## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinetic Energy of the Rigid Body

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinetic Energy of the Rigid Body

We start from the definition of the kinetic energy $T$,
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2,$$ and insert the expression (4.42) for the velocity:
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2+\sum_i m_i\left(\omega \times \mathbf{r}_i\right) \cdot \dot{\mathbf{r}}_0 .$$
The third term is a scalar triple product and can therefore be rewritten as follows:
$$\sum_i m_i \mathbf{r}_i \cdot\left(\dot{\mathbf{r}}_0 \times \omega\right)$$
There are two typical cases for the discussion of the rigid body:

One point of the body remains space-fixed, while the body rotates with the angular velocity $\omega$. Then it appears absolutely reasonable to choose this point as the origin $S$ of $\Sigma$ and in general also as the origin of $\widehat{\Sigma}$. One then speaks of a spinning top for which holds:
$$\mathbf{r}_0=\mathbf{0}, \quad \dot{\mathbf{r}}_0=\mathbf{0}$$

If no point is space-fixed one usually chooses the origin $S$ at the center of mass and that means:
$$\sum_i m_i \mathbf{r}i=\mathbf{0}$$ We see that these two cases, the only relevant ones, both let the third term in (4.43) disappear. We therefore apply from the beginning the kinetic energy in the form: $$T=\frac{1}{2} M \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2=T_T+T{\mathrm{R}}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Properties of the Inertial Tensor

Strictly speaking it is nothing other than a proper extension of the term ‘vector’. By a
tensor of $k$-th rank in an $n$-dimensional space
one understands an $n^k$ number of elements
$$\left(F_{i 1, i_2, \ldots, i_k}\right) ; \quad i_j=1, \ldots, n,$$
which for coordinate rotations transform linearly satisfying certain rules. The elements are called the components of the tensor. They carry $k$ indexes each of which runs from 1 to $n$. The rules are chosen just so that the ‘normal’ vectors are first-rank tensors. One requires that in connection with coordinate rotations a tensor of $k$-th rank transforms itself with respect to all $k$ indexes like a ‘normal’ vector. According to our underlying physical problems of course only the cases $n=1,2,3$ are interesting. Furthermore, in physics we can restrict ourselves to $k=0,1,2$.
$\mathbf{k}=\mathbf{0}:$ scalar: $\quad \bar{x}=x$
$\mathbf{k}=\mathbf{1}$ : vector, $n=3$ components (in the three-dimensional space), for which, according to (1.309), it holds after a coordinate rotation:
$$\bar{x}i=\sum_j d{i j} x_j$$
$\left(d_{i j}\right.$ : components of the rotation matrix (1.307)),
$\mathbf{k}=\mathbf{2}:\left(F_{i j}\right){i, j=1,2,3}: n^2=9$ components with $$\bar{F}{i j}=\sum_{l, m} d_{i l} d_{j m} F_{l m}$$
and so on.
Second-rank tensors can always be written as square matrices. However, in contrast to normal matrices which are represented by collections of elements (numbers), which may behave arbitrarily with coordinate transformations, the above-mentioned transformation behavior is absolutely mandatory for the elements of a tensor.
Why is it necessary that the system of coefficients (4.47) does exhibit tensor properties? The components of the inertial tensor in a given system of coordinates are uniquely determined by the mass distribution of the rigid body. But with a rotation of the system of coordinates the components will change. Furthermore, of course also the components of the angular velocity $\omega$ will undergo a change. However, it is clear that a rotation of the coordinate system should not influence the (measurable) rotational kinetic energy $T_{\mathrm{R}}$.

# 理论力学代写

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinetic Energy of the Rigid Body

$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2$$

$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2+\sum_i m_i\left(\omega \times \mathbf{r}_i\right) \cdot \dot{\mathbf{r}}_0 .$$

$$\sum_i m_i \mathbf{r}_i \cdot\left(\dot{\mathbf{r}}_0 \times \omega\right)$$

$$\mathbf{r}_0=\mathbf{0}, \quad \dot{\mathbf{r}}_0=\mathbf{0}$$

$$\sum_i m_i \mathbf{r} i=\mathbf{0}$$

$$T=\frac{1}{2} M \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2=T_T+T \mathrm{R}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Properties of the Inertial Tensor

$$\left(F_{i 1, i_2, \ldots, i_k}\right) ; \quad i_j=1, \ldots, n$$

$\mathbf{k}=\mathbf{0}$ :标量: $\quad \bar{x}=x$
$\mathbf{k}=\mathbf{1}$ : 矢量， $n=3$ 分量（在三维空间中），根据（1.309)，它在坐标旋转后成立:
$$\bar{x} i=\sum_j d i j x_j$$
$\left(d_{i j}:\right.$ 旋转矩阵的分量 (1.307)),
$\mathbf{k}=\mathbf{2}:\left(F_{i j}\right) i, j=1,2,3: n^2=9$ 组件与
$$\bar{F} i j=\sum_{l, m} d_{i l} d_{j m} F_{l m}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rolling Motion

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rolling Motion

As a further important example of a rigid body with only one rotational degree of freedom we consider the
homogeneous cylinder rolling off an inclined plane
Though the rotation axis is again body-fixed it is not space-fixed. It is shifting in parallel to itself (Fig. 4.11). The velocity of each of the cylinder points is composed by two contributions, a rotational contribution due to the rotation around the cylinder axis during the rolling motion and a translational contribution which is the same for all points of the cylinder and happens in $s$ direction:
$$\dot{\mathbf{r}}i=\dot{\mathbf{r}}{i R}+\dot{\mathbf{r}}{i T}$$ The rotational contribution we have already calculated in (4.8): $$\dot{\mathbf{r}}{i R}=\left(\omega \times \overline{\mathbf{r}}i\right)$$ The translational contribution is obtained from the rolling off condition $$\Delta s=R \Delta \varphi \Longrightarrow\left|\dot{\mathbf{r}}{i T}\right|=|\dot{\mathbf{s}}|=R|\dot{\varphi}|$$
The cylinder shall roll, not slide.
(a) Kinetic Energy
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2=\frac{1}{2} \sum_i m_i\left[\left(\omega \times \overline{\mathbf{r}}_i\right)^2+2 \dot{\mathbf{s}} \cdot\left(\omega \times \overline{\mathbf{r}}_i\right)+\dot{s}^2\right]$$
The mixed term disappears because in a homogeneous cylinder two volume elements located diametrally opposite to the rotation axis have the same mass but rotation velocities are in opposite directions (Fig. 4.12). The sum over all elements is therefore zero. It can of course be shown also by a direct calculation that
$$\sum_i m_i\left(\omega \times \overline{\mathbf{r}}_i\right)=0$$
must hold.

