113年成大海洋碩士班-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:水利及海洋工程
科目:工程數學

解答: $$\textbf{(1)}\;\nabla g =(g_x,g_y,g_z) \Rightarrow \nabla {1\over g}=(-{g_x\over g^2}, -{g_y\over g^2}, -{g_z\over g^2} ) =-{1\over g^2}\nabla g\\ \Rightarrow \nabla(fg)=g\nabla f+f\nabla g \Rightarrow \nabla(f/g)={1\over g}\nabla f+f\nabla{1\over g}={1\over g^2}\cdot g\nabla f+f\cdot (-{1\over g^2}\nabla g) \\={1\over g^2}(g\nabla f-f\nabla g). \bbox[red, 2pt]{QED.} \\\textbf{(2)}\; \text{div}(f\mathbf v) =\frac{\partial  }{\partial x}(f\mathbf v_1) +\frac{\partial  }{\partial y}(f\mathbf v_2) +\frac{\partial  }{\partial z}(f\mathbf v_3)  \\\qquad =\left( f\frac{\partial  }{\partial x}\mathbf v_1 +\mathbf v_1\frac{\partial  }{\partial x} f  \right) + \left(f\frac{\partial  }{\partial y} \mathbf v_2+\mathbf v_2\frac{\partial  }{\partial y} f\right) + \left( f\frac{\partial  }{\partial z}\mathbf v_3 +\mathbf v_3\frac{\partial  }{\partial z}f \right)\\ \qquad = f\left( \frac{\partial  }{\partial x}\mathbf v_1 +\frac{\partial  }{\partial y}\mathbf v_2 +\frac{\partial  }{\partial z} \mathbf v_3\right) + \left( \mathbf v_1\frac{\partial  }{\partial x} f +\mathbf v_2\frac{\partial  }{\partial y} f +\mathbf v_3\frac{\partial  }{\partial z} f \right) \\\qquad =f\;\text{div } \mathbf v + \mathbf v \cdot \nabla f \; \bbox[red, 2pt]{QED}$$

解答: $$\textbf{(1) } A= \begin{bmatrix}1-p& 4\\ 4& 1-p \end{bmatrix}  \Rightarrow \det(A)=(1-p)^2-16 =0 \Rightarrow p=-3,5 \\\qquad \Rightarrow \bbox[red, 2pt]{rank(A)=\begin{cases} 1,& p=5,-3\\ 2,& \text{otherwise}\end{cases}}\\ \textbf{(2) }AA^T=I \Rightarrow \begin{bmatrix}p&-q\\ q& p \end{bmatrix} \begin{bmatrix} p& q\\ -q& p\end{bmatrix} =\begin{bmatrix} p^2+q^2& 0\\ 0 & p^2+q^2\end{bmatrix} =\begin{bmatrix}1& 0\\ 0 & 1 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{p^2+q^2=1}$$

解答: $$\textbf{(1) }y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda= \pm i \Rightarrow y_h= c_1 \cos x+ c_2\sin x\\\qquad y_p= A x\cos x+ Bx\sin x \Rightarrow y_p'=A \cos x-Ax\sin x+B\sin x+ Bx \cos x \\ \qquad \Rightarrow y_p''=-2A\sin x -Ax\cos x+ 2B\cos x - Bx\sin x  \\ \qquad \Rightarrow y_p''+y_p =-2A\sin x+2B\cos x = \cos x \Rightarrow \cases{A=0\\ B=1/2} \Rightarrow y_p= {1\over 2}x\sin x \\ \qquad \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos x+ c_2\sin x+{1\over 2}x\sin x} \\\textbf{(2) }y''+2y'+y=0 \Rightarrow \lambda^2+2 \lambda+1= (\lambda+1)^2=0 \Rightarrow \lambda=-1\Rightarrow y_h=c_1e^{-x} +c_2xe^{-x} \\ \quad \cases{y_1=e^{-x} \\ y_2=xe^{-x}} \Rightarrow W=\begin{vmatrix}y_1& y_2 \\ y_1'& y_2' \end{vmatrix} =e^{-x}\\ \text{Using variations of parameters}, y_p=-e^{-x}\int{xe^{-x} 2x\cos x\over e^{-x}}\,dx +xe^{-x} \int{ e^{-x} 2x \cos x\over e^{-x}}\,dx \\\qquad = (x-1)\sin + \cos x \Rightarrow y= y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-x} +c_2xe^{-x} +(x-1)\sin x + \cos x}$$

解答: $$\textbf{(1) }r(t)=[\cos t, \sin t] \Rightarrow r'(t)=[-\sin t, \cos t] \Rightarrow \mathbf F=[y, -x ]=[\sin t,-\cos t] \\ \Rightarrow \int \mathbf F\cdot dr =\int_0^{2\pi} [\sin t, -\cos t]\cdot [-\sin t, \cos t]\,dt = \int_0^{2\pi} -1\,dt =-2\pi\\ \text{By Green's theorem,}  \int \mathbf F\cdot dr =\iint_R \left( \frac{\partial }{\partial x}(-x)-\frac{\partial }{\partial y}y \right)dA =\iint_R -2\,dA=-2\cdot \pi=-2\pi \bbox[red, 2pt]{QED} \\\textbf{(2) }\mathbf F=[0,x,0] \Rightarrow \text{div }\mathbf F=0\\\text{Using Divergence Theorem,} \iint_S \mathbf F\cdot dS=\iiint_E \text{div }\mathbf F\,dV=\bbox[red, 2pt]{0}$$

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