115年台大轉學考-微積分B詳解

國立臺灣大學 115 學年度轉學生招生考試

科目: 微積分 (B)

Part I: 5 points for each blank

解答:$$ f(x) = (3x)^{2x} \Rightarrow \ln f(x)=2x\ln(3x) \Rightarrow {f'(x)\over f(x)}=2\ln(3x)+2x\cdot {1\over 3x} \cdot 3 \\ \Rightarrow  f'(x) = f(x) [2 \ln(3x) + 2] = (3x)^{2x} [2 \ln(3x) + 2]  \Rightarrow f'(1)=3^2 [2\ln 3+2]= \bbox[red, 2pt]{18(\ln 3+1)}$$

解答:$$x^3+y^3=4xy+1 \Rightarrow 3x^2+3y^2y'=4y+4xy' \Rightarrow y'={4y-3x^2 \over 3y^2-4x} \\ \Rightarrow y'(1,2)={8-3\over 12-4} =\bbox[red, 2pt]{5\over 8}$$

解答:$$\sin x=x-{x^3\over 3!}+{x^5\over 5!}-\cdots \Rightarrow x-\sin x \approx {x^3\over 6 } \\ \cos x=1-{x^2\over 2!}+{x^4\over 4!}-\cdots \Rightarrow 1-\cos x\approx{x^2\over 2}\\ \ln(1-3x)=-3x-{9\over 2}x^2-\cdots \approx-3x \\ \Rightarrow   \lim_{x \to 0} \frac{x - \sin x}{(1 - \cos x)\ln(1 - 3x)} =  \lim_{x \to 0} \frac{\frac{x^3}{6}}{\left(\frac{x^2}{2}\right)(-3x)} = \lim_{x \to 0} \frac{\frac{x^3}{6}}{-\frac{3x^3}{2}}  = \frac{1}{6} \cdot \left(-\frac{2}{3}\right) =\bbox[red, 2pt]{ -\frac{1}{9} }$$

解答:$$f(x) = (2x - 3)e^{-2x} \Rightarrow f'(x)= e^{-2x}(8-4x) \Rightarrow f''(x)=e^{-2x}(8x-20)\\ f''(x)=0 \Rightarrow 8x-20=0 \Rightarrow x={5\over 2} \Rightarrow f(5/2)=2e^{-5} \Rightarrow \text{ inflection point at }\bbox[red, 2pt]{ \left( {5\over 2},2e^{-5} \right)}$$

解答:$$\int_0^{\pi} \sin^2 \theta \cos^2\theta\,d \theta = {1\over 4}\int_0^{\pi} \sin^2(2\theta)\,d\theta ={1\over 8}\int_0^{\pi} (1-\cos (4\theta))\,d\theta ={1\over 8} \left. \left[ \theta-{1\over 4}\sin (4\theta) \right] \right|_0^{\pi} = \bbox[red, 2pt]{ \pi\over 8}$$

解答:$$\cases{u= \ln x\\ dv=dx/\sqrt x} \Rightarrow \cases{du=dx/x\\ v=2\sqrt x} \Rightarrow \int \frac{\ln x}{\sqrt{x}} \, dx = (\ln x)(2\sqrt{x}) - \int (2\sqrt{x})\left(\frac{1}{x}\right) dx \\=2\sqrt x \ln x-2\int{1\over \sqrt x}\,dx = \bbox[red, 2pt]{2\sqrt x\ln x-4\sqrt x+C}$$

解答:$$ V = \int_0^2 \pi (y^3)^2 \, dy = \pi \int_0^2 y^6 \, dy   = \pi \left[ \frac{y^7}{7} \right]_0^2  = \bbox[red, 2pt]{\frac{128\pi}{7}}$$

解答:$${dy\over dx}=x-y \Rightarrow {dy\over dx}+y=x \Rightarrow \text{ integrating factor }I(x)=e^x \Rightarrow {dy\over dx}e^x+ ye^x=xe^x \\ \Rightarrow \left( ye^x \right)'=xe^x \Rightarrow ye^x = \int xe^x \,dx =xe^x-e^x+C \Rightarrow y=x-1+Ce^{-x} \\ \Rightarrow y(0)=-1+C=5 \Rightarrow C=6 \Rightarrow \bbox[red, 2pt]{y=x-1+6e^{-x}}$$