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinematics of the Rigid Body

In our introductory Sect. 4.1 we had already decomposed the general motion of a rigid body into

1. the translation of an arbitrarily chosen point $S$ of the body and
2. the rotation around an axis through this point $S$.
We now introduce two reference systems which are initially both Cartesian:
$\widehat{\Sigma}$ : space-fixed reference system with a space-fixed origin of coordinates $\mathcal{O}$. It is assumed to be an inertial system. Axis : $\hat{\mathbf{e}}\alpha, \alpha=1,2,3$. $\Sigma$ : body-fixed reference system with the body-fixed origin $S$. Axes: $\mathbf{e}\alpha(t), \alpha=$ $1,2,3$

The point $S$ has the position vector $\mathbf{r}0(t)$ as seen from $\widehat{\Sigma}$. Then it holds for the points of the rigid body: $$\begin{array}{ll} \hat{\mathbf{r}}_i(t)=\sum{\alpha=1}^3 \hat{x}{i \alpha}(t) \hat{\mathbf{e}}\alpha & (\text { in } \widehat{\Sigma}), \ \mathbf{r}i(t)=\sum{\alpha=1}^3 x_{i \alpha} \mathbf{e}\alpha(t) & \text { (in } \Sigma) \end{array}$$ with the obvious relation: $$\hat{\mathbf{r}}_i(t)=\mathbf{r}_0(t)+\mathbf{r}_i(t)$$ The coordinates $x{i \alpha}$ in the body-fixed system $\Sigma$ are by the definition of the rigid body time-independent quantities. The position of the rigid body is therewith completely given by the position of $\Sigma$ relative to $\widehat{\Sigma}$.

We are now interested in the velocities of the mass points of the rigid body (Fig. 4.13). These we find rather easily with the general theory of arbitrarily relative to each other moving reference systems that we derived in Sect. 2.2.5. The full time derivative of a vector represented in $\Sigma$ seen from $\widehat{\Sigma}$ can be written as the operator

The first term on the right-hand side plays by definition no role for the rigid body. Thus it remains:
$$\dot{\mathbf{r}}_i=\left(\omega \times \mathbf{r}_i\right)$$
or with (4.40):
$$\dot{\hat{\mathbf{r}}}_i(t)=\dot{\mathbf{r}}_0(t)+\left(\omega \times \mathbf{r}_i\right) .$$
This is an important result. It signifies that at any moment of time the motion of a rigid body can be resolved into the translational motion $\mathbf{r}_0(t)$ of the origin of the body-fixed system and the rotation around the momentary rotation axis $\omega(t)$ where the latter always passes through the origin $S$ of the body-fixed system.

# 理论力学代写

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rolling Motion

，尽管旋转轴同样是物体固定的，但它不是空间固定的。它与自身平行移动（图 4.11)。每个圆柱点的速 度由两个贡献组成，一个是由于在滚动运动期间绕圆柱轴的旋转引起的旋转贡献，另一个是对圆柱的所有 点都相同的平移贡献，并且发生在 $s$ 方向:
$$\dot{\mathbf{r}} i=\dot{\mathbf{r}} i R+\dot{\mathbf{r}} i T$$

$$\dot{\mathbf{r}} i R=(\omega \times \overline{\mathbf{r}} i)$$

$$\Delta s=R \Delta \varphi \Longrightarrow|\dot{\mathbf{r}} i T|=|\dot{\mathbf{s}}|=R|\dot{\varphi}|$$

(a) 动能
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2=\frac{1}{2} \sum_i m_i\left[\left(\omega \times \overline{\mathbf{r}}_i\right)^2+2 \dot{\mathbf{s}} \cdot\left(\omega \times \overline{\mathbf{r}}_i\right)+\dot{s}^2\right]$$

$$\sum_i m_i\left(\omega \times \overline{\mathbf{r}}_i\right)=0$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinematics of the Rigid Body

1. 任意选择点的平移 $S$ 身体的和
2. 通过该点绕轴旋转 $S$.
我们现在介绍两个最初都是笛卡尔的参考系统:
$\widehat{\Sigma}$ : 具有空间固定坐标原点的空间固定参考系 $\mathcal{O}$. 假设它是一个惯性系统。轴: $\hat{e} \alpha, \alpha=1,2,3 . \Sigma$ : 具有身体固定原点的身体固定参考系 $S$. 轴: $\mathrm{e} \alpha(t), \alpha=1,2,3$
重点 $S$ 有位置向量 $\mathbf{r} 0(t)$ 从 $\widehat{\Sigma}$. 然后它适用于刚体的点:
$$\left.\hat{\mathbf{r}}i(t)=\sum \alpha=1^3 \hat{x} i \alpha(t) \hat{\mathbf{e}} \alpha \quad(\text { in } \widehat{\Sigma}), \mathbf{r} i(t)=\sum \alpha=1^3 x{i \alpha} \mathbf{e} \alpha(t) \quad \text { (in } \Sigma\right)$$
具有明显的关系：
$$\hat{\mathbf{r}}_i(t)=\mathbf{r}_0(t)+\mathbf{r}_i(t)$$
我们现在对刚体质点的速度感兴趣（图 4.13) 。这些我们很容易通过我们在第 1 节中导出的任意相对于彼
根据定义，右侧的第一项对刚体没有任何作用。因此它仍然是:
$$\dot{\mathbf{r}}_i=\left(\omega \times \mathbf{r}_i\right)$$
或 $(4.40)$ :
$$\dot{\hat{\mathbf{r}}}_i(t)=\dot{\mathbf{r}}_0(t)+\left(\omega \times \mathbf{r}_i\right)$$
这是一个重要的结果。它表示在任何时刻，刚体的运动都可以分解为平移运动 $\mathbf{r}_0(t)$ 身体固定系统的原点 和绕瞬时旋转轴的旋转 $\omega(t)$ 后者总是通过原点 $S$ 身体固定系统。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS2041

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Central Forces

A force-type of the profile
$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{e}_r$$
is called a ‘central force’. Thus the force is directed along the radial rays which start from the center (origin) of the force (Fig. 2.45). For such forces the angular momentum $\mathbf{L}$, if referred to the force center, is constant, according to (2.245). Central forces in the general form (2.248) are not necessarily conservative. It rather holds:
Central force $\mathbf{F}$ conservative $\Longleftrightarrow \mathbf{F}=f(r) \mathbf{e}_r$.

