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# 数学代写|数学建模代写Mathematical Modeling代考|Motion of a Rocket

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## 数学代写|数学建模代写Mathematical Modeling代考|Motion of a Rocket

As a first idealization, we neglect both gravity and air resistance. A rocket moves forward because of the large supersonic velocity with which gases produced by the burning of the fuel inside the rocket come out of the converging-diverging nozzle of the rocket (Figure 2.8).
Let $m(t)$ be the mass of the rocket at time $t$ and let it move forward with velocity $v(t)$ so that the momentum at time $t$ is $m(t) v(t)$.
In the interval of time $(t, t+\Delta t)$, the mass of the rocket becomes
$$m(t+\Delta t)=m(t)+\frac{d m}{d t} \Delta t+0(\Delta t)$$
Since the rocket is losing mass, $d m / d t$ is negative and the mass of gases $-d m / d t \Delta t$ moves with velocity $u$ relative to the rocket, i.e., with a velocity $v(t+\Delta t)-u$ relative to the Earth so that the total momentum of the rocket and the gases at time $t+\Delta t$ is
$$m(t+\Delta t) v(t+\Delta t)-\frac{d m}{d t} \Delta t(v(t+\Delta t)-u)$$
Since we are neglecting air resistance and gravity, there is no external force on the rocket and as such the momentum is conserved, giving the equation
$$m(t) v(t)=\left(m(t)+\frac{d m}{d t} \Delta t\right)\left(v(t)+\frac{d v}{d t} \Delta t\right)-\frac{d m}{d t} \Delta t(v-u)+0(\Delta t)^2$$
Dividing by $\Delta t$ and proceeding to the limit as $\Delta t \rightarrow 0$, we get
or
$$\begin{gathered} m(t) \frac{d v}{d t}=-u \frac{d m}{d t} \ \frac{d m}{m}=-\frac{1}{u} d v \ \ln \frac{m(t)}{m(0)}=-\frac{v(t)}{u} \end{gathered}$$
assuming that the rocket starts with zero velocity.

## 数学代写|数学建模代写Mathematical Modeling代考|Simple Geometrical Problems

Many geometrical entities can be expressed in terms of derivatives and as such relations between these entities can give rise to differential equations whose solutions will give us a family of curves for which the given relation between geometrical entities is satisfied.
(i) Find curves for which the tangent at a point is always perpendicular to the line joining the point to the origin.
The slope of the tangent is $d y / d x$ and the slope of the line joining the point $(x, y)$ to the origin is $y / x$, and since these lines are given to be orthogonal
$$\frac{d y}{d x}=-\frac{x}{y}$$
integrating
$$x^2+y^2=a^2$$
which represents a family of concentric circles.
(ii) Find curves for which the projection of the normal on the $x$-axis is of constant length.
This condition gives
$$y \frac{d y}{d x}=k$$
integrating
$$y^2=2 k x+A$$
which represents a family of parabolas, all with the same axis and same length of latus rectum.
iii) Find curves for which the tangent makes a constant angle with the radius vector.
Here it is convenient to use polar coordinates, and the conditions of the problem gives
$$r \frac{d \theta}{d r}=\tan \alpha$$
integrating
$$r=A e^{\theta \cot \alpha}$$
which represents a family of equiangular spirals.

## 数学代写|数学建模代写Mathematical Modeling代考|Motion of a Rocket

$$m(t+\Delta t)=m(t)+\frac{d m}{d t} \Delta t+0(\Delta t)$$

$$m(t+\Delta t) v(t+\Delta t)-\frac{d m}{d t} \Delta t(v(t+\Delta t)-u)$$

$$m(t) v(t)=\left(m(t)+\frac{d m}{d t} \Delta t\right)\left(v(t)+\frac{d v}{d t} \Delta t\right)-\frac{d m}{d t} \Delta t(v-u)+0(\Delta t)^2$$

$$\begin{gathered} m(t) \frac{d v}{d t}=-u \frac{d m}{d t} \ \frac{d m}{m}=-\frac{1}{u} d v \ \ln \frac{m(t)}{m(0)}=-\frac{v(t)}{u} \end{gathered}$$

## 数学代写|数学建模代写Mathematical Modeling代考|Simple Geometrical Problems

(i)找到一点的切线总是垂直于该点与原点连接的直线的曲线。

$$\frac{d y}{d x}=-\frac{x}{y}$$

$$x^2+y^2=a^2$$
，这表示一个同心圆族。
(ii)找到法线在$x$轴上的投影长度为常数的曲线。

$$y \frac{d y}{d x}=k$$

$$y^2=2 k x+A$$

iii)找到切线与半径矢量成恒定角的曲线。

$$r \frac{d \theta}{d r}=\tan \alpha$$

$$r=A e^{\theta \cot \alpha}$$

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