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我尝试重写你的（非）MWE，但是我在写到中间部分之后就放弃了...理解你的代码、意图等太耗时了。我希望这个片段能向你展示如何编写数学文本，如何使用提供的不同数学环境amsmath，我非常肯定你的文本的形成可以进一步改进，但是这个任务我留给你了:-)

\documentclass[11pt]{article}
    \usepackage{amsmath}
    \usepackage[margin=25mm,a4paper]{geometry}

    \title{Diwali Assignment 2015}
    \date{2015-11-10}
    \author{Sandeep}

\begin{document}
    \pagenumbering{gobble}
    \maketitle
    \newpage
    \pagenumbering{arabic}

\section{Diwali Assignment 2015}
\label{sec:DiwaliAssignment2015}

\subsection{ATOMIC WASTE DISPOSAL}
\label{sec:ATOMICWASTEDISPOSAL}
Radio active waste is mostly hazardous as it poses potential threats to public health or the environment. Therefore radioactive waste management
is very important and has to be taken seriously before it poses a threat to the survival of mankind.

\emph{So Nuclear Regulatory Commission had disposed  radioactive waste material using tightly sealed drums and then dumped into sea which is of 300 ft. NRC is assured of any leaks in drums.But question arises whether the drum breaks when it hits the ocean floor.Many experiments were conducted on this point and engineers found that the drum breaks when the impact velocity exceeds 40 ft/s}

\paragraph{PROBLEM}
\label{sec:PROBLEM}
The problem is to find the impact velocity and check whether it crosses the limiting velocity to prevent any leakage caused due to impact by using the following data:
    \begin{center}
    \begin{tabular}{rl}
weight of the drum              & 528 lbs  \\
volume of the drum              & 7 $ft^{3}$\\
density of salt water           & 64 lbs\\
drag force of water on Drum     & D = $c \times V \frac{(lb)(s)}{ft}$ where $c=0.08$\\
Acceleration due to gravity     & $32.2 \frac{ft}{s^2}$
    \end{tabular}
    \end{center}
\newpage

\section{Solving The Problem}
\label{sec:SolvingTheProblem}
This model is solved by using \textbf{Newton mechanics}.

\subsection{Newton Mechanics}
\label{sec:NewtonMechanics}
The Newton principle we use in this problem is Newton's second law of motion. Using this law we can state that force acting on any object is equal to mass of that body multiplied by acceleration of that body caused due to force acting on i.e,
    \begin{align}
F   & = Mass \times Acceleration    \notag\\
    & = M \times \frac{d^2 y}{dt^2} \label{eq:1}
    \end{align}
as Acceleration, $a = \frac{d^2 y}{dt^2}$. 
In \eqref{eq:1} $M$, mass of object and $y,t$ denotes displacement and time respectively.

\subsection{Differential Equations}
\paragraph{\textbf{SOLUTION:}}
Let direction of increasing y be downwards then weight of the drum W is positive where as buoyant force B and drag force on drum D is negative
Net force F is given by
    \begin{equation}\label{eq:2}
F = W-B-cV 
    \end{equation}

\subsubsection{Formulating Differential Equation :}
\label{sec:FormulatingDifferentialEquation}
From \eqref{eq:1} and \eqref{eq:2}
    \begin{align}
\frac{d^2 y}{dt^2} 
    & = \frac{1}{M} (W-B-cV)            \notag
    & = \frac{g}{W}(W-B-cV) \label{eq:3}
    \end{align}
where $W$ is mass of drum $\times$ acceleration due to gravity, i.e.,
    \begin{align*}
W & = m \times g = 528 lbs\\
B & = \text{weight of water displaced}\\
  & = \text{volume of drum $\times$ density of salt water}\\
  & = 7 \times 64 = 448 lbs
  \end{align*}
On substituting $V=\frac{dy}{dt}$  in \eqref{eq:3} follows
    \begin{equation}\label{eq:4}
\frac{dV}{dt} + \frac{cg}{W} V = \frac{g}{W}(W-B) 
    \end{equation}

    \subsubsection{Solving Differential Equation}
    \label{sec:SolvingDifferentialEquation}
Equation\eqref{eq:4} can be written as
    \[
\frac{dv}{dt} = \frac{g}{w}(W-B-cV)
    \]
By variable seperation and integrating we get
    \[
\frac{gt}{w} = \int \frac{1}{W-B-cV}dv
    \]
on solving this equation we get
    \[
V(t) = \frac{W-B}{c}[1-\exp(-\frac{cg}{w})t] 
    \quad\text{as } V(0) = 0
    \]
From this the limiting value of $V(t)$ i.e when $t \rightarrow \infty$
    \begin{align*}
Velocity & = \frac{W-B}{c}
         & = \frac{528-448}{0.08} =1 000 \frac{ft}{s} 
    \end{align*}
This velocity is called Terminal Velocity, which is way greater than limiting impact velocity.

\emph{From this,we cannot conclude as terminal velocity of drum $>$ limting impact velocity to avoid breakage.But drum don't break if terminal velocity $<$ maximum impact velocity}
Therefore, we have to write velocity in terms of displacement\\
i.e $V(y)$ instead of $V(t)$.

\subsubsection{To write $V$ in terms of displacement $y$}
From equation \eqref{eq:4} 
    \[
\frac{dv}{dt} = \frac{g}{w}(W-B-cV)
    \]
using chain derivative rule 
    \begin{equation}\label{eq:5}
\frac{dv}{dt}=\frac{dy}{dt}\frac{dv}{dy}
    \end{equation}
substituting in \eqref{eq:4}
    \begin{align*}
\frac{dy}{dt} \times \frac{dv}{dy}
    & = \frac{g}{W}(W-B-cV)
\intertext{as $\frac{dy}{dt} = V$}
    & \Rightarrow V\frac{dv}{dy} = \frac{g}{W}(W-B-cV)\\
    & \Rightarrow \frac{V}{W-B-cV}dv\\
    & =\frac{g}{W}dy \text{ (By variable separation)}
    \end{align*}
By integrating both sides:
    \begin{align*}
\frac{gy}{W} 
    & = \int{\frac{vdv}{W-B-cV}}
\intertext{Taking $-c$ as common and substitute $\frac{W-B}{c} = K$}
    & = -\frac{1}{c}\int{\frac{V-K+K dv}{V-k}} 
        \Rightarrow \frac{gy}{W}\\
    & = -\frac{1}{c}[V-K\ln(V-K)]
\intertext{Therefore}   
\frac{gy}{W} 
    & = \frac{(B-W)\ln(1+\frac{cV}{B-W})-cV}{c^2}
    \end{align*}

etc
\end{document}

您的代码的主要问题已在问题的评论中描述。根据这些问题，我部分更改了您对paragraph和的使用subparagraph，因为它们并非用于强调文本，并尝试在、和的使用上保持一致section。subsection但是subsubsection，由于我对您文档的结构仍不太清楚，因此我可能遗漏了许多内容。

关于使用$$ ... $$：我省略了这一点，因为它不是 LaTeX 符号。我用\[ ... \]或其他 LaTeX/AMS 数学环境替换它们。

要查看我建议的更改后文本的外观，只需编译上述代码。它给出四页文本。

我希望我的重新格式化能够对您有所帮助。

Answer 1