高数强化NO10|导数的应用|切线和法线|导数的物理意义|曲率

一元函数微分学的应用

切线与法线

$$\begin{array}{rl}& \text{设函数}f(x)\text{可导，则曲线}y=f(x)\text{在任意一点的切线斜率等于该点的导数值，}\\ & 也就是说，曲线y=f(x)\text{在}x=x_0\text{处的切线方程可表示为}\\ & y=f^{\prime }(x_0)(x-x_0)+f(x_0).\\ & \text{当}{f}^{\prime }(x_0)\ne 0\text{时，该点的法线方程可表示为}\\ & y=\frac{-1}{f^{\prime }(x_0)}(x-x_0)+f(x_0);\\ & \text{当}{f}^{\prime }(x_0)=0\text{时，该点的法线方程为}\\ & x=x_0.\end{array}$$

单调性与凹凸性

1. 单调性的判定
   $$\begin{array}{rl}& \text{设函数}f(x)\text{在}[a,b]\text{上连续，在}(a,b)\text{内可导.}\\ & \\ & (\text{i})\text{如果在}(a,b)\text{内有}{f}^{\prime }(x)\ge 0,\text{且}{f}^{\prime }(x)\text{不在}(a,b)\text{的任一子区间上恒为零，那么}\\ & \text{函数}f(x)\text{在}[a,b]\text{上单调递增；}\\ & \\ & (\text{ii})\text{如果在}(a,b)\text{内有}{f}^{\prime }(x)\le 0,\text{且}{f}^{\prime }(x)\text{不在}(a,b)\text{的任一子区间上恒为零，那么}\\ & \text{函数}f(x)\text{在}[a,b]\text{上单调递减.}\end{array}$$
2. 凹函数与凸函数的定义
   $$\begin{array}{rl}& \text{设函数}f(x)\text{在区间}I\text{上连续，如果对}I\text{上任意两点}{x}_1,x_2\text{恒有}\\ & \\ & f\left(\frac{x_1+x_2}{2}\right)<\frac{f(x_1)+f(x_2)}{2},\text{则称}f(x)\text{是}I\text{上的凹函数；如果对}I\text{上任意两点}{x}_1,x_2\\ & \text{恒有}f\left(\frac{x_1+x_2}{2}\right)>\frac{f(x_1)+f(x_2)}{2},\text{则称}f(x)\text{是}I\text{上的凸函数.}\\ & \\ & \boxed{【注】}①\text{凹凸函数定义等价的描述方式：设函数}f(x)\text{在区间}I\text{上连续，若对}I\text{上任意不}\\ & \text{同的两点}{x}_1,x_2\text{及}\lambda \in (0,1),\text{恒有}f(\lambda x_1+(1-\lambda )x_2)<\lambda f(x_1)+(1-\lambda )f(x_2),\text{则称函}\\ & \text{数在区间}I\text{上为凹函数；若对}I\text{上任意不同的两点}{x}_1,x_2\text{及}\lambda \in (0,1),\text{恒有}\\ & f(\lambda x_1+(1-\lambda )x_2)>\lambda f(x_1)+(1-\lambda )f(x_2),\text{则称函数在区间}I\text{上为凸函数.}\\ & \\ & ②\text{几何意义：对于凹函数来说，曲线上任意两点的连线都在该曲线的上方，任意一点的}\\ & \text{切线都在该曲线的下方；对于凸函数来说，曲线上任意两点的连线都在该曲线的下方，}\\ & \text{任意一点的切线都在该曲线的上方.}\\ & \\ & ③\text{可导函数的情况下，凹函数的导函数是单调递增的，凸函数的导函数是单调递减的.}\end{array}$$
3. 凹凸性的判定
   $$\begin{array}{rl}& \text{设函数}f(x)\text{在闭区间}[a,b]\text{上连续，在开区间}(a,b)\text{内具有一阶和二阶导数，那么：}\\ & \\ & (\text{i})\text{如果在}(a,b)\text{内有}{f}^{\prime \prime }(x)\ge 0,\text{且}{f}^{\prime \prime }(x)\text{不在}(a,b)\text{的任一子区间上恒为零，那么}\\ & \text{函数}f(x)\text{在}[a,b]\text{上是凹函数；}\\ & \\ & (\text{ii})\text{如果在}(a,b)\text{内有}{f}^{\prime \prime }(x)\le 0,\text{且}{f}^{\prime \prime }(x)\text{不在}(a,b)\text{的任一子区间上恒为零，那么}\\ & \text{函数}f(x)\text{在}[a,b]\text{上是凸函数.}\end{array}$$