解答:$$ \tan^{-1}(u) = u - \frac{u^3}{3} + \frac{u^5}{5} - \frac{u^7}{7} + \dots   \Rightarrow \tan^{-1}(x^2)=  x^2 - \frac{x^6}{3} + \frac{x^{10}}{5} - \dots \\ \Rightarrow \text{The 3rd nonzero term:} \bbox[red, 2pt]{x^{10} \over 5}$$

解答:$$ e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!} \Rightarrow   e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}\\ \Rightarrow  \int_0^1 e^{-x^2} \, dx = \int_0^1 \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} \right) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_0^1 x^{2n} \, dx  = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \cdot \left. \left[ x^{2n+1}\over 2n+1 \right] \right|_0^1 \\= \bbox[red, 2pt]{\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)}}$$

解答:$$r(\theta)=\sin(3\theta) =0 \Rightarrow \theta=0,{\pi\over 3} \Rightarrow \text{area of one piece: } {1\over 2}\int_0^{\pi/3} \sin^2(3\theta)\,d\theta ={1\over 4} \int_0^{\pi/3}(1-\cos (6\theta))\,d\theta \\={1\over 4} \left. \left[ \theta-{1\over 6}\sin (6\theta) \right] \right|_0^{\pi/3} ={\pi\over 12} \Rightarrow \text{ total area: } 3\cdot {\pi\over 12}= \bbox[red, 2pt]{\pi\over 4}$$

解答:$$\cases{ x = \rho\sin\phi\cos\theta\\ y = \rho\sin\phi\sin\theta \\z = \rho\cos\phi } \Rightarrow  \int_{-2}^2 \int_0^{\sqrt{4-x^2}} \int_{\sqrt{x^2+ y^2}}^{\sqrt{8-x^2-y^2}} y\sqrt{x^2+y^2+z^2} \, dz \, dy \, dx \\=   \int_0^\pi \int_0^{\pi/4} \int_0^{2\sqrt{2}} (\rho^2\sin\phi\sin\theta) (\rho^2\sin\phi) \, d\rho \, d\phi \, d\theta =  \int_0^\pi \sin\theta \, d\theta \int_0^{\pi/4} \sin^2\phi \, d\phi \int_0^{2\sqrt{2}} \rho^4 \, d\rho  \\=  2 \times \left( \frac{\pi-2}{8} \right) \times \left( \frac{128\sqrt{2}}{5} \right) = \frac{\pi-2}{4} \times \frac{128\sqrt{2}}{5} = \bbox[red, 2pt]{\frac{32\sqrt{2}(\pi-2)}{5} }$$

Part II: 20 points for each problem.

