107年台聯大轉學考-微積分A3A4A7詳解

 台灣聯合大學系統107學年度學士班轉學生考試

科目：微積分
類組別：A3 A4 A7
甲、填充題：共8題，每題8分，共64分

解答：$$y=f(x)=1-{4\over \pi^2} (\tan^{-1} x)^2 \Rightarrow f'(x)= -{8\over \pi^2} (\tan^{-1} x)\cdot {1\over 1+x^2} =0 \\ \Rightarrow \tan^{-1}x =0 \Rightarrow x=\bbox[red, 2pt]0$$

解答：$$x^2-16 \le 0 \Rightarrow -4\le x\le 4 \Rightarrow \int_a^b (x^2-16)\,dx 的最小值=\int_{-4}^4 (x^2-16)\,dx \\ =\left. \left[ {1\over 3}x^3 -16x\right]\right|_{-4}^4 = \bbox[red, 2pt]{-{256\over 3}}$$

解答：$$令u=e^x \Rightarrow du=e^x\,dx \Rightarrow \int_{-\infty}^\infty {e^x\over 1+e^{2x}}\,dx =\int_0^\infty {1\over 1+u^2}\,d u =\left.\left[ \tan^{-1}x \right]\right|_0^\infty = \bbox[red, 2pt]{\pi \over 2}$$

解答：$$f(x,y)= (x^3+y^3)^{1/3} \Rightarrow f_y(x,y)= y^2(x^3+y^3)^{-2/3} \Rightarrow f_y(0,0)= \bbox[red,2pt]0$$

解答：$$f(x,y)= 4-2x-2y \Rightarrow \cases{f_x= -2\\ f_y=-2} \Rightarrow 欲求之表面積= \iint_D \sqrt{f_x^2 +f_y^2 +1}\,dA \\ =\iint_D 3\,dA = 3\times 1^2\pi\times {1\over 4} =\bbox[red,2pt]{3\pi \over 4}$$

解答：$$r(u,v)= u\mathbf i+v\mathbf j +(u^2+v^2) \mathbf k \Rightarrow z=x^2+y^2 \Rightarrow f(x,y,z)=x^2+y^2-z\\ \Rightarrow \cases{f_x=2x\\ f_y=2y\\ f_z=-1} \Rightarrow \cases{f_x(1,2,5)= 2\\ f_y(1,2,5)= 4\\ f_z(1,2,5)=-1} \Rightarrow 切平面:2(x-1) +4(y-2)-(z-5)=0 \\ \Rightarrow \bbox[red, 2pt]{2x+ 4y-z=5}$$

解答：$$取\cases{x=r\cos \theta\\ y= r\sin \theta} \Rightarrow \int_0^\infty \int_0^\infty {1\over (1+x^2+y^2)^2}\,dx dy = \int_0^{\pi/2} \int_0^\infty {r\over (1+r^2)^2} \,drd\theta \\ 再取u=1+r^2 \Rightarrow du =2rdr,則上式變為\int_0^{\pi/2} \int_1^\infty {1\over 2u^2} \,dud\theta = \int_0^{\pi/2} \left.\left[ -{1\over 2u} \right]\right|_1^\infty \,d\theta \\ =\int_0^{\pi/2} {1\over 2}\,d\theta = \bbox[red,2pt]{\pi \over 4}$$

解答：

$$令\cases{u=x+4y\\ v=x-y} \Rightarrow \cases{x=(u+ 4v)/5\\ y=(u-v)/5} \Rightarrow {\partial(x,y)\over \partial (u,v)} =\begin{vmatrix} 1/5 & 4/5\\ 1/5 & -1/5\end{vmatrix} = -{1\over 5} \\ \Rightarrow \iint_R \sqrt{(x+4y)(x-y)} = -{1\over 5} \int_5^0 \int_0^5  \sqrt{uv}\,dudv =-{1\over 5}\int_5^0 \left.\left[ {2\over 3}u^{3/2}v^{1/2} \right]\right|_0^5\, dv \\=-{2\over 15} 5^{3/2}\int_5^0  v^{1/2}\,dv = {2\over 15} 5^{3/2} \cdot {2\over 3}5^{3/2} ={4\over 45}\cdot 5^3 = \bbox[red, 2pt]{100 \over 9}$$

乙、計算、證明題：共3題，每題12分，共36分

解答：$$\mathbf{a.}\;T(x,y)= 64-2x^2-y^2 \Rightarrow \nabla T=(T_x,T_y)=(-4x,-2y) \Rightarrow \nabla T(-1,2)= (4,-4)\\ \Rightarrow 沿著方向\bbox[red, 2pt]{(4,-4)} 溫度升高最快，其值為||\nabla T(-1,2)|| =\sqrt{4^2+ (-4)^2} = \bbox[red, 2pt]{4\sqrt 2} \\\mathbf{b.}\;路徑為以(-1,2)為起點,向量為(4,-4)的直線即\bbox[red, 2pt]{\{(4t-1,-4t+2)\mid t\in \mathbb R且t\gt 0\}}$$

解答：$$\mathbf{a.}\;\lim_{n\to \infty} \left({3n+2\over n+3}\right)^n =\lim_{n\to \infty} \left({3+2/n\over 1+3/n}\right)^n = \infty  \Rightarrow \bbox[red, 2pt]{發散}\\ \mathbf{b.}\;積分檢定法\text{ integral test: }\int_1^\infty {e^{2/x}\over x^2}\,dx = \left.\left[ -{1\over 2} e^{2/x}\right]\right|_1^\infty =-{1\over 2}+{1\over 2}e^2 \lt \infty \Rightarrow 級數\bbox[red, 2pt]{收斂}$$

解答：$$令\cases{M(x,y)=y^3\\ N(x,y)=27x-x^3} \Rightarrow \cases{M_y= 3y^2\\ N_x=27-3x^2}，依格林定理:\int_C M\,dx+ N\,dy  =\iint_R (N_x-M_y )dA\\ =\iint_R 27-3(x^2+y^2)dA,其中R為圓C所圍區域; 此積分要最大，必須27-3(x^2+y^2)\ge 0 \\ \Rightarrow x^2+y^2 \le 9,即半徑為3的圓,因此該積分變為\int_0^{2\pi}\int_0^3 (27-3r^2) r\,drd\theta =2\pi \left.\left[ {27\over 2}r^2- {3\over 4}r^4 \right] \right|_0^3 \\=2\pi \left({243\over 2} -{243\over 4}\right) = \bbox[red, 2pt]{{243\over 2}\pi}$$

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