113年中山應數系碩士班-微積分詳解

國立中山大學113學年度碩士班招生考試

科目名稱: 微積分【應用數學系】

解答:$$\textbf{(a)}\;\lim_{x \to 1}\frac{x^3-1}{\sqrt x-1} =\lim_{x \to 1}\frac{(x^3-1)'}{(\sqrt x-1)'} = \lim_{x \to 1}\frac{3x^2}{{1\over 2\sqrt x}} = \lim_{x \to 1} 6x^{5/2} =\bbox[red, 2pt] 6 \\ \textbf{(b)}\; \lim_{n\to \infty} \left[{4\over \sqrt{2n^2-1^2}}+  {4\over \sqrt{2n^2-2^2}}+ \cdots +{4\over \sqrt{2n^2-n^2}} \right] \\\qquad = \lim_{n\to \infty} 
 \sum_{k=1}^n {4\over \sqrt{2n^2-k^2}} = \lim_{n\to \infty} \sum_{k=1}^n {4\over n\sqrt{2-(k/n)^2}} = I=\int_0^1 {4\over \sqrt{2-x^2}}\,dx \\\qquad x=\sqrt 2 \cos \theta \Rightarrow dx =-\sqrt 2\sin \theta d\theta \Rightarrow I=\int_{\pi/2}^{\pi/4} {-4\sqrt 2\sin \theta  \over \sqrt 2\sin \theta}\,d\theta =4\cdot {\pi \over 4} =\bbox[red, 2pt]\pi$$

解答:$$f(x,y,z)=\sqrt{xy^3z} +\cos(xy) + z\ln x =x^{1/2}y^{3/2}z^{1/2}+ \cos(xy)+z\ln x\\\Rightarrow \bbox[red, 2pt]{f_x={1\over 2}x^{-1/2}y^{3/2}z^{1/2} -y\sin(xy)+{z\over x}} \\ \Rightarrow f_{xy}=\frac{\partial }{\partial y} f_x \Rightarrow \bbox[red, 2pt]{f_{xy}= {3\over 4}x^{-1/2} y^{1/2}z^{1/2} -\sin(xy) -xy\cos(xy)} \\ \Rightarrow f_{xyz} =\frac{\partial }{\partial z} f_{xy} \Rightarrow \bbox[red, 2pt]{f_{xyz} ={3\over 8} x^{-1/2}y^{1/2}z^{-1/2}}$$

解答:$$\textbf{(a)} \;y=x-3 \Rightarrow dy=dx \Rightarrow I=\int_3^\infty (x-3)^2e^{-x} \,dx = \int_0^\infty y^2e^{-(y+3)}\,dy \\ \quad \cases{u=y^2\\ dv =e^{-(y+3)}dy} \Rightarrow \cases{du =2ydy \\ v= -e^{-(y+3)}} \Rightarrow I= \left. \left[-y^2e^{-(y+3)} \right] \right|_0^\infty +2 \int_0^\infty ye^{-(y+3)}\,dy \\=0+2\int_0^\infty ye^{-(y+3)}\,dy =2 \left. \left[ -ye^{-(y+3) }-e^{-(y+3)} \right] \right|_0^\infty = \bbox[red, 2pt]{2e^{-3}} \\\textbf{(b)}\; \cos x={1-\tan^2(x/2) \over 1+\tan^2(x/2)} \Rightarrow I=\int_0^\pi {1\over 3-\cos(x)}\,dx = \int_0^\pi {1\over 3-{1-\tan^2(x/2) \over 1+\tan^2(x/2)}}\,dx \\ \quad u=\tan(x/2) \Rightarrow x= 2\tan^{-1} u \Rightarrow dx= {2\over 1+u^2}\,du \\\quad \Rightarrow I=\int_0^\infty {1\over 3-{1-u^2\over 1+u^2}} \cdot {2\over 1+u^2} \,du = \int_0^\infty {1\over 1+2u^2}\,du\\ \quad u={v\over \sqrt 2} \Rightarrow du={\sqrt 2 \over 2}dv \Rightarrow I= {\sqrt 2\over 2}\int_0^\infty {1\over 1+v^2}\,dv = {\sqrt 2\over 2}\left. \left[ \tan^{-1} v\right] \right|_0^\infty ={\sqrt 2\over 2}\cdot {\pi\over 2} =\bbox[red, 2pt]{\sqrt 2 \pi \over 4}$$

