106年海洋大學轉學考-微積分詳解

國立臺灣海洋大學106學年度轉學生入學招生考試

考試科目：微積分

解答：$$\mathbf{(a)}\;f(x)={\sin x\over x} \Rightarrow f'(x)={\cos x\over x}-{\sin x\over x^2} \Rightarrow \lim_{x\to \pi/2} f'(x)=0-{4\over \pi^2} =\bbox[red,2pt]{-{4\over \pi^2}}\\ \mathbf{(b)}\; L=\lim_{x\to 0^+} x^{\ln 3\over 1+\ln x} \Rightarrow \ln L =\lim_{x\to 0^+} {\ln 3\over 1+\ln x}\ln x =\ln 3 \lim_{x\to 0^+}{\ln x\over 1+\ln x}=\ln 3 \lim_{x\to 0^+} {1/x\over 1/x} \\\qquad =\ln 3 \Rightarrow L=\bbox[red, 2pt]3$$

解答：$$\mathbf{(a)}\;f(t)={\sqrt{3t+2} \over t} \Rightarrow f'(t)={{3\over 2 t\sqrt{3t+2}}} -{\sqrt{3t+2}\over t^2} ={3t-2(3t+2) \over 2t^2 \sqrt{3t+2}} =\bbox[red,2pt]{-3t-4\over 2t^2 \sqrt{3t+2}} \\\mathbf{(b)}\; y=(\sin x)^x =e^{x\ln (\sin x)} \Rightarrow y'=(\ln (\sin x)+{x \cos x\over \sin x})e^{x\ln (\sin x)} =(\ln (\sin x)+{x \cos x\over \sin x})(\sin x)^x \\\qquad = \bbox[red, 2pt]{(\sin x)^x \ln(\sin x)+ x\cos x(\sin x)^{x-1}} \\\mathbf{(c)}\; g(x)= \sin^3(\cos x) \Rightarrow g'(x)= 3\sin^2(\cos x) \cdot \cos(\cos x) \cdot (-\sin x) \\\qquad =\bbox[red,2pt]{-3\sin x\cos(\cos x) \sin^2(\cos x)}$$

解答：$$3e^{xy}-x=0 \Rightarrow 3(y+xy')e^{xy}-1=0,將\cases{x=3\\ y=0} 代入 \Rightarrow 9y'-1=0 \Rightarrow y'={1\over 9}\\ \Rightarrow 切線方程式：y={1\over 9}(x-3) \Rightarrow \bbox[red,2pt]{x-9y=3}$$

解答：$$\mathbf{(a)}\;f(x)=\ln (e^{\sqrt{2x+1}}) \Rightarrow f(4)=\ln(e^\sqrt{9}) =\bbox[red, 2pt]3 \\\mathbf{(b)}\; \tan \theta={3\over 7} \Rightarrow \sin \theta =\sin (\tan^{-1}{3\over 7})={3\over \sqrt{58}} =\bbox[red,2pt]{ 3\sqrt{58}\over 58}$$

解答：$$外直徑12英尺相當於半徑6英尺，殼厚0.3英吋＝0.3\div 12=0.025英尺\Rightarrow dr= -0.025\\球體積V＝{4\over 3}\pi r^3 \Rightarrow dV= 4\pi r^2dr = 4\pi 6^2\cdot (-0.025) =-3.6\pi,而V(r=6)={4\over 3}\pi 6^3=288\pi \\\Rightarrow 體積近似值＝288\pi -3.6\pi = \bbox[red, 2pt]{284.4\pi}$$

解答：$$\mathbf{(a)}\;f(x,y)=x^2y^3-4y^2 \Rightarrow \nabla f(x,y)=(f_x,f_y) = \bbox[red, 2pt]{(2xy^3,3x^2y^2-8y)} \\\mathbf{(b)}\;z=f(x,y) \Rightarrow dz=f_x\cdot dx + f_y\cdot dy = \bbox[red,2pt]{2xy^3dx+ (3x^2y^2-8y)dy} \\ \mathbf{(c)}\; D_uf(1,2) = \nabla f(1,2)\cdot {\vec u\over |\vec u|} =(16,-4)\cdot ({3\over 5},-{4\over 5}) = \bbox[red,2pt]{64\over 5}$$

解答：$$\mathbf{(a)}\;\int_0^1 \int_0^y xy^2\,dxdy = \int_0^1 [{1\over 2}x^2y^2]|_0^y \,dy = \int_0^1 {1\over 2}y^4\,dy = \left. {1\over 10}y^5 \right|_0^1 =\bbox[red, 2pt]{1\over 10}\\ \mathbf{(b)}\; \iint_R ye^x\,dA = \int_0^4 \int_{y}^{(12-y)/2} ye^x\,dxdy =\int_0^4 ye^{(12-y)/2}-ye^y\, dy \\\qquad =\left.\left[ -2(y+2)e^{(12-y)/2} -(y-1) e^y \right]\right|_0^4  =-15e^4 -(-4e^6+1) =\bbox[red, 2pt]{4e^6-15e^4-1} \\\mathbf{(c)}\; \int_0^1 \int_x^1 e^{y^2}\,dydx =\int_0^1 \int_0^y e^{y^2}\,dxdy = \int_0^1 ye^{y^2}\,dy =\left. \left[ {1\over 2}e^{y^2} \right] \right|_0^1 =\bbox[red, 2pt]{{1\over 2}(e-1)}$$

解答：$$\int x^3e^{-2x}\,dx =-{1\over 2}x^3e^{-2x}+{3\over 2}\int x^2e^{-2x} \,dx=-{1\over 2}x^3e^{-2x}-{3\over 4}x^2 e^{-2x} +{3\over 2}\int xe^{-2x}\,dx\\ =-{1\over 2}x^3e^{-2x}-{3\over 4}x^2 e^{-2x}-{3\over 4}xe^{-2x}-{3\over 8}e^{-2x}+C \Rightarrow \int_0^\infty x^3e^{-2x}\,dx = 0-(-{3\over 8})=\bbox[red, 2pt]{3\over 8}$$

解答：$$x^2-y^2-z^2=1 \Rightarrow x^2=y^2+z^2+1  \\\Rightarrow 與原點距離的平方＝x^2 +y^2+z^2 =f(y,z)= (y^2+z^2+1)+ y^2+z^2= 2y^2+ 2z^2+1 \\ 與原點最近相當於求f(y,z)的最小值；由於\cases{f_y= 4y=0\\ f_z=4z=0} \Rightarrow \cases{y=0\\ z=0} \Rightarrow x^2=1 \Rightarrow x=\pm 1\\ \Rightarrow 與原點最近的點\bbox[red,2pt]{(\pm 1,0,0)}$$

解答：

$$令\cases{u=x+y\\ v=y-x} \Rightarrow \cases{x=(u-v)/2\\ y=(u+v)/2} \Rightarrow {\partial(x,y)\over \partial(u,v)} =\begin{vmatrix} 1/2 & -1/2\\ 1/2 & 1/2\end{vmatrix} ={1\over 2} \\ \Rightarrow \iint_R \left({x-y\over x+y}\right)^4 \,dA =\int_{u=1}^{u=2} \int_{v=-u}^{v=u} ({v\over u})^4\cdot {1\over 2}\,dvdu =\int_1^2 \left.\left[ {1\over 10 u^4}v^5 \right]\right|_{-u}^u\,du =\int_1^2 {1\over 5}u\,du \\ =\left.\left[ {1\over 10}u^2 \right]\right|_1^2 = \bbox[red,2pt]{3\over 10}$$

==================== END ====================

解題僅供參考，其他轉學考歷屆試題及詳解