112年高師大轉學考-微積分詳解

國立高雄師範大學 112 學年度學士班轉學生招生考試

系所別：電機工程學系 二年級
科 目：微積分（全一頁）

◆計算題 （需詳列計算過程）每大題 10 分

解答：$$\mathbf{(a)}\; 2\pi \int_0^2 (x+1)(2x-x^2)\,dx =2\pi \int_0^2 -x^3+x^2+2x\,dx = \left. \left[ -{1\over 4}x^4+{1\over 3}x^3+x^2 \right] \right|_0^2 = \bbox[red, 2pt]{16\pi\over 3}\\ \mathbf{(b)}\;\pi\int_0^4(\sqrt y+1)^2-({y\over 2}+1)^2\,dy= \pi \int_0^4 2\sqrt y-{y^2\over 4} \,dy= \pi \left. \left[ {4\over 3}y^{3/2} -{1\over 12}y^3\right]\right|_0^4 = \bbox[red, 2pt]{16\over 3}\pi$$

解答：$$環表面積(\text{torus area})公式=4\pi^2Rr,依題意R=r ,即\bbox[red, 2pt]{4\pi^2r^2}$$

解答：$$假設\cases{P(x,y,z)在z^2=x^2+y^2\\ Q(4,2,0)} \Rightarrow d=\overline{PQ}= \sqrt{(x-4)^2+(y-2)^2+x^2+y^2}\\ 求d之最小值相當於求d^2=(x-4)^2+(y-2)^2+x^2+y^2之最小值\\ \cases{{d\over dx}d^2=2(x-4)+2x=0 \\{d\over dy}d^2=2(y-2)+2y=0} \Rightarrow \cases{x=2\\ y=1} \Rightarrow z^2=x^2+y^2=5 \Rightarrow z=\pm \sqrt 5\\ \Rightarrow 最接近的點\bbox[red, 2pt]{(2,1,\pm \sqrt 5)}$$

解答：$${d\over dx}\left( x^{3/4}\sqrt{x^2+1}\right)={3\over 4}x^{-1/4} \sqrt{x^2+1}+ {x^{7/4} \over \sqrt{x^2+1}} \\ \Rightarrow y'=\left({3\over 4}x^{-1/4} \sqrt{x^2+1}+ {x^{7/4} \over \sqrt{x^2+1}} \right)\cdot {1\over (3x+2)^5}-{15x^{3/4}\sqrt{x^2+1}\over (3x+2)^6}\\ =\bbox[red, 2pt]{ -39x^3 +14x^2-51x+6\over 4x^{1/4}\sqrt{x^2+1}(3x+2)^6}$$

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

解答：$${d\over dx} {x+1\over \sqrt{x-2}} ={1\over \sqrt{x-2}}-{x+1\over 2(x-2)^{3/2}} ={x-5\over 2(x-2)^{3/2}} \\ \Rightarrow {d\over dx}\ln{x+1\over \sqrt{x-2}} ={\sqrt{x-2} \over x+1} \cdot {x-5\over 2(x-2)^{3/2}} =\bbox[red, 2pt]{x-5\over 2(x+1)(x-2)}$$

解答 ：$$\lim_{h\to 0}{(3+h)^2-9 \over h} =\lim_{h\to 0}{{d\over dh}((3+h)^2-9) \over {d\over dh}h} =\lim_{h\to 0}(2(3+h))=\bbox[red, 2pt]6$$

解答 ：$$f(x,y)=\sin{x\over 1+y} \Rightarrow \bbox[red, 2pt]{\cases{{\partial f\over \partial x}={1\over 1+y} \cos{x \over 1+y} \\{\partial f\over \partial y} =-{x\over (1+y)^2}\cos{x\over 1+y}}}$$

解答 ：$$取\cases{u=\ln x\\ dv ={1\over x}dx} \Rightarrow \cases{du={1\over x}dx \\ v=\ln x} \Rightarrow I=\int_1^e{\ln x\over x}dx = \left. \left [(\ln x)^2\right] \right|_1^e-\int_1^e{\ln x\over x}\,dx \\ \Rightarrow 2I=\left. \left [(\ln x)^2\right] \right|_1^e =1 \Rightarrow I=\bbox[red, 2pt]{1\over 2}$$

解答：$$f(x,y)=\sin(x)+e^{xy} \Rightarrow \cases{f_x=\cos x+ye^{xy} \\ f_y=xe^{xy}} \Rightarrow \nabla f=(f_x,f_y)=\bbox[red, 2pt]{(\cos x+ye^{xy},xe^{xy})}$$
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解題僅供參考,其他歷年試題及詳解