00023 高等数学（工本）2024 年4月真题解析
一、单选设向量\(a=\{-1,0,1\}\)且\(b=\{1,-1,1\}\),则\(2a+b=\)( D )解：\[2a+b=\{-2,0,2\}+\{1,-1,1\}=\{-1,-1,3\}\].设函数\(z=x^2+y^2\),则全微分\(dz\big |_{(1,1)}\)= ( C )解：\[\begin{aligned}dz &=\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial x}\\&=2xdx+2ydy\end{aligned}\\dz\big|_{(1,1)} = 2dx+2dy\]下列微分方程中，可分离变量的微分方程是( A )\[A. \dfrac{dy}{dx}= -\dfrac{y}{x} \quad B.\dfrac{dy}{dx}= e^{xy} \quad C.\dfrac{dy}{dx}= x^2+y^2\quad \dfrac{dy}{dx}= x^2+y^2\]解：A 直接符合可分离变量的微分方程定义常微分方程4. 设级数\(\sum\limits_{n=0}^{\infty} 2x^n\)收敛， 则x的取值可为下列数值中的（ B ）
\[A.1\quad B.\dfrac{1}{2}\quad C.\dfrac{3}{2}\quad D.2\]设积分区域D:\(x^2+y^2=R^2\),则二重积分\(\iint\limits_D(x^2+y^2)dxdy\)=( C )\[A.\dfrac{1}{2}\pi R^2 \quad B.\dfrac{3}{2}\pi R^3 \quad C.\dfrac{1}{2}\pi R^4 \quad D.\pi R^4\]解：
\[\iint\limits_D(x^2+y^2)dxdy=\int_0^{2\pi}\int_0^R R^2Rd_Rd_\theta=\int_0^{2\pi}\dfrac{1}{4}R^4d\theta=\dfrac{1}{2}\pi R^4\]过点\((1,2，-1)\)与\(\dfrac{x-2}{-1}=\dfrac{y+4}{3}=\dfrac{z+1}{1}\)直线平行的直线是（ A ）解：\[直线的方向向量\{-1,3,1\}, 用点向式得直线方程：\\\dfrac{x-1}{-1}=\dfrac{y-2}{3}=\dfrac{z+1}{1}\]极限\(\lim\limits_{x\rightarrow 1\atop y\rightarrow 0} \dfrac{1}{y}sin(xy)\) = ( B )解：\[\lim\limits_{x\rightarrow 1\atop y\rightarrow 0} \dfrac{1}{y}sin(xy)=\lim\limits_{x\rightarrow 1\atop y\rightarrow 0} \dfrac{xsin(xy)}{xy}=\lim\limits_{x\rightarrow 1\atop y\rightarrow 0}x = 1\]设积分区域\(x^2+y^2+z^2\leqslant 10\) 则三重积分\(\iiint (2x-z)dxdydz=\) （ A ）解：根据奇偶性 0TODO级数\(\sum\limits_0^n \dfrac{1}{2^n}=\) ( C )解：\[\sum\limits_0^n \dfrac{1}{2^n}= 1+ \dfrac{1}{2}+ \dfrac{1}{4}+ \dfrac{1}{8}...=2\]设C1，C2是任意常数， 则微分方程\(y^{''}=2x+1\) 的通解y=( D )解：\[y^{'} = x^2+x+C1\\y= \dfrac{1}{3}x^3+\dfrac{1}{2}x^2+C1x+C2\]二、计算题求直线\(\dfrac{x-1}{1}=\dfrac{y-5}{-2}=\dfrac{z+8}{1}\)与直线\(\dfrac{x+1}{1}=\dfrac{y-1}{1}=\dfrac{z}{-2}\)的夹角\(\theta\)解：\[直线的夹角等于方向向量\{1,-2,1\}\{1,1,-2\}的夹角\\cos\theta=\dfrac{\lvert a\times b\rvert}{\lvert a\rvert\lvert