 解答:$$\textbf{(a) }\cases{f(x,y,z)=x+2y+3z\\ g(x)=x^2+y^2+4z^2-4} \Rightarrow \cases{f_x= \lambda g_x \\f_y= \lambda g_y \\f_z= \lambda g_z \\ g=0} \Rightarrow \cases{1=\lambda\cdot 2x\\ 2=\lambda\cdot 2y\\ 3=\lambda\cdot 8z\\ x^2+y^2+4z^2=4} \Rightarrow \cases{x=1/2\lambda\\ y=1/\lambda \\ z=3/8\lambda} \\ \Rightarrow  \left(\frac{1}{2\lambda}\right)^2 + \left(\frac{1}{\lambda}\right)^2 + 4\left(\frac{3}{8\lambda}\right)^2 = 4 \Rightarrow  \frac{29}{16\lambda^2} = 4\Rightarrow  \lambda = \pm \frac{\sqrt{29}}{8} \\ \Rightarrow \cases{\lambda= \sqrt{29}/8 \Rightarrow \cases{x=4/\sqrt{29}\\y=8/\sqrt{29}\\ z=3/\sqrt{29}} \Rightarrow f(4/\sqrt{29},8/\sqrt{29}, 3/\sqrt{29}) =\sqrt{29} \\ \lambda= -\sqrt{29}/8 \Rightarrow \cases{x=-4/\sqrt{29}\\y=-8/\sqrt{29}\\ z=-3/\sqrt{29}} \Rightarrow f(-4/\sqrt{29},-8/\sqrt{29}, -3/\sqrt{29}) =-\sqrt{29} } \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{absolute maximum: }\sqrt{29} \\ \text{absolute minimum: }-\sqrt{29}}} \\\textbf{(b) } g(x,y) = x + 2y + \frac{3}{2}\sqrt{4 - x^2 - y^2} \Rightarrow \cases{g_x=1-3x/2\sqrt{4-x^2-y^2} \\g_y=2-3y/2\sqrt{4-x^2-y^2}}\\ \quad \cases{g_x=0\\ g_y=0} \Rightarrow 3x=3y/2 \Rightarrow y=2x \Rightarrow 2\sqrt{4-x^2-(2x)^2}=3x \Rightarrow x={4\over \sqrt{29}} \Rightarrow y={8\over \sqrt{29}} \\ \quad \Rightarrow \bbox[red, 2pt]{\text{critical point: } \left( {4\over \sqrt{29}},{8\over \sqrt{29}} \right) }\\ \quad\cases{g_{xx}= -{3\over 2} \cdot (4-y^2)/(4-x^2-y^2)^{3/2} \\g_{yy}=  -{3\over 2} \cdot (4-x^2)/(4-x^2-y^2)^{3/2} \\ g_{xy} =-{3\over 2}\cdot xy/(4-x^2-y^2)^{3/2}} \Rightarrow \cases{g_{xx}(4/\sqrt{29}, 8/\sqrt{29}) =-13\sqrt{29}/36 \\ g_{yy} (4/\sqrt{29}, 8/\sqrt{29}) =-25\sqrt{29}/36 \\g_{xy}(4/\sqrt{29}, 8/\sqrt{29}) =-8\sqrt{29}/36}  \\ \quad \Rightarrow D=g_{xx}g_{yy}-(g_{xy})^2  = \left(-\frac{13\sqrt{29}}{36}\right)\left(-\frac{25\sqrt{29}}{36}\right) - \left(-\frac{8\sqrt{29}}{36}\right)^2  = \frac{261 \times 29}{1296} > 0 \\ \quad \Rightarrow \cases{g_{xx}\lt 0 \\ D\gt 0} \Rightarrow \bbox[red, 2pt]{f \left( {4\over \sqrt{29}} , {8\over \sqrt{29}} \right) \text{ is a local minimum}}$$

解答:$$\textbf{(a) }  \text{Let $C$ be a positively oriented (counterclockwise), piecewise-smooth, simple closed curve}\\ \quad \text{in a plane, and let $D$ be the region bounded by $C$. If $P(x,y)$ and $Q(x,y)$ have}\\ \quad \text{continuous partial derivatives on an open region that contains $D$, then:}\\ \qquad \oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \\ \textbf{(b) } \cases{u=x+ 2y\\ v=x-2y} \Rightarrow \cases{x=(u+v)/2\\ y=(y-v)/4} \Rightarrow \frac{\partial(x,y)}{\partial(u,v)} = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} \\\quad = \det \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{4} \end{pmatrix} = \bbox[red, 2pt]{-{1\over 4}} \\ \textbf{(c) } \cases{ P = e^{x^2 - 4y^2}\\ Q = 2e^{x^2 - 4y^2} } \Rightarrow \cases{Q_x=4xe^{x^2-4y^2} \\P_y=-8ye^{-x^2-4y^2}} \Rightarrow \iint_D \left( 4xe^{x^2 - 4y^2} - (-8ye^{x^2 - 4y^2}) \right) dA \\\quad = \iint_D 4(x + 2y)e^{x^2 - 4y^2} dA = \int_{-2}^2 \int_{-2}^2 4u e^{uv} \left(\frac{1}{4}\right) dv \, du = \int_{-2}^2 \int_{-2}^2 u e^{uv} \, dv \, du = \bbox[red, 2pt]0 \\\text{,where} \int_{-2}^2 \left[ e^{uv} \right]_{v=-2}^{v=2} \, du = \int_{-2}^2 (e^{2u} - e^{-2u}) \, du  \text{ and }f(u) = e^{2u} - e^{-2u} \text{ is odd} \\ \qquad \Rightarrow \int_{-2}^2 f(u)\,du =0$$

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解題僅供參考，其他轉學考試題及詳解