It is clear that $\mathbf{F}$ must not depend on $\dot{\mathbf{r}}$ and $t$ to be conservative. For a proof of (2.249) we therefore can restrict ourselves to forces $\mathbf{F}$ of the form:
$$\mathbf{F}=f(\mathbf{r}) \mathbf{e}_r .$$
According to (2.234) the force $\mathbf{F}$ is conservative if and only if the curl of $\mathbf{F}$ vanishes. That we inspect with (1.289):
$$\nabla \times \mathbf{F}=\frac{f(\mathbf{r})}{r} \nabla \times \mathbf{r}+\left[\left(\nabla \frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] .$$
After (1.292) we can exploit $\nabla \times \mathbf{r}=0$, so it remains to require:
$$0 \stackrel{!}{=}\left[\nabla\left(\frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] \text {. }$$
Hence the two vectors in the square bracket have to be parallel. In view of (1.271) and the subsequent discussion one realizes that the gradient vector is orthogonal to the planes $f(\mathbf{r}) / r=$ const. Hence these planes must simultaneously be orthogonal to $\mathbf{r}$. That means, however, that $f(\mathbf{r}) / r$ has to be constant on the surface of a sphere. This is possible only if $f(\mathbf{r})=f(r)$. That proves (2.249)!

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Integration of the Equations of Motion

If the law of conservation of angular-momentum
$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=\text { const }$$
or the energy conservation law
$$E=\frac{m}{2} \dot{\mathbf{r}}^2+V(\mathbf{r})=\mathrm{const}$$
are valid then one speaks of
first integrals of motion
The original equations of motion are always differential equations of second order, the conservation laws, on the other hand, are only of first order. Furthermore, on the basis of the conservation laws a general procedure for the complete solution of the equations of motion can be developed.

We have shown that the angular-momentum conservation law holds if and only if the acting force is a central force:
$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{r}$$
(The trivial case $\mathbf{F} \equiv 0$ shall be excluded!)
If simultaneously the energy conservation law holds then definitely a potential must exist. Hence, the central force is conservative and must be of the form:
$$\mathbf{F}=f(r) \mathbf{r} .$$
Moreover we know that in such a case the potential can depend only on the magnitude of $\mathbf{r}$ :
$$V=V(r) .$$
Therewith we will further evaluate the conservation laws. Because of the constancy of the angular momentum the motion will happen in a fixed plane. Let this be the $x y$ plane.

# 理论力学代写

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Central Forces

$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{e}_r$$

$$\mathbf{F}=f(\mathbf{r}) \mathbf{e}_r .$$

$$\nabla \times \mathbf{F}=\frac{f(\mathbf{r})}{r} \nabla \times \mathbf{r}+\left[\left(\nabla \frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] .$$

$$0 \stackrel{!}{=}\left[\nabla\left(\frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] \text {. }$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Integration of the Equations of Motion

$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=\text { const }$$

$$E=\frac{m}{2} \dot{\mathbf{r}}^2+V(\mathbf{r})=\text { const }$$

$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{r}$$
（琐碎的情况 $\mathbf{F} \equiv 0$ 应被排除!）

$$\mathbf{F}=f(r) \mathbf{r} .$$

$$V=V(r) .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYC90007

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Angular Momentum and Torque

If we multiply the basic dynamical equation (2.43) vectorially by $\mathbf{r}$,
$$m(\mathbf{r} \times \ddot{\mathbf{r}})=(\mathbf{r} \times \mathbf{F}),$$ then there appears on the left-hand side the time-derivative of an important physical quantity:
$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=(\mathbf{r} \times \mathbf{p}) \quad \text { angular momentum } .$$
Since both position $\mathbf{r}$ and momentum $\mathbf{p}$ are polar vectors the resulting $\mathbf{L}$ must be an axial vector oriented perpendicularly to the plane spanned by $\mathbf{r}$ and $\mathbf{p}$. With the further definition,
$$\mathbf{M}=(\mathbf{r} \times \mathbf{F}) \quad \text { torque (moment) },$$
it follows from (2.241):
$$\frac{d}{d t} \mathbf{L}=\mathbf{M} .$$
This equation represents the angular-momentum law:
The time rate of the change of angular momentum is equal to the applied torque. If the torque is identical to zero then this theorem becomes the
Law of Conservation of Angular-Momentum
$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Law of Conservation of Angular-Momentum

$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$
There are two possibilities for getting $\mathbf{M}=\mathbf{0}$ :
1) $\mathbf{F} \equiv \mathbf{0} \quad$ (trivial case) ,
2) $\mathbf{F} \uparrow \uparrow \mathbf{r} \quad$ (central field) .
$(2.246)$
Case (1) is identical to the uniform straight-line motion of the mass point:
$$\dot{\mathbf{r}}=\mathbf{v}=\text { const . }$$
At first glance it appears astonishing that a uniform straight-line movement possesses any, even if constant, angular momentum. In Fig. $2.44 \mathbf{L}$ is perpendicular to the plane of the paper with the magnitude $m v d$. Only if the reference point (origin of coordinates) lies on the straight line then $\mathbf{L}$ indeed disappears. That gives evidence that the angular momentum is not at all a genuine particle property, but rather depends on the choice of the reference point.
A shift of the origin of coordinates by the constant vector a,
$$\mathbf{r}^{\prime}=\mathbf{r}+\mathbf{a} ; \quad \dot{\mathbf{r}}^{\prime}=\dot{\mathbf{r}} \Longrightarrow \mathbf{p}^{\prime}=\mathbf{p}$$
means for the angular momentum:
$$\mathbf{L}^{\prime}=\left(\mathbf{r}^{\prime} \times \mathbf{p}^{\prime}\right)=(\mathbf{r} \times \mathbf{p})+(\mathbf{a} \times \mathbf{p})=\mathbf{L}+(\mathbf{a} \times \mathbf{p}) .$$
If $\mathbf{L}$ is constant then $\mathbf{L}^{\prime}$ is also constant only if simultaneously the conservation of momentum also holds $\mathbf{p}=$ const. Furthermore, it does not necessarily follow from $\mathbf{L}=\mathbf{0}$ that also $\mathbf{L}^{\prime}=\mathbf{0}$. In general that is indeed not the case.

