115台聯大轉學考A2-微積分詳解

台灣聯合大學系統 115 學年度學士班轉學生考試

科目：微積分 類組別： A2

解答:$$\begin{array}{l} & X & Y & X^2 &   XY\\\hline & 0& 60& 0 & 0\\ & 2& 74& 4& 148\\ &4& 90& 16& 360\\ & 6&106 & 36& 636\\\hline \sum& 12& 330& 56& 1144\end{array} \Rightarrow m= {n \sum XY-(\sum X)(\sum Y)\over n(\sum X^2)-(\sum X)^2} = {4\cdot 1144-12\cdot 330\over 4\cdot 56-12^2}={77\over 10} \\ \Rightarrow b={\sum Y-m(\sum X) \over n}={330-{77\over 10}\cdot 12\over 4}={297\over 5} \Rightarrow y=mx+b \Rightarrow \bbox[red, 2pt]{y={77\over 10}x+{297\over 5}}$$

解答:$$\lim_{x\to 0}{e^{x^2}-1\over 1-\cos x} = \lim_{x\to 0}{{d\over dx}(e^{x^2}-1)\over {d\over dx}(1-\cos x)} = \lim_{x\to 0}{2xe^{x^2}\over \sin x} = \lim_{x\to 0}{{d\over dx}2xe^{x^2}\over {d\over dx}\sin x} = \lim_{x\to 0}{2e^{x^2}+ 4x^2e^{x^2}\over \cos x} = \bbox[red, 2pt]2$$

解答:$$y=x^2-2x+3 \Rightarrow y'=2x-2 \\ \text{Let the point of tangency on the parabola be }(a,a^2-2a+3). \\\text{Then the slope }m={a^2-2a+3-(-1) \over a-2}  ={a^2-2a+4\over a-2}= y'(a)=(2a-2) \\ \Rightarrow (2a-2)(a-2)=a^2-2a+4 \Rightarrow a(a-4)=0 \Rightarrow \cases{a=0 \Rightarrow m=-2 \Rightarrow y=-2(x-2)-1\\ a=4\Rightarrow m=6 \Rightarrow y=6(x-2)-1} \\\Rightarrow \text{tangent lines:} \bbox[red, 2pt]{\cases{y=-2x+ 3\\ y=6x-13}}$$

解答:$$u=x^2 \Rightarrow du=2x\,dx \Rightarrow x\,dx={1\over 2}du \Rightarrow \int_0^1 x^2 e^{x^2}\cdot x\,dx = \int_0^1 {1\over 2}ue^u \,du \\= {1\over 2} \left. \left[ ue^u-e^u \right] \right|_0^1 = \bbox[red, 2pt]{1\over 2} $$

解答:$$x^2 y^3-y^2+x^3y=1 \Rightarrow 2xy^3+ 3x^2y^2y'-2yy'+3x^2y+x^3y'=0 \\ \Rightarrow y'={2xy^3+3x^2y\over 2y-3x^2y^2-x^3} \Rightarrow y'(1,1)= \bbox[red, 2pt]{-{5\over 2}}$$

解答:$$\cases{f(x,y,z)= x^2+y^2+z^3 \\g(x,y,z)=3x+2y+z-80} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y \\f_z= \lambda g_z\\ g=0} \Rightarrow \cases{2x= 3\lambda\\ 2y=2\lambda\\ 3z^2=\lambda \\ 3x+2y+z=80} \\ \Rightarrow \cases{x=9z^2/2\\ y=3z^2\\ z=z} \Rightarrow {27z^2\over 2}+6z^2+z=80 \Rightarrow 39z^2+2z-160=0 \Rightarrow (z-2)(39z+80)=0 \\ \Rightarrow z=2 \Rightarrow \cases{x=18\\ y=12\\ z=2} \Rightarrow f(18,12,2)= \bbox[red, 2pt]{476}$$

解答:$$y=(\ln x)^x  \Rightarrow \ln y=x \ln (\ln x) \Rightarrow {y'\over y}=\ln(\ln (x))+x\cdot {1/x\over \ln x} =\ln(\ln(x) )+{1\over \ln x} \\ \Rightarrow y'=y\ln(\ln(x))+{y\over \ln x} \Rightarrow y'(e,y(e)) =y'(e,1)=0+1= \bbox[red, 2pt]1$$