极值点与拐点

1. 极值点的定义
   $$\begin{array}{rl}& \text{设函数}f(x)\text{在点}{x}_0\text{的某邻域}U(x_0)\text{内有定义，如果对任意的}x\in \mathring{U}(x_0),\text{有}\\ & \\ & f(x_0)<f(x)(\text{或}f(x_0)>f(x)),\text{则称}f(x_0)\text{是函数}f(x)\text{的一个极小值（或极大值）}.\end{array}$$
2. 极值点的判定
   $$\begin{array}{rl}& \text{a.必要条件：设函数}f(x)\text{在}{x}_0\text{处可导，并在}{x}_0\text{处取得极值，那么}{f}^{\prime }(x_0)=0.\\ & \\ & \text{b.第一充分条件：设函数}f(x)\text{在}{x}_0\text{处连续，并在}{x}_0\text{的某去心邻域}\mathring{U}(x_0,\delta )\text{内可导：}\\ & (\text{i})\text{若}x\in (x_0-\delta ,x_0)\text{时，}{f}^{\prime }(x)>0,\text{而}x\in (x_0,x_0+\delta )\text{时，}{f}^{\prime }(x)<0,\text{则}f(x)\text{在}{x}_0\\ & \text{处取得极大值；}\\ & (\text{ii})\text{若}x\in (x_0-\delta ,x_0)\text{时，}{f}^{\prime }(x)<0,\text{而}x\in (x_0,x_0+\delta )\text{时，}{f}^{\prime }(x)>0,\text{则}f(x)\text{在}\\ & x_0\text{处取得极小值；}\\ & (\text{iii})\text{若}x\in \mathring{U}(x_0,\delta )\text{时，}{f}^{\prime }(x)\text{符号保持不变，则}f(x)\text{在}{x}_0\text{处没有极值.}\\ & \\ & \text{c.第二充分条件：设函数}f(x)\text{在}{x}_0\text{处存在二阶导数且}{f}^{\prime }(x_0)=0,\text{那么：}\\ & (\text{i})\text{若}{f}^{\prime \prime }(x_0)>0,\text{则}f(x)\text{在}{x}_0\text{处取得极小值；}\\ & (\text{ii})\text{若}{f}^{\prime \prime }(x_0)<0,\text{则}f(x)\text{在}{x}_0\text{处取得极大值.}\end{array}$$
3. 拐点的定义
   $$连续函数凹凸性的分界点称为拐点.$$
4. 拐点的判定
   $$\begin{array}{rl}& \text{a.必要条件：设函数}f(x)\text{在}{x}_0\text{处二阶可导，并且}(x_0,f(x_0))\text{是它的拐点，那么}\\ & f^{\prime \prime }(x_0)=0.\\ & \\ & \text{b.第一充分条件：设函数}f(x)\text{在}{x}_0\text{处连续，并在}{x}_0\text{的某去心邻域}\mathring{U}(x_0,\delta )\text{内二阶可导，}\\ & \text{若在点}{x}_0\text{两侧函数的二阶导数异号，则点}(x_0,f(x_0))\text{为函数}f(x)\text{的拐点；若在点}{x}_0\text{两}\\ & \text{侧函数的二阶导数同号，则点}(x_0,f(x_0))\text{不是函数}f(x)\text{的拐点.}\\ & \\ & \text{c.第二充分条件：设函数}f(x)\text{在}{x}_0\text{处三阶可导，若在点}{x}_0\text{处}{f}^{\prime \prime }(x_0)=0,\text{且有}\\ & f^{\prime \prime \prime }(x_0)\ne 0,\text{则点}(x_0,f(x_0))\text{为函数}f(x)\text{的拐点.}\end{array}$$

