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2024年9月2日 星期一

113年成大資源工程碩士班-工程數學詳解

 國立成功大學113學年度碩士班招生考試

系所:資源工程學系
科目:工程數學

解答:(a)dydx=y+x2yy=x2exyyex=x2ex(exy)=x2exexy=x2exdx=ex(x2+2x+2)+c1y=(x2+2x+2)+c1exy(0)=2+c1=1c1=3y=3ex(x2+2x+2)(b)y+y=0yh=c1cosx+c2sinxyp=Axcosx+Bxsinxyp=Acosx+BsinxAxsinx+Bxcosxyp=2Asinx+2BcosxAxcosxBxsinxyp+yp=2Asinx+2Bcosx=cos(x){A=0B=1/2yp=12xsinxy=yh+yp=c1cosx+c2sinx+12xsinxy=(12c1)sinx+c2cosx+12xcosx{y(0)=c1=0y(0)=c2=0y=12xsinx
解答:(a)L{exp(3t2)}=0e3t2estdt=e20e(3s)tdt=e2[13se(3s)t]|0=e2s3(b)L1{s+2(s+1)s}=L1{2s1s+1}=L1{2s}L1{1s+1}=2et


解答:(a-1)A=[1112112112112111]R2R1R2,R3R1R3,R42R1R4[1112001101010113]R1R3R1,R1+R4R4[1013001101010014]R4+R2R4[1013001101010005]R2R3,R4=R4/(5)[1013010100110001]R1R3R1,R2+R4[1004010000110001]R14R4R1,R3+R4R3[1000010000100001]=I4rref(A)=[1000010000100001](a-2)rref(A)=I4rank(A)=4nullity(A)=4rank(A)=0(b)B=[1111]det
解答:\textbf{(a)}\;\vec F=3x^2y^3\vec i-2y^3z^2\vec j+x^3z^2\vec k \Rightarrow \text{curl }\vec F=\begin{vmatrix}\vec i & \vec j& \vec k \\{\partial \over \partial x} & {\partial \over \partial y} &{\partial \over \partial z} \\\vec F_1 & \vec F_2 & \vec F_3\end{vmatrix} =\begin{vmatrix}\vec i & \vec j& \vec k \\{\partial \over \partial x} & {\partial \over \partial y} &{\partial \over \partial z} \\3x^2y^3 & -2y^3z^2 & x^3z^2\end{vmatrix} \\ =0\vec i+ 0\vec j+0\vec k -9x^2y^2\vec k-3x^2z^2 \vec j+4y^3z \vec i  \Rightarrow \text{curl }\vec F(2,-1,1) =\bbox[red, 2pt]{-4\vec i-12\vec j-36\vec k} \\\textbf{(b)}\; \cases{f(x,y, z)={xy^2\over z^3} \Rightarrow \nabla f =({y^2\over z^3},{2xy\over z^3}, -{3xy^2\over z^4} ) \\ \vec n= 2\vec i+2\vec j+ 1\vec k \Rightarrow{\vec n\over |\vec n|}={2\over 3}\vec i+ {2\over 3}\vec j+{1\over 3}\vec k}   \\ \Rightarrow \nabla f\cdot {\vec n\over |\vec n|} ={2\over3}\cdot {y^2\over z^3} +{2\over 3}\cdot {2xy\over z^3 } -{1\over3}\cdot {3xy^2\over z^4}\\ \Rightarrow \nabla f(3,2,1)\cdot {\vec n\over |\vec n|}  ={8\over 3}+8-12 = \bbox[red, 2pt]{-{4 \over 3}}
\textbf{(c)}\;\text{Let }M(x,y) \text{ and }N(x,y) \text{ be continuous and have continuous partial derivatives } \\\text{in a region }R \text{ and on its boundary }C. \text{ Then, for both simply-connected and }\\\text{ multiply-connected }R.\\ \oint_C (Mdx+ Ndy)= \displaystyle \iint_R \left( {\partial N\over \partial x} -{\partial M\over \partial y}\right)\,dxdy
(d) Let S be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. Also let \vec F be a vector field then,\int_C \vec F\cdot d\vec r= \iint_S \text{curl }\vec F\cdot d\vec S
解答:\textbf{(a)}\;a_0= {1\over 2}\int_{-1}^1 f(x)  \,dx = {1\over 2}\int_{0}^1   1\,dx  ={1\over 2} \\ a_n= \int_0^1\cos(n\pi x)\,dx =\left.\left[{1\over n\pi} \sin(n\pi x) \right] \right|_0^1=0\\ b_n= \int_0^1 \sin(n\pi x)\,dx =\left.\left[-{1\over n\pi} \cos(n\pi x) \right] \right|_0^1= {1\over n\pi}(1-(-1)^n) \\ \Rightarrow f(x)=\bbox[red, 2pt]{{1\over 2}+\sum_{n=1}^\infty  {1\over n\pi}(1-(-1)^n) \sin(n\pi x)} \\ \textbf{(b)}\;\mathcal F(\omega) =\int_{-\infty}^\infty f(t)e^{-j\omega t}\,dt =\int_0^\infty ke^{-at} e^{-j\omega t}\,dt = k\int_0^\infty   e^{-(a+j\omega) t}\,dt =k \left. \left[-{1\over a+j\omega} e^{-(a+j\omega) t}\right] \right|_0^\infty \\= \bbox[red, 2pt]{k\over a+j\omega}

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解題僅供參考,其他碩士班歷年試題及詳解

2 則留言:

  1. 第三題的(b)eigenvalues已求錯,故eigenvectors 也不對.
    第四題的(b)方向導數不對,因為那單位向量是(2/3,2/3,1/3)

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