臺灣綜合大學系統112學年度學士班轉學生聯合招生
科目名稱：微積分A
解答：$$\lim_{x \rightarrow 4}{4-x\over 2-\sqrt x} =\lim_{x \rightarrow 4}{\frac{\text{d} }{\text{d}x}(4-x)\over \frac{\text{d} }{\text{d}x}(2-\sqrt x)} = \lim_{x\to 4}{ -1\over-{1\over 2\sqrt x} } = \lim_{x\to 4} 2\sqrt x= \bbox[red, 2pt]4$$
解答：$$令\cases{P(t,\sqrt t) \in y=\sqrt x\\ Q(1,0)} \Rightarrow \overline{PQ}=\sqrt{(t-1)^2+t} =\sqrt{t^2-t+1}\\ 令f(t)=t^2-t+1 \Rightarrow f'(t)=2t-1=0 \Rightarrow f(t={1\over 2})為最小值 \Rightarrow P\bbox[red, 2pt]{({1\over 2},{\sqrt 2\over 2})}$$
解答：$$f(x)=\int^{e^x}_2 \sqrt{1+t^2}\,dt \Rightarrow f'(x)=e^x \sqrt{1+e^{2x}}\\ 又f(x)=0 \Rightarrow e^x=2 \Rightarrow x=\ln 2 \Rightarrow f'(\ln 2)=2\sqrt{1+4} =2\sqrt 5\\ \Rightarrow (f^{-1})'(0)={1\over f'(\ln 2)} ={1\over 2\sqrt 5} =\bbox[red, 2pt]{\sqrt 5\over 10}$$
解答：$$r=\sin^3(\theta/3) \Rightarrow {dr\over d\theta} =3\sin^2(\theta /3) \cos(\theta/3) \cdot {1\over 3}=\sin^2 {\theta\over 3}\cos{\theta \over 3}\\ 因此曲線長=\int_0^\pi \sqrt{r^2+({dr\over d\theta})^2}\,d\theta =\int_0^\pi \sqrt{ \sin^6 {\theta\over 3}+ \sin^4{\theta\over 3} \cos^2 {\theta\over 3}}\, d\theta =\int_0^\pi \sqrt{\sin^4 {\theta\over 3}}\, d\theta \\ =\int_0^\pi \sin^2{\theta \over 3}\, d\theta ={1\over 2}\int_0^\pi 1-\cos{2\theta \over 3}\, d\theta ={1\over 2} \left. \left[ \theta-{3\over 2}\sin{2\theta \over 3}\right] \right|_0^\pi =\bbox[red, 2pt]{{1\over 2}\pi -{3\over 8}\sqrt 3}$$
解答：$$令a_n={(-1)^n 2^n\over \sqrt n}x^n \Rightarrow \lim_{n\to \infty}\left|{a_{n+1}\over a_n} \right| = \lim_{n\to \infty}\left|{(-1)^{n+1} 2^{n+1} x^{n+1 }\over \sqrt {n+1}} \cdot {\sqrt n\over (-1)^n 2^n x^n} \right| \\=\lim_{n\to \infty}{2\sqrt n |x|\over \sqrt{n+1}} =\lim_{n\to \infty} 2|x| \lt 1 \Rightarrow |x| \lt {1\over 2} \Rightarrow 收斂半徑=\bbox[red, 2pt]{1\over 2}\\ 又\cases{ x={1\over 2} \Rightarrow \sum_{n=1}^\infty a_n =\sum_{ n=1}^\infty {(-1)^n \over \sqrt n} 收斂(依交錯級數審斂法) \\ x=-{1\over 2} \Rightarrow \sum_{n=1}^\infty a_n =\sum_{ n=1}^\infty {1\over \sqrt n}發散({1\over \sqrt n}\ge {1\over n} 發散)} \Rightarrow 收斂區間： \bbox[red, 2pt]{(-{1\over 2},{1\over 2}]}$$
解答：$$r(t)=\cos 3t\mathbf i+ \sin 3t\mathbf j+4t \mathbf k \Rightarrow r'(t)=-3\sin 3t\mathbf i+ 3\cos 3t\mathbf j+4\mathbf k\\ \Rightarrow \text{unit tangent vector }T(t)={r'(t) \over \Vert r'(t)) \Vert} ={1\over \sqrt{9+16}}(-3\sin 3t\mathbf i+ 3\cos 3t\mathbf j+4\mathbf k) \\ =\bbox[red, 2pt]{-{3 \over 5}\sin 3t\mathbf i+ {3 \over 5} \cos 3t\mathbf j+{4 \over 5}\mathbf k} \Rightarrow T'(t)=-{9\over 5}\cos 3t\mathbf i -{9\over 5} \sin 3t\mathbf j+ 0\mathbf k\\ \Rightarrow \text{unit normal vector }N(t)= {T'(t) \over \Vert T'(t)\Vert} ={5\over 9}T'(t)= \bbox[red, 2pt]{-\cos 3t\mathbf i- \sin 3t\mathbf j +0\mathbf k}$$