解答:$$\text{Change the order of integration: }\int_0^1 \int_y^1 {y\over x^3+1}\,dx\,dy =\int_0^1 \int_0^x {y\over x^3+1}\,dy\,dx ={1\over 2}\int_0^1 {x^2\over x^3+1}\,dx \\=  {1\over 2} \left. \left[ {1\over 3}\ln(x^3+1) \right] \right|_0^1 = \bbox[red, 2pt]{{1\over 6}\ln 2}$$

解答:$$a_n= {n!(x+1)^n\over 3^n} \Rightarrow \lim_{n\to \infty} \left| {a_{n+1} \over a_n}\right| =\lim_{n\to \infty} \left| {(n+1)!(x+1)^{n+1} \over  3^{n+1}} \cdot {3^n\over n!(x+1)^n}\right| = \lim_{n\to \infty} \left| {n+1\over 3}(x+1) \right| \\=\infty, \text{ for }x\ne -1 \Rightarrow \text{The radius of convergence is  }\bbox[red, 2pt]0.$$

解答:$$y'(x)={\sin(x+y)+\sin(x-y) \over \sin(x+y)- \sin(x-y)} ={\sin x\cos y+\sin y\cos x+\sin x\cos y-\sin y\cos x\over \sin x\cos y+\sin y\cos x-\sin x\cos y+\sin y\cos x} \\={2\sin x\cos y\over 2\sin y\cos x} =\tan x\cdot \cot y \Rightarrow \tan y\,dy=\tan x\,dx \Rightarrow \int \tan y\,dy= \int \tan x\,dx \\ \Rightarrow -\ln|\cos y|= -\ln |\cos x| +C_1 \Rightarrow \cos y= C_2\cos x\\ y(0)={\pi\over 3} \Rightarrow \cos {\pi\over 3}=C_2\cos 0 \Rightarrow C_2={1\over 2} \Rightarrow \cos y={1\over 2}\cos x \Rightarrow \cos y(\pi/2)={1\over 2}\cos {\pi\over 2}=0 \\ \Rightarrow y({\pi\over 2})= \bbox[red, 2pt]{\pi\over 2}$$

乙) 計算題：共 2 題，每題 10 分

解答:$$V=\pi r^2 h =12\pi \Rightarrow h={12\over r^2} \\ \text{Let $k$ be the cost per square centimeter of the material for the base.} \\\text{Therefore, the cost per square centimeter of the material for the sides is $\frac{2}{3}k$.}\\\text{Now,}\cases{\text{area of the base: }\pi r^2\\ \text{area of the side :}2\pi rh} \Rightarrow \cases{\text{cos of the base: }k\pi r^2\\ \text{cost of the sides: }{2\over 3}k \cdot 2\pi rh={4\over 3}k \pi rh} \\ \Rightarrow \text{The total cost function is: }C(r,h)= k\pi r^2+{4\over 3}k \pi r h =k \pi r^2+{4\over 3}k\pi r \cdot {12\over r^2} =k\pi r^2+{16 k \pi\over r} \\=k\pi \left( r^2+{16\over r} \right) \Rightarrow \text{Let }f(r)=r^2+{16\over r} \Rightarrow f'(r)=2r-{16\over r^2}=0 \Rightarrow r^3=8 \Rightarrow r=2 \\  \Rightarrow h ={12\over 4}=3 \Rightarrow \text{To minimize the construction cost, the dimensions of the cup should be a }\\ \bbox[red, 2pt]{\text{radius of 2 cm and a height of 3 cm.}}$$

 

解答:$$x^3+y^3=9xy \Rightarrow 3x^2{dx \over dt}+3y^2 {dy\over dt}=9 \left( y{dx\over dt}+x{dy\over dt} \right) \Rightarrow x^2{dx\over dt}+y^2 {dy\over dt}= 3y{dx\over dt}+ 3x{dy\over dt} \\ \Rightarrow 2^2\cdot 3+4^2{dy\over dt}+3\cdot 4\cdot 3+3\cdot 2{dy\over dt} \Rightarrow {dy\over dt}= \bbox[red, 2pt]{12\over 5}$$

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解題僅供參考，其他轉學考試題及詳解