115年台大碩士班-微積分A詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 微積分(A)

解答:$$\textbf{(a) }L= {\sin x\over x} \Rightarrow \lim_{x\to 0^+} L=\lim_{x\to 0^+}{{d\over dx}\sin x\over {d\over dx}x}=  \lim_{x\to 0^+}{\cos x\over 1} =1 \Rightarrow \lim_{x\to 0^+}\sqrt L =\bbox[red, 2pt]1 \\\textbf{(b) }{d\over dx}(2x)^\pi =2^\pi{d\over dx}x^\pi = \bbox[red, 2pt]{2^\pi\pi x^{\pi-1}} \\\textbf{(c) }{d\over dx}2^{-x} ={d\over dx}e^{-x \ln 2} = -\ln 2 e^{-x\ln 2} = \bbox[red, 2pt]{-\ln 2\cdot 2^{-x}} \\\textbf{(d) } {d^2\over dx^2} \sin^{-1}x ={d\over dx}{1\over \sqrt{1-x^2}} =-{1\over 2}(1-x^2)^{-3/2}(-2x) = \bbox[red, 2pt]{x\over (1-x^2)^{3/2}} \\\textbf{(e) }\int \sec^2 x\,dx = \bbox[red, 2pt]{\tan x+C} \\\textbf{(f) }{d\over dx}\int_1^{x^2} f(\sqrt t)\,dt = f(\sqrt{x^2})\cdot {d\over dx}x^2= \bbox[red, 2pt]{2xf(\sqrt {x^2})} \\\textbf{(g) } \cases{u=x\\ dv=e^{-x}dx} \Rightarrow \cases{du=dx\\ v=-e^{-x}} \Rightarrow \int_1^\infty xe^{-x}\,dx = \left. \left[ -xe^{-x} \right] \right|_1^\infty +\int e^{-x}\,dx = \left. \left[ -xe^{-x}-e^{-x} \right] \right|_0^\infty \\\qquad =0+(e^{-1}+e^{-1}) = \bbox[red, 2pt]{2\over e} \\\textbf{(h) }\lim_{x\to \infty} \left( \sqrt{x+\sqrt x}-\sqrt x \right) =\lim_{x\to \infty} {\left( \sqrt{x+\sqrt x}-\sqrt x \right) \left( \sqrt{x+\sqrt x}+\sqrt x \right)\over \sqrt{x+\sqrt x}+\sqrt x} \\\quad =\lim_{x\to \infty} {\sqrt x \over\sqrt{x+\sqrt x}+\sqrt x} =\lim_{x\to \infty} {1 \over\sqrt{1+1/\sqrt x}+1} = \bbox[red, 2pt]{1\over 2} \\\textbf{(i) } \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt n \cdot \sqrt{n+k}} =\lim_{n\to \infty} \sum_{k=1}^n {1\over n \cdot \sqrt{1+k/n}} =\int_0^1 {1\over \sqrt{1+x}}\,dx = \left. \left[ 2\sqrt{1+x} \right] \right|_0^1 \\\quad = \bbox[red, 2pt]{2(\sqrt 2-1)} \\\textbf{(j) }{1\over 1-u} = \sum_{n=0}^\infty u^n \Rightarrow {d\over du} \left( {1\over 1-u} \right) ={1\over (1-u)^2} = \sum_{n=1}^\infty nu^{n-1}  \\ \quad \Rightarrow {1\over (1-3x)^2} =\sum_{n=1}^\infty n(3x)^{n-1} =\sum_{k=0}^\infty \bbox[red, 2pt]{(k+1)3^k}x^k \text{ for }|3x|\lt 1\Rightarrow |x|\lt \bbox[red, 2pt]{1\over 3}$$

解答:$$(y-1){dy\over dx}+x(y-1)^2 =0 \Rightarrow \int{1\over y-1}\,dy = \int -x\,dx \Rightarrow \ln(y-1)=-{1\over 2}x^2+c_1 \\ \Rightarrow y-1=c_2e^{-x^2/2} \Rightarrow y=1+c_2e^{-x^2/2} \\\textbf{(a) } y(0)=1=1+c_2 \Rightarrow c_2=0 \Rightarrow \bbox[red, 2pt]{y=1} \\\textbf{(b) }y(0)=0=1+c_2 \Rightarrow c_2=-1 \Rightarrow \bbox[red, 2pt]{y=1-e^{-x^2/2}}$$