解答:$$\sqrt{1+x} =(1+x)^{1/2} =\sum_{k=0}^\infty {1/2 \choose k}  x^k = \bbox[red, 2pt]{\sum_{k=0}^\infty {2k\choose k} {(-1)^{k+1}\over 2^{2k}(2k-1)}x^k }$$

解答:$$y'+3y=e^{2x} \Rightarrow 積分因子I(x)=e^{\int 3\,dx} =e^{3x} \Rightarrow I(x)y'+3yI(x)=e^{2x}I(x) \\ \Rightarrow e^{3x}y'+ 3e^{3x} y=e^{5x} \Rightarrow \left( e^{3x}y \right)'=e^{5x} \Rightarrow e^{3x} y={1\over 5}e^{5x}+c_1 \Rightarrow y={1\over 5}e^{2x}+ c_1e^{-3x} \\ \text{Initial value }y(0)=1 \Rightarrow 1={1\over 5}+c_1 \Rightarrow c_1={4\over 5} \Rightarrow \bbox[red, 2pt]{y={1\over 5}e^{2x}+ {4\over 5}e^{-3x}}$$

解答:$$f(x(r,\theta),y(r,\theta)),\text{ where }\cases{x(r,\theta) =r\cos \theta\\ y(r,\theta )= r\sin \theta} \Rightarrow \cases{f_r=f_xx_r+ f_yy_r =f_x\cos \theta+f_y\sin \theta \\ f_\theta=f_xx_\theta+ f_yy_\theta =-f_x r\sin \theta+ f_y r\cos \theta} \\ \Rightarrow f_{rr} = (f_x\cos \theta+f_y\sin \theta)_r = (f_{xx}x_r+ f_{xy} y_r) \cos \theta+ (f_{yx}x_r+ f_{yy}y_r) \sin \theta \\\qquad = (f_{xx}\cos \theta+ f_{xy}\sin \theta) \cos \theta+ (f_{yx}\cos \theta+ f_{yy}\sin \theta) \sin \theta\\\qquad = f_{xx} \cos^2\theta +2f_{xy}\cos \theta \sin \theta+f_{yy} \sin^2 \theta\\ \Rightarrow f_{\theta\theta}= (-f_x r\sin \theta+ f_y r\cos \theta)_\theta \\\qquad =(-f_{xx}x_\theta -f_{xy}y_\theta)r\sin \theta-f_xr\cos \theta+ (f_{yx}x_\theta+ f_{yy}y_\theta) r\cos \theta -f_yr \sin \theta \\\qquad =(f_{xx} r\sin \theta-f_{xy} r\cos \theta) r\sin \theta-f_xr\cos \theta+ (-f_{yx} r\sin \theta+ f_{yy} r\cos \theta) r\cos \theta-f_y r\sin \theta \\\qquad = f_{xx}r^2\sin^2 \theta-2f_{xy}r^2 \cos \theta\sin \theta+ f_{yy}r^2\cos^2 \theta-f_xr\cos \theta-f_y r\sin \theta \\ \Rightarrow f_{rr}+{1\over r^2}f_{\theta\theta }=f_{xx}(\cos^2\theta+ \sin^2 \theta)+ f_{yy}(\sin^2\theta+ \cos^2\theta)-{1\over r}(f_x \cos \theta+f_y \sin \theta)\\ \qquad = f_{xx}+f_{yy}+{1\over r}f_r \\ \Rightarrow f_{rr}+{1\over r^2}f_{\theta\theta}- {1\over r} f_r=f_{xx}+f_{yy}= 0 \Rightarrow f_{rr}+{1\over r^2}f_{\theta\theta}- {1\over r} f_r=0. \bbox[red, 2pt]{QED}$$

解答:$$\text{Given }\frac{\text{d}V}{\text{d}t} =120,  V={4\over 3}\pi r^3 \Rightarrow \frac{\text{d}V}{\text{d}t}= \frac{\text{d}V}{\text{d}r} \frac{\text{d}r}{\text{d}t} =4\pi r^2 \frac{\text{d}r}{\text{d}t} \\ \Rightarrow 120 =4\pi \cdot 30^2\cdot  \frac{\text{d}r}{\text{d}t} \Rightarrow \frac{\text{d}r}{\text{d}t}= \bbox[red, 2pt]{{1\over 30\pi} \text{cm/s}}$$
 

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