112年清華科技管理碩士班-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: 科技管理研究所(乙組)
考試科目: 微積分

解答: $$\textbf{(a)}\; \lim_{x \to 0} \frac{(\sin(a+2x)-2\sin(a+x)+ \sin a}{x^2} =\lim_{x \to 0} \frac{\sin(a+2x)-2\sin(a+x)+ \sin a)'}{(x^2)'} \\=\lim_{x \to 0} \frac{2\cos(a+2x)-2\cos(a+x) }{2x} = \lim_{x \to 0} \frac{ (\cos(a+2x)-\cos(a+x))' }{(x)'} \\=\lim_{x \to 0} (-2\sin(a+2x)+\sin(a+x)) = \bbox[red, 2pt]{-\sin a}\\ \textbf{(b)}\; L=\left( 1+{2\over n}+ {3\over n^2}\right)^n =\left(  {n^2+2n+3\over n^2}\right)^n \Rightarrow \ln L=n\ln\left(  {n^2+2n+3\over n^2}\right)\\\qquad \Rightarrow \lim_{n\to \infty} \ln L= \lim_{n\to \infty}\frac{\ln\left(  {n^2+2n+3\over n^2}\right)}{1\over n} =\lim_{n\to \infty} \frac{\left(\ln\left(  {n^2+2n+3\over n^2}\right) \right)'}{({1\over n})'} =\lim_{n\to \infty} \frac{{n^2\over n^2+2n+3} \cdot {-2n-6\over n^3}}{-1\over n^2} \\\qquad =\lim_{n\to \infty}= \frac{n(2n+6)}{n^2+2n+3} =2 \Rightarrow \lim_{n\to \infty} L=\bbox[red, 2pt]{e^2} \\\textbf{(c)}\; \int_1^{10} \frac{\log_{10} x}{x}\,dx ={1\over \ln 10}\int_1^{10} \frac{\ln x}{x}\,dx = {1\over \ln 10} \left. \left[{1\over 2}(\ln x)^2 \right] \right|_1^{10} ={1\over \ln 10}\cdot {1\over 2}(\ln 10)^2 =\bbox[red, 2pt]{{1\over 2}\ln 10} \\\textbf{(d)}\; \int_0^{\pi/2} \int_y^{\pi/2} {\sin x\over x}\,dxdy =\int_0^{\pi/2} \int_0^x {\sin x\over x}\,dydx =\int_0^{\pi/2} \sin x\,dx =\bbox[red, 2pt]1 \\ \textbf{(e)}\; \cases{f(x,y)=x^{3/4}y^{1/4} \\ g(x,y)=x+y-100} \Rightarrow \cases{f_x=\lambda g_x \\f_y= \lambda g_y\\ g=0} \Rightarrow \cases{{3\over 4}x^{-1/4}y^{1/4}= \lambda\\ {1\over 4}x^{3/4}y^{-3/4} =\lambda} \Rightarrow {3y\over x}=1 \Rightarrow x=3y \\ \qquad \Rightarrow 3y+y=100 \Rightarrow y=25\Rightarrow x=75 \Rightarrow \max(f) =f(75,25) = \bbox[red, 2pt]{75^{3/4}25^{1/4}} \\\textbf{(f)}\; y'+2y=e^{-x} \Rightarrow e^{2x}y'+2e^{2x} y=e^x \Rightarrow (e^{2x}y)'=e^x \Rightarrow e^{2x} y=e^x+C \\\qquad \Rightarrow  y=e^{-x}+Ce^{-2x} \Rightarrow y(0)=1+C=0 \Rightarrow C=-1 \Rightarrow \bbox[red, 2pt]{y=e^{-x}-e^{-2x}}  \\\textbf{(g)}\ \text{The circumference of the circle center }A(2,0)\text{ around the y-axis is equal to }4\pi, \\\qquad \text{and the area of circle is: }\pi  , \text{then the volume of the torus is: }4\pi \times \pi=\bbox[red, 2pt]{4\pi^2}$$

