IS_Math-2020-02

IS Math-2022-02

题目来源：Problem 2 日期：2024-07-06 题目主题: Math-Calculus-二元积分

具体题目

Let $\mathbf{p}=(p(t),q(t))(t\in [a,b])$ be a smooth curve in the $xy$-plane. The length ${\ell }_{a^{\prime },b^{\prime }}$ of p from time $t=a^{\prime }$ to $b^{\prime }$ is defined by

$${\ell }_{a^{\prime },b^{\prime }}={\int }_{a^{\prime }}^{b^{\prime }}\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}t}\right)}^2}\mathrm{d}t,$$

and the total length ${\ell }_{a,b}$ of $\mathbf{p}$ is denoted by $L$. Suppose that $\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}=(0,0)$ never holds. Let variable $s=s(t)$ be defined as the length ${\ell }_{a,t}$ of $\mathbf{p}$ from time $a$ to $t$. Then the curve $\mathbf{p}$ is parametrized by the variable $s\in [0,L]$, where $s$ is also referred to as time. Answer the following questions.

1. Show that the following equation holds:

$$\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2}=1.$$

1. Let $\theta =\theta (s)$ be the angle between the tangent vector $\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}s}=\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)$ of $\mathbf{p}$ at time $s$ and the $x$-axis. Show that the following equation holds:

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\frac{\mathrm{d}\theta }{\mathrm{d}s}.$$

In the following, suppose that $\mathbf{p}$ is a smooth closed curve and is the boundary of a convex set $K$, where $\mathbf{p}$ rounds around $K$ in the counterclockwise direction.

1. Explain that $\frac{\mathrm{d}\theta }{\mathrm{d}s}\ge 0$ holds for any time $s$.

2. A point $\mathbf{x}=(x,y)$ that is not contained in $K$ is uniquely represented as

$$\mathbf{x}=\mathbf{p}(s)+r\mathbf{u}(s)$$

by time $s\in [0,L]$ and the distance $r$ between $\mathbf{p}$ and $K$, where $\mathbf{u}(s)$ is the unit normal vector of $\mathbf{p}$ at the time $s$ with direction toward the outside of $K$. Show that the following equation holds for such an $\mathbf{p}=(x,y)$:

$$\left|\det \left(\begin{array}{cc}\frac{\partial x}{\partial s}& \frac{\partial x}{\partial r}\\ \frac{\partial y}{\partial s}& \frac{\partial y}{\partial r}\end{array}\right)\right|=1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}.$$

3. For a nonnegative real number $D$, let $K_D$ be the set of points whose distance from $K$ is at most $D$. Show that the area $A_D={\iint }_{K_D}\mathrm{d}x\mathrm{d}y$ of $K_D$ is represented as

$$A_D=A+LD+\pi D^2$$

by the area $A$ of $K$ and the total length $L$ of $\mathbf{p}$.

正确解答

1. Prove $\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2}=1$

Since $s(t)$ is the arc length parameter, we have

$$s(t)={\int }_a^t\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}u}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}u}\right)}^2}\mathrm{d}u$$

Differentiating both sides with respect to $t$, we get

$$\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}t}\right)}^2}$$

By the chain rule,

$$\frac{\mathrm{d}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}s}\frac{\mathrm{d}s}{\mathrm{d}t}$$

Thus,

$$\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}p}{\mathrm{d}s}\frac{\mathrm{d}s}{\mathrm{d}t},\frac{\mathrm{d}q}{\mathrm{d}t}=\frac{\mathrm{d}q}{\mathrm{d}s}\frac{\mathrm{d}s}{\mathrm{d}t}$$

Squaring and adding,

$${\left(\frac{\mathrm{d}p}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}t}\right)}^2={\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)}^2$$

Since $\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}t}\right)}^2}$, we have

$${\left(\frac{\mathrm{d}p}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}t}\right)}^2={\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2$$

So,

$$\sqrt{{\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2}=1$$

2. Prove $\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\frac{\mathrm{d}\theta }{\mathrm{d}s}$

Let $\mathbf{T}=\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)$ be the unit tangent vector. The angle $\theta (s)$ between $\mathbf{T}$ and the $x$-axis satisfies

$$\mathbf{T}=\left(\cos \theta ,\sin \theta \right)$$

Differentiating $\mathbf{T}$ with respect to $s$, we get

$$\frac{\mathrm{d}\mathbf{T}}{\mathrm{d}s}=\left(\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2},\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}\right)=\left(-\sin \theta \frac{\mathrm{d}\theta }{\mathrm{d}s},\cos \theta \frac{\mathrm{d}\theta }{\mathrm{d}s}\right)$$