The second possibility for $\mathbf{M}=\mathbf{0}$ in (2.246) shall be discussed in a separate section.

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Angular Momentum and Torque

$$m(\mathbf{r} \times \ddot{\mathbf{r}})=(\mathbf{r} \times \mathbf{F}),$$

$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=(\mathbf{r} \times \mathbf{p}) \quad \text { angular momentum } .$$

$$\mathbf{M}=(\mathbf{r} \times \mathbf{F}) \quad \text { torque }(\text { moment }),$$

$$\frac{d}{d t} \mathbf{L}=\mathbf{M}$$

$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Law of Conservation of Angular-Momentum

$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$

$$\mathbf{r}^{\prime}=\mathbf{r}+\mathbf{a} ; \quad \dot{\mathbf{r}}^{\prime}=\dot{\mathbf{r}} \Longrightarrow \mathbf{p}^{\prime}=\mathbf{p}$$

$$\mathbf{L}^{\prime}=\left(\mathbf{r}^{\prime} \times \mathbf{p}^{\prime}\right)=(\mathbf{r} \times \mathbf{p})+(\mathbf{a} \times \mathbf{p})=\mathbf{L}+(\mathbf{a} \times \mathbf{p})$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS3020

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Classical Particle Paths

Our very general considerations already permit us to draw far reaching conclusions about possible particle paths. Since $T$ is non-negative it follows from (2.213):
classically allowed region of motion : $E \geq V(x)$,
(2.215)
classically forbidden region of motion: $E<V(x)$,
(2.216)
classical turning points : $E=V(x)$.
The supplement classical is important since the above statement has to be commented on when dealing with the all-embracing quantum theory.
Examples
(a) Harmonic oscillator:
Because of (2.215) it is to be expected that an oscillatory motion takes place between the two turning points $\pm x_0$. The distance between $E=E_0$ and $V(x)$ is a measure for the velocity of the mass point (Fig. 2.35). At the turning points the velocity of the particle is zero. The direction of motion reverses.
(b) General state dependence of potential:
For $x \leq x_1$ no movement is possible, and so is the case between $x_2$ and $x_3$, also. Between $x_1$ and $x_2$ an oscillatory behavior takes place, whilst a particle coming from $+\infty$ is reflected at $x_3$ (Fig. 2.36). Possible equilibrium positions of the particle are those points where no forces act. Obviously these are the extremal values of the potential $V$ :
$$F=0=-\frac{d V}{d x} \Longleftrightarrow V \text { extremal } .$$
In case of a maximum the particle is in an unstable equilibrium. The smallest position change lets it fall down the potential wall. In case of a minimum the particle finds itself in a stable equilibrium.

Finally we add a remark about the dimension, which is the same for $T, W, V$ and $E$ :
$$[E]=k g m^2 s^{-2}=\text { Joule . }$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Work, Power, and Energy

We start with the term ‘work’ which has to be generalized for arbitrary force fields,
$$\mathbf{F}=\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t)$$
in analogy and compared to (2.204). To produce an infinitesimal displacement $d \mathbf{r}$ the work
$$\delta W=-\mathbf{F} \cdot d \mathbf{r}$$
has to be invested. The sign convention is the same as explained after (2.205). The symbol $\delta$ is chosen consciously since this expression does not necessarily represent a total differential as we will see in the following. Here it merely denotes an infinitesimally small quantity.
For finite pathways (Fig. 2.40) it holds:
$$W_{21}=-\int_{P_1}^{P_2} \mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) \cdot d \mathbf{r} .$$

This quantity normally depends on:
1) force field $\mathbf{F}$,
2) endpoints $P_1, P_2$,
3) path $C$,
4) temporal course of movement.
If $\mathbf{F}=\mathbf{F}(\mathbf{r})$ then of course point 4) becomes meaningless, i.e. $W_{21}$ depends only on the shape of the path and no longer on the temporal course of motion of the mass point along the trajectory. The integration in (2.220) represents a socalled curvilinear (line) integral. One evaluates such line integrals by tracing them back, in some way, to normal Riemann-integrals. That can be done with the parametrization of the space curve $C$ introduced in Sect. 1.4.1 (Fig. 2.41). The parameter $\alpha$ can but need not necessarily be the time $t$ :
$$\begin{gathered} C: \mathbf{r}=\mathbf{r}(\alpha) ; \quad \alpha_1 \leq \alpha \leq \alpha_2 ; \ d \mathbf{r}=\frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha . \end{gathered}$$
Therewith Eq. (2.220) can also be written as follows:
$$W_{21}=-\int_{\alpha_1}^{\alpha_2} \mathbf{F}\left(\mathbf{r}_{,} \dot{\mathbf{r}}, t\right) \cdot \frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Classical Particle Paths

(2.215)

(2.216)

(a) 谐振子:

(b) 势能的一般状态依赖性:

$$F=0=-\frac{d V}{d x} \Longleftrightarrow V \text { extremal } .$$

$$[E]=k \mathrm{~km}^2 \mathrm{~s}^{-2}=\text { Joule } .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Work, Power, and Energy

$$\mathbf{F}=\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t)$$

$$\delta W=-\mathbf{F} \cdot d \mathbf{r}$$

$$W_{21}=-\int_{P_1}^{P_2} \mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) \cdot d \mathbf{r} .$$

