臺灣綜合大學系統 115 學年度學士班轉學生聯合招生考試
科目名稱 微積分 C
解答:$$\lim_{x\to 0} {\int_{3x}^{x^2} \ln(1+t^2) \over x^2-3x} =\lim_{x\to 0} {{d\over dx}\int_{3x}^{x^2} \ln(1+t^2) \over {d\over dx}(x^2-3x)} = \lim_{x\to 0} {2x\ln(1+x^4) -3\ln(1+9x^2)\over 2x-3} \\={0-0\over -3} =\bbox[red, 2pt]0$$
解答:$$\textbf{(a) }{d\over dx}x^3 +{d\over dx} y^3= {d\over dx}6xy \Rightarrow 3x^2+3y^2y' =6y+6xy' \Rightarrow y'={6y-3x^2\over 3y^2-6x} ={2y-x^2\over y^2-2x} \\ \qquad \Rightarrow \text{slope }m= y'(3,3) ={6-9\over 9-6}=-1 \Rightarrow \text{ tangent line:} y-3=-1(x-3) \Rightarrow \bbox[red, 2pt]{x+y=6} \\\textbf{(b) }\text{Given: }\cases{\text{rate of change of volume:} \frac{dV}{dt} = 2 \\ \text{dimensions of the tank:} \cases{\text{height }H = 6\\ \text{radius} R = 3} \\ \text{instantaneous height to evaluate:} h = 2 } \\ \text{Because the water forms a smaller cone inside the tank, the ratio of its radius to its height }\\ \text{remains constant by similar triangles: } {r\over h}={R\over H}={3\over 6}={1\over 2} \Rightarrow r={h\over 2} \\ \Rightarrow V={1\over 3}\pi r^2 h= {1\over 3}\pi \left( {h\over 2} \right)^2 h={1\over 12}\pi h^3 \Rightarrow {dV\over dt} ={1\over 4}\pi h^2\cdot {dh\over dt} \Rightarrow 2={1\over 4}\pi \cdot 2^2\cdot {dh\over dt} \\ \Rightarrow {dh\over dt}= \bbox[red, 2pt]{2\over \pi}$$
解答:$$\textbf{(a) } I= \int_0^{\pi/2} {\sin^3 x\over \sin^3x +\cos^3 x} \,dx =\int_0^{\pi/2} {\sin^3 \left( {\pi\over 2}-x \right)\over \sin^3 \left( {\pi\over 2}-x \right) +\cos^3 \left( {\pi\over 2}-x \right)} \,dx \\\qquad =\int_0^{\pi/2} {\cos^3 x\over \cos^3x +\sin^3 x} \,dx\Rightarrow I+I =\int_0^{\pi/2} {\sin^3 x+ \cos^3 x\over \sin^3x +\cos^3 x} \,dx = \int_0^{\pi/2}1\,dx \\ \qquad \Rightarrow 2I= {\pi\over 2} \Rightarrow I= \bbox[red, 2pt]{\pi\over 4}\\ \textbf{(b) } \text{Let }\int_0^{\pi} f(x)\sin x\,dx =I_1 \text{ and }\int_0^{\pi} f''(x) \sin x\,dx = I_2, \\\qquad\text{ then }I=\int_0^{\pi}\left[ f(x)+f''(x) \right]\sin x\,dx = I_1+I_2=3\\ \qquad \cases{u= \sin x \\ dv= f''(x)\,dx } \Rightarrow \cases{du= \cos x\,dx \\ v=f'(x)} \Rightarrow I_2=\left. \left[ \sin x\cdot f'(x) \right] \right|_0^{\pi} -\int_0^{\pi} f'(x)\cos x\,dx\\\qquad =0-\int_0^{\pi} f'(x)\cos x\,dx = -\int_0^{\pi} f'(x)\cos x\,dx \\ \qquad \cases{u=\cos \\ dv=f'(x)\,dx} \Rightarrow \cases{du=-\sin x\,dx\\ v=f(x)} \Rightarrow I_2 =- \left( \left. \left[ \cos x\cdot f(x) \right] \right|_0^{\pi}+\int_0^{\pi} f(x) \sin x\,dx\right) \\\qquad =f(\pi)+f(0)-I_1 \Rightarrow I_2= f(\pi)+f(0)-I_1 \Rightarrow I=I_1+I_2 = f(\pi)+f(0)=3 \\ \qquad \Rightarrow f(\pi) =3-f(0)= \bbox[red, 2pt]2$$