Notice that

$$\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)\cdot \left(-\sin \theta \frac{\mathrm{d}\theta }{\mathrm{d}s},\cos \theta \frac{\mathrm{d}\theta }{\mathrm{d}s}\right)=0$$

Therefore,

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

Step-by-Step Derivation

1. Tangent Vector and Angle:

   Let the curve $\mathbf{p}(s)$ be parametrized by arc length $s$. The tangent vector to the curve at $s$ is given by:

$$\mathbf{T}(s)=\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)$$

The angle $\theta (s)$ is defined as the angle between the tangent vector and the $x$-axis. Thus, we can express the tangent vector in terms of $\theta (s)$:

$$\mathbf{T}(s)=\left(\cos \theta (s),\sin \theta (s)\right)$$

1. First Derivative of the Tangent Vector:

   The derivative of the tangent vector with respect to $s$ gives the change in the direction of the tangent vector, which is related to the curvature of the curve:

$$\frac{\mathrm{d}\mathbf{T}}{\mathrm{d}s}=\left(\frac{\mathrm{d}}{\mathrm{d}s}\cos \theta (s),\frac{\mathrm{d}}{\mathrm{d}s}\sin \theta (s)\right)$$

Using the chain rule, this becomes:

$$\frac{\mathrm{d}\mathbf{T}}{\mathrm{d}s}=\left(-\sin \theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s},\cos \theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}\right)$$

1. Relating the Derivatives:

   Now, express the components of the tangent vector in terms of $p$ and $q$:

$$\mathbf{T}(s)=\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)$$

Therefore,

$$\frac{\mathrm{d}p}{\mathrm{d}s}=\cos \theta (s)\text{and}\frac{\mathrm{d}q}{\mathrm{d}s}=\sin \theta (s)$$

1. Second Derivative of the Curve:

   Next, consider the second derivatives of $p$ and $q$:

$$\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)=\frac{\mathrm{d}}{\mathrm{d}s}(\cos \theta (s))=-\sin \theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}$$ $$\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}=\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)=\frac{\mathrm{d}}{\mathrm{d}s}(\sin \theta (s))=\cos \theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

1. Combining the Results:

   Substitute the expressions for $\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}$ and $\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}$ into the given equation:

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}$$

Using the relationships $\frac{\mathrm{d}p}{\mathrm{d}s}=\cos \theta (s)$ and $\frac{\mathrm{d}q}{\mathrm{d}s}=\sin \theta (s)$, we get:

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}=\cos \theta (s)\cdot \cos \theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}={\cos }^2\theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}$$ $$\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\sin \theta (s)\cdot (-\sin \theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s})=-{\sin }^2\theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

Therefore,

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}={\cos }^2\theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s}-(-{\sin }^2\theta (s)\frac{\mathrm{d}\theta }{\mathrm{d}s})=({\cos }^2\theta (s)+{\sin }^2\theta (s))\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

1. Simplifying Using the Pythagorean Identity:

   Recall that ${\cos }^2\theta (s)+{\sin }^2\theta (s)=1$:

$$({\cos }^2\theta (s)+{\sin }^2\theta (s))\frac{\mathrm{d}\theta }{\mathrm{d}s}=1\cdot \frac{\mathrm{d}\theta }{\mathrm{d}s}=\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

Thus, we have shown that:

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\frac{\mathrm{d}\theta }{\mathrm{d}s}.$$

3. Explain $\frac{\mathrm{d}\theta }{\mathrm{d}s}\ge 0$ for any time $s$

Since $\mathbf{p}$ is a smooth closed curve that is the boundary of a convex set $K$ and rounds around $K$ in the counterclockwise direction, the angle $\theta (s)$ increases as we move along the curve. The convexity of $K$ ensures that the curvature is non-negative, leading to $\frac{\mathrm{d}\theta }{\mathrm{d}s}\ge 0$.

Notice: The requirement of the problem is to “Explain” rather than to “Prove” this statement. Therefore, a qualitative explanation based on the properties of convex curves is sufficient.

