IS Math-2016-02

题目来源：Problem 2 日期：2024-08-12 题目主题: Math-Calculus-Variation of Surface Area by Rotation

解题思路

本题探讨了一个平面曲线绕 $x$ 轴旋转所形成的柱体表面的面积问题。通过旋转曲线 $y(x)$ 关于 $x$ 轴，我们需要计算其表面积，并应用变分法中的 Euler-Lagrange 方程来进一步分析该表面积函数。该题目涉及到了计算表面积的公式推导，Euler-Lagrange 方程的应用，以及用常数 $c$ 表示的具体曲线方程。

Solution

Question 1

To prove that the surface area $S$ of the cylindrical object formed by the rotation of the curve $y(x)$ about the $x$-axis is given by:

$$S=2\pi {\int }_{-1}^{1}F(y,y^{\prime })\mathrm{d}x,$$

where

$$F(y,y^{\prime })=y\sqrt{1+(y^{\prime }{)}^2},$$

we start by recalling the formula for the surface area of a solid of revolution generated by rotating a curve $y(x)$ about the $x$-axis:

$$S=2\pi {\int }_a^{b}y(x)\sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{d}x.$$

In this problem, the curve $y(x)$ is defined from $x=-1$ to $x=1$, and the surface area $S$ can be written as:

$$S=2\pi {\int }_{-1}^{1}y(x)\sqrt{1+{\left(y^{\prime }(x)\right)}^2}\mathrm{d}x,$$

where $y^{\prime }=\frac{\mathrm{d}y}{\mathrm{d}x}$ is the first derivative of $y$ with respect to $x$.

Comparing this with the given expression:

$$S=2\pi {\int }_{-1}^{1}F(y,y^{\prime })\mathrm{d}x,$$

we can identify the function $F(y,y^{\prime })$ as:

$$F(y,y^{\prime })=y\sqrt{1+(y^{\prime }{)}^2}.$$

This completes the proof for the first part.

Question 2

We are asked to prove that the following relation holds:

$$F-y^{\prime }\frac{\partial F}{\partial y^{\prime }}=c,\text{\hspace{1em}(2.4)}$$

where $c$ is a constant. To do this, we need to consider the Euler-Lagrange equation:

$$\frac{\partial F}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y^{\prime }}=0.\text{\hspace{1em}(2.3)}$$

Step 1: Compute the Total Derivative of $F$

We start by considering the total derivative of $F(y,y^{\prime })$ with respect to $x$. Since $F$ is a function of both $y$ and $y^{\prime }$, which themselves are functions of $x$, the total derivative is given by:

$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial y}\cdot \frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial F}{\partial y^{\prime }}\cdot \frac{\mathrm{d}{y}^{\prime }}{\mathrm{d}x}.$$

Since $\frac{\mathrm{d}y}{\mathrm{d}x}=y^{\prime }$ and $\frac{\mathrm{d}{y}^{\prime }}{\mathrm{d}x}=y^{\prime \prime }$, this becomes:

$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial y}\cdot y^{\prime }+\frac{\partial F}{\partial y^{\prime }}\cdot y^{\prime \prime }.$$

Step 2: Apply the Euler-Lagrange Equation

From the Euler-Lagrange equation, we know:

$$\frac{\partial F}{\partial y}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial F}{\partial y^{\prime }}\right).$$

Substituting this into the expression for $\frac{\mathrm{d}F}{\mathrm{d}x}$, we get:

$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial F}{\partial y^{\prime }}\right)\cdot y^{\prime }+\frac{\partial F}{\partial y^{\prime }}\cdot y^{\prime \prime }.$$

This expression can be simplified using the product rule:

$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}\right).$$

Step 3: Derive the Conservation Law

Since the total derivative $\frac{\mathrm{d}F}{\mathrm{d}x}$ is equal to $\frac{\mathrm{d}}{\mathrm{d}x}\left(y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}\right)$, we can write:

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(F-y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}\right)=0.$$

This equation indicates that the quantity inside the derivative, $F-y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}$, does not change with $x$. Therefore, it must be a constant. Denoting this constant by $c$, we have:

$$F-y^{\prime }\frac{\partial F}{\partial y^{\prime }}=c.$$

Question 3

We need to express a differential equation satisfied by the curve $y(x)$ using $y$, $y^{\prime }$, and the constant $c$. From the previous result in Question 2, we know that:

$$F-y^{\prime }\frac{\partial F}{\partial y^{\prime }}=y\sqrt{1+(y^{\prime }{)}^2}-y^{\prime }\frac{\partial }{\partial y^{\prime }}(y\sqrt{1+(y^{\prime }{)}^2})=\frac{y}{\sqrt{1+(y^{\prime }{)}^2}}=c.$$

