高等数学（第七版）同济大学 习题3-7 个人解答

高等数学（第七版）同济大学 习题3-7

 

$\begin{array}{rlr}& 1.求椭圆4x^2+y^2=4在点(0,2)处的曲率。& \end{array}$

解：

$\begin{array}{rlr}& 对椭圆方程两边求导，得8x+2y{y}^{\prime }=0，y^{\prime }=-\frac{4x}{y}，y^{\prime \prime }=-\frac{16}{y^3}，当x=0时，y^{\prime }=0，y^{\prime \prime }=-2，& \\ & & \\ & 由曲率公式K=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}，得椭圆在点(0,2)处得曲率K=2.& \end{array}$

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$\begin{array}{rlr}& 2.求曲线y=lnsecx在点(x,y)处的曲率及曲率半径。& \end{array}$

解：

$\begin{array}{rlr}& y^{\prime }=\frac{secxtanx}{secx}=tanx，y^{\prime \prime }=se{c}^{2}x，由曲率公式K=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}，得K=\frac{se{c}^{2}x}{(1+ta{n}^{2}x{)}^{\frac{3}{2}}}=|cosx|，& \\ & & \\ & 曲率半径\rho =\frac{1}{K}=|secx|.& \end{array}$

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$\begin{array}{rlr}& 3.求抛物线y=x^2-4x+3在其顶点处的曲率及曲率半径。& \end{array}$

解：

$\begin{array}{rlr}& 抛物线在顶点处曲率最大，顶点为(2,-1)，y^{\prime }=2x-4，y^{\prime \prime }=2，由曲率公式K=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}，& \\ & & \\ & 得抛物线在顶点(2,-1)处的曲率K=\frac{2}{(1+0^2{)}^{\frac{3}{2}}}=2，曲率半径\rho =\frac{1}{K}=\frac{1}{2}.& \end{array}$

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$\begin{array}{rlr}& 4.求曲线x=aco{s}^{3}t，y=asi{n}^{3}t在t=t_0相应的点处的曲率。& \end{array}$

解：

$\begin{array}{rlr}& \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3asi{n}^{2}tcost}{-3aco{s}^{2}tsint}=-tant，\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{-se{c}^{2}t}{-3aco{s}^{2}tsint}=\frac{1}{3asintco{s}^{4}t}.& \\ & & \\ & 曲线在t=t_0处的曲率K=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=\frac{\left|\frac{1}{3asintco{s}^{4}t}\right|}{[1+(-tant{)}^2{]}^{\frac{3}{2}}}=\frac{2}{|3asin(2t_0)|}.& \end{array}$

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$\begin{array}{rlr}& 5.对数曲线y=lnx上哪一点处的曲率半径最小？求出该点处的曲率半径。& \end{array}$

解：

$\begin{array}{rlr}& y^{\prime }=\frac{1}{x}，y^{\prime \prime }=-\frac{1}{x^2}，曲线曲率K=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=\frac{\left|-\frac{1}{x^2}\right|}{{\left[1+{\left(\frac{1}{x}\right)}^2\right]}^{\frac{3}{2}}}=\frac{x}{(1+x^2{)}^{\frac{3}{2}}}，曲率半径\rho =\frac{(1+x^2{)}^{\frac{3}{2}}}{x}& \\ & & \\ & 因{\rho }^{\prime }=\frac{(1+x^2{)}^{\frac{1}{2}}(2x^2-1)}{x^2}，令{\rho }^{\prime }=0，得x=\frac{\sqrt{2}}{2}，x=-\frac{\sqrt{2}}{2}，& \\ & & \\ & 当0<x<\frac{\sqrt{2}}{2}时，{\rho }^{\prime }<0，\rho 在\left(0,\frac{\sqrt{2}}{2}\right)上单调减少，& \\ & & \\ & 当\frac{\sqrt{2}}{2}<x<+\infty 时，{\rho }^{\prime }>0，\rho 在\left[\frac{\sqrt{2}}{2},+\infty \right)上单调增加，& \\ & & \\ & 因此\rho 在x=\frac{\sqrt{2}}{2}处取得极小值，又因驻点唯一，极小值就是最小值，所以最小曲率半径为\rho =\frac{{\left(1+\frac{1}{2}\right)}^{\frac{3}{2}}}{\frac{\sqrt{2}}{2}}=\frac{3\sqrt{3}}{2}.& \end{array}$

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$\begin{array}{rlr}& 6.证明曲线y=ach\frac{x}{a}在点(x,y)处的曲率半径为\frac{y^2}{a}.& \end{array}$

解：

$\begin{array}{rlr}& y^{\prime }=sh\frac{x}{a}，y^{\prime \prime }=\frac{1}{a}ch\frac{x}{a}，曲线在点(x,y)处的曲率为K=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=\frac{\left|\frac{1}{a}ch\frac{x}{a}\right|}{{\left(1+s{h}^2\frac{x}{a}\right)}^{\frac{3}{2}}}=\frac{1}{ac{h}^2\frac{x}{a}}，& \\ & & \\ & 曲率半径为\rho =\frac{1}{K}=ac{h}^2\frac{x}{a}=\frac{y^2}{a}.& \end{array}$

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$\begin{array}{rlr}& 7.一飞机沿抛物线路径y=\frac{x^2}{10000}（y轴铅直向上，单位为m）做俯冲飞行。在坐标原点O处的飞机的速度& \\ & & \\ & 为v=200m/s。飞行员体重G=70kg。求飞机俯冲至最低点即原点O处时坐椅对飞行员的反力。& \end{array}$

