115年高師大電機轉學考-微積分詳解

 國立高雄師範大學115 學年度暑假轉學生招生考試

系所別： 電機工程學系二年級
科 目： 微積分（全二頁）

一、選擇題 (【不需要】列出計算過程，只需要回答選項) (每題 4 分)

解答:$$\lim_{x \to 2} \left( \frac{1}{x-2} - \frac{1}{x^2-3x+2} \right) = \lim_{x \to 2} \left( \frac{x-1}{(x-2)(x-1)} - \frac{1}{(x-2)(x-1)} \right) \\= \lim_{x \to 2} {x-2\over (x-2)(x-1)} = \lim_{x \to 2} {1\over  x-1}=1，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x) = \frac{x^2+3x-1}{2x+1} \Rightarrow f'(x)={2x+3\over 2x+1}-{(x^2+3x-1)\cdot 2 \over (2x+1)^2} \Rightarrow f'(2)={7\over 5}-{18\over 25}={17\over 25}，故選\bbox[red, 2pt]{(A)}$$

解答:$$g(\theta) = e^\theta(\tan\theta - \theta) \Rightarrow g'(\theta)= e^\theta(\tan\theta - \theta)+e^\theta( \sec^2\theta - 1)\\ \Rightarrow g'(\pi/4)= e^{\pi/4}(1-{\pi\over 4})+ e^{\pi/4}(2-1) =e^{\pi/4} \left( 2-{\pi\over 4} \right) =k e^{\pi/4} \Rightarrow k=2-{\pi\over 4}，故選\bbox[red, 2pt]{(B)}$$

解答:$$f(x)= \log_{10} (a+\sin x) = {\ln (a+\sin x) \over \ln 10} \Rightarrow f'(x)={\cos x \over (a+\sin x) \ln 10} \Rightarrow f'(0)={1\over a\ln 10}={1\over 3\ln 10} \\ \Rightarrow a=3，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)=(x-2)^{2/3}(7-x) \Rightarrow f'(x)={2\over 3}(x-2)^{-1/3}(7-x)-(x-2)^{2/3}= {20-5x\over 3(x-2)^{1/3}} \\ \cases{20-5x=0\\ x-2=0} \Rightarrow \cases{x=4\\ x=2} \Rightarrow 4+2=6，故選\bbox[red, 2pt]{(C)}$$

解答:$$ \lim_{x \to \infty} \frac{\ln x}{x^k} = \lim_{x \to \infty} \frac{\frac{1}{x}}{k x^{k-1}} = \lim_{x \to \infty} \frac{1}{k x^k} = 0 \Rightarrow k\gt 0，故選\bbox[red, 2pt]{(C)}$$

解答:

$$\cases{A(x,0)\\ B(-x,0)\\ C(-x,12-x^2)\\D(x,12-x^2)} \Rightarrow \text{the area of the rectangle }ABCD= \overline{AB}\times \overline{AD} =g(x)=2x\cdot(12-x^2) \\ \Rightarrow g'(x)=2(12-x^2)+2x(-2x)=24-6x^2=0 \Rightarrow x=2 \Rightarrow g(2)=4\cdot 8=32，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)=x^2-4 \Rightarrow f'(x)=2x\\ \text{Newton's Method: }x_{n+1}=x_n-{f(x_n) \over f'(x_n)} \Rightarrow x_1= 2.5=x_0-{f(x_0)\over f'(x_0)} =x_0-{x_0^2-4\over 2x_0} \\ \Rightarrow x_0^2-5x_0+4=0 \Rightarrow (x_0-4)(x_0-1)=0 \Rightarrow x_0=4,1，故選\bbox[red, 2pt]{(B)}$$

解答:$$u=(x-1)^2 \Rightarrow du =2(x-1)\,dx \Rightarrow \int (x-1)e^{(x-1)^2} \,dx = \int {1\over 2}e^u\,du ={1\over 2}e^{(x-1)^2}+C \\ F(2)-F(1)=  \int_0^2 (x - 1)e^{(x-1)^2} dx - \int_0^1 (x - 1)e^{(x-1)^2} dx = \int_1^2 (x - 1)e^{(x-1)^2} dx = \left. \left[ {1\over 2}e^{(x-1)^2} \right] \right|_1^2\\= {1\over 2}\left( e-1 \right)，故選\bbox[red, 2pt]{(A)}$$

解答:$$A=\int_a^b v(t)\,dt \text{ may be positive or negative} \Rightarrow A= \text{ displacement} \\ B= \int_a^b |v(t)|\,dt \ge 0 \Rightarrow B= \text{total distance traveled}\\，故選\bbox[red, 2pt]{(B)}$$