To derive the differential equation, we start by squaring both sides of Equation (3.1) to eliminate the square root:

$${\left(\frac{y}{\sqrt{1+(y^{\prime }{)}^2}}\right)}^2=c^2,$$

which simplifies to:

$$\frac{y^2}{1+(y^{\prime }{)}^2}=c^2.$$

Multiplying both sides by $(1+(y^{\prime }{)}^2)$ gives:

$$y^2=c^2(1+(y^{\prime }{)}^2).$$

Expanding the right-hand side:

$$y^2=c^2+c^2(y^{\prime }{)}^2.$$

Rearranging this equation to express it in terms of $y$ and $y^{\prime }$ gives us the differential equation:

$$y^2-c^2=c^2(y^{\prime }{)}^2,\text{\hspace{1em}(3.2)}$$

or equivalently:

$$(y^{\prime }{)}^2=\frac{y^2-c^2}{c^2}.$$

This is the differential equation that the curve $y(x)$ satisfies.

Question 4

Step 1: Start with the Differential Equation

We have already derived the differential equation from Question 3:

$$(y^{\prime }{)}^2=\frac{y^2-c^2}{c^2}.$$

Taking the square root on both sides:

$$y^{\prime }=\frac{\sqrt{y^2-c^2}}{c}.$$

Note: Since $c$ is an arbitrary constant, the "$\pm$" can be absorbed by $c$.

This can be rewritten as:

$$\frac{\mathrm{d}y}{\sqrt{y^2-c^2}}=\frac{\mathrm{d}x}{c}.$$

Step 2: Separate the Variables and Integrate

Integrating the left side:

$$\int \frac{\mathrm{d}y}{\sqrt{y^2-c^2}}=\frac{1}{c}\int \mathrm{d}x.$$

The integral on the right side is straightforward:

$$\frac{1}{c}\int \mathrm{d}x=\frac{x}{c}+C_1,$$

where $C_1$ is a constant of integration.

For the left side, recall the standard integral form:

$$\int \frac{\mathrm{d}y}{\sqrt{y^2-a^2}}=\ln \left|y+\sqrt{y^2-a^2}\right|+C_2,$$

where $a$ is a constant. Here, $a=c$, so:

$$\ln \left|y+\sqrt{y^2-c^2}\right|=\frac{x}{c}+C_3,$$

where $C_3=C_1-C_2$ is the combined constant of integration.

Let’s revisit the solution and correctly solve for $y(x)$ and determine the range of $c$ based on the boundary conditions.

Step 3: Solve for $y(x)$

To solve for $y(x)$, exponentiate both sides:

$$y+\sqrt{y^2-c^2}=e^{\frac{x}{c}+C_3}.$$

Let $K=e^{C_3}$, then:

$$y+\sqrt{y^2-c^2}=K{e}^{\frac{x}{c}}.$$

Step 4: Isolate $y(x)$

To isolate $y$, subtract $\sqrt{y^2-c^2}$ from both sides and then square both sides:

$$\sqrt{y^2-c^2}=K{e}^{\frac{x}{c}}-y.$$

Squaring both sides gives:

$$y^2-c^2={\left(K{e}^{\frac{x}{c}}-y\right)}^2.$$

Expanding the square on the right:

$$y^2-c^2=K^2{e}^{\frac{2x}{c}}-2Ky{e}^{\frac{x}{c}}+y^2.$$

The $y^2$ terms cancel out, rearranging this equation to solve for $y$:

$$y=\frac{K{e}^{\frac{x}{c}}}{2}+\frac{c^2}{2K{e}^{\frac{x}{c}}}.$$

Thus, the general solution for $y(x)$ is:

$$y(x)=\frac{K{e}^{\frac{x}{c}}}{2}+\frac{c^2}{2K{e}^{\frac{x}{c}}}.$$

Step 5: Apply Boundary Conditions

Given the curve passes through points $A(-1,2)$ and $B(1,2)$, we start with the equation for $y(x)$:

$$y(x)=\frac{K{e}^{\frac{x}{c}}}{2}+\frac{c^2}{2K{e}^{\frac{x}{c}}}.$$

Now, substitute the coordinates of the points $A(-1,2)$ and $B(1,2)$.

