國立臺灣大學 115 學年度轉學生招生考試
科目: 微積分 (C)
解答:$$\textbf{(a) }y=(\sin x)^x \Rightarrow \ln y=x\ln (\sin x) \Rightarrow {y'\over y}=\ln (\sin x)+ x\cdot {\cos x\over \sin x} \Rightarrow y'=y \left( \ln(\sin x)+x\cot x \right) \\ \quad y'= \bbox[red, 2pt]{(\sin x)^x(\ln( \sin x)+ x\cot x)} \\ \textbf{(b) }y= \arctan(2x) \Rightarrow y'={2\over 1+(2x)^2} ={2\over 1+4x^2} \Rightarrow y''=-{2\cdot 8x\over (1+4x^2)^2} = \bbox[red, 2pt]{-{16x\over (1+4x^2)^2}} \\ \textbf{(c) }\cases{u=\ln x\\ dv=dx/x^2} \Rightarrow \cases{du=dx/x\\ v=-1/x} \Rightarrow \int \frac{\ln x}{x^2} dx = (\ln x)\left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) dx \\\quad = -\frac{\ln x}{x} + \int x^{-2} dx= \bbox[red, 2pt]{-\frac{\ln x}{x} -{1\over x}+C} \\\textbf{(f) } \lim_{x \to 0} \frac{\sin x - x \cos x}{x^3} = \lim_{x \to 0} \frac{{d\over dx}(\sin x - x \cos x)}{ {d\over dx} x^3} = \lim_{x \to 0} \frac{ x\sin x}{3x^2} = {1\over 3}\lim_{x \to 0} \frac{ \sin x}{x} =\bbox[red, 2pt]{1\over 3} \\ \textbf{(g) } \lim_{x \to \infty} x(\sqrt{x^2 + 1} - x) = \lim_{x \to \infty} x \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1} + x} \\\quad = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x^2}} + 1} =\bbox[red, 2pt]{1\over 2} \\\textbf{(h) } \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{n^2 + k^2} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{1 + \left(\frac{k}{n}\right)^2} \cdot \frac{1}{n} = \int_0^1 \frac{1}{1 + x^2} dx = \left. \left[ \tan^{-1}x \right] \right|_0^1 =\bbox[red, 2pt]{\pi\over 4} \\ \textbf{(i,j) } \ln(1+u) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} u^k \quad \text{for } -1 < u \le 1 \Rightarrow \ln(1+2x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} (2x)^k \\\quad = \sum_{k=1}^{\infty} \bbox[red, 2pt]{\frac{(-1)^{k-1} 2^k}{k} }x^k\quad \text{for } -1 < 2x \le 1 \Rightarrow -{1\over 2}\lt x\le {1\over 2} \Rightarrow |x| \lt \bbox[red, 2pt]{1\over 2}$$
解答:$$\textbf{(a) }\text{surface: }F(x,y,z)=f(x,y)-z=0 \Rightarrow \nabla F=(f_x,f_y,-1) \\ \text{Given } \mathbf{n_1} = -4\mathbf{i} - 2\mathbf{j} - 2\mathbf{k} = 2(-2\mathbf{i} - \mathbf{j} - \mathbf{k}) \Rightarrow \nabla f=-2\mathbf i-\mathbf j \text{(top-right to bottom-left)}\\ \Rightarrow \bbox[red, 2pt]{\cases{\textbf{(2a-i) }Q\\ \textbf{(2a-ii) } -2\mathbf i-\mathbf j }} \\ \textbf{(b) } \mathbf{n_2} = \alpha\mathbf{i} + \beta\mathbf{j} - \mathbf{k} \Rightarrow \nabla f(P)= \alpha \mathbf i+ \beta \mathbf j \Rightarrow P \text{ is located to the lower right of }M \\ \quad \Rightarrow \bbox[red, 2pt]{\cases{\textbf{(2b-i) }\text{positive} \\ \textbf{(2b-ii) }\text{negative}}} \\ \textbf{(c) }\text{The function value increases outward from point $M$} \\ \Rightarrow f(x_0, y_0) \text{ is a local minimum.} \Rightarrow \bbox[red, 2pt]{\textbf{(2c-i) }\text{local minimum}} \Rightarrow f_{xx}\gt 0 \Rightarrow \bbox[red, 2pt]{\textbf{(2c-ii) }\text{positive}} \\\textbf{(d) } \text{At $t_0$, the curve passes through the local minimum point $M$} \\ \quad \text{Because $\nabla f(M)$=0, $g'(t_0)=0$. Then }g(t) \text{ is a local minimum } \\ \text{At $t_1$, the curve $C$ is tangent to the outermost contour line at $P$, and the curve} \\ \qquad \text{trajectory remains inside the contour line before and after this point.} \\ \qquad \text{This indicates that as the curve moves, the function value reaches a local maximum} \\ \qquad \text{at $P$, hence there is a local extremum at $t_1$} \\ \text{At $t_2$, the curve $C$ directly crosses the countour line where $Q$ is located and moves} \\\qquad \text{ towards the center $M$. The function value decreases monotonically during this }\\ \qquad \text{ crossing, and there are no instances where the tangent slope is zero; That is,}\\ \qquad \text{no local extremum at $t_2$.