109年台綜大轉學考-微積分B詳解

臺灣綜合大學系統109學年度學士班轉學生聯合招生考試

科目名稱：微積分B

解答： $$\mathbf{(a)}\;\lim_{x\to 2}{x+4\over x-7}={2+4 \over 2-7} = \bbox[red,2pt]{-{6\over 5}}\\ \mathbf{(b)}\; \lim_{x\to 0} {4\ln(1-7x) \over x} =\lim_{x\to 0} {(4\ln(1-7x))' \over (x)'} =\lim_{x\to 0} {-28/(1-7x) \over 1} = -28 \\ \quad \Rightarrow \lim_{x\to 0} (1-7x)^{4/x} =\lim_{x\to 0} e^{4 \ln(1-7x)/x } =\bbox[red,2pt]{e^{-28}}$$

解答： $$\mathbf{(a)}\;\cases{u=x \Rightarrow du=dx\\ dv =\sec^2x\,dx \Rightarrow v=\tan x} \Rightarrow \int_0^{\pi/4} x\sec^2 x\,dx = \left.\left[x\tan x \right]\right|_0^{\pi/4}-\int_0^{\pi/4} \tan x\,dx\\ = \left.\left[ x\tan x+\ln|\cos x| \right]\right|_0^{\pi/4} ={\pi\over 4}+ \ln {\sqrt 2\over 2} =\bbox[red, 2pt]{{\pi\over 4}-{1\over 2}\ln 2}\\\mathbf{(b)}\;令u=x-3 \Rightarrow \int_4^5 {x-3\over \sqrt{x^2-6x+13}} \,dx = \int_4^5 {x-3\over \sqrt{(x-3)^2+4}} \,dx = \int_1^2 {u\over \sqrt{u^2+4}}\,du \\=\left.\left[ \sqrt{u^2+4} \right]\right|_1^2 = \bbox[red, 2pt]{2\sqrt 2-\sqrt 5}$$

解答： $$u(x,y)=(e^x-{y\over 6}) \int_{-2}^{2x} \sqrt{4-t^2}\,dt \Rightarrow {\partial u\over \partial x} =e^x\int_{-2}^{2x} \sqrt{4-t^2}\,dt + 2(e^x-{y\over 6}) \sqrt{4-4x^2}\\ \Rightarrow {\partial u\over \partial x}|_{(0,1)} =\int_{-2}^{0} \sqrt{4-t^2}\,dt + 4(1-{1\over 6})= \bbox[red, 2pt]{\pi +{10\over 3}}$$

解答： $$h(s,t)={t\over 4s}-t^2+{\tan s\over \ln s} \Rightarrow u(x,y)=h(x^2+y^2,3x-4y)= {3x-4y\over 4(x^2+y^2)} -(3x-4y)^2+ {\tan (x^2+y^2) \over \ln(x^2+y^2)}\\ \Rightarrow {\partial u\over \partial y} ={-4\over 4(x^2+y^2)} -{(3x-4y)(2y)\over 4(x^2+y^2)^2} +8(3x-4y)+ {\sec^2(x^2+y^2)\cdot (2y)\over \ln(x^2+y^2)} -{\tan(x^2+y^2) \over  (\ln(x^2+y^2))^2}\cdot {2y\over x^2+y^2} \\ \Rightarrow \left.{\partial u\over \partial y} \right|_{(2,0)} =-{1\over 4}-0+48+0+ 0 =\bbox[red, 2pt]{191\over 4}$$

解答：

$$y=\sin(x^2) \Rightarrow x^2=\sin^{-1}y \Rightarrow 繞y軸旋轉體積=\int_0^{\sin (1)} \pi\sin^{-1}y\,dy =\pi \left.\left[ x \sin^{-1}x +\sqrt{1-x^2}\right] \right|_0^{\sin (1)} \\ =\pi (\sin(1)+ \sqrt{1- \sin^2 (1)}-1) = \bbox[red, 2pt]{\pi(\sin 1+ \cos 1-1)}$$