二、計算題

解答:$$\textbf{(1) } x^2 + xy + y^2 = 7 \Rightarrow  \frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(7) \Rightarrow 2x+y+x{dy\over dx}+2y{dy\over dx}=0 \\\qquad  \Rightarrow (x+2y){dy\over dx}=-(2x+y) \Rightarrow {dy\over dx}=\bbox[red, 2pt]{-{2x+y\over x+2y}} \\ \textbf{(2) } \text{ slope }m= {dy\over dx}(1,2)=-{2+2\over 1+4}=-{4\over 5} \Rightarrow \text{tangent line: }y=-{4\over 5}(x-1)+2 \\\qquad \Rightarrow 5y=-4x+4+10 \Rightarrow 4x+5y=14 \Rightarrow \cases{a=4\\ b=5\\ c=14} \Rightarrow a+b+c= \bbox[red, 2pt]{23}$$

解答:$$x=t \Rightarrow \cases{x^2=2y \Rightarrow y=t^2/2\\ 3z=xy \Rightarrow z=xy/3=t^3/6} \Rightarrow C: r(t)= \left( t,{t^2\over 2},{t^3\over 6} \right)  \Rightarrow r'(t)= \left( 1,t,{t^2\over 2} \right)\\ \Rightarrow \vert r'(t) \vert=  \sqrt{1 + t^2 + \frac{t^4}{4}} =1+{t^2\over 2}\\ (0,0,0) \to (6,18,36) \Rightarrow r(0)\to r(6) \Rightarrow L= \int_0^6 |r'(t)|\,dt =  \int_{0}^{6} \left(1 + \frac{t^2}{2}\right) dt   = \left[ t + \frac{t^3}{6} \right]_{0}^{6} =\bbox[red, 2pt]{42}$$

解答:$$ \int_0^{1/2} \int_{\sqrt{3}y}^{\sqrt{1-y^2}} xy^2 \,dx\,dy =  \int_0^{1/2}  \left[ \frac{1}{2}x^2 y^2 \right]_{x=\sqrt{3}y}^{x=\sqrt{1-y^2}}   \,dy =\int_0^{1/2} \frac{1}{2} (y^2 - 4y^4) \,dy  = \frac{1}{2} \left[ \frac{y^3}{3} - \frac{4y^5}{5} \right]_0^{1/2} \\ = \frac{1}{2} \left( \frac{1}{24} - \frac{1/8}{5} \right) =\bbox[red, 2pt]{1\over 120}$$

解答:$$\cases{A(0,0) \\B(1,1) \\C(2,0) \\D(1,-1)} \Rightarrow \cases{\overleftrightarrow{AB}: y-x=0\\ \overleftrightarrow{BC}: y+x=2\\ \overleftrightarrow{CD}: y-x=-2\\ \overleftrightarrow{DA}: y+x=0} \Rightarrow \cases{u=x+y\\ v=x-y} \Rightarrow \cases{0\le  u\le 2\\ 0\le v\le 2} \Rightarrow \cases{x=(u+v)/2\\ y=(u-v)/2} \\ \Rightarrow  J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}   = -\frac{1}{2} \Rightarrow dA= |J|du\,dv ={1\over 2}du\,dv \\ \Rightarrow \iint_R xy \,dA = \iint_S \left( \frac{u^2 - v^2}{4} \right) \left( \frac{1}{2} \right) \,du\,dv = \frac{1}{8} \int_0^2 \int_0^2 (u^2 - v^2) \,du\,dv \\=\frac{1}{8} \int_0^2 \left( \frac{8}{3} - 2v^2 \right) \,dv  = \frac{1}{8} \left[ \frac{8}{3}v - \frac{2v^3}{3} \right]_0^2 =\bbox[red, 2pt]0$$

解答:$$S: x^2+y^2+ z^2=1 \Rightarrow \vec n=(x,y,z) \Rightarrow \vec F\cdot \vec n=\vec F\cdot (x,y,z) = 2x+2y+z^2 \\ \Rightarrow \vec F=(2,2,z) \Rightarrow \text{div }\vec F=0+0+1=1 \Rightarrow  \iint_S \vec{F} \cdot \vec{n} \, dS = \iiint_E \text{div} \vec{F} \, dV = \iiint_E 1\,dV= \bbox[red, 2pt]{{4\over 3}\pi}$$

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解題僅供參考，其他轉學考試題及詳解