115年海洋大學運輸碩士班-微積分詳解

 國立臺灣海洋大學115學年度碩士班考試入學招生考試

考試科目： 微積分
學系組名稱： 運輸科學系碩士班不分組

解答:$$L = \left(1 + \frac{1}{x}\right)^x \Rightarrow \ln L = x\ln \left( 1+{1\over x} \right) \Rightarrow  \lim_{x\to\infty} \ln L = \lim_{x\to\infty} \frac{\ln \left(1 + \frac{1}{x}\right)}{\frac{1}{x}} \\= \lim_{x\to\infty} \frac{{d\over dx}\ln \left(1 + \frac{1}{x}\right)}{{d\over dx}\frac{1}{x}}   = \lim_{x\to\infty} \frac{1}{1 + \frac{1}{x}} =1 \Rightarrow \lim_{x\to \infty }L=e^1=\bbox[red, 2pt]e$$

解答:$$x^2=2x \Rightarrow x=0,2 \Rightarrow V= \int_0^2 2\pi x(2x-x^2)\,dx = 2\pi \left. \left[ {2\over 3}x^3 -{1\over 4}x^4\right] \right|_0^2 =\bbox[red, 2pt]{8\pi\over 3}$$

解答:

$$f(x)=0 \Rightarrow x^3=1 \Rightarrow x=1 \Rightarrow \begin{cases} f\le 0,& -2\le x\le 1\\ f\ge 0, & 1\le x\le 2\end{cases} \\ \Rightarrow \text{area }A  = \int_{-2}^{1} -(3x^3 - 3) \,dx + \int_{1}^{2} (3x^3 - 3) \,dx =  \left[ -\frac{3}{4}x^4 + 3x \right]_{-2}^{1}+  \left[ \frac{3}{4}x^4 - 3x \right]_{1}^{2} \\={81\over 4}+ {33\over 4} = \bbox[red, 2pt]{57\over 2}$$

解答:$$\cases{x=0 \\ y=\pi } \Rightarrow x^2\cos^2 y-\sin y= 0\cdot (-1)^2-0=0 \Rightarrow (0,\pi) \text{ lies on the curve }\bbox[red, 2pt]{QED.} \\ x^2\cos^2 y-\sin y= 0 \Rightarrow  \frac{d}{dx}(x^2 \cos^2 y) - \frac{d}{dx}(\sin y) = 0\\ \Rightarrow  2x \cos^2 y - 2x^2 \sin y \cos y \frac{dy}{dx} - \cos y \frac{dy}{dx} = 0\\ \Rightarrow 2(0) \cos^2(\pi) - 2(0)^2 \sin(\pi) \cos(\pi) \frac{dy}{dx} - \cos(\pi) \frac{dy} {dx} = 0 \Rightarrow  \frac{dy}{dx} = 0  \Rightarrow\bbox[red, 2pt]{ \cases{\text{tangent line: }y=\pi \\ \text{normal line: }x=0}}$$

解答:$$ f(x) = ax^3 + bx^2 + cx + d \Rightarrow  f'(x) = 3ax^2 + 2bx + c \\f\text{ has a local maximum at (0,0)} \Rightarrow \cases{f(0)=0 \\f'(0)=0} \Rightarrow \cases{d=0\\ c=0} \Rightarrow \cases{f(x)=ax^3+bx^2\\f'(x)=3ax^2+2bx} \\ f\text{ has a local minimum at (1,-1)} \Rightarrow  \cases{f(1)=-1\\ f'(1)=0} \Rightarrow \cases{a+b=-1 \\3a+2b=0} \Rightarrow  \bbox[red, 2pt]{\cases{a=2\\ b=-3 \\ c=0\\d=0}}$$

解答:$$  k(z) = \frac{1-z}{2z} \Rightarrow  k'(z) = \lim_{h \to 0} \frac{k(z+h) - k(z)}{h}   = \lim_{h \to 0} \frac{\frac{1-(z+h)}{2(z+h)} - \frac{1-z}{2z}}{h} \\ = \lim_{h \to 0} \frac{1}{h} \left( \frac{1-z-h}{2(z+h)} - \frac{1-z}{2z} \right) = \lim_{h \to 0} \frac{1}{h} \left( \frac{-h}{2z(z+h)} \right)  = \frac{-1}{2z(z+0)} =\bbox[red, 2pt]{-{1\over 2z^2}}$$

解答:$$$$

解答:$$f(x,y) =\log_y x= {\ln x\over \ln y} \Rightarrow \bbox[red, 2pt]{{\partial f\over \partial x} ={1\over x\ln y}}\\ \Rightarrow  {\partial f\over \partial y}=  \ln x \cdot \left( -\frac{1}{(\ln y)^2} \cdot \frac{1}{y} \right)\Rightarrow \bbox[red, 2pt]{{\partial f\over \partial y} =-{\ln x\over y(\ln y)^2} }$$

解答:$$\textbf{(a) }u=x^2+6 \Rightarrow du =2x\,dx \Rightarrow I=  \int x^3 \sqrt{x^2 + 6} \,dx = \int x^2 \sqrt{x^2 + 6} \cdot x \,dx =  \int (u - 6) \sqrt{u} \cdot \frac{1}{2} \,du \\=  \frac{1}{2} \int (u^{\frac{3}{2}} - 6u^{\frac{1}{2}}) \,du =  \frac{1}{2} \left( \frac{2}{5} u^{\frac{5}{2}} - 6 \cdot \frac{2}{3} u^{\frac{3}{2}} \right) + C = \frac{1}{5} u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + C  \\= \bbox[red, 2pt]{\frac{1}{5}(x^2 + 6)^{\frac{5}{2}} - 2(x^2 + 6)^{\frac{3}{2}} + C } \\ \textbf{(b) } \cases{u=xe^x \\dv=dx/(x+1)^2} \Rightarrow \cases{du =e^x+xe^x\\ v=-1/(x+1)} \\\Rightarrow  \int \frac{xe^x}{(x+1)^2} \,dx = (xe^x)\left(-\frac{1}{x+1}\right) - \int \left(-\frac{1}{x+1}\right) e^x(x+1) \,dx  = -\frac{xe^x}{x+1} + \int e^x \,dx \\ = \frac{-xe^x + e^x(x+1)}{x+1} + C = \bbox[red, 2pt]{{e^x\over x+1}+C}$$

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