曲率

1. 曲率
   $$\text{曲线}y=f(x)\text{在点}(x,y)\text{处的曲率为}K=\frac{|y^{\prime \prime }|}{{\left(1+(y^{\prime }{)}^2\right)}^{\frac{3}{2}}}.$$
2. 曲率半径
   $$\rho =\frac{1}{K}$$
3. 曲率圆
   $$\begin{array}{rl}& \text{在}M(x_0,y_0)\text{点的法线上取曲线凹向一侧的一点}D,\text{使得}DM=\rho ,\text{则以点}D\text{为圆心，}\\ & \rho \text{为半径的圆称之为曲线在点}M\text{处的曲率圆.}\\ & \\ & \text{经计算可得，曲率圆的方程为}(x-\alpha {)}^2+(y-\beta {)}^2={\rho }^2,\text{其中}\\ & \alpha =x_0-\frac{y^{\prime }\left[1+(y^{\prime }{)}^2\right]}{y^{\prime \prime }},\\ & \beta =y_0+\frac{1+(y^{\prime }{)}^2}{y^{\prime \prime }}.\end{array}$$

多元函数微分学的应用

无条件极值

1. 多元函数极值的定义
   $$\begin{array}{rl}& \text{设点}{P}_0\text{是函数}z=f(x,y)\text{的定义域}D\text{的内点，若存在}{P}_0\text{的邻域}U(P_0),\text{使得对任意异}\\ & \text{于}{P}_0\text{的点}P\in U(P_0),\text{都有}f(P)<f(P_0),\text{则称函数}z=f(x,y)\text{在点}{P}_0\text{有极大值，}\\ & \text{点}{P}_0\text{称为函数}z=f(x,y)\text{的极大值点.类似地，可以定义极小值与极小值点.极大值与极}\\ & \text{小值统称为极值，极大值点与极小值点统称为极值点.}\end{array}$$
2. 多元函数极值的判定
   $$\begin{array}{rl}& \text{a.必要条件：设函数}z=f(x,y)\text{在}(x_0,y_0)\text{点具有偏导数，且在该点取得极值，则有}\\ & f_x^{\prime }(x_0,y_0)=0,f_y^{\prime }(x_0,y_0)=0.\\ & \\ & \text{b.充分条件:设函数}z=f(x,y)\text{在}(x_0,y_0)\text{点的某邻域内具有连续的一阶及二阶偏导数，}\\ & \text{又设}{f}_x^{\prime }(x_0,y_0)=0,f_y^{\prime }(x_0,y_0)=0.\text{令}{f}_{xx}^{\prime \prime }(x_0,y_0)=A,f_{xy}^{\prime \prime }(x_0,y_0)=B,\\ & f_{yy}^{\prime \prime }(x_0,y_0)=C.\\ & \\ & (\text{i})\text{若}AC-B^2>0,\text{则函数}z=f(x,y)\text{在}(x_0,y_0)\text{点具有极值.且当}A>0\text{时取得极}\\ & \text{小值；当}A<0\text{时取得极大值.}\\ & (\text{ii})\text{若}AC-B^2<0,\text{则函数}z=f(x,y)\text{在}(x_0,y_0)\text{点不取极值.}\\ & (\text{iii})\text{若}AC-B^2=0,\text{则函数}z=f(x,y)\text{在}(x_0,y_0)\text{点可能取极值，也可能不取极值.}\end{array}$$

