114年成大水利及海洋碩士班-工程數學詳解

 國立成功大學114學年度碩士班招生考試

系所:水利及海洋工程學系
科目:工程數學

解答: $$\textbf{(1) }\nabla =(\frac{\partial }{\partial x},\frac{\partial }{\partial y}, \frac{\partial }{\partial z}) \Rightarrow \nabla \left({f\over g} \right) =  {\nabla f\over g}+f \nabla({1\over g}) ={g\nabla f\over g^2} -{f\nabla g\over g^2} ={1\over g^2}(g\nabla f -f\nabla g) \quad \bbox[red, 2pt]{QED} \\ \mathbf{(2)} \cases{f=f(x,y,z) \\ \mathbf v=(v_1,v_2,v_2)} \Rightarrow   div(f\mathbf v) =div((fv_1, fv_2, fv_3)) ={\partial \over \partial x}(fv_1) +{\partial \over \partial y}(fv_2) +{\partial \over \partial z}(fv_3) \\ =v_1{\partial \over \partial x}f + f{\partial \over \partial x}v_1 +v_2{\partial \over \partial y}f + f{\partial \over \partial y}v_2 + v_3{\partial \over \partial z}f + f{\partial \over \partial z}v_3  \\=\left(v_1{\partial \over \partial x}f + v_2{\partial \over \partial y}f + v_3{\partial \over \partial z}f \right) + f \left( {\partial \over \partial x}v_1+{\partial \over \partial y}v_2 + {\partial \over \partial z}v_3\right)= \mathbf v\cdot \nabla f+f \,div \mathbf v \\ \Rightarrow div(f\mathbf v) =f\,div \mathbf v+\mathbf v\cdot \nabla f \qquad \bbox[red, 2pt]{\text{QED}}$$

解答: $$\mathbf{(1) }\det(A)=(1-p)^2-16 =(p-5)(p+3) =0 \Rightarrow p=-3,5 \Rightarrow \bbox[red, 2pt] {\begin{cases} rank(A)=1 &  p=-3,5\\ rank(A)=2& \text{otherwise}\end{cases}} \\\mathbf{(2) }A=\begin{bmatrix} p&-q\\ q&p\end{bmatrix} \Rightarrow A^T= \begin{bmatrix} p& q\\ -q&p\end{bmatrix} \Rightarrow AA^T =\begin{bmatrix} p^2+q^2&0\\ 0& p^2+q^2\end{bmatrix} =I \Rightarrow \bbox[red, 2pt]{p^2+q^2=1}$$

解答: $$\mathbf{(1) }y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow y_h= c_1\cos x+ c_2\sin x\\\qquad y_p= Ax\cos x+Bx\sin x \Rightarrow y_p'=A\cos x-Ax\sin x+B\sin x+ Bx\cos x \\ \qquad \Rightarrow y_p''=-2A\sin x -Ax\cos x+ 2B\cos x -Bx\sin x \\\qquad \Rightarrow y_p''+y_p=-2A\sin x+2B\cos x= \cos x \Rightarrow \cases{-2A=0 \\2B=1} \Rightarrow \cases{A=0\\ B=1/2} \Rightarrow y_p= {1\over 2}x\sin x\\ \qquad \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos x+c_2\sin x+ {1\over 2}x\sin x}\\ \mathbf{(2) }y''+ 2y'+y=0 \Rightarrow \lambda^2+ 2\lambda+1=0 \Rightarrow (\lambda+1)^2=0 \Rightarrow \lambda= -1 \Rightarrow y_h= c_1e^{-x}+ c_2xe^{-x} \\\qquad y_p= A\cos x+ B\sin x +Cx\cos x+ Dx\sin x \\\qquad \Rightarrow y_p'= (-A+D)\sin x+(B+C)\cos x -Cx\sin x+  Dx\cos x\\ \qquad \Rightarrow y_p''=(-A+ 2D) \cos x-(B+ 2C)\sin x-Cx\cos x -Dx\sin x \\\qquad \Rightarrow y_p''+2 y_p'+y_p \\\qquad \quad =(2B+2C+2D) \cos x+ (-2A-2C+2D)\sin x+2D x\cos x-2Cx\sin x= 2x\cos x \\ \qquad \Rightarrow \cases{B+C+D=0\\ -2A-2C+2D=0\\C=0\\D=1} \Rightarrow \cases{A=1\\B=-1\\ C=0\\ D=1} \Rightarrow y_p=\cos x-\sin x+ x\sin x \\\qquad \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y =  c_1e^{-x}+ c_2xe^{-x}+\cos x-\sin x+ x\sin x }$$

解答: $$\mathbf{(1) } \cases{x=\cos \theta\\ y=\sin \theta} \Rightarrow \cases{dx=-\sin \theta d\theta\\ dy=\cos \theta d\theta} \Rightarrow \oint_C \mathbf F\cdot d\mathbf r = \oint_C ydx-xdy =\int_0^{2\pi} (-\sin^2 \theta-\cos^2\theta)\,d\theta \\\quad = \int_0^{2\pi} -1\,d\theta =-2\pi \cdots(1)\\\quad \text{By Green Theorem, }\cases{P(x,y)=y\\ Q(x,y)=-x} \Rightarrow \oint_C ydx-xdy =\iint( Q_x-P_y)\,dA = \iint -2\,dA\\ \quad =(-2)\cdot 1^2\pi =-2\pi \cdots(2)\\\qquad (1)=(2)=-2\pi \qquad \bbox[red, 2pt]{\text{QED}}\\ \mathbf{(2) } \mathbf F=[0,x,0] \Rightarrow div \,\mathbf F=0 \\\quad  \text{By Divergence Theorem of Gauss}, \iint_S \mathbf F\cdot \mathbf n\,d A = \iiint_T div \mathbf \,F\,dV =\bbox[red, 2pt]0$$

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

解題僅供參考，碩士班歷年試題及詳解