解答: $$f(x)={x+\cos x \over 1+\sin x} \Rightarrow f'(x)={-x\cos x\over (1+\sin x)^2} =0 \Rightarrow x=0,\pi/2 , \text{ for }x\in [0,\pi] \\ \Rightarrow \cases{f(0)=1\\ f(\pi/2)= \pi/4\\ f(\pi)=\pi-1} \Rightarrow \bbox[red, 2pt]{\cases{\text{abs max: } \pi-1\\ \text{abs min: }\pi/4}}$$

解答: $$\frac{\text{d}S}{\text{d}t} =k(D-S) \Rightarrow S'+kS=kD \Rightarrow e^{kt} S'+k e^{kt}S = kDe^{kt} \Rightarrow \left(e^{kt}S\right)'=kDe^{kt} \\ \Rightarrow e^{kt}S=\int kDe^{kt}\,dt = De^{kt}+C \Rightarrow S=D+Ce^{-kt} \\ \Rightarrow S(0)=D+C=S_0 \Rightarrow C=S_0-D \Rightarrow \bbox[red, 2pt]{S=D+(S_0-D)e^{-kt}}$$

解答: $$f(x,y)=x^3-3x^2+y^2 \Rightarrow \cases{f_x=3x^2-6x\\ f_y=2y} \Rightarrow \cases{f_{xx}=6x-6\\ f_{xy}=0\\ f_{yy}=2}\\ \Rightarrow d(x,y)=f_{xx} f_{yy}-(f_{xy})^2 =12x-12\\ \cases{f_x=0 \\f_y=0} \Rightarrow \cases{3x(x-2)=0 \Rightarrow x=0,2\\ y=0} \Rightarrow \cases{d(0,0)=0\\ d(2,0)=12 \gt 0} \Rightarrow f_{xx}(2,0)=6\gt 0\\ \text{Additionally, } \cases{f(x,0)=x^3-3x^2 \text{ has a local maximum at }(0,0)\\ f(0,y)=y^2 \text{has a local minimum at }(0,0)} \\\Rightarrow (0,0)\text{ is a saddle point of }f\\ \text{In summary} \bbox[red, 2pt]{\cases{ f \text{ has a local minimum at }(2,0) \\ f\text{ has a saddle point at }(0,0)}}$$

解答: $$\textbf{(a)}\; \frac{\partial f}{\partial x}(0,0)= \lim_{h \to 0} \cfrac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0} {0-0\over h} =0 \Rightarrow \bbox[red,2pt]{ \frac{\partial f}{\partial x}(0,0) \text{ exists}} \\\frac{\partial f}{\partial y}(0,0) =\lim_{h\to 0} {f(0,0+h)-f(0,0)\over h} =\lim_{h\to 0} {h\sin{1\over h^2}-0 \over h} =\lim_{h\to 0}\sin{1\over h^2} \text{ Not exists} \\ \qquad \Rightarrow \frac{\partial f}{\partial y}(0,0) \bbox[red, 2pt]{\text{ does NOT exists}}\\ \textbf{(b)}\; -|y| \le \left| y\sin {1\over x^2+y^2}\right| \le |y| \Rightarrow \lim_{(x,y)\to (0,0)} -|y| \le \lim_{(x,y)\to (0,0)} \left| y\sin {1\over x^2+y^2}\right| \le \lim_{(x,y) \to (0,0)}|y| \\ \quad \Rightarrow 0\le \lim_{(x,y)\to (0,0)} \left| y\sin {1\over x^2+y^2}\right|\le 0 \Rightarrow \lim_{(x,y)\to (0,0)} \left| y\sin {1\over x^2+y^2}\right|=0 \Rightarrow f \text{  is continuous at} (0,0) \\\quad \Rightarrow \bbox[red, 2pt]{\text{Yes}}$$

 

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