### 数学代写|微分几何代写Differential Geometry代考|MATH 464

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微分几何代写Differential Geometry代考|Frames Associated to Coordinate Systems

Many problems in introductory mechanics involve finding the trajectory of a particle under the influence of various forces and/or subject to certain constraints. The first approach uses the coordinate functions and describes the trajectory as
$$\vec{r}(t)=(x(t), y(t), z(t))=x(t) \vec{\imath}+y(t) \vec{\jmath}+z(t) \vec{k} .$$
Newton’s equations of motion then lead to differential equations in the three coordinate functions $x(t), y(t)$, and $z(t)$. The velocity function is the derivative, namely
\begin{aligned} \vec{r}^{\prime}(t) &=\frac{d}{d t}(x(t) \vec{\imath})+\frac{d}{d t}(y(t) \vec{\jmath})+\frac{d}{d t}(z(t) \vec{k}) \ &=x^{\prime}(t) \vec{\imath}+x(t) \frac{d}{d t}(\vec{\imath})+y^{\prime}(t) \vec{\jmath}+y(t) \frac{d}{d t}(\vec{\jmath})+z^{\prime}(t) \vec{k}+z(t) \frac{d}{d t}(\vec{k}) \ &=x^{\prime}(t) \vec{\imath}+y^{\prime}(t) \vec{\jmath}+z^{\prime}(t) \vec{k} \end{aligned}
because $\frac{d}{d t} \vec{\imath}=0, \frac{d}{d t} \vec{\jmath}=0$, and $\frac{d}{d t} \vec{k}=0$. This last remark shows that the frame $(\vec{\imath}, \vec{\jmath}, \vec{k})$ associated to the Cartesian coordinate systems is a constant frame.

As we discuss variable frames, we introduce a nice way to describe the rate of change of a variable frame. Suppose that $\left{\vec{u}{1}, \vec{u}{2}, \vec{u}{3}\right}$ is a basis of $\mathbb{R}^{3}$ and let $\vec{a}$ and $\vec{b}$ be two other vectors with components $\vec{a}=a{1} \vec{u}{1}+a{2} \vec{u}{2}+a{3} \vec{u}_{3}$ and $\vec{b}=$ $b_{1} \vec{u}{1}+b{2} \vec{u}{2}+b{3} \vec{u}{3}$. Assuming that all vectors are column vectors, we can write these component definitions of $\vec{a}$ and $\vec{b}$ in the matrix expression $$\left(\begin{array}{ll} \vec{a} & \vec{b} \end{array}\right)=\left(\begin{array}{lll} \vec{u}{1} & \vec{u}{2} & \vec{u}{3} \end{array}\right)\left(\begin{array}{ll} a_{1} & b_{1} \ a_{2} & b_{2} \ a_{3} & b_{3} \end{array}\right)$$
Using this notation, we can express the relationships $\frac{d}{d t} \vec{\imath}=0, \frac{d}{d t} \vec{\jmath}=0$, and $\frac{d}{d t} \vec{k}=0$ by
$$\frac{d}{d t}\left(\begin{array}{lll} \vec{\imath} & \vec{\jmath} & \vec{k} \end{array}\right)=\left(\begin{array}{lll} \vec{\imath} & \vec{\jmath} & \vec{k} \end{array}\right)\left(\begin{array}{lll} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{array}\right) .$$

## 数学代写|微分几何代写Differential Geometry代考|Frames Associated to Trajectories

In the study of trajectories, whether in physics or geometry, it is often convenient to use a frame that is different from the Cartesian frame. Changing types of frames sometimes makes difficult integrals tractable or makes certain difficult differential equations manageable. In the particular context of special relativity, one talks about a momentarily comoving reference frame, abbreviated to MCRF. [50]