解答： $$e^x = 1+x+{1\over 2!}x^2+{1\over 3!}x^3+\cdots +{1\over k!}x^k+\cdots\\ \Rightarrow xe^x =x+x^2 +{1\over 2!}x^3+ {1\over 3!}x^4+\cdots +{1\over k!}x^{k+1}+\cdots \\ \Rightarrow \int xe^x\,dx = \int \left( x+x^2 +{1\over 2!}x^3+ {1\over 3!}x^4+\cdots +{1\over k!}x^{k+1}+\cdots \right)\,dx\\ \Rightarrow f(x)=xe^x-e^x +C= {1\over 2}x^2+{1\over 3}x^3+ {1\over 4\cdot 2! }x^4+ {1\over 5\cdot 3! }x^5+ \cdots +{1 \over (k+2)k!}x^{k+2}+\cdots \\ \Rightarrow f(0)=-1+C=0 \Rightarrow C=1 \Rightarrow f(1)= \bbox[red, 2pt]1=\sum_{k=0}^\infty {1\over (k+2)k!}$$

解答： $$假設圓心O，則小狗坐標為P(\cos (0.1t),\sin (0.1t)),其中t為時間單位(分鐘)\\ 小狗每分鐘走0.1公里，\stackrel{\frown}{AD} ={3\over 4}\pi公里需時{3\over 4}\pi \times 10={15\over 2}\pi 分鐘\\ 因此h=\overline{PA}= \sqrt{(\cos(0.1t)-1)^2 +\sin^2 (0.1t)} =\sqrt{2-2\cos (0.1t)}\\ \Rightarrow h'= {0.1\sin(0.1t)\over \sqrt{2-2\cos(0.1t)}} \Rightarrow h'({15\over 2}\pi) = {0.1\sin({3\over 4}\pi)\over \sqrt{2-2\cos({3\over 4}\pi)}} = \bbox[red,2pt]{ \sqrt 2\over 20\sqrt{2+\sqrt 2}}$$

解答： $$f(x)=x^5+2x^3+ 2x \Rightarrow f'(x)=5x^4+ 6x^2+ 2 \Rightarrow \cases{f(1)=5 \\ f'(1)=13}\\f^{-1}(f(x))= x \Rightarrow  {d\over dx} f^{-1}(f(x))=1 \Rightarrow (f^{-1})'(f(x)) f'(x)=1 \Rightarrow (f^{-1})'(5) f'(1)=1 \\ \Rightarrow (f^{-1})'(5) =1/f'(1) = \bbox[red, 2pt]{1/13}$$

解答：

$$積分區域為四分之一環形，如上圖著色區域；\\因此取\cases{x=r\cos \theta\\ y=r\sin \theta}，則原式=\int_0^{\pi/2} \int_1^3 re^{r^2}\,drd\theta = \int_0^{\pi/2}\left.\left[ {1\over 2}e^{r^2}\right]\right|_1^3\,d\theta =\bbox[red, 2pt]{ {\pi\over 4}\left( e^8-e\right)}$$

解答： $$\cases{f(x,y,z)= e^{xyz}\\ g(x,y,z)=x^3-y^3+z^3-24} \Rightarrow \cases{f_x =\lambda g_x \\ f_y =\lambda g_y \\f_z =\lambda g_z \\ g=0} \Rightarrow \cases{yz e^{xyz}= \lambda (3x^2) \cdots(1) \\xz e^{xyz}= \lambda (-3y^2) \cdots(2) \\ xy e^{xyz}= \lambda (3z^2) \cdots(3) }\\ 因此\cases{(1)/(2) \Rightarrow y/x=-x^2/y^2\\ (2)/(3) \Rightarrow z/y=-y^2/z^2 \\ (3)/(1)\Rightarrow x/z=z^2/x^2} \Rightarrow \cases{y^3=-x^2 \\ z^3=x^3} 代入g=0 \Rightarrow 3x^3-24=0 \Rightarrow x=2\\ \Rightarrow \cases{y=-2\\ z=2} \Rightarrow f(2,-2,2)= \bbox[red, 2pt]{1/e^8 為極小值}$$

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