### 数学代写|常微分方程代写ordinary differential equation代考|MATH211

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• 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

## 数学代写|常微分方程代写ordinary differential equation代考|The Simplest Falling Object Model

The Earth’s gravity is the most obvious force acting on our falling object. Checking a convenient physics text, we find that the force of the Earth’s gravity acting on an object of mass $m$ is given by
$$F_{\text {grav }}=-g m \quad \text { where } \quad g=9.8\left(\text { meters } / \text { second }^2\right) .$$
Of course, the value for $g$ is an approximation and assumes that the object is not too far above the Earth’s surface. It also assumes that we’ve chosen “up” to be the positive direction (hence the negative sign).

For this model, let us suppose the Earth’s gravity, $F_{\text {grav }}$, is the only significant force involved. Assuming this (and keeping in mind that we are measuring distance in meters and time in seconds), we have
$$F=F_{\text {grav }}=-9.8 \mathrm{~m}$$
in the ” $F=m a$ ” equation. In particular, equation $\left(1.2^{\prime}\right)$ becomes
$$-9.8 m=m \frac{d^2 y}{d t^2} .$$
The mass conveniently divides out, leaving us with
$$\frac{d^2 y}{d t^2}=-9.8 .$$
Taking the indefinite integral with respect to $t$ of both sides of this equation yields
$$\begin{array}{rlrl} \int \frac{d^2 y}{d t^2} d t & =\int-9.8 d t \ \hookrightarrow \quad \int \frac{d}{d t}\left(\frac{d y}{d t}\right) d t & =\int-9.8 d t \ \hookrightarrow & \frac{d y}{d t}+c_1 & =-9.8 t+c_2 \ \hookrightarrow & \frac{d y}{d t} & =-9.8 t+c \end{array}$$
where $c_1$ and $c_2$ are the arbitrary constants of integration and $c=c_2-c_1$. This gives us our formula for ${ }^{d y / d t}$ up to an unknown constant $c$. But recall that the initial velocity is zero.
$$\left.\frac{d y}{d t}\right|_{t=0}=v(0)=0 .$$

## 数学代写|常微分方程代写ordinary differential equation代考|A Better Falling Object Model

The above model does not take into account the resistance of the air to the falling object – a very important force if the object is relatively light or has a parachute. Let us add this force to our model. That is, for our ” $F=m a$ ” equation, we’ll use
$$F=F_{\text {grav }}+F_{\text {air }}$$
where $F_{\text {grav }}$ is the force of gravity discussed above, and $F_{\text {air }}$ is the force due to the air resistance acting on this particular falling body.

Part of our problem now is to determine a good way of describing $F_{\text {air }}$ in terms relevant to our problem. To do that, let us list a few basic properties of air resistance that should be obvious to anyone who has stuck their hand out of a car window:

1. The force of air resistance does not depend on the position of the object, only on the relative velocity between it and the surrounding air. So, for us, $F_{\text {air }}$ will just be a function of $v$, $F_{\text {air }}=F_{\text {air }}(v)$. (This assumes, of course, that the air is still – no up-or downdrafts – and that the density of the air remains fairly constant throughout the distance this object falls.)
2. This force is zero when the object is not moving, and its magnitude increases as the speed increases (remember, speed is the magnitude of the velocity). Hence, $F_{\mathrm{air}}(v)=0$ when $v=0$, and $\left|F_{\text {air }}(v)\right|$ gets bigger as $|v|$ gets bigger.
3. Air resistance acts against the direction of motion. This means that the direction of the force of air resistance is opposite to the direction of motion. Thus, the sign of $F_{\text {air }}(v)$ will be opposite that of $v$.

While there are many formulas for $F_{\text {air }}(v)$ that would satisfy the above conditions, common sense suggests that we first use the simplest. That would be
$$F_{\mathrm{air}}(v)=-\gamma v$$ where $\gamma$ is some positive value. The actual value of $\gamma$ will depend on such parameters as the object’s size, shape, and orientation, as well as the density of the air through which the object is moving. For any given object, this value could be determined by experiment (with the aid of the equations we will soon derive).

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|The Simplest Falling Object Model

$$F=F_{\text {grav }}=-9.8 \mathrm{~m}$$

$$-9.8 m=m \frac{d^2 y}{d t^2} .$$

$$\frac{d^2 y}{d t^2}=-9.8 .$$

$$\int \frac{d^2 y}{d t^2} d t=\int-9.8 d t \hookrightarrow \int \frac{d}{d t}\left(\frac{d y}{d t}\right) d t=\int-9.8 d t \hookrightarrow \frac{d y}{d t}+c_1=-9.8 t+c_2 \hookrightarrow \frac{d y}{d t}$$

$$\left.\frac{d y}{d t}\right|_{t=0}=v(0)=0$$

## 数学代写|常微分方程代写ordinary differential equation代考|A Better Falling Object Model

$$F=F_{\text {grav }}+F_{\text {air }}$$

1. 空气阻力的大小与物体的位置无关，只与物体与周围空气的相对速度有关。所以，对我们来说， $F_{\text {air }}$ 将只是一个函数 $v, F_{\text {air }}=F_{\text {air }}(v)$. (当然，这是假设空气是静止的一一没有上升气流或下 降气流一一并且在这个物体下落的整个距离内空气的密度保持相当恒定。)
2. 当物体不动时这个力为零，它的大小随㸔速度的增加而增加（记住，速度是速度的大小）。因此， $F_{\text {air }}(v)=0$ 什么时候 $v=0$ ，和 $\left|F_{\text {air }}(v)\right|$ 变大为 $|v|$ 变大。
3. 空气阻力与运动方向相反。这意味着空气阻力的方向与运动方向相反。因此，标志 $F_{\text {air }}(v)$ 将与 $v$.
虽然有很多公式 $F_{\text {air }}(v)$ 满足上述条件，常识建议我们首先使用最简单的。那将是
$$F_{\text {air }}(v)=-\gamma v$$
在哪里 $\gamma$ 是一些正值。的实际价值 $\gamma$ 将取决于物体的大小、形状和方向等参数，以及物体移动时空气的密 度。对于任何给定的物体，这个值可以通过实验来确定（借助于我们很快就会推导出来的方程式）。

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## MATLAB代写

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