113年台綜大轉學考-工程數學D37詳解

 臺灣綜合大學系統113學年度學士班轉學生聯合招生考試

科目名稱:工程數學
類組代碼:D37

解答: $$\bbox[cyan,2pt]{題目有疑義}, 只有dx沒有dy$$

$$\bbox[cyan,2pt]{題目有疑義}, 只有dx沒有dy$$

$$\bbox[cyan,2pt]{題目有疑義}, 只有dx沒有dy$$

解答: $$\mathbf F(x,y,z)=5yz \mathbf i+x^2z \mathbf j+3x^3\mathbf k \Rightarrow \text{curl }\mathbf F =\begin{vmatrix} \mathbf i &\mathbf j &\mathbf k \\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\\mathbf F_1 & \mathbf F_2 &\mathbf F_3 \end{vmatrix} =\begin{vmatrix} \mathbf i &\mathbf j &\mathbf k \\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 5yz & x^2z & 3x^3 \end{vmatrix} \\= 0+2xz \mathbf k+5y \mathbf j -5z\mathbf k-9x^2 \mathbf j-x^2\mathbf i =(-x^2,-9x^2+5y,2xz-5z), 故選\bbox[red, 2pt]{(E)}$$

解答: $$\mathbf F(x,y,z)=ye^{5xy} \mathbf i+x^2 \sin yz \mathbf j+\cos xz^3\mathbf k \Rightarrow  \text{div }\mathbf F=\frac{\partial }{\partial x} \mathbf F_1 + \frac{\partial }{\partial y} \mathbf F_2 + \frac{\partial }{\partial z} \mathbf F_3\\  =5y^2 e^{5xy} +x^2z \cos yz- 3xz^2 \sin xz^3, 故選\bbox[red, 2pt]{(C)}$$

解答: $$y=c_1e^{-x} +c_2e^x \Rightarrow y'=-c_1e^{-x}+ c_2e^x \Rightarrow \cases{y(0) =c_1+c_2=0\\ y'(0) =-c_1+c_2=1} \Rightarrow \cases{c_1=-1/2\\ c_2=1/2} \\ \Rightarrow y=-{1\over 2}e^{-x}+{1\over 2}e^x, 故選\bbox[red, 2pt]{(B)}$$

解答: $$L\{f(t)\} =\int_0^\infty f(t)e^{-st}\,dt =\int_0^1 -e^{-st}\,dt +\int_1^\infty e^{-st}\, dt = \left. \left[{1\over s} e^{-st}\right] \right|_0^1 + \left. \left[ -{1\over s}e^{-st} \right] \right|_1^\infty \\= {1\over s}\left( e^{-s}-1\right) -{1\over s}\left(0-e^{-s} \right) ={2\over s}e^{-s} -{1\over s},故選\bbox[red, 2pt]{(E)}$$

解答: $$\cases{x_1-x_2+3x_3=-1\\ x_1-3x_2+4x_3=5\\ x_1-x_2+6x_3=2} \Rightarrow \begin{bmatrix} 1 &-1  &3  \\1 & -3 & 4 \\1 &-1  & 6\end{bmatrix} \begin{bmatrix}  x\\ y\\z\end{bmatrix} = \begin{bmatrix}  -1\\ 5 \\2 \end{bmatrix} \\ A=\begin{bmatrix} 1 &-1  &3  \\1 & -3 & 4 \\1 &-1  & 6\end{bmatrix} \Rightarrow rref(A) = \left[\begin{matrix}1 & -1 & 3 \\0 & -2 & 1 \\0 & 0 & 3 \end{matrix} \right] \Rightarrow rank(A) =3 \Rightarrow \text{ unique solution} ,故選\bbox[red, 2pt]{(D)}$$

解答: $$AK_1 =\begin{bmatrix} -2\\ 9 \\-1\end{bmatrix} \ne \lambda K_1, \forall \lambda \in \mathbb R\\ AK_2= \begin{bmatrix} 5\\ 10 \\9\end{bmatrix} \ne \lambda K_2, \forall \lambda \in \mathbb R\\ AK_3=\begin{bmatrix} 3\\ 12 \\9\end{bmatrix} = 3 K_3 \Rightarrow K_3 \text{ is a eigenvector} \\ 故選\bbox[red, 2pt]{(E)}$$

解答: $$A  =\begin{bmatrix}0& 1 \\-1&0\end{bmatrix}  \Rightarrow \det(A-\lambda I)=\lambda^2+1=0 \Rightarrow \lambda=\pm i \\ \Rightarrow A=\begin{bmatrix}-i& i \\ 1& 1\end{bmatrix} \begin{bmatrix}i& 0 \\0& -i\end{bmatrix} \begin{bmatrix}i/2& 1 /2\\-i/2& 1/2\end{bmatrix} =\begin{bmatrix}i& -i \\ 1& 1\end{bmatrix} \begin{bmatrix}-i& 0 \\0& i\end{bmatrix} \begin{bmatrix}-i/2& 1 /2\\i/2& 1/2\end{bmatrix} ,故選\bbox[red, 2pt]{(CD)}$$

解答: $$選\bbox[red, 2pt]{(E)}$$

解答: $$\left[ \begin{array}{rr|rr}1 & 5 & 1 & 0\\2 & 4 & 0 & 1 \end{array}\right] \xrightarrow{R_2-2R_1\to R_2}\left[ \begin{array}{rr|rr}1 & 5 & 1 & 0\\0 & -6 & -2 & 1 \end{array}\right] \xrightarrow{R_2/(-6) \to R_2} \left[\begin{array}{rr|rr}1 & 5 & 1 & 0\\0 & 1 & \frac{1}{3} & - \frac{1}{6} \end{array}\right] \\ \xrightarrow{R_1-5R_2\to R_1} \left[\begin{array}{rr|rr}1 & 0 & - \frac{2}{3} & \frac{5}{6}\\0 & 1 & \frac{1}{3} & - \frac{1}{6} \end{array}\right] \Rightarrow A=\begin{bmatrix}  - \frac{2}{3} & \frac{5}{6}\\  \frac{1}{3} & - \frac{1}{6} \end{bmatrix},故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{P(x,y)= x+2y\\ Q(x,y)= 2x-y} \Rightarrow P_y=2=Q_x \Rightarrow \text{independent} \\ \Phi(x,y)=\int (x+2y)\,dx = \int(2x-y)dy \Rightarrow \Phi(x,y)={1\over 2}x^2+2xy-{1\over 2}y^2+c_1 \\ \Rightarrow \int_{(1,0)}^{(3,2)} P\,dx+Q\,dy=\Phi(3,2)-\Phi(1,0)=14,故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{u=x^2-y^2\\ v=xy} \Rightarrow {\partial(u,v)\over \partial(x,y)} =\begin{vmatrix}2x& -2y\\ y& x \end{vmatrix} =2x^2+2y^2=2(x^2+y^2),故選\bbox[red, 2pt]{(C)}$$

解答: $$\iint_R (x^2+y^2)\sin xy\,dA = \int_{1}^9 \int_{-2}^2 {1\over 2} \sin v\,dv du=0,故選\bbox[red, 2pt]{(D)}$$

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解題僅供參考，轉學考歷年試題及詳解