4. Show $\left|\det \left(\begin{array}{cc}\frac{\partial x}{\partial s}& \frac{\partial x}{\partial r}\\ \frac{\partial y}{\partial s}& \frac{\partial y}{\partial r}\end{array}\right)\right|=1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}$

Since $\mathbf{p}(s)$ moves counterclockwise around $K$, the outward unit normal vector $\mathbf{u}(s)$ can be expressed as:

$$\mathbf{u}(s)=\left(\frac{\mathrm{d}q}{\mathrm{d}s},-\frac{\mathrm{d}p}{\mathrm{d}s}\right)$$

First, calculate the partial derivatives of $\mathbf{x}$ with respect to $s$ and $r$:

$$\frac{\partial \mathbf{x}}{\partial s}=\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}s}+r\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}s}$$ $$\frac{\partial \mathbf{x}}{\partial r}=\mathbf{u}(s)$$

Where $\mathbf{p}(s)=(p(s),q(s))$ and $\mathbf{u}(s)=\left(\frac{\mathrm{d}q}{\mathrm{d}s},-\frac{\mathrm{d}p}{\mathrm{d}s}\right)$. The derivatives are:

$$\frac{\partial \mathbf{p}}{\partial s}=\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)$$ $$\frac{\partial \mathbf{u}}{\partial s}=\left(\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2},-\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}\right)$$

Thus,

$$\frac{\partial \mathbf{x}}{\partial s}=\left(\frac{\mathrm{d}p}{\mathrm{d}s},\frac{\mathrm{d}q}{\mathrm{d}s}\right)+r\left(\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2},-\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}\right)$$

We also have:

$$\frac{\partial \mathbf{x}}{\partial r}=\left(\frac{\mathrm{d}q}{\mathrm{d}s},-\frac{\mathrm{d}p}{\mathrm{d}s}\right)$$

The Jacobian matrix $J$ is:

$$J=\left(\begin{array}{cc}\frac{\partial x}{\partial s}& \frac{\partial x}{\partial r}\\ \frac{\partial y}{\partial s}& \frac{\partial y}{\partial r}\end{array}\right)=\left(\begin{array}{cc}\frac{\mathrm{d}p}{\mathrm{d}s}+r\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}& \frac{\mathrm{d}q}{\mathrm{d}s}\\ \frac{\mathrm{d}q}{\mathrm{d}s}-r\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}& -\frac{\mathrm{d}p}{\mathrm{d}s}\end{array}\right)$$

To find the determinant, we expand:

$$\left|\det J\right|=\left|\det \left(\begin{array}{cc}\frac{\mathrm{d}p}{\mathrm{d}s}+r\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}& \frac{\mathrm{d}q}{\mathrm{d}s}\\ \frac{\mathrm{d}q}{\mathrm{d}s}-r\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}& -\frac{\mathrm{d}p}{\mathrm{d}s}\end{array}\right)\right|$$

Using the determinant formula for a $2\times 2$ matrix, we get:

$$\det J=\left(\frac{\mathrm{d}p}{\mathrm{d}s}+r\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}\right)\left(-\frac{\mathrm{d}p}{\mathrm{d}s}\right)-\left(\frac{\mathrm{d}q}{\mathrm{d}s}-r\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}\right)\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)$$

Expanding this, we get:

$$=\left(\frac{\mathrm{d}p}{\mathrm{d}s}\left(-\frac{\mathrm{d}p}{\mathrm{d}s}\right)+r\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}\left(-\frac{\mathrm{d}p}{\mathrm{d}s}\right)\right)-\left(\frac{\mathrm{d}q}{\mathrm{d}s}\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)-r\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)\right)$$ $$=-{\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2-r\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2+r\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}$$

Using the result from the second question, we know:

$$\frac{\mathrm{d}p}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}q}{\mathrm{d}{s}^2}-\frac{\mathrm{d}q}{\mathrm{d}s}\frac{{\mathrm{d}}^{2}p}{\mathrm{d}{s}^2}=\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

So,

$$=-\left({\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2\right)-r\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

Since ${\left(\frac{\mathrm{d}p}{\mathrm{d}s}\right)}^2+{\left(\frac{\mathrm{d}q}{\mathrm{d}s}\right)}^2=1$,

$$=-1-r\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

Since $r$ is the distance between $\mathbf{p}$ and $K$, it is non-negative, and $\frac{\mathrm{d}\theta }{\mathrm{d}s}\ge 0$ for a convex curve, we can conclude to:

$$\left|\det J\right|=1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

5. Show $A_D=A+LD+\pi D^2$

To rigorously prove the formula for the area $A_D$ of the set $K_D$, which consists of all points within a distance $D$ from the convex set $K$, we will use the result from the fourth question and the concept of the Jacobian determinant.

First, recall the fourth question’s result, which states:

$$\left|\det \left(\begin{array}{cc}\frac{\partial x}{\partial s}& \frac{\partial x}{\partial r}\\ \frac{\partial y}{\partial s}& \frac{\partial y}{\partial r}\end{array}\right)\right|=1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

We will use this Jacobian determinant to transform the integration from the $(x,y)$ coordinates to the $(s,r)$ coordinates.