} \\ \Rightarrow \bbox[red, 2pt]{\textbf{(2d) }t_0,t_1} \\\textbf{(e) } \text{Since $\mathbf{r}'(t_1)$ is parallel to the tangent to the contour line, and}\\\qquad \text{$\nabla f(P)$ is perpendicular to the contour line, these two vectors are orthogonal .} \\ \quad \text{The dot product of mutually perpendicular vectors is zero.} \\ \Rightarrow \bbox[red, 2pt]{\textbf{(2e) }zero}$$
解答:$$x{dy\over dx}+2y=e^{-x^2} \Rightarrow {dy\over dx}+{2\over x} y={e^{-x^2}\over x} \Rightarrow \text{ integrating factor }I(x)=e^{\int (2/x)\,dx} =x^2 \\ \Rightarrow x^2{dy\over dx}+2xy=xe^{-x^2} \Rightarrow (x^2y)' =xe^{-x^2} \Rightarrow x^2y= \int xe^{-x^2}\,dx =-{1\over 2}e^{-x^2}+C \\ \Rightarrow y=-{1\over 2x^2}e^{-x^2}+{C\over x^2} \Rightarrow y(1)=-{1\over 2}e^{-1}+C= 0 \Rightarrow C={1\over 2}e^{-1} \\ \Rightarrow \bbox[red, 2pt]{y=-{1\over 2x^2}e^{-x^2}+{1\over 2x^2 }e^{-1}}$$
解答:$$\textbf{(a) }\text{Changing the order of the integrating, } \int_0^1 \int_{y^2}^1 y e^{x^2} dx dy= \int_0^1 \int_0^{\sqrt{x}} y e^{x^2} dy dx \\= \int_0^1e^{x^2} \left. \left[ {1\over 2} y^2\right] \right|_0^{\sqrt x} \,dx = \int_0^1 \frac{1}{2} x e^{x^2} dx = \left. \left[ {1\over 4}e^{x^2} \right] \right|_0^1 = \bbox[red, 2pt]{{1\over 4}(e-1)} \\ \textbf{(b)} \cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iiint_E \sqrt{x^2 + y^2} dV = \int_0^{2\pi} \int_0^1 \int_0^{3-r^2} r^2 \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^1 (3r^2-r^4)\,dr\,d\theta \\=\int_0^{2\pi} \left. \left[ r^3-{1\over 5}r^5 \right] \right|_0^1\,d\theta=\int_0^{2\pi} {4\over 5}\,d\theta={4\over 5}\cdot 2\pi= \bbox[red, 2pt]{8\pi \over 5}$$
解答:$$\textbf{(a) } f(x,y) = x^2 - 4xy + y^3 \Rightarrow \nabla f=(f_x,f_y)=(2x-4y,-4x+3y^2) \Rightarrow \nabla f(2,1)=(0,-5) \\ \quad \text{from (2,1) to (0,0) }\Rightarrow \vec v=(-2,-1) \Rightarrow |\vec v|=\sqrt 5 \Rightarrow \vec e={\vec v\over |\vec v|}= \left( -{2\over \sqrt 5},-{1\over \sqrt 5} \right) \\ \quad \Rightarrow D_{\vec e} f(2,1) =\nabla f(2,1) \cdot \vec e= (0,-5) \cdot \left( -{2\over \sqrt 5},-{1\over \sqrt 5} \right) = \bbox[red, 2pt]{\sqrt 5} \\ \textbf{(b) } \cases{f_x=0\\ f_y=0} \Rightarrow \cases{2x-4y=0\\ -4x+3y^2 =0}\Rightarrow x=2y \Rightarrow -4\cdot(2y)+3y^2=0 \Rightarrow y(3y-8)=0\\ \quad \Rightarrow \cases{y=0 \Rightarrow x=0\\ y=8/3 \Rightarrow x=16/3} \Rightarrow \bbox[red, 2pt]{\text{ critical points:} (0,0), \left( {16\over 3},{8\over 3} \right)} \\ \cases{f_{xx}=2\\ f_{yy}=6y \\f_{xy} =-4} \Rightarrow D(x,y) = f_{xx}f_{yy} - (f_{xy})^2 = 12y - 16\\\quad \Rightarrow \cases{D(0,0)=-16\lt 0\\ D(16/3,8/3)=16\gt 0 \text{ and }f_{xx}=2 \gt 0 } \Rightarrow \bbox[red, 2pt]{\cases{\text{saddle point:(0,0)} \\ \text{local minimum at }(16/3,8/3)}} \\ \textbf{(c) }\cases{g(x,y)=f_x=2x-4y \\ h(x,y)=2x^2+y^2-1} \Rightarrow \cases{g_x = \lambda h_x\\ g_y= \lambda h_y \\ h=0} \Rightarrow \cases{2=\lambda\cdot 4x\\ -4=\lambda\cdot 2y \\ 2x^2+y^2=1} \Rightarrow y=-4x \\ \quad \Rightarrow 2x^2+(-4x)^2=1 \Rightarrow 18x^2=1 \Rightarrow \cases{x= 1/3\sqrt 2 \Rightarrow y=-4/3\sqrt 2 \\ x=-1/3\sqrt2 \Rightarrow y= 4/3\sqrt 2} \\\quad \Rightarrow \cases{g(1/3\sqrt 2, -4/3\sqrt 2)= 3\sqrt 2\\ g(-1/3\sqrt 2, 4/3\sqrt 2) =-3\sqrt 2} \Rightarrow \bbox[red, 2pt]{\cases{\text{absolute maximum: }3\sqrt 2\\ \text{absolute minimum: }-3\sqrt 2}}$$
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