解答:$$a_n = \frac{(-1)^{n-1}(x-2)^n}{n \cdot 3^n} \Rightarrow L=\lim_{n\to \infty} \left| a_{n+1} \over a_n\right| =\lim_{n \to \infty} \left\vert{} \frac{\frac{(-1)^n(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}}}{\frac{(-1)^{n-1}(x-2)^n}{n \cdot 3^n}} \right\vert{} = \lim_{n \to \infty} \left\vert{} (x-2) \cdot \frac{1}{3} \cdot \frac{n}{n+1} \right\vert{} \\={1\over 3}|x-2|\lt 1 \Rightarrow -1\lt x\lt 5\\ x=5 \Rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \Rightarrow \text{ alternating harmonic series} \Rightarrow \text{ convergent} \\x=-1\Rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{n} = -\sum_{n=1}^{\infty} \frac{1}{n} \Rightarrow \text{ negative of the harmonic series }\Rightarrow \text{divergent} \\ \Rightarrow \text{ interval of convergence: }\bbox[red, 2pt]{(-1,5]}$$
解答:$$\textbf{(a) }g(x)={1\over 1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n} \Rightarrow g'(x)=-{2x\over (1+x^2)^2 } = \sum_{n=1}^{\infty} (-1)^n (2n)x^{2n-1} \\ \qquad \Rightarrow f(x)={x\over (1+x^2)^2} = -\frac{1}{2} \sum_{n=1}^{\infty} (-1)^n (2n)x^{2n-1} = \sum_{n=1}^{\infty} (-1)^{n-1} n x^{2n-1}= \bbox[red, 2pt]{\sum_{n=0}^{\infty} (-1)^{n } (n+1) x^{2n+1} } \\\text{ And radius of Convergence: }R=\bbox[red, 2pt]1 \\\textbf{(b) }2n+1=7 \Rightarrow n=3 \Rightarrow \text{ coefficient }=(-1)^3\cdot (3+1)=-4 \Rightarrow {f^{(7)}(0) \over 7!}=-4 \\\qquad \Rightarrow f^{(7)}(0)=-4\cdot 7! =\bbox[red, 2pt]{-20160}$$
解答:$$\textbf{(a) }\cases{f(x,y,z)= x^4+y^4 +z^4\\ g(x,y,z)=x^2+y^2+z^2 -1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ f_z= \lambda g_z\\ g=0} \Rightarrow \cases{4x^3=\lambda(2x) \\ 4y^3=\lambda(2y) \\ 4z^3=\lambda(2z) \\x^2+y^2+z^2=1}\\ \qquad \Rightarrow (x,y,z)=\cases{(\pm 1,0,0), (0,\pm 1,0), (0,0,\pm 1) \Rightarrow f=1\\ (\pm 1/\sqrt 2, \pm 1/\sqrt 2,0), (\pm 1/\sqrt 2, 0,\pm 1/\sqrt 2), (0,\pm 1/\sqrt 2, \pm 1/\sqrt 2 ) \Rightarrow f=1/2 \\ (\pm 1/\sqrt 3,\pm 1/\sqrt 3,\pm 1/\sqrt 3) \Rightarrow f=1/3 } \\ \qquad\Rightarrow \bbox[red, 2pt]{\cases{\text{max:}1\\ \text{min:}1/3}} \\ \textbf{(b) }\cases{f_x= 4x^3=0\\ f_y=4y^3=0\\ f_z=4z^3=0} \Rightarrow \text{ a single critical point at the origin $(0, 0, 0)$} \Rightarrow f(0,0,0)=0 \\ \qquad \text{From part(a), we have } \cases{\text{Maximum on boundary: 1} \\ \text{Minimum on boundary:}1/3} \Rightarrow \bbox[red, 2pt] {\cases{\text{The absolute maximum value is $1$} \\ \text{The absolute minimum value is $0$}}} $$