条件极值和拉格朗日乘数法

$$\begin{array}{rl}& \text{模型：求函数}z=f(x,y)\text{在约束条件}\phi (x,y)=0\text{下的极值点.}\\ & \text{方法：}\\ & (\text{i})\text{作拉格朗日函数}L(x,y,\lambda )=f(x,y)+\lambda \phi (x,y);\\ & (\text{ii})\text{求拉格朗日函数的驻点，即解方程组}\{\begin{array}{l}{f}_x^{\prime }(x,y)+\lambda {\phi }_x^{\prime }(x,y)=0\\ f_y^{\prime }(x,y)+\lambda {\phi }_y^{\prime }(x,y)=0\\ \phi (x,y)=0\end{array};\\ & (\text{iii})\text{根据实际条件判断所求出的点是极大值点还是极小值点.}\\ & \text{此方法还可以推广到变量多于两个，约束条件多于一个的情形.}\\ & \text{如求函数}u=f(x,y,z)\text{在约束条件}\phi (x,y,z)=0,\psi (x,y,z)=0\text{下的极值点，只需要}\\ & \text{把拉格朗日函数改为}L(x,y,z,\lambda ,\mu )=f(x,y,z)+\lambda \phi (x,y,z)+\mu \psi (x,y,z)\text{即可.}\end{array}$$

导数的几何意义和物理意义

切线与法线

$$\begin{array}{rl}& \{\begin{array}{l}\text{求方程（导数）}\{\begin{array}{l}\text{直角坐标（导数计算6种题型）}\\ \text{极坐标（转化为参数方程求导）}\end{array}\\ \text{相切的充要条件：}\{\begin{array}{l}f(x_0)=g(x_0)\\ f^{\prime }(x_0)=g^{\prime }(x_0)\end{array}\end{array}\end{array}$$

$$\begin{array}{rl}& \text{解：对数螺线}\rho =e^{\theta }\text{的直角坐标方程为}\\ & \{\begin{array}{l}x=\rho \cos \theta =e^{\theta }\cos \theta \\ y=\rho \sin \theta =e^{\theta }\sin \theta \end{array}\\ & \text{则}\frac{dy}{dx}=\frac{y^{\prime }(\theta )}{x^{\prime }(\theta )}=\frac{e^{\theta }\sin \theta +e^{\theta }\cos \theta }{e^{\theta }\cos \theta -e^{\theta }\sin \theta }\\ & \\ & {\frac{dy}{dx}|}_{\theta =\frac{\pi }{2}}=-1,\left(e^{\frac{\pi }{2}},\frac{\pi }{2}\right)\text{对应的直角坐标点为}\left(0,e^{\frac{\pi }{2}}\right)\\ & \text{直角坐标条件为}y=-x+e^{\frac{\pi }{2}}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{本题的难点在于对极坐标方程求导数，核心是要知道}\rho \text{是关于}\theta \text{的函数，只}\\ & \text{要遇到极坐标表示的方程，一律利用}\{\begin{array}{l}x=\rho (\theta )\cos \theta ,\\ y=\rho (\theta )\sin \theta ,\end{array}\text{转化为参数方程.}\end{array}$$