解：

$\begin{array}{rlr}& y^{\prime }=\frac{2x}{10000}=\frac{x}{5000}，y^{\prime \prime }=\frac{1}{5000}，& \\ & & \\ & 抛物线在坐标原点的曲率半径为\rho =\frac{1}{K}=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=5000，& \\ & & \\ & 向心力F_1=\frac{m{v}^2}{\rho }=\frac{70\times 200^2}{5000}=560N，& \\ & & \\ & 飞机俯冲至最低点即原点O处时坐椅对飞行员的反力等于飞行员离心力和飞行员重量对坐椅压力的和，& \\ & & \\ & F=mg+F_1=70\times 9.8+560=1246N.& \end{array}$

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$\begin{array}{rlr}& 8.汽车连同载重共5t，在抛物线拱桥上行驶，速度为21.6km/h，桥的跨度为10m，拱地矢高位0.25m& \\ & & \\ & （图3-37）。求汽车越过桥顶时对桥地压力。& \end{array}$

解：

$\begin{array}{rlr}& 设拱桥抛物线方程为y=a{x}^2，因抛物线过点(5,0.25)，得a=\frac{y}{x^2}=\frac{0.25}{25}=0.01，y^{\prime }=2ax，y^{\prime \prime }=2a，& \\ & & \\ & 则y^{\prime }=0，y^{\prime \prime }=0.02，\rho =\frac{1}{K}=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=50，& \\ & & \\ & 所以，汽车越过桥顶时对桥地压力为F=mg-\frac{m{v}^2}{\rho }=5\times 10^3\times 9.8-\frac{5\times 10^3\times {\left(\frac{21.6\times 10^3}{3600}\right)}^2}{50}=45400N.& \end{array}$

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$\begin{array}{rlr}& 9.求曲线y=lnx在与x轴交点处的曲率圆方程。& \end{array}$

解：

$\begin{array}{rlr}& 方程组\{\begin{array}{l}y=lnx，\\ \\ y=0，\end{array}得曲线与x轴交点为(1,0)，y^{\prime }=\frac{1}{x}，y^{\prime \prime }=-\frac{1}{x^2}，所以当x=1时，y^{\prime }=1，y^{\prime \prime }=-1，& \\ & & \\ & 设曲线在点(1,0)处得曲率中心为(\alpha ,\beta )，则& \\ & & \\ & \alpha =\left[x-\frac{y^{\prime }(1+y^{\prime 2})}{y^{\prime \prime }}\right]=1-\frac{1\cdot (1+1^2)}{-1}=3，\beta =\left[y+\frac{1+y^{\prime 2}}{y^{\prime \prime }}\right]=0+\frac{1+1^2}{-1}=-2，& \\ & & \\ & 曲率半径\rho =\frac{1}{K}=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=\sqrt{8}，曲率圆方程为(\xi -3{)}^2+(\eta +2{)}^2=8.& \end{array}$

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$\begin{array}{rlr}& 10.求曲线y=tanx在点\left(\frac{\pi }{4},1\right)处的曲率圆方程。& \end{array}$

解：

$\begin{array}{rlr}& y^{\prime }=se{c}^{2}x，y^{\prime \prime }=2se{c}^{2}xtanx，当x=\frac{\pi }{4}时,y^{\prime }=2，y^{\prime \prime }=4，& \\ & & \\ & 设曲线在点\left(\frac{\pi }{4},1\right)处的曲率中心的坐标为(\alpha ,\beta )，则& \\ & & \\ & \alpha =x-\frac{y^{\prime }(1+y^{\prime 2})}{y^{\prime \prime }}=\frac{\pi }{4}-\frac{2(1+4)}{4}=\frac{\pi -10}{4}，\beta =y+\frac{1+y^{\prime 2}}{y^{\prime \prime }}=1+\frac{1+4}{4}=\frac{9}{4}.& \\ & & \\ & 曲率半径\rho =\frac{1}{K}=\frac{|y^{\prime \prime }|}{(1+y^{\prime 2}{)}^{\frac{3}{2}}}=\frac{\sqrt{125}}{4}，& \\ & & \\ & 曲率圆方程为{\left(\xi -\frac{\pi -10}{4}\right)}^2+{\left(\eta -\frac{9}{4}\right)}^2=\frac{125}{16}.& \end{array}$

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$\begin{array}{rlr}& 11.求抛物线y^2=2px的渐屈线方程。& \end{array}$

解：

$\begin{array}{rlr}& 对抛物线方程两边求导，2y{y}^{\prime }=2p，二阶求导，y^{\prime 2}+y{y}^{\prime \prime }=0，得y^{\prime }=\frac{p}{y}，y^{\prime \prime }=-\frac{p^2}{y^3}，& \\ & & \\ & 抛物线渐屈线方程为& \\ & & \\ & \{\begin{array}{l}\alpha =x-\frac{y^{\prime }(1+y^{\prime 2})}{y^{\prime \prime }}=x-\frac{\frac{p}{y}\left[1+{\left(\frac{p}{y}\right)}^2\right]}{-\frac{p^2}{y^3}}=\frac{3y^2}{2p}+p，\\ \\ \beta =y+\frac{1+y^{\prime 2}}{y^{\prime \prime }}=y+\frac{1+{\left(\frac{p}{y}\right)}^2}{-\frac{p^2}{y^3}}=-\frac{y^3}{p^2}\end{array}& \\ & & \\ & 消去y得渐屈线方程为27p{\beta }^2=8(\alpha -p{)}^3& \end{array}$