解答:$$\textbf{(a) }f(x,y,z)= x^2z-y^2z+xy \Rightarrow \nabla f=(f_x,f_y,f_z) =(2xz+y,-2yz+x, x^2-y^2) \\ \qquad \Rightarrow \nabla f(P) =\nabla f(1,-1,2)= \bbox[red, 2pt]{(3, 5,0)} \\ \textbf{(b) } u={v\over |v|}={(1,2,-2)\over 3} = \left( {1\over 3},{2\over 3},-{2\over 3} \right) \Rightarrow D_uf(P) = \nabla f(P)\cdot u =(3,5,0) \cdot\left( {1\over 3},{2\over 3},-{2\over 3} \right) \\\qquad =1+{10\over 3}+0= \bbox[red, 2pt]{13\over 3} \\\textbf{(c) }|\nabla f(P)|= \sqrt{3^2+5^2+0^2} = \sqrt{34} \Rightarrow u_{max}={\nabla f(P) \over |\nabla f(P)|} = \left( {3\over \sqrt{34}}, {5\over \sqrt{34}},0 \right) \\ \qquad \Rightarrow \bbox[red, 2pt]{\cases{\text{The maximum directional derivative is} \sqrt{34} \\\text{The corresponding unit direction vector is} \left\langle \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}}, 0 \right\rangle}}$$
解答:$$\cases{x= r\cos \theta\\ y=r\sin \theta} \Rightarrow \int_{-2}^0 \int_0^{\sqrt{4-x^2}} e^{x^2+y^2}\,dy\,dx = \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{2} e^{r^2} r \,dr\,d\theta = \int_{\frac{\pi}{2}}^{\pi} {1\over 2}\left( e^4-1 \right)\,d\theta \\={1\over 2}\left( e^4-1 \right)\cdot {\pi\over 2} =\bbox[red, 2pt] {{\pi\over 4} (e^4-1)}$$
解答:$$\textbf{(a) }\cases{P(x,y) = 1 \\Q(x,y) = x + e^{\sin y}} \Rightarrow \cases{P_y=0\\ Q_x=1} \Rightarrow \oint_C P\,dx + Q\,dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA= \iint_D 1\,dA \\ \qquad ={1\over 2}\cdot 1\cdot 2\pi=\bbox[red, 2pt]\pi \quad (D: \text{ upper half of the ellipse}) \\ \textbf{(b) } \cases{C_1: \text{line segment from $(-1,0)$ to $(1,0)$} \\ \text{$C_2$: The upper arc of the ellipse from $(1,0)$ to $(-1,0)$}} \Rightarrow \oint_C = \int_{C_1}+ \int_{C_2}= \pi\\ \qquad C_1: \cases{y=0 \Rightarrow dy=0\\ x:-1\to 1} \Rightarrow \int_{C_1} dx+(x+e^{\sin y})\,dy = \int_{-1}^1 1\,dx +(x+e^{\sin y})\cdot 0 =2 \\ \qquad \Rightarrow 2+ \int_{C_2}=\pi \Rightarrow \int_{C_2}= \pi-2 \Rightarrow \int_{\gamma}=-\int_{C_2}= \bbox[red, 2pt]{2-\pi}$$
解答:$$ F(x,y,z) = (x + \sin(yz), y^3 + e^z, z^3 + \ln(x^2 + y^2 + 1)) =(P,Q,R) \\ \Rightarrow \iint_S F\cdot dS = \iiint_V (\nabla\cdot F)\,dV = \iiint_V (P_x+Q_y+ R_z)\,dV = \iiint_V ( 1+3y^2+3z^2)\,dV \\=\iiint_V 1\,dV+ 3 \iiint_V (y^2+z^2)\,dV={4\over 3}\pi \cdot 3^3+2 \iiint_V (x^2+y^2+z^2)\, dV =36\pi+ 2\iiint_V \rho^2 \,dV\\ =36\pi +2 \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{3} (\rho^2)(\rho^2 \sin\phi) \,d\rho \,d\phi \,d\theta =36\pi+ 2 \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{\pi} \sin\phi \,d\phi \right) \left( \int_{0}^{3} \rho^4 \,d\rho \right) \\=36\pi+2\cdot 2\pi\cdot 2\cdot {243\over 5}= \bbox[red, 2pt] {{2124\over 5}\pi}$$
解題僅供參考，其他轉學考試題及詳解
沒有留言:
張貼留言