1) 力场 $\mathbf{F}$,
2) 端点 $P_1, P_2$ ，
3) 路径 $C$ ，
4) 时间运动过程。

$$C: \mathbf{r}=\mathbf{r}(\alpha) ; \quad \alpha_1 \leq \alpha \leq \alpha_2 ; d \mathbf{r}=\frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha$$

$$W_{21}=-\int_{\alpha_1}^{\alpha_2} \mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) \cdot \frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYSICS2532

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Inverse Matrix

Definition 1.6.7 $A=\left(a_{i j}\right)$ is a given $(n \times n)$-matrix. Then one denotes as its inverse matrix
$$A^{-1}=\left(\left(a^{-1}\right){i j}\right)$$ just the $(n \times n)$-matrix, for which holds: $$A^{-1} A=A A^{-1}=\mathbb{1} .$$ Theorem 1.6.2 $A^{-1}$ exists only when $\operatorname{det} A \neq 0$. The elements are then found by: $$\left(a^{-1}\right){i j}=\frac{U_{j i}}{\operatorname{det} A} .$$
(Note the order of the indexes!)

Proof Let $\widehat{A}=\left(\alpha_{i j}=U_{j i}\right)$ be an $(n \times n)$-matrix. With the expansion theorems (1.327) and (1.332) we find:
\begin{aligned} &\operatorname{det} A=\sum_j a_{i j} U_{i j}=\sum_j a_{i j} \alpha_{j i}=(A \cdot \widehat{A}){i i}, \ &\operatorname{det} A=\sum_i a{i j} U_{i j}=\sum_i \alpha_{j i} a_{i j}=(\widehat{A} \cdot A){i j} . \end{aligned} The diagonal elements of the product matrices $A \cdot \widehat{A}$ and $\widehat{A} \cdot A$ are thus all identical to $\operatorname{det} A$. What about the non-diagonal elements? With (1.336) one finds: $$(A \cdot \widehat{A}){i j}=\sum_k a_{i k} \alpha_{k j}=\sum_k a_{i k} U_{j k}=0 \quad \text { for } i \neq j$$
It follows that $A \cdot \widehat{A}$ and $\widehat{A} \cdot A$ are diagonal matrices with
$$A \cdot \widehat{A}=\widehat{A} \cdot A=\operatorname{det} A \cdot \mathbb{1} \text {. }$$
With $\operatorname{det} A \neq 0$ and by comparison with (1.337) the theorem is proved:
$$\frac{\widehat{A}}{\operatorname{det} A}=A^{-1} \Longleftrightarrow \frac{U_{j i}}{\operatorname{det} A}=\left(a^{-1}\right)_{i j}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rotation Matrix

We remember the question which came up in connection with (1.321). Under what conditions an arbitrary matrix $D$ based in a given $\operatorname{CONS}\left{\mathbf{e}i\right}$ is a rotation matrix? At first it must satisfy the orthonormality relations (1.308) and (1.316): \begin{aligned} &\sum_m d{i m} d_{j m}=\delta_{i j}, \ &\sum_m d_{m i} d_{m j}=\delta_{i j} . \end{aligned}
What is more, the new basis system $\left{\overline{\mathbf{e}}{\mathbf{j}}\right}$ originating from the original system $\left{\mathbf{e}_i\right}$ by rotation shall again be a right-handed trihedron, i.e. (1.342) must also be valid for the $\overline{\mathbf{e}}{\mathbf{j}}$. That is not yet guaranteed by the conditions (1.308) and (1.316). For instance, if we replace in the $i$-th row of $D$ the $d_{i j}$ by $\left(-d_{i j}\right)$, the orthonormality relations will still be valid. On the other hand, however, according to (1.305) $\left{\overline{\mathbf{e}}_i\right}$ transfers into $\left(-\overline{\mathbf{e}}_i\right)$. Thus the right-handed trihedron becomes a left-handed one. However, we notice with (1.305):
\begin{aligned} \overline{\mathbf{e}}1 \cdot\left(\overline{\mathbf{e}}_2 \times \overline{\mathbf{e}}_3\right) &=\sum{m, n, p} d_{1 m} d_{2 n} d_{3 p} \mathbf{e}m \cdot\left(\mathbf{e}_n \times \mathbf{e}_p\right)=\ &=\sum{m, n, p} \varepsilon_{m n p} d_{1 m} d_{2 n} d_{3 p}=\operatorname{det} D \end{aligned}
That means that besides the orthonormality of rows and columns a rotation matrix $D$ still must fulfill:
$$\operatorname{det} D=1$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Inverse Matrix

$$A^{-1}=\left(\left(a^{-1}\right) i j\right)$$

$$A^{-1} A=A A^{-1}=1 .$$

$$\left(a^{-1}\right) i j=\frac{U_{j i}}{\operatorname{det} A} .$$
(注意索引的顺序!)

$\operatorname{det} A=\sum_j a_{i j} U_{i j}=\sum_j a_{i j} \alpha_{j i}=(A \cdot \widehat{A}) i i, \quad \operatorname{det} A=\sum_i a i j U_{i j}=\sum_i \alpha_{j i} a_{i j}=(\widehat{A} \cdot A) i j .$

$$(A \cdot \widehat{A}) i j=\sum_k a_{i k} \alpha_{k j}=\sum_k a_{i k} U_{j k}=0 \quad \text { for } i \neq j$$

$$A \cdot \widehat{A}=\widehat{A} \cdot A=\operatorname{det} A \cdot 1 .$$

$$\frac{\widehat{A}}{\operatorname{det} A}=A^{-1} \Longleftrightarrow \frac{U_{j i}}{\operatorname{det} A}=\left(a^{-1}\right)_{i j}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rotation Matrix

loperatorname{CONS}\left{\mathbf{e}i\right } } \text { 是旋转矩阵? 首先它必须满足正交关系 (1.308) 和 (1.316) : }
$$\sum_m d i m d_{j m}=\delta_{i j}, \quad \sum_m d_{m i} d_{m j}=\delta_{i j} .$$