解答:$$\textbf{(a) }\int_0^1 \int_{\sqrt y}^1 e^{x^3}\,dx\,dy =\int_0^1 \int_{0}^{x^2} e^{x^3}\,dy\,dx = \int_0^1 x^2e^{x^3}\,dx = \left. \left[ {1\over 3}e^{x^3} \right] \right|_0^1= \bbox[red, 2pt]{{1\over 3}(e-1)} \\\textbf{(b) }x^2+2y^2 \le z\le 2 \Rightarrow I= \iiint_P y^2\,dV= \iint_D \left( \int_{x^2+2y^2}^2 y^2\,dz \right) \,dA= \iint_D (2y^2-x^2y^2-2y^4)\,dA \\ D=\{(x,y) \mid x^2+2y^2 \le 2\} \Rightarrow {x^2\over 2}+y^2\le1 \Rightarrow \cases{x=\sqrt 2r \cos \theta\\ y=r\sin \theta} \Rightarrow |J|= \begin{Vmatrix} \sqrt 2\cos \theta& -\sqrt 2r\sin \theta\\ \sin \theta& r\cos \theta\end{Vmatrix} \\=\sqrt 2 r \Rightarrow I =\int_0^{2\pi} \int_0^1(r^2\sin^2\theta)\cdot 2(1-r^2)\cdot \sqrt 2r\,dr\,d\theta =2\sqrt 2 \left( \int_0^{2\pi} \sin^2\theta\,d\theta \right) \left( \int_0^1(r^3-r^5)\,dr \right) \\=2\sqrt 2\cdot (\pi) \cdot ({1\over 12}) = \bbox[red, 2pt]{\sqrt 2\pi\over 6}$$

解答:$$\textbf{(a) }\nabla f(a,b,c)= (f_x,f_y,f_z) =(b^2c^3, 2abc^3, 3ab^2c^2) \parallel \bbox[red, 2pt]{({1\over a}, {2\over b},{3\over c})} \\\textbf{(b) }f(x,y,z) =xy^2z^3 \ge 0, \text{ where }x,y,z\ge 0 \Rightarrow \min(f)=\bbox[red, 2pt] 0 \\\textbf{(c) }\cases{f(x,y,z) =xy^2z^3\\ g(x,y,z)=x^2+y^2+z^2-6} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ f_z = \lambda g_z\\ g=0} \Rightarrow \cases{y^2z^3= \lambda(2x) \\ 2xyz^3=\lambda(2y) \\3xy^2z^2=\lambda(2z) \\ x^2+y^2+z^2=6} \Rightarrow \cases{{y\over 2x}={x\over y} \\ {{2z \over 3y}}={y\over z}} \\\quad \Rightarrow \cases{y^2=2x^2\\2z^2=3y^2} \Rightarrow \cases{y^2=2x^2\\ z^2=3x^2} \Rightarrow x^2+y^2+z^2=6x^2=6 \Rightarrow x=1 \Rightarrow \cases{y=\sqrt 2\\ z=\sqrt 3}\\\quad  \Rightarrow f(1,\sqrt 2,\sqrt 3) = \bbox[red, 2pt]{6\sqrt 3}$$

解答:$$\textbf{(a) }{\partial \over \partial x} \left( {x\over (x^2+y^2+z^2)^{3/2}} \right) ={-2x^2+y^2+z^2\over (x^2+y^2+z^2)^{5/2}},{\partial \over \partial y} \left( {y\over (x^2+y^2+z^2)^{3/2}} \right) ={ x^2-2y^2+z^2\over (x^2+ y^2+ z^2)^{5/2}}, \\\quad {\partial \over \partial z} \left( {z\over (x^2+y^2+ z^2)^{3 /2}} \right) ={ x^2+y^2-2z^2\over (x^2+y^2+z^2)^{5/2}}, \\ \Rightarrow \text{div }E={-2x^2+y^2+z^2\over (x^2+y^2+z^2)^{5/2}} +{ x^2-2y^2+z^2\over (x^2+ y^2+ z^2)^{5/2}} +{ x^2+y^2-2z^2\over (x^2+y^2+z^2)^{5/2}} =\bbox[red, 2pt]0 \\\textbf{(b) }\iint_{\partial V} E\cdot dS = \iiint_V div(E) \,dV=0 \\ \text{The boundary }\partial V \text{ consists of }S \text{ (oriented upward/outward) and }Q \text{ (oriented downward/inward)} \\\text{relative  to }V. \Rightarrow \iint_SE \cdot n_{up}dS -\iint_QE \cdot n_{up} dS=0 \Rightarrow  \iint_Q E\cdot dS= \iint_S E\cdot dS\\ =\iint 1dS = \text{ Area of unit hemisphere}= \bbox[red, 2pt]{2\pi}\\ \bbox[cyan,2pt]{\text{singular point: }(0,0,0)}$$

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解題僅供參考，碩士班歷年試題及詳解