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\large 
\underline
{MA2160 \hspace{5mm}Calculus II with Technology \hspace{50mm} Ch. 11-2}

Today, we analyze the graph of the family of solutions.  We call this graphing the slope field 


\Large
Goal
\large
\begin{enumerate}
\item Define a slope field
\item Generate a slope field
\item Graph possible solutions using a slope field.

\end{enumerate}

What is a slope field?  A slope field is a graph on a coordinate plane of the values of $y'$ given an equation for $y'$.  Given the differential equation $y'=-\frac{x}{y}$, the slope field generated is the following graph:
\begin{center}
\includegraphics[scale=1]{xover-ysf.pdf} 
\end{center}

How is this generated?  By taking various points, we calculate the slope using the differential equation, and draw a line accordingly.
\newpage

For the following differential equations, graph the corresponding slope fields.

\begin{enumerate}
\item $y'=y$
\begin{center}
\includegraphics[scale=1]{blankgraph.pdf} 
\end{center}
\vfill

\item $y'=x$
\begin{center}
\includegraphics[scale=1]{blankgraph.pdf} 
\end{center}
\vfill

\item $y'=2-y$
\begin{center}
\includegraphics[scale=1]{blankgraph.pdf} 
\end{center}
\vfill
\end{enumerate}
\newpage
What good are slope fields?  They help us in two respects.  One, we can see a graph of what potential solutions look like; two, if we have a starting point, we can trace the solution and possibly determine the equation for a particular solution.  Take for example our original slope field.
\begin{center}
\includegraphics[scale=.7]{xover-ysf.pdf} 
\end{center}
It seems obvious that the solution to this equation is a circle, which when we differentiate $x^2+y^2=r^2$, we find $y'=-\frac{x}{y}$.

With the following initial conditions, we can determine the actual equation:
\begin{enumerate}
\item $y(0)=4$; we get the graph of a circle with radius 4; hence the equation $x^2+y^2=16$
\vfill
\item $y(2)=0$; we get the graph of a circle with radius 2; hence the equation $x^2+y^2=4$
\vfill
\item $y(0)=-3$; we get the graph of a circle with radius 3; hence the equation $x^2+y^2=9$
\end{enumerate}

On our three previous examples sketch the solution for the first when $y(0)=1$, the second when $y(0)=0$, and the third when $y(0)=-1$
\newpage

What about the following differential equation $y'=\frac{x}{y}$ and its accompanying slope field:
\begin{center}
\includegraphics[scale=.7]{sfxovery.pdf} 
\end{center}

If you think this looks like a hyperbola, you would be correct, as the solution of this equation is either $x^2-y^2=r^2$ or $y^2-x^2=r^2$.

Trace the solution which pass through each of the following points:
\begin{enumerate}
\item (1,0)
\vspace{5mm}
\item (0,1)
\vspace{5mm}
\item (-3,1)
\vspace{5mm}
\item (2,-1)
\vspace{5mm}
\item (1,1)
\end{enumerate}
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