Veft{\overline{\mathbf{e}}_i\right } } \text { 䉽入 } ( – \overline { \mathbf { e } } _ { i } ) \text { . 因此右手三面体变成左手三面体。但是，我们注意到 (1.305): }
$$\overline{\mathbf{e}} 1 \cdot\left(\overline{\mathbf{e}}2 \times \overline{\mathbf{e}}_3\right)=\sum m, n, p d{1 m} d_{2 n} d_{3 p} \mathbf{e} m \cdot\left(\mathbf{e}n \times \mathbf{e}_p\right)=\quad \sum m, n, p \varepsilon{m n p} d_{1 m} d_{2 n} d_{3 p}=\operatorname{det} D$$

$\operatorname{det} D=1$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS2041

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Calculation Rules for Matrices

Let us first agree upon what we want to understand as the sum of two matrices:
Definition 1.6.3 If $A=\left(a_{i j}\right), B=\left(b_{i j}\right)$ are two $(m \times n)$-matrices then the sum is given as the matrix $C=A+B=\left(c_{i j}\right)$ with the elements:
$$c_{i j}=a_{i j}+b_{i j}, \quad \forall i, j .$$
$C$ is again a $(m \times n)$-matrix.
Example
\begin{aligned} &A=\left(\begin{array}{lll} 6 & 3 & 0 \ 1 & 4 & 5 \end{array}\right) \ &B=\left(\begin{array}{lll} 1 & 3 & 5 \ 2 & 4 & 6 \end{array}\right) \Longrightarrow C=A+B=\left(\begin{array}{ccc} 7 & 6 & 5 \ 3 & 8 & 11 \end{array}\right) . \end{aligned}
The so defined addition is obviously commutative as well as associative.
The next step concerns the multiplication of a matrix by a real number:
Definition 1.6.4 If $A=\left(a_{i j}\right)$ is a $(m \times n)$-matrix then the matrix $\lambda A(\lambda \in \mathbb{R})$ is to understand as the $(m \times n)$-matrix:
$$\lambda A=\left(\lambda a_{i j}\right) .$$
Hence each matrix element is multiplied by $\lambda$
Example
$$3\left(\begin{array}{ccc} 5 & -3 & 1 \ 0 & 2 & -1 \end{array}\right)=\left(\begin{array}{ccc} 15 & -9 & 3 \ 0 & 6 & -3 \end{array}\right) \text {. }$$
We know from normal vectors, which represent nothing else than special matrices, namely $(n \times 1)$ – and $(1 \times n)$-matrices, respectively, that they can be multiplicatively connected in form of scalar products. That is generalized correspondingly for matrices.

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Transformation of Coordinates

Let $\Sigma, \bar{\Sigma}$ be two systems of coordinates specified by the orthonormal basis vectors (Fig. 1.72):
$\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ and $\overline{\mathbf{e}}_1, \overline{\mathbf{e}}_2, \overline{\mathbf{e}}_3$, respectively
Translations are relatively uninteresting. We therefore assume that the origins of $\Sigma$ and $\bar{\Sigma}$ coincide. Let us now consider an arbitrarily chosen position vector $\mathbf{r}$ :
$$\begin{array}{ll} \mathbf{r}=\left(x_1, x_2, x_3\right) \text { in } \Sigma & {[\mathbf{r}(\Sigma)]} \ \mathbf{r}=\left(\bar{x}_1, \bar{x}_2, \bar{x}_3\right) \text { in } \bar{\Sigma} & {[\mathbf{r}(\bar{\Sigma})]} \end{array}$$
Let us presume that the elements $x_i$ in $\Sigma$ are known while the elements $\bar{x}_j$ in $\bar{\Sigma}$ are to be determined. $\mathbf{r}$ itself is of course independent of the special choice of the system of coordinates, both with respect to direction as well as magnitude.

Therefore:
$$\sum_{j=1}^3 x_j \mathbf{e}j=\sum{j=1}^3 \bar{x}j \overline{\mathbf{e}}_j$$ The basis vectors $\overline{\mathbf{e}}_j$ can be represented in $\Sigma$ : $$\overline{\mathbf{e}}_j=\sum_k d{j k} \mathbf{e}k$$ We determine the expansion coefficients $d{j k}$ by scalar multiplication of this equation by $\mathbf{e}m$ : $$d{j m}=\overline{\mathbf{e}}j \cdot \mathbf{e}_m=\cos \varphi{j m}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Calculation Rules for Matrices

$$c_{i j}=a_{i j}+b_{i j}, \quad \forall i, j .$$
$C$ 又是一个 $(m \times n)$-矩阵。

$$\lambda A=\left(\lambda a_{i j}\right) .$$

$$3\left(\begin{array}{lllll} 5 & -3 & 10 & 2 & -1 \end{array}\right)=\left(\begin{array}{llllll} 15 & -9 & 30 & 6 & -3 \end{array}\right) .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Transformation of Coordinates

$\mathbf{e}1, \mathbf{e}_2, \mathbf{e}_3$ 和 $\overline{\mathbf{e}}_1, \overline{\mathbf{e}}_2, \overline{\mathbf{e}}_3$ ，分别 翻译比较无趣。因此，我们假设 $\Sigma$ 和 $\bar{\Sigma}$ 重合。现在让我们考虑一个任意选择的位置向量 $\mathbf{r}$ : $$\mathbf{r}=\left(x_1, x_2, x_3\right) \text { in } \Sigma \quad[\mathbf{r}(\Sigma)] \mathbf{r}=\left(\bar{x}_1, \bar{x}_2, \bar{x}_3\right) \text { in } \bar{\Sigma} \quad[\mathbf{r}(\bar{\Sigma})]$$ 我们假设元素 $x_i$ 在 $\Sigma$ 已知元素 $\bar{x}_j$ 在 $\bar{\Sigma}$ 有待确定。 $\mathbf{r}$ 它本身当然独立于坐标系统的特殊选择，无论是方向还是幅度。 所以： $$\sum{j=1}^3 x_j \mathbf{e} j=\sum j=1^3 \bar{x} j \overline{\mathbf{e}}_j$$

$$\overline{\mathbf{e}}_j=\sum_k d j k \mathbf{e} k$$

$$d j m=\overline{\mathbf{e}} j \cdot \mathbf{e}_m=\cos \varphi j m$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS3020