Step 1: Set Up the Integral in $(x,y)$ Coordinates

The area of $A_D$ outside of $K$, $A_D^{\prime }$ , can be expressed as a double integral over the region $K_D$:

$$A_D^{\prime }={\iint }_{K_D}\mathrm{d}x\mathrm{d}y$$

Step 2: Change of Variables to $(s,r)$ Coordinates

We use the parameterization $\mathbf{x}=\mathbf{p}(s)+r\mathbf{u}(s)$, where $s\in [0,L]$ and $r\in [0,D]$. The Jacobian determinant for the transformation is $1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}$. Therefore, we can rewrite the area integral as:

$$A_D^{\prime }={\int }_0^L{\int }_0^D\left(1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}\right)\mathrm{d}r\mathrm{d}s$$

Step 3: Evaluate the Inner Integral

Evaluate the inner integral with respect to $r$:

$${\int }_0^D\left(1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}\right)\mathrm{d}r={\int }_0^{D}1\mathrm{d}r+{\int }_0^{D}r\frac{\mathrm{d}\theta }{\mathrm{d}s}\mathrm{d}r$$

The first term is straightforward:

$${\int }_0^{D}1\mathrm{d}r=D$$

For the second term:

$${\int }_0^{D}r\frac{\mathrm{d}\theta }{\mathrm{d}s}\mathrm{d}r=\frac{\mathrm{d}\theta }{\mathrm{d}s}{\int }_0^{D}r\mathrm{d}r=\frac{\mathrm{d}\theta }{\mathrm{d}s}{\left[\frac{r^2}{2}\right]}_0^D=\frac{\mathrm{d}\theta }{\mathrm{d}s}\cdot \frac{D^2}{2}$$

Combining these, we get:

$${\int }_0^D\left(1+r\frac{\mathrm{d}\theta }{\mathrm{d}s}\right)\mathrm{d}r=D+\frac{D^2}{2}\frac{\mathrm{d}\theta }{\mathrm{d}s}$$

Step 4: Evaluate the Outer Integral

Next, integrate with respect to $s$:

$$A_D^{\prime }={\int }_0^L\left(D+\frac{D^2}{2}\frac{\mathrm{d}\theta }{\mathrm{d}s}\right)\mathrm{d}s$$

This can be split into two integrals:

$$A_D^{\prime }=D{\int }_0^L\mathrm{d}s+\frac{D^2}{2}{\int }_0^L\frac{\mathrm{d}\theta }{\mathrm{d}s}\mathrm{d}s$$

The first integral is straightforward:

$$D{\int }_0^L\mathrm{d}s=D\cdot L=LD$$

For the second integral, we note that integrating $\frac{\mathrm{d}\theta }{\mathrm{d}s}$ over one complete loop gives the total change in $\theta$, which is $2\pi$ for a simple closed curve that winds around the convex set $K$ once:

$${\int }_0^L\frac{\mathrm{d}\theta }{\mathrm{d}s}\mathrm{d}s=\theta (L)-\theta (0)=2\pi$$

So, the second term becomes:

$$\frac{D^2}{2}\cdot 2\pi =\pi D^2$$

Step 5: Include the Area $A$ of the Convex Set $K$

In addition to the area contributed by the strip of width $D$ around $K$ and the corner areas, we need to add the area $A$ of the convex set $K$ itself. Therefore, the total area $A_D$ is:

$$A_D=A+A_D^{\prime }=A+LD+\pi D^2$$

This confirms the formula for the area of $K_D$.

知识点

Jacobi行列式 二元积分

Jacobi行列式 Jacobi Determinant

二重积分 Double Integral

解题技巧和信息

1. Arc Length Parameterization: Use the definition of arc length to reparametrize curves.
2. Tangent and Normal Vectors: Understand the geometric interpretations of these vectors and their derivatives.
3. Convexity: Utilize the properties of convex sets and curves for inequality proofs.
4. Jacobian Determinant: Apply the Jacobian determinant to change variables in integrals.
5. Geometric Decomposition: Decompose complex shapes into simpler parts to find areas.

重点词汇

1. Arc length parameterization - 弧长参数化
2. Tangent vector - 切向量
3. Normal vector - 法向量
4. Convex set - 凸集
5. Determinant - 行列式
6. Parametrization - 参数化
7. Strip - 带状区域

参考资料

1. Calculus of Variations by I. M. Gelfand and S. V. Fomin, Chap. 1-3
2. Convex Analysis by R. Tyrrell Rockafellar, Chap. 5
3. Introduction to Differential Geometry by Luther Pfahler Eisenhart, Chap. 2