For Point $A(-1,2)$

$$2=\frac{K}{2}\left(e^{-\frac{1}{c}}+\frac{c^2}{K^2}{e}^{\frac{1}{c}}\right).$$

Multiplying both sides by 2 gives:

$$4=K{e}^{-\frac{1}{c}}+\frac{c^2}{K}{e}^{\frac{1}{c}}.$$

For Point $B(1,2)$

$$2=\frac{K}{2}\left(e^{\frac{1}{c}}+\frac{c^2}{K^2}{e}^{-\frac{1}{c}}\right).$$

Again, multiplying both sides by 2 gives:

$$4=K{e}^{\frac{1}{c}}+\frac{c^2}{K}{e}^{-\frac{1}{c}}.$$

Step 6: Solve the System of Equations

Now, let’s solve these two equations simultaneously.

Adding equations:

$$4+4=K\left(e^{-\frac{1}{c}}+e^{\frac{1}{c}}\right)+\frac{c^2}{K}\left(e^{\frac{1}{c}}+e^{-\frac{1}{c}}\right).$$

Using the identity $e^{\alpha }+e^{-\alpha }=2\cosh (\alpha )$:

$$8=\left(K+\frac{c^2}{K}\right)2\cosh \left(\frac{1}{c}\right).$$

Dividing by $2$:

$$4=\left(K+\frac{c^2}{K}\right)\cosh \left(\frac{1}{c}\right).$$

Now subtract equations on both sides:

$$0=K\left(e^{\frac{1}{c}}-e^{-\frac{1}{c}}\right)-\frac{c^2}{K}\left(e^{\frac{1}{c}}-e^{-\frac{1}{c}}\right).$$

Using the identity $e^{\alpha }-e^{-\alpha }=2\sinh (\alpha )$:

$$0=\left(K-\frac{c^2}{K}\right)2\sinh \left(\frac{1}{c}\right).$$

Since $2\sinh \left(\frac{1}{c}\right)\ne 0$, this implies:

$$K=\frac{c^2}{K},$$

or equivalently:

$$K^2=c^2.$$

Step 7: Determine the Value of $c$

Substituting $K^2=c^2$ back:

$$4=2c\cosh \left(\frac{1}{c}\right).$$

Dividing both sides by 2:

$$2=c\cosh \left(\frac{1}{c}\right).$$

This is the equation satisfied by $c$.

知识点

旋转曲面变分法Euler-Lagrange方程微分方程

难点思路

1. 理解旋转曲面的表面积公式及其推导过程
2. 掌握 Euler-Lagrange 方程的应用及其在变分问题中的意义
3. 推导并理解由 Euler-Lagrange 方程得出的守恒律
4. 将守恒律转化为描述曲线 y(x) 的微分方程
5. 解决非线性微分方程并应用边界条件确定参数

解题技巧和信息

1. 旋转曲面的表面积计算涉及到曲线长度公式的应用,需要熟悉微积分中的弧长公式
2. Euler-Lagrange 方程是变分法中的核心工具,它将变分问题转化为微分方程问题
3. 在处理守恒律时,注意识别不随 x 变化的量,这通常暗示着一个常数 c 的存在
4. 解非线性微分方程时,变量分离法往往是一个有效的策略
5. 在应用边界条件时,要注意方程的对称性,可能会简化计算过程
6. 处理包含双曲函数的方程时,熟悉双曲函数的性质和恒等式会很有帮助

重点词汇

• Surface Area of Revolution 旋转曲面的表面积
• Variational Problem 变分问题
• Euler-Lagrange Equation Euler-Lagrange 方程
• Conservation Law 守恒律
• Nonlinear Differential Equation 非线性微分方程
• Boundary Conditions 边界条件
• Hyperbolic Functions 双曲函数

参考资料

1. Calculus of Variations by I. M. Gelfand and S. V. Fomin
2. Advanced Engineering Mathematics by Erwin Kreyszig, Chapter on Calculus of Variations
3. Mathematical Methods for Physicists by George B. Arfken and Hans J. Weber
4. Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo
5. A Course in Modern Mathematical Physics by Peter Szekeres, Chapter on Variational Principles’