In the study of plane curves, it is common to use the frame ${\vec{T}, \vec{U}}$ to study the local properties of a plane curve $\vec{x}(t)$. (See [5, Chapter 1].) The vector $\vec{T}(t)$ is the unit tangent vector $\vec{T}(t)=\vec{x}^{\prime}(t) /\left|\vec{x}^{\prime}(t)\right|$, and the unit normal vector $\vec{U}(t)$, is the result of rotating $\vec{T}(t)$ by $\pi / 2$ in the counterclockwise direction. This is a moving frame that is defined in terms of a given regular curve $\vec{x}(t)$ and, at $t=t_{0}$, is viewed as based at the point $\vec{x}\left(t_{0}\right)$. To compare with applications in physics, it is important to note that the ${\vec{T}, \vec{U}}$ frame is not the same as the polar coordinate frame $\left{\vec{e}{r}, \vec{e}{\theta}\right}$. From Equation (2.4) (and ignoring the $z$-component), we know that
$$\vec{e}{T}=(\cos \theta, \sin \theta) \quad \text { and } \quad \vec{e}{\theta}=(-\sin \theta, \cos \theta) .$$
Assuming that $x, y, r$, and $\theta$ are functions of $t$ and since $x=r \cos \theta$ and $y=r \sin \theta$, we have
$$\vec{x}^{\prime}(t)=\left(x^{\prime}(t), y^{\prime}(t)\right)=\left(r^{\prime} \cos \theta-r \theta^{\prime} \sin \theta, r^{\prime} \sin \theta+r \theta^{\prime} \cos \theta\right)=r^{\prime} \vec{e}{r}+r \theta^{\prime} \vec{e}{\theta} .$$
We then calculate the speed function to be
$$s^{\prime}(t)=\left|\vec{x}^{\prime}(t)\right|=\sqrt{\left(r^{\prime}\right)^{2}+r^{2}\left(\theta^{\prime}\right)^{2}}$$
and find the unit tangent and unit normal vectors to be
\begin{aligned} &\vec{T}=\frac{1}{\sqrt{\left(r^{\prime}\right)^{2}+r^{2}\left(\theta^{r}\right)^{2}}}\left(r^{\prime} \vec{e}{r}+r \theta^{\prime} \vec{e}{\theta}\right), \ &\vec{U}=\frac{1}{\sqrt{\left(r^{\prime}\right)^{2}+r^{2}\left(\theta^{r}\right)^{2}}}\left(-r \theta^{\prime} \vec{e}{r}+r^{\prime} \vec{e}{\theta}\right) . \end{aligned}

## 数学代写|微分几何代写Differential Geometry代考| Frames Associated to Coordinate Systems

$$\vec{r}(t)=(x(t), y(t), z(t))=x(t) \vec{\imath}+y(t) \vec{\jmath}+z(t) \vec{k} .$$

$$\vec{r}^{\prime}(t)=\frac{d}{d t}(x(t) \vec{\imath})+\frac{d}{d t}(y(t) \vec{\jmath})+\frac{d}{d t}(z(t) \vec{k}) \quad=x^{\prime}(t) \vec{\imath}+x(t) \frac{d}{d t}(\vec{\imath})+y^{\prime}(t) \vec{\jmath}+y(t) \frac{d}{d t}(\vec{\jmath})+z^{\prime}(t)$$

## 数学代写|微分几何代写Differential Geometry代考| Frames Associated to Trajectories

$$\vec{e} T=(\cos \theta, \sin \theta) \quad \text { and } \quad \vec{e} \theta=(-\sin \theta, \cos \theta) .$$

$$\vec{x}^{\prime}(t)=\left(x^{\prime}(t), y^{\prime}(t)\right)=\left(r^{\prime} \cos \theta-r \theta^{\prime} \sin \theta, r^{\prime} \sin \theta+r \theta^{\prime} \cos \theta\right)=r^{\prime} \vec{e} r+r \theta^{\prime} \vec{e} \theta$$

$$s^{\prime}(t)=\left|\vec{x}^{\prime}(t)\right|=\sqrt{\left(r^{\prime}\right)^{2}+r^{2}\left(\theta^{\prime}\right)^{2}}$$

$$\vec{T}=\frac{1}{\sqrt{\left(r^{\prime}\right)^{2}+r^{2}\left(\theta^{r}\right)^{2}}}\left(r^{\prime} \vec{e} r+r \theta^{\prime} \vec{e} \theta\right), \quad \vec{U}=\frac{1}{\sqrt{\left(r^{\prime}\right)^{2}+r^{2}\left(\theta^{r}\right)^{2}}}\left(-r \theta^{\prime} \vec{e} r+r^{\prime} \vec{e} \theta\right) .$$

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