$$\begin{array}{rl}& \{\begin{array}{l}f(1)=1^2-1=0\\ f^{\prime }(1)=(2x-1){|}_{x=1}=1\end{array}①\text{因为}f(x)\text{在}x=1\text{可导，故}{f}_{-}^{\prime }(1)=f^{\prime }(1)\\ & \\ & \text{解：}\underset{n\to \infty }{\lim }nf\left(\frac{n}{n+2}\right)=\underset{n\to \infty }{\lim }\frac{f\left(\frac{n}{n+2}\right)-f(1)}{\frac{n}{n+2}-1}\cdot \left(\frac{n}{n+2}-1\right)\cdot n=-2f_{-}^{\prime }(1)\\ & \stackrel{\text{①}}{=}-2f^{\prime }(1)=-2\cdot 1=-2\\ & \\ & n\to \infty \text{时，}\frac{n}{n+2}\to 1^{-}\end{array}$$
$$\begin{array}{rl}& \text{改题}\\ & \text{设曲线}f(x)=\{\begin{array}{ll}1-x& x\ge 1f_{+}^{\prime }(1)=-1\\ x-1& x<1f_{-}^{\prime }(1)=1\end{array}\text{则，}\\ & \underset{n\to \infty }{\lim }nf\left(\frac{n}{n+2}\right)=\underline{}\\ & \\ & \underset{n\to \infty }{\lim }nf\left(\frac{n}{n+2}\right)=-2f_{-}^{\prime }(1)=-2.\\ & \\ & \text{②求}\underset{n\to \infty }{\lim }n\cdot f\left(\frac{n+2}{n}\right)=-2f_{+}^{\prime }(1)=-2\cdot (-1)=2.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{本题的关键是记住两条曲线}y=f(x)\text{与}y=g(x)\text{在点}x=x_0\text{处相切的充要条}\\ & \text{件为}f(x_0)=g(x_0)\text{且}{f}^{\prime }(x_0)=g^{\prime }(x_0).\end{array}$$
$$\begin{array}{rl}& \text{关于单侧极限的总结：}\\ & \text{若}\underset{x\to 1^{+}}{\lim }f(x)=1,\underset{x\to 1^{-}}{\lim }f(x)=-1,\text{则}\\ & ①\underset{x\to (-1{)}^{+}}{\lim }f(-x)=-1\\ & ②\underset{x\to 1^{-}}{\lim }f\left(\frac{1}{x}\right)=1\\ & ③\underset{x\to 0}{\lim }f(\cos x)=-1(\cos x\le 1,\cos x\to 1^{-})\\ & ④\underset{n\to \infty }{\lim }f\left(\frac{n+1}{n}\right)=-1\left(\frac{n+1}{n}<1,\frac{n+1}{n}\to 1^{-}\right)\\ & ⑤\underset{x\to 0^{+}}{\lim }f\left(\frac{\sin x}{x}\right)=-1\left(x\to 0^{+},\sin x<x,\frac{\sin x}{x}<1,\frac{\sin x}{x}\to 1^{-}\right)\\ & ⑥\underset{x\to 0^{-}}{\lim }f\left(\frac{\sin x}{x}\right)=-1\left(x\to 0^{-},\sin x>x,\frac{\sin x}{x}<1,\frac{\sin x}{x}\to 1^{-}\right)\\ & \underset{x\to 0}{\lim }f\left(\frac{\sin x}{x}\right)=-1\\ & ⑦\underset{x\to 1}{\lim }f(x^2-2x+2)=1\left(x\to 1,x^2-2x+2=(x-1{)}^2+1\ge 1,x^2-2x+2\to 1^{+}\right)\end{array}$$

导数的物理意义

$$\begin{array}{rl}& \text{解：设在}t\text{时刻，}P\text{点的坐标为}\left(x_0,\frac{4}{9}{x}_0^2\right)\\ & S=\frac{1}{2}\left(1+\frac{4}{9}{x}_0^2\right)\cdot x_0-{\int }_0^{x_0}\frac{4}{9}{u}^{2}du\\ & =\frac{1}{2}{x}_0+\frac{2}{9}{x}_0^3-{\int }_0^{x_0}\frac{4}{9}{u}^{2}du\\ & \\ & \frac{dS}{dt}=\frac{dS}{d{x}_0}\cdot \frac{d{x}_0}{dt}=\left(\frac{1}{2}+\frac{2}{3}{x}_0^2-\frac{4}{9}{x}_0^2\right)\cdot \frac{d{x}_0}{dt}\\ & \\ & {\frac{dS}{dt}|}_{x_0=3}={\left(\frac{1}{2}+\frac{2}{9}{x}_0^2\right)|}_{x_0=3}\cdot {\frac{d{x}_0}{dt}|}_{x_0=3}\\ & =\frac{5}{2}\cdot 4=10\end{array}$$
$$\begin{array}{rl}& 改题\\ & \text{已知曲线}L:x=\frac{3}{2}\sqrt{y},\text{若}P\text{运动到点}(3,4)\text{时}\\ & \text{沿}y\text{轴正方向的速度是}4,\text{求此时}S\text{关于时间}t\text{的变化率.}\\ & \\ & \text{解：设在}t\text{时刻，}P\text{点坐标为}\left(\frac{3}{2}\sqrt{y},y\right)\\ & S=\frac{1}{2}(1-y)\cdot \frac{\sqrt{3}}{2}y+{\int }_0^y\frac{\sqrt{3}}{2}udu\\ & =\frac{\sqrt{3}}{4}y-\frac{\sqrt{3}}{4}{y}^2+{\int }_0^y\frac{\sqrt{3}}{2}udu\\ & \\ & \frac{dS}{dt}=\frac{dS}{dy}\cdot \frac{dy}{dt}=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}y+\frac{\sqrt{3}}{2}\right)\cdot \frac{dy}{dt}\\ & \\ & {\frac{dS}{dt}|}_{y=4}=\frac{\sqrt{3}}{4}{\frac{dy}{dt}|}_{y=4}=\sqrt{3}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{本题求的是面积关于时间的变化率，求解的关键是利用题目所给条件列出面}\\ & \text{积}S\text{和时间}t\text{关系式，然后直接求导即可得到变化率.}\\ & 2.\text{本题的难点有两个，第一个是正确理解“}P\text{点沿}x\text{轴正方向的速度是}4\text{”实质上是}\\ & \frac{dx}{dt}=4;\text{第二个是对函数关系的梳理.}\end{array}$$