b\rvert}=\dfrac{\lvert1-2-2\rvert}{\sqrt{6}\sqrt{6}}=\dfrac{1}{2}\\\theta=\dfrac{\pi}{3}\]求平面\(x-2y+2z-3=0\)的法向量的方向余弦解：\[法向量为\{1，-2，2\}\\方向余弦： 向量与x,y,z轴的夹角的余弦\\x=\{1,0,0\} \quad cos\alpha=\dfrac{1}{\sqrt{9}\sqrt{1}} = \dfrac{1}{3} \\y=\{0,1,0\} \quad cos\beta=\dfrac{-2}{\sqrt{9}\sqrt{1}} = \dfrac{-2}{3} \\z=\{0,0,1\} \quad cos\gamma=\dfrac{2}{\sqrt{9}\sqrt{1}} = \dfrac{2}{3}\]求曲面\(z=4-x^2-y^2\)在点\((1,1,2)\)的切平面方程解：偏导的应用\[F(x)=4-x^2-y^2-z\\{-2x,-2y,-1}法线的方向向量为{-2,-2,-1}\\点法式：切平面为-2（x-1）-2(y-1)-(z-2)=0 \quad \Rightarrow 2（x-1）+2(y-1)+(z-2)=0\]求函数\(u=2xy-z^2\)在\(（2，-1，1）\)处的梯度解：偏导的应用\[\dfrac{\partial u}{\partial x} = 2y\\\dfrac{\partial u}{\partial y} = 2x\\\dfrac{\partial u}{\partial z} = -2z\\gradu(2,-1,1)=\{-2,4,-2\}\]求\(f(x,y)=4(x-y)-x^2-y^2\) 的极值解：偏导的应用\[f_x=4-2x=0 \quad\Rightarrow\quad x=2\\f_y=-4-2y=0 \quad\Rightarrow\quad y=-2\\则驻点为（2，-2）\\A=f_{xx}=-2\quad B=f_{xy}=0 \quad C=f_{yy}=-2\\\Delta=B^2-AC< 0 且A0\)时\(\dfrac{1+n}{1+n^2} >\dfrac{n}{n^2}= \dfrac{1}{n}\)已知\(\sum\limits_0^\infty\dfrac{1}{n} 发散（调和级数）\)所以\(\sum\limits_0^\infty\dfrac{1+n}{1+n^2}\)的发散求微分方程的\(y^{''}-2^{'}y-3y=0\)通解解： 特征根法\[特征方程 r^2-2r-3= 0\\特征根： r_1= 3 \quad r_2=-1\\通解y= C_1e^{3x}+C_2e^{-1x}\]三、综合题求幂级数\(\sum\limits_0^\infty\dfrac{2^n}{2n+1} x^n\)的收敛半径和收敛区间解：\[\rho=\lim\limits_{x\rightarrow \infty}\lvert\dfrac{a_{n+1}}{a_n}\rvert=\lim\limits_{x\rightarrow \infty}\dfrac{4n+2}{2n+3} = 2\\所以收敛半径为R=\dfrac{1}{\rho}=\dfrac{1}{2}\\x=\dfrac{1}{2} 时\sum\limits_0^\infty\dfrac{2^n}{2n+1} x^n 发散\\x=-\dfrac{1}{2} 时 \sum\limits_0^\infty\dfrac{2^n}{2n+1} x^n 发散\\所以收敛区间为(-\dfrac{1}{2},\dfrac{1}{2})\]计算对坐标曲面积分\(I=\iint \limits_\Sigma(x^2+y^2+z^2-1)dxdy\), 其中\(\Sigma\) 是半球面\(\sqrt{3-x^2-y^2}\)解：\[z=\sqrt{3-x^2-y^2}\\\Sigma在x0y 平面的投影x^2+y^2=3\\\Sigma法向量与z轴夹角小于\dfrac{\pi}{2}\\\begin{aligned}I &=\iint \limits_\Sigma(x^2+y^2+z^2-1)dxdy\\&=\iint\limits_{D_{xy}} (3-1)dxdy\\&=2\int_0^{2\pi}d\theta\int_0^{\sqrt{3}}rdr\\&=6\pi\end{aligned}\]