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

With the aid of the partial derivative we have the possibility to find out how a field alters as we proceed along one of the axis of coordinates. We want to investigate now how a scalar field changes along an arbitrary (!) space direction e, i.e. we are interested in the term
\begin{aligned} &\Delta \varphi=\varphi(\mathbf{r}+\Delta \mathbf{r})-\varphi(\mathbf{r}), \ &\Delta \mathbf{r}=\left(\Delta x_1, \Delta x_2, \Delta x_3\right) \uparrow \uparrow \mathbf{e} . \end{aligned}
If $\Delta \mathbf{r}$ were, e.g., parallel to the 1-axis then for sufficiently small changes $\Delta \mathbf{r}=$ $\Delta x_1 \mathbf{e}_1$ we would have:
$$\Delta \varphi=\frac{\partial \varphi}{\partial x_1} \Delta x_1 \quad\left[\varphi(\mathbf{r}+\Delta \mathbf{r})=\varphi\left(x_1+\Delta x_1, x_2, x_3\right)\right] .$$
This presumption is not in general fulfilled (Fig. 1.71). It is, however, possible to realize it by a proper rotation of the coordinate axes. The physical field $\varphi$ is of course not affected by such a redefinition of the axes directions. We execute the rotation in such a way that the new 1 axis coincides with the e direction. Then we must have:
$$\Delta \varphi=\frac{\partial \varphi}{\partial \bar{x}_1} \Delta \bar{x}_1 .$$
We now can express $\Delta \mathbf{r}$ in the new and old system of coordinates, respectively, as follows:
$$\Delta \mathbf{r}=\Delta \bar{x}_1 \overline{\mathbf{e}}_1=\Delta x_1 \mathbf{e}_1+\Delta x_2 \mathbf{e}_2+\Delta x_3 \mathbf{e}_3 .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Divergence and Curl

The gradient, nabla operator introduced in the last section acts exclusively on scalar fields $\varphi$, while the resulting gradient field $\operatorname{grad} \varphi=\nabla \varphi$ is itself a vector. An obvious question then is whether it is possible to apply the nabla operator $\nabla$, formally defined in (1.269) as vector-differential operator, also to vectors. The answer is yes! There are again two kinds of application, similar to the previously discussed multiplicative connection of two ordinary vectors, one in the sense of a scalar product, the other in the sense of a vector product.

Definition 1.5.6 Let a $(\mathbf{r}) \equiv\left(a_1(\mathbf{r}), a_2(\mathbf{r}), a_3(\mathbf{r})\right)$ be a continuously differentiable vector field.
Then one calls
$$\sum_{j=1}^3 \frac{\partial a_j}{\partial x_j} \equiv \operatorname{div} \mathbf{a}(\mathbf{r}) \equiv \nabla \cdot \mathbf{a}(\mathbf{r})$$
the divergence (the source field) of $\mathbf{a}(\mathbf{r})$.
By this definition, to a given vector field $\mathbf{a}(\mathbf{r})$ a new scalar field diva(r) is assigned. The illustrative interpretation of div $\mathbf{a}(\mathbf{r})$ as a source field of $\mathbf{a}(\mathbf{r})$ will become understandable later by some examples from physics.
The reader should prove as an exercise the following calculation rules:
\begin{aligned} \operatorname{div}(\mathbf{a}+\mathbf{b}) &=\operatorname{diva}+\operatorname{divb}, \ \operatorname{div}(\gamma \mathbf{a}) &=\gamma \operatorname{diva} ; \quad \gamma \in \mathbb{R}, \ \operatorname{div}(\varphi \mathbf{a}) &=\varphi \operatorname{diva}+\mathbf{a} \cdot \operatorname{grad} \varphi \end{aligned}
( $\varphi$ : scalar field; a: vectorial field).

## 理论力学代写

$$\Delta \varphi=\varphi(\mathbf{r}+\Delta \mathbf{r})-\varphi(\mathbf{r}), \quad \Delta \mathbf{r}=\left(\Delta x_1, \Delta x_2, \Delta x_3\right) \uparrow \uparrow \mathbf{e} .$$

$$\Delta \varphi=\frac{\partial \varphi}{\partial x_1} \Delta x_1 \quad\left[\varphi(\mathbf{r}+\Delta \mathbf{r})=\varphi\left(x_1+\Delta x_1, x_2, x_3\right)\right] .$$

$$\Delta \varphi=\frac{\partial \varphi}{\partial \bar{x}_1} \Delta \bar{x}_1 .$$

$$\Delta \mathbf{r}=\Delta \bar{x}_1 \overline{\mathbf{e}}_1=\Delta x_1 \mathbf{e}_1+\Delta x_2 \mathbf{e}_2+\Delta x_3 \mathbf{e}_3$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Divergence and Curl

$$\sum_{j=1}^3 \frac{\partial a_j}{\partial x_j} \equiv \operatorname{div} \mathbf{a}(\mathbf{r}) \equiv \nabla \cdot \mathbf{a}(\mathbf{r})$$

$$\operatorname{div}(\mathbf{a}+\mathbf{b})=\operatorname{diva}+\operatorname{divb}, \operatorname{div}(\gamma \mathbf{a})=\gamma \operatorname{diva} ; \quad \gamma \in \mathbb{R}, \operatorname{div}(\varphi \mathbf{a})=\varphi \operatorname{diva}+\mathbf{a} \cdot \operatorname{grad} \varphi$$
( $\varphi$ : 标量场；a：矢量场)。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS2041

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Elements of Integral Calculus

The technique of ‘differentiation’, which we discussed in the previous section, follows the scope of work:
given: $\quad y=f(x)$
finding: $\quad f^{\prime}(x)=\frac{d f}{d x}:$ ‘derivation’,
The reverse program, namely
given: $\quad f^{\prime}(x)=\frac{d f}{d x}$
finding: $\quad y=f(x)$
leads to the technique of ‘integration’. Consider for example
$$f^{\prime}(x)=c=\text { const }$$
then we remember according to (1.77) that
$$y=f(x)=c \cdot x$$
fulfills the condition $f^{\prime}(x)=c$.
Definition $F(x)$ is the ‘antiderivative (primitive function)’ of $f(x)$, if it holds:
$$F^{\prime}(x)=f(x) \quad \forall x .$$
In this connection the above example means:
$$f(x) \equiv c \quad \curvearrowright \quad F(x)=c \cdot x+d .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Theorem of Calculus

