\pi/3)= \bbox[red, 2pt]{{3\over 4}\sqrt 3 r^2}$$

解答: $$I=\int {1\over x^4+4}\,dx =\int \left({1/4-x/8\over x^2-2x+2}+{1/4+x/8 \over x^2+2x+2} \right)\,dx = I_1+ I_2,\\\qquad \text{where }I_1= {1\over 8}\int{2-x\over x^2-2x+2} \,dx ,I_2={1\over 8}\int {2+x \over x^2+2x+2}\,dx \\ I_1= {1\over 8}\int \left({1-x\over x^2-2x+2} +{1\over x^2-2x+2} \right)\,dx ={1\over 8} \left(-{1\over 2}\ln(x^2-2x+2) +\tan^{-1}(x-1) \right)+c_1 \\\text{By the same way, }I_2={1\over 8} \left({1\over 2}\ln(x^2+2x+2) +\tan^{-1}(x+1) \right)+c_2 \\ \Rightarrow I=I_1+ I_2 = {1\over 16} (\ln(x^2+2x+2)- \ln(x^2-2x+2)) +{1\over 8}(\tan^{-1}(x-1)+ \tan^{-1}(x+1))+c_3 \\= \bbox[red, 2pt]{{1\over 16} \ln{x^2+2x+2\over x^2-2x+2} +{1\over 4} \tan^{-1}{x\over 2-x^2}+C}$$

解答: $$I=\int  \sqrt{1+(y')^2}\,dx =\int \sqrt{1+\left({2\over 3}x^{-1/3}\right)^2}\,dx =\int  \sqrt{1+{4\over 9}x^{-2/3}}\,dx\\ \text{Let }u=x^{-1/3}, \text{ then }du =-{1\over 3}x^{-4/3}dx =-{1\over 3}u^4dx \Rightarrow dx= -3{du\over u^4} \Rightarrow I=-3\int  {\sqrt{1+{4\over 9}u^2}\over u^4} du\\ \text{Let }u={3\over 2}\sinh v \Rightarrow du ={3\over 2}\cosh v\,dv \Rightarrow I= -3\int  {\cosh v \over {81 \over 16}\sinh^4 v} \cdot {3\over 2}\cosh v\,dv \\ \quad = -{8\over 9}\int  {\cosh^2 v\over \sinh^4 v}\,dv = -{8\over 9}\int {1\over \tanh^4 v \cosh^2 v}\,dv \\ \text{Let }w=\tanh v \Rightarrow dw =\sec^2 vdv \Rightarrow I=-{8\over 9}\int {1\over w^4}dw = {8\over 27}w^{-3}+C ={x\over 27}(9+4x^{-2/3})^{3/2}+C \\ \Rightarrow \int_1^8 \sqrt{1+(y')^2}\,dx = {8\over 27}10^{3/2} -{1\over 27}13^{3/2} = \bbox[red, 2pt]{{80\over 27}\sqrt{10}-{13\over 27}\sqrt{13}}$$

解答: $$\frac{d y}{dx}=x^2y\cos x \Rightarrow {1\over y}dy =x^2 \cos x\,dx \Rightarrow \ln y=(x^2-2) \sin x+2x\cos x+c_1 \\\Rightarrow y=c_2e^{(x^2-2) \sin x+2x\cos x} \Rightarrow y(0)=c_2=2 \Rightarrow c_2=2 \Rightarrow \bbox[red, 2pt]{y=2e^{(x^2-2) \sin x+2x\cos x}}$$

解答: $$\textbf{(A)}\; \cos \theta ={\vec a \cdot \vec b\over |\vec a||\vec b|} ={\sqrt 3\over 2\cdot 1} \Rightarrow \theta =\bbox[red, 2pt]{\pi\over 6} \\\textbf{(B)} \;\vec a\times (\vec c\times \vec d) =(1,0,\sqrt 3)\times ((-2,3,4)\times (3,-1,1)) =(1,0,\sqrt 3)\times (7,14,-7) \\ \qquad = \bbox[red, 2pt]{(-14\sqrt 3,7+7\sqrt 3,14)}$$

解答: $$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_\Omega xy^2 \sqrt{x^2+y^2} \,dA  =\int_{\pi/4}^{\pi/2} \int_0^\sqrt 3 r^3\cos\theta \sin^2\theta\cdot r\cdot rdrd\theta \\=\int_{\pi/4}^{\pi/2} \int_0^\sqrt 3 r^5\cos\theta \sin^2\theta drd\theta =\int_{ \pi/4}^{\pi/2} {9\over 2} \cos\theta \sin^2\theta\,d\theta = \left. \left[ {3\over 2} \sin^3 \theta \right] \right|_{\pi/4}^{\pi/2} = \bbox[red, 2pt]{{3\over 2}(1-{\sqrt 2\over 4})}$$

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