解答：$$令F(x,y,z)=x+2y+3z-\sin(xyz) \Rightarrow \cases{F_x=1-yz\cos(xyz) \\ F_y=2-xz \cos(xyz) \\ F_z=3-xy\cos(xyz)} \\ \Rightarrow \nabla F(2,-1,0)=(F_x(2,-1,0),F_y(2,-1,0),F_z(2,-1,0)) =(1,2,5) \\ \Rightarrow \cases{\text{tangent plane: } 1(x-2)+2(y+1)+5(z-0)=0 \Rightarrow \bbox[red, 2pt]{x+2y-5x+2=0} \\\text{normal line: } \bbox[red, 2pt]{{x-2\over 1}={y+1\over 2}={ z\over 5}}}$$
解答：
$$\cases{x=\sqrt{1-y^2} \Rightarrow x^2+y^2=1\\ x=\sqrt 3y} \Rightarrow 積分區域為夾角30^\circ的扇形區域，見上圖\\ 取\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \int_0^{1/2} \int_{\sqrt 3y}^{\sqrt{1-y^2}} x^2y\,dxdy =\int_0^{\pi /6} \int_0^1r^2\cos^2 \theta\cdot r\sin\theta\cdot r\,drd\theta \\\int_0^{\pi /6} \int_0^1 r^4\cos^2\theta \sin \theta \,drd\theta ={1\over 5} \int_0^{\pi/6} \cos^2\theta \sin\theta \,d\theta ={1\over 5} \left.\left[ -{1\over 3}\cos^3 \theta \right] \right|_0^{\pi/6} \\={1\over 5}\cdot {1\over 3}(1-{3\sqrt 3\over 8}) =\bbox[red, 2pt]{{1\over 15}-{\sqrt 3\over 40}}$$
解答：$$取\cases{x(t)=\cos t\\ y(t)= \sin t},0\le t\le 2\pi \Rightarrow \cases{dx = -\sin tdt\\ dy=\cos t dt\\ ye^{xy}= \sin te^{\sin(2t)/2} \\ xe^{xy}= \cos te^{\sin(2t)/2}} \\ \Rightarrow \int_C \mathbf F\cdot d\mathbf r =\int_0^{2\pi } \sin te^{\sin(2t)/2}\cdot (-\sin tdt)+ \cos te^{\sin(2t)/2}\cdot \cos tdt \\=\int_0^{2\pi } e^{\sin(2t)/2}(\cos^2t-\sin^2 t)\,dt=\int_0^{2\pi } e^{ \sin(2t)/2}\cos(2t)\,dt =\left. \left[ e^{\sin(2t)/2} \right] \right|_0^{2\pi}= \bbox[red, 2pt]0$$
解答：$$利用散度定理來求解，div \mathbf F= \frac{\partial }{\partial x}F_1 + \frac{\partial }{\partial y}F_2 + \frac{\partial }{\partial z}F_3 =z^2+ x^2+ y^2=1 \\ \Rightarrow \iint_S \mathbf F\cdot d\mathbf S =\iiint_V div \mathbf F\,dV =\iiint_V 1\,dV=半球體積=\bbox[red, 2pt]{{2\over 3}\pi}$$
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解答：$$取\cases{x(t)=\cos t\\ y(t)= \sin t},0\le t\le 2\pi \Rightarrow \cases{dx = -\sin tdt\\ dy=\cos t dt\\ ye^{xy}= \sin te^{\sin(2t)/2} \\ xe^{xy}= \cos te^{\sin(2t)/2}} \\ \Rightarrow \int_C \mathbf F\cdot d\mathbf r =\int_0^{2\pi } \sin te^{\sin(2t)/2}\cdot (-\sin tdt)+ \cos te^{\sin(2t)/2}\cdot \cos tdt \\=\int_0^{2\pi } e^{\sin(2t)/2}(\cos^2t-\sin^2 t)\,dt=\int_0^{2\pi } e^{ \sin(2t)/2}\cos(2t)\,dt =\left. \left[ e^{\sin(2t)/2} \right] \right|_0^{2\pi}= \bbox[red, 2pt]0$$
解答：$$利用散度定理來求解，div \mathbf F= \frac{\partial }{\partial x}F_1 + \frac{\partial }{\partial y}F_2 + \frac{\partial }{\partial z}F_3 =z^2+ x^2+ y^2=1 \\ \Rightarrow \iint_S \mathbf F\cdot d\mathbf S =\iiint_V div \mathbf F\,dV =\iiint_V 1\,dV=半球體積=\bbox[red, 2pt]{{2\over 3}\pi}$$
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