We consider the definite integral over a continuous function $f(t)$, but now with variable upper limit:
$$F(x)=\int_{a}^{a} f(t) d t \quad \text { ‘area function’ }$$

The area under the curve $f(t)$ in this case is not constant but a function of $x$ (Fig. 1.22). If the upper bound of integration is shifted by $\Delta x$ the area will change by:
$$\Delta F=F(x+\Delta x)-F(x)=\int_{a}^{x+\Delta x} f(t) d t-\int_{a}^{x} f(t) d t=\int_{x}^{x+\Delta x} f(t) d t$$
In the last step we have used the rule (1.114). Without explicit proof we accept the important
‘mean value theorem of integral calculus’
This theorem implies:
$$\exists \hat{x} \in[x, x+\Delta x] \text { with } \Delta F=\Delta x \cdot f(\hat{x}) .$$
Although not exactly proven the theorem appears rather plausible according to Fig. 1.23. So we can further conclude:
$$F^{\prime}(x)=\lim {\Delta x \rightarrow 0} \frac{\Delta F}{\Delta x}=\lim {\Delta x \rightarrow 0} f(\hat{x})=f(x) .$$
Thus after (1.105), the area function is the antiderivative of $f(x)$ ! Furthermore, the equivalence of the definitions (1.105) and (1.110) for the antiderivative, which remained unsettled in Sect. $1.2 .1$, is now settled.
‘fundamental theorem of calculus’
$$\frac{d}{d x} F(x) \equiv \frac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Elements of Integral Calculus

$$f^{\prime}(x)=c=\text { const }$$

$$y=f(x)=c \cdot x$$

$$F^{\prime}(x)=f(x) \quad \forall x .$$

$$f(x) \equiv c \quad \curvearrowright \quad F(x)=c \cdot x+d .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Theorem of Calculus

$$F(x)=\int_{a}^{a} f(t) d t \quad \text { ‘area function’ }$$

$$\Delta F=F(x+\Delta x)-F(x)=\int_{a}^{x+\Delta x} f(t) d t-\int_{a}^{x} f(t) d t=\int_{x}^{x+\Delta x} f(t) d t$$

$$\exists \hat{x} \in[x, x+\Delta x] \text { with } \Delta F=\Delta x \cdot f(\hat{x}) .$$

$$F^{\prime}(x)=\lim \Delta x \rightarrow 0 \frac{\Delta F}{\Delta x}=\lim \Delta x \rightarrow 0 f(\hat{x})=f(x) .$$

1.2.1，现已解决。
‘微积分基本定理’
$$\frac{d}{d x} F(x) \equiv \frac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYC90007

statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Differential Quotient

The ‘slope (gradient)’ of a straight line is the quotient of ‘height difference’ $\Delta y$ and ‘base line’ $\Delta x$ (see Fig. 1.10). For the gradient angle $\alpha$ we obviously have:
$$\tan \alpha=\frac{\Delta y}{\Delta x} .$$
Analogously one defines the slope (gradient) of an arbitrary function $f(x)$ at a point $P$ (see Fig. 1.11). The secant $\overline{P Q}$ has the increase
$$\frac{\Delta y}{\Delta x}=\tan \alpha^{\prime}$$
One dennotês
$$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
as ‘difference quotient’. If we now shift the point $Q$ along the curve towards the point $P$ then the increase of the secant becomes the increase of the tangent on the curve $f(x)$ at $P$ (broken line in Fig. 1.11),
$$\tan \alpha=\lim {\alpha^{\prime} \rightarrow \alpha} \tan \alpha^{\prime}=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$$
and one arrives at the ‘differential quotient’
$$\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \equiv \frac{d y}{d x} .$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Taylor Expansion

Occasionally it is unavoidable for a physicist to digress from rigorous mathematical exactness in order to come by adopting some ‘reasonable’ mathematical simplifications to concrete physical results. In this respect, the so-called ‘Taylor expansion (series)’ of a mathematical function $y=f(x)$ represents a very important and frequently used auxiliary means. We assume that this function possesses arbitrarily many continuous derivatives at $x=x_{0}$. Then the following power series expansion is valid what is explicitly proved as Exercise $1.1 .9$ :
\begin{aligned} f(x) &=f\left(x_{0}\right)+\frac{f^{\prime}\left(x_{0}\right)}{1 !}\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\ldots \ &=\sum_{n=0}^{\infty} \frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n} \ f^{(n)}\left(x_{0}\right) &=\left.f^{(n)}(x)\right|{x=x{0}} . \end{aligned}
The assumption $\left|x-x_{0}\right|<1$ guarantees the convergence of the series. Then one can assume that the terms of the series become smaller and smaller with increasing index $n$, so that it should be allowed, in the sense of a controlled approximation, to cut the series after a finite number of summands. The error can strictly be estimated as will be demonstrated in Sect. $1.2$ of volume 3 .

However, the Taylor expansion can also be used for the derivation of exact series as is shown by the following examples:
$1 .$
$$f(x)=\frac{1}{1+x} ; x_{0}=0 ;|x|<1$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Differential Quotient

$$\tan \alpha=\frac{\Delta y}{\Delta x} .$$

$$\frac{\Delta y}{\Delta x}=\tan \alpha^{\prime}$$

$$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

$$\tan \alpha=\lim \alpha^{\prime} \rightarrow \alpha \tan \alpha^{\prime}=\lim \Delta x \rightarrow 0 \frac{\Delta y}{\Delta x}$$

$$\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \equiv \frac{d y}{d x}$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Taylor Expansion

$$f(x)=f\left(x_{0}\right)+\frac{f^{\prime}\left(x_{0}\right)}{1 !}\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\ldots \quad=\sum_{n=0}^{\infty} \frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n} f^{(n)}\left(x_{0}\right)$$

$$f(x)=\frac{1}{1+x} ; x_{0}=0 ;|x|<1$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。