曲率

$$\begin{array}{rl}& \{\begin{array}{l}\text{公式: 曲率}K=\frac{|y^{\prime \prime }|}{{\left(1+(y^{\prime }{)}^2\right)}^{\frac{3}{2}}},\rho =\frac{1}{K},\\ \text{几何意义}\{\begin{array}{l}①\text{曲率: 弯曲程度越大, 曲率就越大}\\ ②\text{曲线抱着曲率圆.}\end{array}\end{array}\end{array}$$

$$\begin{array}{rl}& \text{解: }\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}=\frac{2t+4}{2t}=1+\frac{2}{t},{\frac{dy}{dx}|}_{t=1}=3\\ & \\ & \frac{d^{2}y}{d{x}^2}=\frac{d\left(1+\frac{2}{t}\right)}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\left(-\frac{2}{t^2}\right)\cdot \frac{1}{2t}=-\frac{1}{t^3},{\frac{d^{2}y}{d{x}^2}|}_{t=1}=-1\\ & \\ & K=\frac{|-1|}{{\left(1+3^2\right)}^{\frac{3}{2}}}=\frac{1}{10\sqrt{10}},R=\frac{1}{K}=10\sqrt{10},\text{选}(C)\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{本题只要写出曲率公式，以及曲率半径}R=\frac{1}{K},\text{再根据参数方程求出一阶导}\\ & \text{数和二阶导数，代入到公式中求解即可.}\end{array}$$

$$\begin{array}{rl}& \text{解: 因为}{f}_i(x)(i=1,2)\text{且有二阶连续导数，}{f}_i^{\prime \prime }(x_0)>0,(i=1,2)\\ & \text{则由保号性可知，在}{x}_0\text{的附近，}{f}_i^{\prime \prime }(x)<0,i=1,2,\\ & \text{曲线}{f}_i(x)(i=1,2)\text{是凸的，弯曲程度越大，曲率越大.}\\ & \text{故可得}g(x)\ge f_2(x)\ge f_1(x),\text{选}(A)\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{对于选择题，考生可以采用图形法进行快速求解.在本题中，根据曲率越大，曲}\\ & \text{线的弯曲程度越大的性质，以及}{f}_1(x)\text{与}{f}_2(x)\text{均为凸函数，画出两个函数以及公切线}\\ & \text{图形，从而快速得到答案.}\end{array}$$

选B
$$\begin{array}{rl}& \boxed{【小结】}\text{本题的关键在于考生需要准确理解曲率圆的定义及几何意义，应用图形法的时}\\ & \text{候应当准确画出函数，切线以及曲率圆的图形.}\\ & \text{曲率的考点虽然简单，但是有两个层次的要求，第一是记公式，第二，是理解几何意义。}\\ & \text{分别是曲率和曲率圆的几何意义。}\end{array}$$